The solvability for a nonlinear degenerate hyperbolic–parabolic coupled system arising from nematic liquid crystals

3.1 The problem in a partial hodograph plane

We introduce the new independent variables

(3.5) $$ \begin{align} \tau=\theta(t,x),\quad y=x. \end{align} $$

It is easy to calculate the Jacobian of this transformation

$$ \begin{align*} J:=\frac{\partial (\tau, y)}{\partial (t,x)}=\tau_ty_x-\tau_xy_t=\theta_t=\frac{\widetilde{R}+\widetilde{S}}{2}-\tau-\widetilde{v}+\theta_0, \end{align*} $$

which, together with (3.1) and (3.3), gives $J|_{t=0}=\theta _0\geq \underline {\theta }>0$ . Furthermore, it acquires that

(3.6) $$ \begin{align} \partial _t=\bigg(\frac{\widetilde{R}+\widetilde{S}}{2}-\theta-\widetilde{v}+\theta_0\bigg)\partial _\tau,\quad \partial _x=\frac{\widetilde{R}-\widetilde{S}}{2\tau}\partial _\tau +\partial _y. \end{align} $$

From system (3.2) and the relations in (3.6), we can obtain a closed quasilinear hyperbolic system in terms of $(\widetilde {R},\widetilde {S},t)$

(3.7) $$ \begin{align} \left\{ \begin{array}{l} \widetilde{R}_\tau-\frac{\tau}{f+\widetilde{S}} \widetilde{R}_y=\frac{\widetilde{R}-\widetilde{S}}{2\tau} +\frac{1}{f+\widetilde{S}}\cdot\frac{(\widetilde{R}-\widetilde{S})^2}{4\tau} -\frac{\widetilde{R}-\widetilde{S}}{f+\widetilde{S}} +\frac{\theta_{0}'}{f+\widetilde{S}}\tau-\frac{w(t,y)}{f+\widetilde{S}}\tau,\\[10pt] \widetilde{S}_\tau+\frac{\tau}{f+\widetilde{R}} \widetilde{S}_y=\frac{\widetilde{S}-\widetilde{R}}{2\tau} +\frac{1}{f+\widetilde{R}}\cdot\frac{(\widetilde{R}-\widetilde{S})^2}{4\tau} +\frac{\widetilde{R}-\widetilde{S}}{f+\widetilde{R}} -\frac{\theta_{0}'}{f+\widetilde{R}}\tau+\frac{w(t,y)}{f+\widetilde{R}}\tau,\\[10pt] t_\tau=\frac{1}{\frac{\widetilde{R}+\widetilde{S}}{2}-\tau-\widetilde{v}(t,y)+\theta_0(y)}, \end{array} \right. \end{align} $$

where $f=\theta _0(y)-\widetilde {v}(t,y)-\tau $ . Obviously, system (3.7) has a clear regularity-singularity structure. The initial values of $(\widetilde {R},\widetilde {S},t)$ are

(3.8) $$ \begin{align} \widetilde{R}(0,y)=\widetilde{S}(0,y)=t(0,y)=0. \end{align} $$

It is easily seen that the three eigenvalues of system (3.7) are

(3.9) $$ \begin{align} \lambda_-=-\frac{\tau}{f+\widetilde{S}},\ \ \lambda_+=\frac{\tau}{f+\widetilde{R}},\ \ \lambda_0=0. \end{align} $$

Moreover, the three characteristics passing through a point $(\xi ,\eta )$ are defined by

(3.10) $$ \begin{align} \left\{ \begin{array}{l} \displaystyle \frac{\mathrm{d}y_\pm(\tau;\xi,\eta)}{\mathrm{ d}\tau}=\lambda_\pm(\tau,y,t,\widetilde{R},\widetilde{S})(\tau,y_\pm(\tau;\xi,\eta)),\\ y_\pm(\xi;\xi,\eta)=\eta, \end{array} \right. \quad y_0(\tau;\xi,\eta)=\eta. \end{align} $$

Suppose that $\widetilde {M}$ is a sufficiently large positive constant determined later. We first choose a sufficiently small positive constant $\widetilde {\delta }_1\leq 1$ such that

(3.11) $$ \begin{align} \widetilde{\delta}_1\leq\frac{1}{2}\underline{\theta},\quad \widetilde{M}\cdot\widetilde{\delta}_1\leq 1,\quad \bigg(2+\frac{2\widehat{M}_0}{\underline{\theta}}\bigg)\widetilde{\delta}_1\leq\frac{1}{2}\underline{\theta}. \end{align} $$

We use the notation $\widetilde {\Sigma }(\widetilde {\delta }_1)$ to denote the function class which incorporates all continuously vector functions $\textbf {F}=(\widetilde {R},\widetilde {S},t)^T:\ [0,\widetilde {\delta }_1]\times \mathbb {R}\rightarrow \mathbb {R}^{3}$ satisfying the following properties:

(3.12) $$ \begin{align} \begin{array}{l} \mathrm{(P_1)}: (\widetilde{R}, \widetilde{S}, t)(\tau, y)\ \mathrm{are\ continuous\ on}\ [0,\widetilde{\delta}_1]\times\mathbb{R}; \\ \mathrm{(P_2)}: (\widetilde{R}, \widetilde{S}, t)(0,y)=\textbf{0},\ \forall\ y\in\mathbb{R}; \\ \displaystyle \mathrm{(P_3)}: |\widetilde{R}(t,y)|,|\widetilde{S}(t,y)|\leq \widetilde{M}\tau^2,\ \frac{\tau}{2\overline{\theta}}\leq t(\tau,y)\leq \frac{2\tau}{\underline{\theta}}\leq1, \\ \qquad \ \ \ \mathrm{for\ all}\ (\tau,y)\in[0,\widetilde{\delta}_1]\times\mathbb{R}. \end{array} \end{align} $$

If $(\widetilde {R}, \widetilde {S}, t)$ is an arbitrary element in $\widetilde {\Sigma }(\widetilde {\delta }_1)$ , we see by (3.11) and (3.12) that

(3.13) $$ \begin{align} \displaystyle f+\widetilde{R}=&\theta_0(y)-\widetilde{v}(t,y)-\tau +\widetilde{R}\geq \underline{\theta} -\widehat{M}_0t-\tau -|\widetilde{R}| \nonumber \\ \displaystyle \geq &\underline{\theta} -\widehat{M}_0\cdot\frac{2\tau}{\underline{\theta}}-\tau -\widetilde{M}\tau^2\geq \underline{\theta} -\bigg(\frac{2\widehat{M}_0}{\underline{\theta}}+ 1+\widetilde{M}\widetilde{\delta}_1\bigg)\tau \geq\frac{1}{2}\underline{\theta} \end{align} $$

and

(3.14) $$ \begin{align} \displaystyle f+\widetilde{R}=&\theta_0(y)-\widetilde{v}(t,y)-\tau +\widetilde{R}\leq \overline{\theta} +\widehat{M}_0t+\tau +|\widetilde{R}| \nonumber \\ \displaystyle &\leq \overline{\theta} +\widehat{M}_0\cdot\frac{2\tau}{\underline{\theta}}+\tau +\widetilde{M}\tau^2 \nonumber \\ &\leq \overline{\theta} +\bigg(\frac{2\widehat{M}_0}{\underline{\theta}}+ 1+\widetilde{M}\widetilde{\delta}_1\bigg)\tau \leq\overline{\theta}+\frac{1}{2}\underline{\theta}\leq2\overline{\theta}. \end{align} $$

Similarly, one has

(3.15) $$ \begin{align} \frac{1}{2}\underline{\theta}\leq f+\widetilde{S} \leq2\overline{\theta} \end{align} $$

and

(3.16) $$ \begin{align} \frac{1}{2}\underline{\theta}\leq\frac{\widetilde{R}+\widetilde{S}}{2}-\tau-\widetilde{v}(t,y)+\theta_0(y)\leq \overline{\theta}+ \frac{1}{2}\underline{\theta} \leq2\overline{\theta}. \end{align} $$

Furthermore, we have by (3.9) and (3.13)–(3.15)

(3.17) $$ \begin{align} \frac{\tau}{2\overline{\theta}}\leq -\lambda_-, \lambda_+\leq \frac{2\tau}{\underline{\theta}}, \end{align} $$

from which and (3.10) one achieves

(3.18) $$ \begin{align} |y_+(\tau;\xi,\eta)-y_-(\tau;\xi,\eta)|\leq2\int_{0}^\xi\frac{2\tau}{\underline{\theta}}\ \mathrm{ d}\tau=\frac{2}{\underline{\theta}}\xi^2. \end{align} $$

3.2 The iterative sequence

For any $(\xi ,\eta )\in [0.\widetilde {\delta }_1]\times \mathbb {R}$ , integrating the differential system (3.7) along the characteristic curves $y=y_i(\tau ;\xi ,\eta )\ (i=\pm ,0)$ defined in (3.10), we utilize the boundary conditions (3.8) to gain a system of integral equations

(3.19) $$ \begin{align} \left\{ \begin{array}{l} \displaystyle \widetilde{R}(\xi,\eta)=\int_{0}^\xi\bigg\{\frac{\widetilde{R}-\widetilde{S}}{2\tau} +A\cdot\frac{(\widetilde{R}-\widetilde{S})^2}{4\tau} \\[6pt] \displaystyle \qquad \qquad \quad -A (\widetilde{R}-\widetilde{S}) +A\theta_{0}'\tau-Aw(t,y)\tau \bigg\}(\tau, y_-(\tau;\xi,\eta))\ \mathrm{d}\tau,\\[6pt] \displaystyle \widetilde{S}(\xi,\eta)=\int_{0}^\xi\bigg\{\frac{\widetilde{S}-\widetilde{R}}{2\tau} +B\cdot\frac{(\widetilde{R}-\widetilde{S})^2}{4\tau} \\[6pt] \displaystyle \qquad \qquad \quad +B (\widetilde{R}-\widetilde{S}) -B\theta_{0}'\tau+Bw(t,y)\tau \bigg\}(\tau, y_+(\tau;\xi,\eta))\ \mathrm{d}\tau,\\[6pt] \displaystyle t(\xi,\eta)=\int_{0}^\xi C(\tau, y_0(\tau;\xi,\eta))\ \mathrm{d}\tau, \end{array} \right. \end{align} $$

where

(3.20) $$ \begin{align} A=\frac{1}{f+\widetilde{S}},\quad B=\frac{1}{f+\widetilde{R}},\quad C=\frac{1}{\frac{\widetilde{R}+\widetilde{S}}{2}-\tau-\widetilde{v}(t,y)+\theta_0(y)}. \end{align} $$

One can apply the integral system (3.19) to construct the iterative sequences. For any $(\tau ,y)\in [0,\widetilde {\delta }_1]\times \mathbb {R}$ , set

(3.21) $$ \begin{align} \widetilde{R}^{(0)}(\tau,y)= \widetilde{S}^{(0)}(\tau,y)=0, \quad t^{(0)}(\tau,y)=\frac{\tau}{2\overline{\theta}}, \end{align} $$

from which we define the characteristic curves $y=y_{i}^{(0)}(\tau )=:y_{i}^{(0)}(\tau ;\xi ,\eta )\ (i=\pm ,0)$ for $\tau \in [0,\xi ]$ as

(3.22) $$ \begin{align} \left\{ \begin{array}{l} \displaystyle \frac{\mathrm{d}y_{\pm}^{(0)}(\tau)}{\mathrm{ d}\tau}=\lambda_\pm(\tau,y,t^{(0)},\widetilde{R}^{(0)},\widetilde{S}^{(0)})(\tau,y_{\pm}^{(0)}(\tau)),\\[6pt] y_{\pm}^{(0)}(\xi)=\eta, \end{array} \right. \quad y_{0}^{(0)}(\tau)=\eta. \end{align} $$

Thus, the functions $(\widetilde {R}^{(1)}, \widetilde {S}^{(1)}, t^{(1)})(\tau , y)$ can be defined as

(3.23) $$ \begin{align} \left\{ \begin{array}{l} \displaystyle \widetilde{R}^{(1)}(\xi,\eta)=\int_{0}^\xi\bigg\{\frac{\widetilde{R}^{(0)}-\widetilde{S}^{(0)}}{2\tau} +A^{(0)}\cdot\frac{(\widetilde{R}^{(0)}-\widetilde{S}^{(0)})^2}{4\tau} \\[6pt] \displaystyle \quad -A^{(0)}(\widetilde{R}^{(0)}-\widetilde{S}^{(0)}) +A^{(0)}\theta_{0}'\tau-A^{(0)}w(t^{(0)},y)\tau \bigg\}(\tau, y_{-}^{(0)}(\tau))\ \mathrm{d}\tau,\\[6pt] \displaystyle \widetilde{S}^{(1)}(\xi,\eta)=\int_{0}^\xi\bigg\{\frac{\widetilde{S}^{(0)}-\widetilde{R}^{(0)}}{2\tau} +B^{(0)}\cdot\frac{(\widetilde{R}^{(0)}-\widetilde{S}^{(0)})^2}{4\tau} \\[6pt] \displaystyle \quad +B^{(0)} (\widetilde{R}^{(0)}-\widetilde{S}^{(0)}) -B^{(0)}\theta_{0}'\tau+B^{(0)}w(t^{(0)},y)\tau \bigg\}(\tau, y_{+}^{(0)}(\tau))\ \mathrm{d}\tau,\\[6pt] \displaystyle t^{(1)}(\xi,\eta)=\int_{0}^\xi C^{(0)}(\tau, y_{0}^{(0)}(\tau))\ \mathrm{d}\tau, \end{array} \right. \end{align} $$

where

$$ \begin{align*} &A^{(0)}=\frac{1}{f^{(0)}+\widetilde{S}^{(0)}},\ \ B^{(0)}=\frac{1}{f^{(0)}+\widetilde{R}^{(0)}},\\ &C^{(0)}=\frac{1}{\frac{\widetilde{R}^{(0)}+\widetilde{S}^{(0)}}{2}-\tau-\widetilde{v}(t^{(0)},y)+\theta_0(y)}, \end{align*} $$

and $f^{(0)}=\theta _0(y)-\widetilde {v}(t^{(0)},y)-\tau $ . After obtaining the functions $(\widetilde {R}^{(k)}, \widetilde {S}^{(k)}, t^{(k)})(\tau ,y)$ , one gets the characteristic curves $y=y_{i}^{(k)}(\tau )=:y_{i}^{(k)}(\tau ;\xi ,\eta )\ (i=\pm ,0)$ by solving the following ODE equations:

(3.24) $$ \begin{align} \left\{ \begin{array}{l} \displaystyle \frac{\mathrm{d}y_{\pm}^{(k)}(\tau)}{\mathrm{ d}\tau}=\lambda_\pm(\tau,y,t^{(k)},\widetilde{R}^{(k)},\widetilde{S}^{(k)})(\tau,y_{\pm}^{(k)}(\tau)),\\[6pt] y_{\pm}^{(k)}(\xi)=\eta, \end{array} \right. \quad y_{0}^{(k)}(\tau)=\eta. \end{align} $$

Hence, we determine the functions $(\widetilde {R}^{(k+1)}, \widetilde {S}^{(k+1)}, t^{(k+1)})(\tau ,y)$ by the following relations:

(3.25) $$ \begin{align} \left\{ \begin{array}{l} \displaystyle \widetilde{R}^{(k+1)}(\xi,\eta)=\int_{0}^\xi\bigg\{\frac{\widetilde{R}^{(k)}-\widetilde{S}^{(k)}}{2\tau} +A^{(k)}\cdot\frac{(\widetilde{R}^{(k)}-\widetilde{S}^{(k)})^2}{4\tau} \\[6pt] \displaystyle \ -A^{(k)}(\widetilde{R}^{(k)}-\widetilde{S}^{(k)}) +A^{(k)}\theta_{0}'\tau-A^{(k)}w(t^{(k)},y)\tau \bigg\}(\tau, y_{-}^{(k)}(\tau))\ \mathrm{d}\tau,\\\\[-6pt] \displaystyle \widetilde{S}^{(k+1)}(\xi,\eta)=\int_{0}^\xi\bigg\{\frac{\widetilde{S}^{(k)}-\widetilde{R}^{(k)}}{2\tau} +B^{(k)}\cdot\frac{(\widetilde{R}^{(k)}-\widetilde{S}^{(k)})^2}{4\tau} \\[6pt] \displaystyle \ +B^{(k)}(\widetilde{R}^{(k)}-\widetilde{S}^{(k)}) -B^{(k)}\theta_{0}'\tau+B^{(k)}w(t^{(k)},y)\tau \bigg\}(\tau, y_{+}^{(k)}(\tau))\ \mathrm{d}\tau,\\[6pt] \displaystyle t^{(k+1)}(\xi,\eta)=\int_{0}^\xi C^{(k)}(\tau, y_{0}^{(k)}(\tau))\ \mathrm{d}\tau, \end{array} \right. \end{align} $$

where

$$ \begin{align*} &A^{(k)}=\frac{1}{f^{(k)}+\widetilde{S}^{(k)}},\ \ B^{(k)}=\frac{1}{f^{(k)}+\widetilde{R}^{(k)}},\\ &C^{(k)}=\frac{1}{\frac{\widetilde{R}^{(k)}+\widetilde{S}^{(k)}}{2}-\tau-\widetilde{v}(t^{(k)},y)+\theta_0(y)}, \end{align*} $$

and $f^{(k)}=\theta _0(y)-\widetilde {v}(t^{(k)},y)-\tau $ .

We next choose the constants $\widetilde {M}$ and $\widetilde {\delta }\leq \widetilde {\delta }_1$ to verify the uniform convergence of the sequences $(\widetilde {R}^{(k)}, \widetilde {S}^{(k)}, t^{(k)})(\tau ,y)$ .

3.3 Several lemmas

It first follows by the detailed expressions of $A, B, C$ in (3.20) that

(3.26) $$ \begin{align} \begin{array}{l} \displaystyle A_{\widetilde{S}}=-A^2,\quad B_{\widetilde{R}}=-B^2,\quad C_{\widetilde{R}}=C_{\widetilde{S}}=-\frac{1}{2}C^2, \\ \displaystyle A_{t}=A^2\widetilde{v}_t(t, y), \quad A_{y}=-A^2[\theta_{0}'-w(t, y)], \quad B_{t}=B^2\widetilde{v}_t(t, y), \\ \displaystyle B_{y}=-B^2[\theta_{0}'-w(t, y)],\quad C_{t}=C^2\widetilde{v}_t(t,y), \quad C_{y}=C^2[w(t,y)-\theta_{0}']. \end{array} \end{align} $$

If $(\widetilde {R}, \widetilde {S}, t)(\tau ,y)\in \widetilde {\Sigma }(\widetilde {\delta }_1)$ , we have by (3.13)–(3.16) and (3.26)

(3.27) $$ \begin{align} \begin{array}{c} \displaystyle |A|,|B|\leq\frac{2}{\underline{\theta}},\quad \frac{1}{2\overline{\theta}}\leq C\leq\frac{2}{\underline{\theta}},\quad |A_{\widetilde{S}}|,|B_{\widetilde{R}}|,|C_{\widetilde{R}}|,|C_{\widetilde{S}}| \leq\bigg(\frac{2}{\underline{\theta}}\bigg)^2, \\[8pt] \displaystyle |A_t|,|B_t|,|C_t|\leq \bigg(\frac{2}{\underline{\theta}}\bigg)^2\widehat{M}_0,\ |A_y|,|B_y|,|C_y|\leq \bigg(\frac{2}{\underline{\theta}}\bigg)^2\bigg(|\theta_{0}'|+\frac{2}{\underline{\theta}}\widehat{M}_0\tau\bigg). \end{array} \end{align} $$

Denote $\overline {\theta }_0=\max \{\max _{z\in \mathbb {R}}|\theta _{0}'(z)|, \max _{z\in \mathbb {R}}|\theta _{0}"(z)|, \max _{z\in \mathbb {R}}|\theta _{0}"'(z)|\}$ and

(3.28) $$ \begin{align} \overline{K}=\max\bigg\{1, \frac{2}{\underline{\theta}}, \frac{4}{\underline{\theta}^2}, \frac{8}{\underline{\theta}^3}, \overline{\theta}_1, \frac{2\overline{\theta}_0}{\underline{\theta}}, \frac{4\overline{\theta}_0}{\underline{\theta}^2}, \frac{8\overline{\theta}_{0}^2}{\underline{\theta}^2}\bigg\}, \end{align} $$

which along with (3.27) give

(3.29) $$ \begin{align} \begin{array}{c} \displaystyle |A|,|B|,|A\theta_{0}'|,|B\theta_{0}'|,|A_{\widetilde{S}}|,|B_{\widetilde{R}}|, |C_{\widetilde{R}}|,|C_{\widetilde{S}}| \leq \overline{K}, \\ \displaystyle |A_t|,|B_t|,|C_t| \leq \overline{K}\widehat{M}_0, \quad |Aw|, |Bw|\leq \overline{K}\widehat{M}_0\tau,\\ |A_y|,|B_y|,|C_y|, |(A\theta_{0}')_y|,|(B\theta_{0}')_y|\leq \overline{K}(1+\widehat{M}_0\tau). \end{array} \end{align} $$

We now choose $\widetilde {M}, \overline {M}$ , and $\widetilde {\delta }$ satisfying

(3.30) $$ \begin{align} \widetilde{M}=3\overline{M},\quad \overline{M}=16\overline{K}, \quad \widetilde{\delta}=\min\bigg\{ \frac{1}{16\overline{M}}, \frac{1}{16\overline{K}\widehat{M}_{0}}, \frac{1}{\widehat{M}_{0}^2}\bigg\}\leq\widetilde{\delta}_1, \end{align} $$

such that there hold

(3.31) $$ \begin{align} \begin{array}{l} \displaystyle \overline{M}\widetilde{\delta}\leq\frac{1}{16},\quad \overline{K}\widehat{M}_{0}\widetilde{\delta}\leq\frac{1}{16},\quad \widehat{M}_0\sqrt{\widetilde{\delta}}\leq1,\quad e^{\overline{M}\widetilde{\delta}^2}\leq \exp\bigg\{\frac{1}{64}\bigg\}\leq\frac{16}{15}. \end{array} \end{align} $$

Hence, if $(\widetilde {R}, \widetilde {S}, t)(\tau ,y)\in \widetilde {\Sigma }(\widetilde {\delta })$ , one obtains

(3.32) $$ \begin{align} \begin{array}{r} \displaystyle \frac{1}{2\overline{\theta}}\leq C\leq\frac{2}{\underline{\theta}}, \quad \widetilde{\delta}|A_t|,\widetilde{\delta}|B_t|,\widetilde{\delta}|C_t|\leq \frac{1}{16}, \quad |Aw|, |Bw|\leq \overline{K}\leq \frac{\overline{M}}{16},\\[8pt] \displaystyle |A|,|B|,|A\theta_{0}'|,|B\theta_{0}'|,|A_{\widetilde{S}}|,|B_{\widetilde{R}}|, |C_{\widetilde{R}}|,|C_{\widetilde{S}}| \leq \overline{K}\leq \frac{\overline{M}}{16}, \\[8pt] \displaystyle |A_y|,|B_y|,|C_y|, |(A\theta_{0}')_y|,|(B\theta_{0}')_y|\leq 2\overline{K}\leq \frac{\overline{M}}{8}. \end{array} \end{align} $$

Thanks to (3.31) and (3.32), we have the following lemma.

Lemma 3.1 Let the sequences $(\widetilde {R}^{(k)}, \widetilde {S}^{(k)}, t^{(k)})$ be defined in (3.25). For all $k\geq 1$ , the following inequalities

(3.33) $$ \begin{align} \begin{array}{l} \big|\widetilde{R}^{(k)}(\xi,\eta)\big|, \big|\widetilde{S}^{(k)}(\xi,\eta)\big|\leq \overline{M}\xi^2\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j,\ \ \frac{\xi}{2\overline{\theta}}\leq t^{(k)}(\xi,\eta)\leq \frac{2\xi}{\underline{\theta}}, \\ \big|\widetilde{R}^{(k)}(\xi,\eta)-\widetilde{S}^{(k)}(\xi,\eta)\big| \leq \overline{M}\xi^2\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j \end{array} \end{align} $$

hold in $[0,\widetilde {\delta }]\times \mathbb {R}$ .

Proof We show this lemma by the standard argument of induction. That is, we first check that each inequality in (3.33) is true for $n=1$ , then assume that they all hold for $n=k$ and establish (3.33) for $n=k+1$ .

Obviously, the functions $(\widetilde {R}^{(0)},\widetilde {S}^{(0)},t^{(0)})(\xi ,\eta )$ defined in (3.21) are in $\widetilde {\Sigma }(\widetilde {\delta })$ , then we find by (3.32) that

(3.34) $$ \begin{align} \frac{1}{2\overline{\theta}}\leq C^{(0)}\leq\frac{2}{\underline{\theta}},\quad |I^{(0)}|,|I^{(0)}\theta_{0}'|,|I^{(0)}w(t^{(0)},y)|\leq \frac{\overline{M}}{16}, \end{align} $$

for $I=A,B$ . It concludes by (3.23) and (3.34) that

(3.35) $$ \begin{align} \big|\widetilde{R}^{(1)}(\xi,\eta)\big|&\leq \int_{0}^\xi\bigg\{ |A^{(0)}\theta_{0}'|\tau+|A^{(0)}w(t^{(0)},y)|\tau \bigg\}\ \mathrm{d}\tau \nonumber \\ &\leq \int_{0}^\xi\frac{\overline{M}}{8}\tau\ \mathrm{d}\tau=\frac{\overline{M}}{16}\xi^2\leq \overline{M}\xi^2\displaystyle\sum_{j=0}^1\bigg(\frac{2}{3}\bigg)^j. \end{align} $$

The estimate (3.35) is also true for $\widetilde {S}^{(1)}(\xi ,\eta )$ . For the variable $t^{(1)}(\xi ,\eta )$ , one arrives at

(3.36) $$ \begin{align} \frac{1}{2\overline{\theta}}\xi\leq t^{(1)}(\xi,\eta)=\int_{0}^\xi C^{(0)}(\tau, y_{0}^{(0)}(\tau))\ \mathrm{ d}\tau\leq \frac{2}{\underline{\theta}}\xi. \end{align} $$

Furthermore, applying (3.23) and (3.34) again acquires

(3.37) $$ \begin{align} &\big|\widetilde{R}^{(1)}(\xi,\eta)-\widetilde{S}^{(1)}(\xi,\eta)\big| \nonumber \\ &\leq \int_{0}^\xi\bigg\{ |A^{(0)}\theta_{0}'|\tau+|A^{(0)}w(t^{(0)},y)|\tau + |B^{(0)}\theta_{0}'|\tau+|B^{(0)}w(t^{(0)},y)|\tau \bigg\}\ \mathrm{d}\tau \nonumber \\ &\leq \int_{0}^\xi\frac{\overline{M}}{4}\tau \ \mathrm{d}\tau =\frac{\overline{M}}{8}\xi^2\leq \overline{M}\xi^2\displaystyle\sum_{j=0}^1\bigg(\frac{2}{3}\bigg)^j. \end{align} $$

We combine (3.35)–(3.37) to achieve (3.33) for $n=1$ . In addition, the functions $(\widetilde {R}^{(1)}, \widetilde {S}^{(1)}, t^{(1)})(\xi ,\eta )$ satisfy

(3.38) $$ \begin{align} \big|\widetilde{R}^{(1)}(\xi,\eta)\big|, \big|\widetilde{S}^{(1)}(\xi,\eta)\big|\leq 3\overline{M}\xi^2=\widetilde{M}\xi^2,\quad \frac{1}{2\overline{\theta}}\xi\leq t^{(1)}(\xi,\eta)\leq \frac{2}{\underline{\theta}}\xi, \end{align} $$

which means that $(\widetilde {R}^{(1)}, \widetilde {S}^{(1)}, t^{(1)})(\xi ,\eta )\in \widetilde {\Sigma }(\widetilde {\delta })$ .

Suppose that all inequalities in (3.33) hold for $n=k$ . Thus,

(3.39) $$ \begin{align} \big|\widetilde{R}^{(k)}(\xi,\eta)\big|, \big|\widetilde{S}^{(k)}(\xi,\eta)\big|\leq \overline{M}\xi^2\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j\leq 3\overline{M}\xi^2=\widetilde{M}\xi^2, \end{align} $$

from which we see that $(\widetilde {R}^{(k)}, \widetilde {S}^{(k)}, t^{(k)})(\xi ,\eta )\in \widetilde {\Sigma }(\widetilde {\delta })$ . Therefore, one has by (3.32)

(3.40) $$ \begin{align} \frac{1}{2\overline{\theta}}\leq C^{(k)}\leq\frac{2}{\underline{\theta}},\quad |I^{(k)}|,|I^{(k)}\theta_{0}'|,|I^{(k)}w(t^{(k)},y)|\leq \frac{\overline{M}}{16}, \end{align} $$

for $I=A,B$ . In view of (3.25) and (3.40), we employ the induction assumptions to obtain

(3.41) $$ \begin{align} \big|\widetilde{R}^{(k+1)}(\xi,\eta)\big|&\leq \int_{0}^\xi\bigg\{\frac{1}{2}\overline{M}\tau \displaystyle\sum_{j=0}^k\bigg(\frac{2}{3}\bigg)^j +\frac{\overline{M}}{16}\cdot\frac{1}{4}\bigg(\overline{M}\displaystyle\sum_{j=0}^k\bigg(\frac{2}{3}\bigg)^j\bigg)^2\tau^3 \nonumber \\ \displaystyle & \qquad \quad +\frac{\overline{M}}{16}\cdot\overline{M}\tau^2 \displaystyle\sum_{j=0}^k\bigg(\frac{2}{3}\bigg)^j +\frac{\overline{M}}{16}\tau +\frac{\overline{M}}{16}\tau \bigg\}\ \mathrm{d}\tau \nonumber \\ &\leq \overline{M}\xi^2\bigg\{\frac{1}{16} +\bigg(\frac{1}{4}+\frac{3(\overline{M}\widetilde{\delta})^2}{256} +\frac{\overline{M}\widetilde{\delta}}{48}\bigg)\displaystyle\sum_{j=0}^k\bigg(\frac{2}{3}\bigg)^j\bigg\} \nonumber \\ &\leq \overline{M}\xi^2\bigg\{1 +\frac{2}{3}\cdot\displaystyle\sum_{j=0}^k\bigg(\frac{2}{3}\bigg)^j\bigg\} =\overline{M}\xi^2\displaystyle\sum_{j=0}^{k+1}\bigg(\frac{2}{3}\bigg)^j. \end{align} $$

The above estimate (3.41) also holds for $\widetilde {S}^{(k+1)}(\xi ,\eta )$ . Moreover, it is easy to find by (3.40) that

(3.42) $$ \begin{align} \frac{1}{2\overline{\theta}}\xi\leq t^{(k+1)}(\xi,\eta)=\int_{0}^\xi C^{(k)}(\tau, y_{0}^{(k)}(\tau))\ \mathrm{ d}\tau\leq \frac{2}{\underline{\theta}}\xi. \end{align} $$

For the term $|\widetilde {R}^{(k+1)}(\xi ,\eta )-\widetilde {S}^{(k+1)}(\xi ,\eta )|$ , we proceed by using (3.25) and (3.40) again

(3.43) $$ \begin{align} & \big|\widetilde{R}^{(k+1)}(\xi,\eta)- \widetilde{S}^{(k+1)}(\xi,\eta)\big| \nonumber \\ &\leq \int_{0}^\xi\bigg\{\overline{M}\tau \displaystyle\sum_{j=0}^k\bigg(\frac{2}{3}\bigg)^j +\frac{\overline{M}}{16}\cdot\frac{1}{2}\bigg(\overline{M}\displaystyle\sum_{j=0}^k\bigg(\frac{2}{3}\bigg)^j\bigg)^2\tau^3 \nonumber \\ \displaystyle & \qquad +\frac{\overline{M}}{16}\cdot\overline{M}\tau^2 \displaystyle\sum_{j=0}^k\bigg(\frac{2}{3}\bigg)^j +\frac{\overline{M}}{4}\tau \bigg\}\ \mathrm{d}\tau \nonumber \\ &\leq \overline{M}\xi^2\bigg\{\frac{1}{8} +\bigg(\frac{1}{2}+\frac{3}{128} +\frac{1}{48}\bigg)\displaystyle\sum_{j=0}^k\bigg(\frac{2}{3}\bigg)^j\bigg\} \leq \overline{M}\xi^2\displaystyle\sum_{j=0}^{k+1}\bigg(\frac{2}{3}\bigg)^j. \end{align} $$

Combining (3.41)–(3.43) ends the proof of the lemma.

By virtue of Lemma 3.1, the functions $(\widetilde {R}^{(k)}, \widetilde {S}^{(k)}, t^{(k)})(\xi ,\eta )$ are in the space $\widetilde {\Sigma }(\widetilde {\delta })$ for each $k\geq 0$ . Thus, the estimates in (3.32) are true for the functions $(\widetilde {R}^{(k)}, \widetilde {S}^{(k)}, t^{(k)})\ (k\geq 0)$ . To derive the uniform convergence of the iterative sequences $(\widetilde {R}^{(k)}, \widetilde {S}^{(k)}, t^{(k)})$ , one needs the properties of the sequences $(\widetilde {R}_{\eta }^{(k)}, \widetilde {S}_{\eta }^{(k)}, t_{\eta }^{(k)})(\xi ,\eta )$ . We differentiate system (3.25) with respect to $\eta $ to calculate

(EQ.a) $$ \begin{align} \displaystyle &\widetilde{R}_{\eta}^{(k+1)}(\xi,\eta)=\int_{0}^\xi\bigg\{\frac{\widetilde{R}_{y}^{(k)}-\widetilde{S}_{y}^{(k)}} {2\tau} + A^{(k)}\cdot\frac{(\widetilde{R}^{(k)}-\widetilde{S}^{(k)}) (\widetilde{R}_{y}^{(k)}-\widetilde{S}_{y}^{(k)})}{2\tau} \nonumber \\ \displaystyle &\ -A^{(k)}(\widetilde{R}_{y}^{(k)}-\widetilde{S}_{y}^{(k)}) +A_{11}^{(k)}A_{\widetilde{S}}^{(k)}\widetilde{S}_{y}^{(k)} +[A_{11}^{(k)}A_{t}^{(k)}-A^{(k)}w_t\tau]t_{y}^{(k)} \nonumber \\ \displaystyle & \ + A_{11}^{(k)}A_{y}^{(k)} +A^{(k)}(\theta_{0}"-w_y)\tau \bigg\}\frac{\partial y_{-}^{(k)}(\tau)}{\partial \eta}(\tau, y_{-}^{(k)}(\tau))\ \mathrm{d}\tau, \end{align} $$

where

(3.44) $$ \begin{align} \begin{array}{l} \displaystyle A_{11}^{(k)}=\frac{(\widetilde{R}^{(k)}-\widetilde{S}^{(k)})^2}{4\tau} -(\widetilde{R}^{(k)}-\widetilde{S}^{(k)}) +\theta_{0}'\tau-w(t^{(k)},y)\tau,\\[6pt] A_{\widetilde{S}}^{(k)}=-(A^{(k)})^2,\quad A_{t}^{(k)}=(A^{(k)})^2\widetilde{v}_t(t^{(k)}, y), \\ \displaystyle A_{y}^{(k)}=-(A^{(k)})^2[\theta_{0}'-w(t^{(k)}, y)], \end{array} \end{align} $$

(EQ.b) $$ \begin{align} \displaystyle &\widetilde{S}_{\eta}^{(k+1)}(\xi,\eta)=\int_{0}^\xi\bigg\{\frac{\widetilde{S}_{y}^{(k)}-\widetilde{R}_{y}^{(k)}} {2\tau} + B^{(k)}\cdot\frac{(\widetilde{R}^{(k)}-\widetilde{S}^{(k)})(\widetilde{R}_{y}^{(k)}-\widetilde{S}_{y}^{(k)})}{2\tau} \nonumber \\ \displaystyle &\ +B^{(k)}(\widetilde{R}_{y}^{(k)}-\widetilde{S}_{y}^{(k)}) +B_{11}^{(k)}B_{\widetilde{R}}^{(k)}\widetilde{R}_{y}^{(k)} +[B_{11}^{(k)}A_{t}^{(k)}+B^{(k)}w_t\tau]t_{y}^{(k)} \nonumber \\ &\ \displaystyle + B_{11}^{(k)}B_{y}^{(k)} -B^{(k)}(\theta_{0}"-w_y)\tau \bigg\}\frac{\partial y_{+}^{(k)}(\tau)}{\partial \eta}(\tau, y_{+}^{(k)}(\tau))\ \mathrm{d}\tau, \end{align} $$

where

(3.45) $$ \begin{align} \begin{array}{l} \displaystyle B_{11}^{(k)}=\frac{(\widetilde{R}^{(k)}-\widetilde{S}^{(k)})^2}{4\tau} +(\widetilde{R}^{(k)}-\widetilde{S}^{(k)}) -\theta_{0}'\tau+w(t^{(k)},y)\tau,\\[6pt] B_{\widetilde{R}}^{(k)}=-(B^{(k)})^2,\quad B_{t}^{(k)}=(B^{(k)})^2\widetilde{v}_t(t^{(k)}, y), \\ \displaystyle B_{y}^{(k)}=-(B^{(k)})^2[\theta_{0}'-w(t^{(k)}, y)], \end{array} \end{align} $$

and

(EQ.c) $$ \begin{align} t_{\eta}^{(k+1)}(\xi,\eta)=\int_{0}^\xi \bigg\{ C_{\widetilde{R}}^{(k)}\widetilde{R}_{y}^{(k)} + C_{\widetilde{S}}^{(k)}\widetilde{S}_{y}^{(k)} +C_{t}^{(k)}t_{y}^{(k)} + C_{y}^{(k)} \bigg\}(\tau, \eta)\ \mathrm{ d}\tau, \end{align} $$

where

(3.46) $$ \begin{align} \begin{array}{l} \displaystyle C_{\widetilde{R}}^{(k)}=C_{\widetilde{S}}^{(k)}=-\frac{1}{2}(C^{(k)})^2, \\[6pt] \displaystyle C_{t}^{(k)}=(C^{(k)})^2\widetilde{v}_t(t^{(k)},y), \quad C_{y}^{(k)}=(C^{(k)})^2[w(t^{(k)},y)-\theta_{0}']. \end{array} \end{align} $$

The functions $\frac {\partial y_{\pm }^{(k)}(\tau )}{\partial \eta }$ in (EQ.a) and (EQ.b) are

(3.47) $$ \begin{align} \frac{\partial y^{(k)}_\pm}{\partial \eta}(\tau)=\exp\bigg\{\int_{\xi}^\tau\frac{\partial \lambda_{\pm}^{(k)}}{\partial y}(s,y_{\pm}^{(k)}(s))\ \mathrm{d}s\bigg\}, \end{align} $$

where

(3.48) $$ \begin{align} \begin{array}{l} \displaystyle \frac{\partial \lambda_{-}^{(k)}}{\partial y}=-s\big[A_{\widetilde{S}}^{(k)}\widetilde{S}_{y}^{(k)} +A_{t}^{(k)}t_{y}^{(k)} + A_{y}^{(k)}\big], \\[8pt] \displaystyle \frac{\partial \lambda_{+}^{(k)}}{\partial y}=s\big[B_{\widetilde{R}}^{(k)}\widetilde{R}_{y}^{(k)} +B_{t}^{(k)}t_{y}^{(k)} + B_{y}^{(k)}\big]. \end{array} \end{align} $$

Due to $(\widetilde {R}^{(k)}, \widetilde {S}^{(k)}, t^{(k)})\in \widetilde {\Sigma }(\widetilde {\delta })$ , for $k\geq 1$ , it suggests by Lemma 3.1 and (3.27), (3.30), and (3.32) that

(3.49) $$ \begin{align} &|A_{11}^{(k)}|,|B_{11}^{(k)}|\leq\frac{9}{4}\overline{M}^2\tau^3+\frac{3}{2}\overline{M}\tau^2 +(|\theta_{0}'|+|w|)\tau \nonumber \\ &\leq 4\overline{M}\tau^2 +\bigg(|\theta_{0}'|+\frac{2}{\underline{\theta}}\widehat{M}_0\widetilde{\delta}\bigg)\tau, \end{align} $$

and then

(3.50) $$ \begin{align} \begin{array}{l} \displaystyle |A_{11}^{(k)}A_{\widetilde{S}}^{(k)}|\leq 4\overline{M}\tau^2 \cdot\frac{\overline{M}}{16} +\bigg(\frac{2}{\underline{\theta}}\bigg)^2\cdot \bigg(|\theta_{0}'|+\frac{2}{\underline{\theta}}\widehat{M}_0\widetilde{\delta}\bigg)\tau \leq \frac{\overline{M}^2}{4}\tau^2 +\frac{\overline{M}}{8}\tau, \\[6pt] \displaystyle |A_{11}^{(k)}A_{t}^{(k)}-A^{(k)}w_t\tau|\leq \bigg\{4\overline{M}\tau^2 +\bigg(|\theta_{0}'|+\frac{2}{\underline{\theta}}\widehat{M}_0\widetilde{\delta}\bigg)\tau\bigg\} \cdot\bigg(\frac{2}{\underline{\theta}}\bigg)^2\widehat{M}_{0} +\frac{2}{\underline{\theta}}\widehat{M}_{0}\tau \\[6pt] \displaystyle \leq \widehat{M}_{0}\tau\bigg\{\frac{2}{\underline{\theta}} +\frac{16\overline{M}\tau}{\underline{\theta}^2} + \bigg(\frac{2}{\underline{\theta}}\bigg)^2 \bigg(|\theta_{0}'|+\frac{2}{\underline{\theta}}\widehat{M}_0\widetilde{\delta}\bigg)\bigg\}\leq 4\overline{K}\widehat{M}_{0}\widetilde{\delta}\leq\frac{1}{4}, \\[6pt] \displaystyle |A_{11}^{(k)}A_{y}^{(k)}|\leq \bigg\{4\overline{M}\tau^2 +\bigg(|\theta_{0}'|+\frac{2}{\underline{\theta}}\widehat{M}_0\widetilde{\delta}\bigg)\tau\bigg\} \bigg(\frac{2}{\underline{\theta}}\bigg)^2 \bigg(|\theta_{0}'|+\frac{2}{\underline{\theta}}\widehat{M}_0\widetilde{\delta}\bigg) \\[8pt] \displaystyle \leq 4\overline{M}\tau^2\cdot\frac{\overline{M}}{8} +\frac{\overline{M}}{8}\tau\leq \frac{\overline{M}}{4}\tau, \\[6pt] \displaystyle |A^{(k)}(\theta_{0}"-w_y)|\leq \frac{2}{\underline{\theta}}\bigg(|\theta_{0}"| +\widehat{M}_{0}\sqrt{\widetilde{\delta}}\bigg)\leq \frac{\overline{M}}{8}. \end{array} \end{align} $$

We obtain analogously

(3.51) $$ \begin{align} \begin{array}{l} \displaystyle |B_{11}^{(k)}B_{\widetilde{R}}^{(k)}|\leq \frac{\overline{M}^2}{4}\tau^2 +\frac{\overline{M}}{8}\tau, \quad |B_{11}^{(k)}B_{t}^{(k)}+B^{(k)}w_t\tau|\leq \frac{1}{4}, \\[8pt] \displaystyle |B_{11}^{(k)}B_{y}^{(k)}|\leq \frac{\overline{M}}{4}\tau, \qquad \qquad \quad |B^{(k)}(\theta_{0}"-w_y)|\leq \frac{\overline{M}}{8}. \end{array} \end{align} $$

Furthermore, there also hold

(3.52) $$ \begin{align} |C_{\widetilde{R}}^{(k)}|,|C_{\widetilde{S}}^{(k)}|\leq\frac{\overline{M}}{16}, \quad |C_{t}^{(k)}|\leq \overline{K}\widehat{M}_0, \quad |C_{y}^{(k)}|\leq \frac{\overline{M}}{8}, \end{align} $$

and

(3.53) $$ \begin{align} \bigg|\frac{\partial \lambda_{\pm}^{(k)}}{\partial y}\bigg|\leq s\bigg\{\frac{\overline{M}}{16}(|\widetilde{R}_{y}^{(k)}|+|\widetilde{S}_{y}^{(k)}|) +\overline{K}\widehat{M}_{0}|t_{y}^{(k)}| +\frac{\overline{M}}{8} \bigg\}. \end{align} $$

For the sequences $(\widetilde {R}_{\eta }^{(k)}, \widetilde {S}_{\eta }^{(k)}, t_{\eta }^{(k)})(\xi ,\eta )$ , we have the following lemma.

Lemma 3.2 For all $k\geq 1$ , the following inequalities

(3.54) $$ \begin{align} &\big|\widetilde{R}_{\eta}^{(k)}(\xi,\eta)\big|, \big|\widetilde{S}_{\eta}^{(k)}(\xi,\eta)\big|\leq \overline{M}\xi^2\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j,\ \big|t_{\eta}^{(k)}(\xi,\eta)\big|\leq \overline{M}\xi\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j, \nonumber \\ &\big|\widetilde{R}_{\eta}^{(k)}(\xi,\eta)-\widetilde{S}_{\eta}^{(k)}(\xi,\eta)\big| \leq \overline{M}\xi^2\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j \end{align} $$

hold in $[0,\widetilde {\delta }]\times \mathbb {R}$ .

Proof The proof is also based on the standard argument of induction. According to

$$ \begin{align*}(\widetilde{R}^{(0)},\widetilde{S}^{(0)},t^{(0)})(\xi,\eta) \in\widetilde{\Sigma}(\widetilde{\delta}), \end{align*} $$

we get by (3.53) and (3.47)

(3.55) $$ \begin{align} \bigg|\frac{\partial \lambda_{\pm}^{(0)}}{\partial y}\bigg|\leq s\cdot\frac{\overline{M}}{8},\quad \bigg|\frac{\partial y^{(0)}_\pm}{\partial \eta}(\tau)\bigg|\leq\exp\bigg\{\int_{0}^\xi\frac{\overline{M}}{8}s\ \mathrm{ d}s\bigg\}\leq\exp\bigg(\frac{\overline{M}\xi^2}{16}\bigg). \end{align} $$

One combines (EQ.a), (3.50), and (3.55) to acquire

(3.56) $$ \begin{align} &\big|\widetilde{R}_{\eta}^{(1)}(\xi,\eta)\big|\leq\int_{0}^\xi\bigg\{ | A_{11}^{(0)}A_{y}^{(0)}| +|A^{(0)}(\theta_{0}"-w_y)|\tau \bigg\}\bigg|\frac{\partial y_{-}^{(0)}(\tau)}{\partial \eta}\bigg|\ \mathrm{ d}\tau \nonumber \\ &\leq \int_{0}^\xi\bigg\{ \frac{\overline{M}}{4}\tau +\frac{\overline{M}}{8}\tau \bigg\}\exp\bigg(\frac{\overline{M}\xi^2}{16}\bigg)\ \mathrm{d}\tau\leq \overline{M}\xi^2\cdot\frac{3}{16}\exp\bigg(\frac{\overline{M}\widetilde{\delta}^2}{16}\bigg) \nonumber \\ &\leq \overline{M}\xi^2\displaystyle\sum^{1}_{j=0} \bigg(\frac{2}{3}\bigg)^j. \end{align} $$

The estimate in (3.56) is also valid for the function $\widetilde {S}_{\eta }^{(1)}(\xi ,\eta )$ . For the function $t_{\eta }^{(1)}(\xi ,\eta )$ , it concludes by (EQ.c) and (3.52) that

(3.57) $$ \begin{align} \big|t_{\eta}^{(1)}(\xi,\eta)\big|\leq\int_{0}^\xi |C_{y}^{(0)}|(\tau, \eta)\ \mathrm{d}\tau\leq \int_{0}^\xi \frac{\overline{M}}{8}\ \mathrm{d}\tau \leq \overline{M}\xi\displaystyle\sum^{1}_{j=0} \bigg(\frac{2}{3}\bigg)^j. \end{align} $$

Moreover, for the term $|\widetilde {R}_{\eta }^{(1)}(\xi ,\eta )-\widetilde {S}_{\eta }^{(1)}(\xi ,\eta )|$ , it is easy to check that

(3.58) $$ \begin{align} \big|\widetilde{R}_{\eta}^{(1)}(\xi,\eta)-\widetilde{S}_{\eta}^{(1)}(\xi,\eta)\big|\leq \overline{M}\xi^2\cdot\frac{3}{8}\exp\bigg(\frac{\overline{M}\widetilde{\delta}^2}{16}\bigg) \leq \overline{M}\xi^2\displaystyle\sum^{1}_{j=0} \bigg(\frac{2}{3}\bigg)^j, \end{align} $$

which along with (3.56) and (3.57) give (3.54) for $n=1$ .

Let all the inequalities in (3.54) hold for $n=k$ . Then

(3.59) $$ \begin{align} |\widetilde{R}_{y}^{(k)}|,|\widetilde{S}_{y}^{(k)}|\leq 3\overline{M}\xi^2,\quad |t_{y}^{(k)}|\leq3\overline{M}\xi. \end{align} $$

Combining (3.53), (3.47), and (3.59) yields

(3.60) $$ \begin{align} \bigg|\frac{\partial \lambda_{\pm}^{(k)}}{\partial y}\bigg|\leq s\bigg\{\frac{\overline{M}}{16}\cdot6\overline{M}s^2 +\overline{K}\widehat{M}_{0}\cdot3\overline{M}s +\frac{\overline{M}}{8} \bigg\} \end{align} $$

and then

(3.61) $$ \begin{align} \bigg|\frac{\partial y^{(k)}_\pm}{\partial \eta}(\tau)\bigg|&\leq \exp\bigg\{\int_{0}^\xi s\bigg(\frac{\overline{M}}{16}\cdot6\overline{M}s^2 +\overline{K}\widehat{M}_{0}\cdot3\overline{M}s +\frac{\overline{M}}{8} \bigg)\ \mathrm{d}s\bigg\} \nonumber \\ &\leq \exp\bigg\{\overline{M}\widetilde{\delta}^2\bigg(\frac{\overline{M}\widetilde{\delta}^2}{8} +\overline{K}\widehat{M}_0\widetilde{\delta} +\frac{1}{16}\bigg)\bigg\}\leq e^{\overline{M}\widetilde{\delta}^2}. \end{align} $$

Utilizing the induction assumptions, we have by (EQ.a), (3.50), and (3.61)

(3.62) $$ \begin{align} &\big|\widetilde{R}_{\eta}^{(k+1)}(\xi,\eta)\big|\leq \int_{0}^\xi\bigg\{\frac{1}{2}\overline{M}\tau\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j +\frac{\overline{M}}{16}\cdot\frac{3\overline{M}\tau}{2}\cdot\overline{M}\tau^2\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j \nonumber \\ &\quad +\frac{\overline{M}}{16}\cdot \overline{M}\tau^2\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j+\bigg(\frac{\overline{M}^2\tau^2}{4} +\frac{\overline{M}\tau}{8}\bigg)\cdot \overline{M}\tau^2\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j \nonumber \\ &\quad +\frac{1}{4}\cdot \overline{M}\tau \displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j +\frac{\overline{M}}{4}\tau +\frac{\overline{M}}{8}\tau \bigg\}e^{\overline{M}\widetilde{\delta}^2}\ \mathrm{ d}\tau \nonumber \\ &\leq \overline{M}\xi^2\bigg\{ \frac{3}{16} +\bigg(\frac{1}{4} +\frac{(\overline{M}\widetilde{\delta})^2}{32} +\frac{\overline{M}\widetilde{\delta}}{32} +\bigg(\frac{(\overline{M}\widetilde{\delta})^2}{4}+\frac{\overline{M}\widetilde{\delta}}{8}\bigg) \cdot\frac{\widetilde{\delta}}{3} +\frac{1}{8} \bigg)\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j\bigg\}e^{\overline{M}\widetilde{\delta}^2} \nonumber \\ &\leq \overline{M}\xi^2\bigg\{ \frac{15}{32} +\bigg[\frac{1}{4} +\frac{(\overline{M}\widetilde{\delta})^2}{32} +\frac{\overline{M}\widetilde{\delta}}{32} +\bigg(\frac{(\overline{M}\widetilde{\delta})^2}{4}+\frac{\overline{M}\widetilde{\delta}}{8}\bigg) \cdot\frac{\widetilde{\delta}}{3} +\frac{1}{32} \bigg]\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j\bigg\}e^{\overline{M}\widetilde{\delta}^2} \nonumber \\ &\leq \overline{M}\xi^2\bigg\{ \frac{15}{32} +\bigg[\frac{9}{32} +\frac{1}{32\times16} +\frac{1}{32\times4} +\bigg(\frac{1}{4\times16}+\frac{1}{8\times4}\bigg) \cdot\frac{1}{3} \bigg]\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j\bigg\}e^{\overline{M}\widetilde{\delta}^2}\nonumber \\ &\leq \overline{M}\xi^2\bigg\{ \frac{15}{32}e^{\overline{M}\widetilde{\delta}^2} +\bigg(\frac{5}{16}e^{\overline{M}\widetilde{\delta}^2}\bigg)\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j\bigg\} \leq \overline{M}\xi^2\displaystyle\sum^{k+1}_{j=0} \bigg(\frac{2}{3}\bigg)^j. \end{align} $$

The above estimate also holds for the function $\widetilde {S}_{\eta }^{(k+1)}(\xi ,\eta )$ . For the term $|\widetilde {R}_{\eta }^{(k+1)}(\xi ,\eta )-\widetilde {S}_{\eta }^{(k+1)}(\xi ,\eta )|$ , we can perform the process as in (3.62) to obtain

(3.63) $$ \begin{align} \big|\widetilde{R}_{\eta}^{(k+1)}(\xi,\eta)- \widetilde{S}_{\eta}^{(k+1)}(\xi,\eta)\big| &\leq \overline{M}\xi^2\bigg\{ \frac{15}{16}e^{\overline{M}\widetilde{\delta}^2} +\frac{5}{8}e^{\overline{M}\widetilde{\delta}^2}\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j\bigg\}, \end{align} $$

which, together with the fact $15e^{\overline {M}\widetilde {\delta }^2}\leq 16$ by (3.31), leads to

(3.64) $$ \begin{align} \big|\widetilde{R}_{\eta}^{(k+1)}(\xi,\eta)- \widetilde{S}_{\eta}^{(k+1)}(\xi,\eta)\big| \leq \overline{M}\xi^2\displaystyle\sum^{k+1}_{j=0} \bigg(\frac{2}{3}\bigg)^j. \end{align} $$

For the function $t_{\eta }^{(k+1)}(\xi ,\eta )$ , it follows by (EQ.c), (3.52), and the induction assumptions that

(3.65) $$ \begin{align} \big|t_{\eta}^{(k+1)}(\xi,\eta)\big|&\leq \int_{0}^\xi \bigg\{ \frac{\overline{M}}{16}\cdot \overline{M}\tau^2\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j \times 2 +\overline{K}\widehat{M}_0\cdot \overline{M}\tau \displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j +\frac{\overline{M}}{8} \bigg\} \ \mathrm{d}\tau \nonumber \\ &= \overline{M}\xi\bigg\{\frac{1}{8} +\bigg(\frac{\overline{M}\widetilde{\delta}^2}{24} +\frac{\overline{K}\widehat{M}_0\widetilde{\delta}}{2}\bigg)\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j \bigg\} \nonumber \\ &\leq \overline{M}\xi\bigg\{1 +\bigg(\frac{1}{24} +\frac{1}{2}\bigg)\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j \bigg\} \leq \overline{M}\xi\displaystyle\sum^{k+1}_{j=0} \bigg(\frac{2}{3}\bigg)^j. \end{align} $$

We combine (3.62), (3.64), and (3.65) to complete the proof of the lemma.

By means of Lemmas 3.1 and 3.2, one achieves the following lemma.

Lemma 3.3 For all $k\geq 0$ , the following inequalities

(3.66) $$ \begin{align} \begin{array}{r} \displaystyle \big|\widetilde{R}^{(k+1)}(\xi,\eta)-\widetilde{R}^{(k)}(\xi,\eta)\big| \leq \overline{M}\xi^2\displaystyle\bigg(\frac{2}{3}\bigg)^k, \\[8pt] \displaystyle \big|\widetilde{S}^{(k+1)}(\xi,\eta)-\widetilde{S}^{(k)}(\xi,\eta)\big| \leq \overline{M}\xi^2\displaystyle\bigg(\frac{2}{3}\bigg)^k, \\[8pt] \displaystyle \big|t^{(k+1)}(\xi,\eta)-t^{(k)}(\xi,\eta)\big| \leq \overline{M}\xi\displaystyle\bigg(\frac{2}{3}\bigg)^k \end{array} \end{align} $$

hold in $[0,\widetilde {\delta }]\times \mathbb {R}$ .

Proof We use again the argument of induction to verify the lemma. It is obvious to see by the functions $(\widetilde {R}^{(0)},\widetilde {S}^{(0)},t^{(0)})(\xi ,\eta )$ in (3.21) and (3.33) that the inequalities in (3.66) are valid for $n=1$ . Suppose that each inequality in (3.66) holds for $n\leq k-1$ . We shall show that they are true for $n=k$ .

We first estimate the term $|t^{(k+1)}(\xi ,\eta )-t^{(k)}(\xi ,\eta )|$ . In view of (3.25) and (3.32), we apply the induction assumptions and the fact $y_{0}^{k}(\tau )\equiv \eta $ to find that

(3.67) $$ \begin{align} &\big|t^{(k+1)}(\xi,\eta)-t^{(k)}(\xi,\eta)\big|\leq \int_{0}^\xi \big|C^{(k)}(\tau, \eta)-C^{(k-1)}(\tau, \eta)\big|\ \mathrm{d}\tau \nonumber \\ &\leq \int_{0}^\xi \bigg\{|C_{\widetilde{R}}|\cdot\big|\widetilde{R}^{(k)}(\tau,\eta) -\widetilde{R}^{(k-1)}(\tau,\eta)\big|+ |C_{\widetilde{S}}|\cdot\big|\widetilde{S}^{(k)}(\tau,\eta) -\widetilde{S}^{(k-1)}(\tau,\eta)\big| \nonumber \\ &\qquad \quad + |C_{t}|\cdot\big|t^{(k)}(\tau,\eta) -t^{(k-1)}(\tau,\eta)\big|\bigg\}\ \mathrm{d}\tau \nonumber \\ &\leq \int_{0}^\xi \bigg\{\frac{\overline{M}}{16}\cdot \overline{M}\tau^2\bigg(\frac{2}{3}\bigg)^{k-1}\times2 +\frac{1}{16\widetilde{\delta}}\cdot \overline{M}\tau \bigg(\frac{2}{3}\bigg)^{k-1}\bigg\}\ \mathrm{d}\tau \nonumber \\ &\leq \bigg(\frac{\overline{M}\widetilde{\delta}^2}{24}+\frac{1}{32}\bigg)\cdot\overline{M}\xi\bigg(\frac{2}{3}\bigg)^{k-1} \leq \overline{M}\xi\bigg(\frac{2}{3}\bigg)^{k}. \end{align} $$

To derive the difference between $\widetilde {R}^{(k+1)}(\xi ,\eta )$ and $\widetilde {R}^{(k)}(\xi ,\eta )$ , we need to first obtain the estimate of $|y^{(k)}_-(\tau )-y^{(k-1)}_-(\tau )|$ . By virtue of (3.24), it suggests that for $\tau \in [0,\xi ]$ ,

(3.68) $$ \begin{align} &y^{(k)}_-(\tau)-\int_{\tau}^{\xi}\frac{s}{f^{(k)}+\widetilde{S}^{(k)}}(s,y^{(k)}_-(s))\ \mathrm{d}s \nonumber \\ =&\eta =y^{(k-1)}_-(\tau)-\int_{\tau}^{\xi}\frac{s}{f^{(k-1)}+\widetilde{S}^{(k-1)}}(s,y^{(k-1)}_-(s))\ \mathrm{ d}s, \end{align} $$

from which we see by Lemma 3.1 and (3.15) that

(3.69) $$ \begin{align} \big|y^{(k)}_-(\tau)-y^{(k-1)}_-(\tau)\big|&\leq \int_{\tau}^{\xi}\frac{T_{1}^{(k)}+T_{2}^{(k)}+T_{3}^{(k)}}{|f^{(k)} +\widetilde{S}^{(k)}|\cdot|f^{(k-1)}+\widetilde{S}^{(k-1)}|}\cdot s\ \mathrm{d}s \nonumber \\ &\leq \frac{4}{\underline{\theta}^2} \int_{0}^{\xi}(T_{1}^{(k)}+T_{2}^{(k)}+T_{3}^{(k)})s\ \mathrm{d}s, \end{align} $$

where

(3.70) $$ \begin{align} \begin{array}{l} T_{1}^{(k)}=\big|\theta_0(y^{(k)}_-(s))-\theta_0(y^{(k-1)}_-(s))\big|, \\ T_{2}^{(k)}=\big|\widetilde{S}^{(k)}(s,y^{(k)}_-(s))-\widetilde{S}^{(k-1)}(s,y^{(k-1)}_-(s))\big|,\\ T_{3}^{(k)}=\big|\widetilde{v}\big(t^{(k)}(s, y^{(k)}_-(s)), y^{(k)}_-(s)\big) -\widetilde{v}\big(t^{(k-1)}(s, y^{(k-1)}_-(s)), y^{(k-1)}_-(s)\big)\big|. \end{array} \end{align} $$

For the term $T_{1}^{(k)}$ , one easily finds by the mean value theorem that

(3.71) $$ \begin{align} T_{1}^{(k)} \leq \max|\theta_{0}'|\cdot\big|y^{(k)}_-(s)-y^{(k-1)}_-(s)\big| \leq \overline{\theta}_0\big|y^{(k)}_-(s)-y^{(k-1)}_-(s)\big|. \end{align} $$

For the term $T_{2}^{(k)}$ , we have by Lemma 3.2 and the induction assumptions

(3.72) $$ \begin{align} T_{2}^{(k)} &\leq \big|\widetilde{S}^{(k)}(s,y^{(k)}_-(s))-\widetilde{S}^{(k)}(s,y^{(k-1)}_-(s))\big| \nonumber \\ &+\big|\widetilde{S}^{(k)}(s,y^{(k-1)}_-(s))-\widetilde{S}^{(k-1)}(s,y^{(k-1)}_-(s))\big| \nonumber \\ &\leq 3\overline{M}s^2 \big|y^{(k)}_-(s)-y^{(k-1)}_-(s)\big| +\overline{M}s^2\bigg(\frac{2}{3}\bigg)^{k-1}. \end{align} $$

Similarly, it concludes that for the term $T_{3}^{(k)}$ ,

(3.73) $$ \begin{align} T_{3}^{(k)} &\leq \big|\widetilde{v}\big(t^{(k)}(s, y^{(k)}_-(s)), y^{(k)}_-(s)\big) -\widetilde{v}\big(t^{(k)}(s, y^{(k)}_-(s)), y^{(k-1)}_-(s)\big)\big| \nonumber \\ &\ +\big|\widetilde{v}\big(t^{(k)}(s, y^{(k)}_-(s)), y^{(k-1)}_-(s)\big) -\widetilde{v}\big(t^{(k-1)}(s, y^{(k-1)}_-(s)), y^{(k-1)}_-(s)\big)\big| \nonumber \\ &\leq \widehat{M}_0t^{(k)}(s, y^{(k)}_-(s))\cdot\big|y^{(k)}_-(s)-y^{(k-1)}_-(s)\big| \nonumber \\ & +\widehat{M}_0\{\big|t^{(k)}(s, y^{(k)}_-(s)) -t^{(k)}(s, y^{(k-1)}_-(s))\big| \nonumber \\ &+\big|t^{(k)}(s, y^{(k-1)}_-(s)) -t^{(k-1)}(s, y^{(k-1)}_-(s))\big|\} \nonumber \\ &\leq \widehat{M}_0\frac{2s}{\underline{\theta}}\cdot\big|y^{(k)}_-(s)-y^{(k-1)}_-(s)\big| \nonumber \\ &+\widehat{M}_0\bigg\{3\overline{M}s\cdot \big|y^{(k)}_-(s)-y^{(k-1)}_-(s)\big| +\overline{M}s\bigg(\frac{2}{3}\bigg)^{k-1}\bigg\} \nonumber \\ = & \bigg(\frac{2\widehat{M}_0 s}{\underline{\theta}} +3\widehat{M}_0\overline{M}s\bigg)\big|y^{(k)}_-(s)-y^{(k-1)}_-(s)\big| + \widehat{M}_0\overline{M}s\bigg(\frac{2}{3}\bigg)^{k-1}. \end{align} $$

Putting (3.71)–(3.73) into (3.69) yields for $\tau \in [0,\xi ]$ ,

(3.74) $$ \begin{align} &\big|y^{(k)}_-(\tau)-y^{(k-1)}_-(\tau)\big| \nonumber \\ &\leq \frac{4}{\underline{\theta}^2} \int_{0}^{\xi}\bigg\{\bigg( \overline{\theta}_0 +3\overline{M}s^2+ \frac{2\widehat{M}_0 s}{\underline{\theta}} +3\widehat{M}_0\overline{M}s\bigg)\big|y^{(k)}_-(s)-y^{(k-1)}_-(s)\big| \nonumber \\ &\qquad \qquad +\bigg(\overline{M}s^2 +\widehat{M}_0\overline{M}s\bigg)\bigg(\frac{2}{3}\bigg)^{k-1}\bigg\}\cdot s\ \mathrm{d}s \nonumber \\ &\leq \int_{0}^{\xi}\bigg\{\bigg( \overline{K}\widetilde{\delta} +3\overline{K}\overline{M}\widetilde{\delta}^3+ \overline{K}\widehat{M}_0\widetilde{\delta}^2 +3\overline{K}\widehat{M}_0\overline{M}\widetilde{\delta}^2\bigg)\big|y^{(k)}_-(s)-y^{(k-1)}_-(s)\big| \nonumber \\ &\qquad \qquad +2\overline{K}\widehat{M}_0\overline{M}s^2\bigg(\frac{2}{3}\bigg)^{k-1}\bigg\}\ \mathrm{d}s \nonumber \\ &\leq \int_{0}^{\xi}\frac{1}{4}\big|y^{(k)}_-(s)-y^{(k-1)}_-(s)\big|\ \mathrm{d}s + +\frac{2}{3}\overline{K}\widehat{M}_0\overline{M}\xi^3\bigg(\frac{2}{3}\bigg)^{k-1}. \end{align} $$

Denote

$$ \begin{align*}d_{-}^{(k)}=\max_{\tau\in[0,\xi]}\big|y_{-}^{(k)}(\tau)-y_{-}^{(k-1)}(\tau)\big|. \end{align*} $$

Then we obtain by (3.74)

$$ \begin{align*}d_{-}^{(k)}\leq \frac{1}{4}d_{-}^{(k)} +\frac{2}{3}\overline{K}\widehat{M}_0\overline{M}\xi^3\bigg(\frac{2}{3}\bigg)^{k-1}, \end{align*} $$

which indicates that

(3.75) $$ \begin{align} d_{-}^{(k)}\leq \overline{K}\widehat{M}_0\overline{M}\xi^3\bigg(\frac{2}{3}\bigg)^{k-1}. \end{align} $$

Analogously, one gets

(3.76) $$ \begin{align} \max_{\tau\in[0,\xi]}\big|y_{+}^{(k)}(\tau)-y_{+}^{(k-1)}(\tau)\big|\leq \overline{K}\widehat{M}_0\overline{M}\xi^3\bigg(\frac{2}{3}\bigg)^{k-1}. \end{align} $$

We now estimate the difference between $\widetilde {R}^{(k+1)}(\xi ,\eta )$ and $\widetilde {R}^{(k)}(\xi ,\eta )$ . Applying (3.25) gives

(3.77) $$ \begin{align} &\big|\widetilde{R}^{(k+1)}(\xi,\eta)-\widetilde{R}^{(k)}(\xi,\eta)\big| \nonumber \\ &\leq \int_{0}^\xi\bigg\{T_{4}^{(k)}+T_{5}^{(k)}+T_{6}^{(k)}+(T_{7}^{(k)}+T_{8}^{(k)})\tau \bigg\}\ \mathrm{d}\tau, \end{align} $$

where

(3.78) $$ \begin{align} \begin{array}{l} \displaystyle T_{4}^{(k)}=\bigg|\frac{\widetilde{R}^{(k)}-\widetilde{S}^{(k)}}{2\tau}(\tau, y_{-}^{(k)}(\tau)) -\frac{\widetilde{R}^{(k-1)}-\widetilde{S}^{(k-1)}}{2\tau}(\tau, y_{-}^{(k-1)}(\tau))\bigg|, \\[8pt] \displaystyle T_{5}^{(k)}=\bigg|A^{(k)}\frac{(\widetilde{R}^{(k)}-\widetilde{S}^{(k)})^2}{4\tau}(\tau, y_{-}^{(k)}(\tau)) \\[8pt] \qquad \qquad \displaystyle -A^{(k-1)}\frac{(\widetilde{R}^{(k-1)}-\widetilde{S}^{(k-1)})^2}{4\tau}(\tau, y_{-}^{(k-1)}(\tau))\bigg|,\\[8pt] \displaystyle T_{6}^{(k)}=\bigg|A^{(k)}(\widetilde{R}^{(k)}-\widetilde{S}^{(k)})(\tau, y_{-}^{(k)}(\tau)) \\[8pt] \qquad \qquad \displaystyle -A^{(k-1)}(\widetilde{R}^{(k-1)}-\widetilde{S}^{(k-1)})(\tau, y_{-}^{(k-1)}(\tau))\bigg|, \\[8pt] \displaystyle T_{7}^{(k)}=\bigg|A^{(k)}\theta_{0}'(\tau, y_{-}^{(k)}(\tau)) -A^{(k-1)}\theta_{0}'(\tau, y_{-}^{(k-1)}(\tau))\bigg|,\\[8pt] \displaystyle T_{8}^{(k)}=\bigg|A^{(k)}w(t^{(k)},y)(\tau, y_{-}^{(k)}(\tau)) -A^{(k-1)}w(t^{(k-1)},y)(\tau, y_{-}^{(k-1)}(\tau))\bigg|. \end{array} \end{align} $$

Thanks to Lemma 3.2 and (3.75), one employs the induction assumptions to arrive at

(3.79) $$ \begin{align} T_{4}^{(k)}&\leq \frac{|\widetilde{R}^{(k)}(\tau, y_{-}^{(k)}(\tau))-\widetilde{R}^{(k-1)}(\tau, y_{-}^{(k-1)}(\tau))|}{2\tau} \nonumber \\ &+\frac{|\widetilde{S}^{(k)}(\tau, y_{-}^{(k)}(\tau))-\widetilde{S}^{(k-1)}(\tau, y_{-}^{(k-1)}(\tau))|}{2\tau} \nonumber \\ &\leq 3\overline{M}\tau d_{-}^{(k)}+\overline{M}\tau\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ &\leq \big(1+3\overline{K}\widehat{M}_0\overline{M}\widetilde{\delta}^3\big)\overline{M}\tau\bigg(\frac{2}{3}\bigg)^{k-1}\leq \frac{97}{96}\overline{M}\tau\bigg(\frac{2}{3}\bigg)^{k-1}. \end{align} $$

Recalling the expressions of $A^{(k)}$ in (3.25) and $T_{1,2,3}^{(k)}$ in (3.70) achieves

(3.80) $$ \begin{align} &\big|A^{(k)}(\tau, y_{-}^{(k)}(\tau)) -A^{(k-1)}(\tau, y_{-}^{(k-1)}(\tau))\big|\leq \frac{4}{\underline{\theta}^2}(T_{1}^{(k)}+T_{2}^{(k)}+T_{3}^{(k)}) \nonumber \\ &\leq \frac{4}{\underline{\theta}^2}\bigg\{ \bigg(\overline{\theta}_0+3\overline{M}\xi^2 +\frac{2\widehat{M}_0 \xi}{\underline{\theta}} +3\widehat{M}_0\overline{M}\xi\bigg)d_{-}^{(k)} + \bigg( \overline{M}\xi^2 + \widehat{M}_0\overline{M}\xi\bigg)\bigg(\frac{2}{3}\bigg)^{k-1} \bigg\} \nonumber \\ &\leq \frac{4}{\underline{\theta}^2}\bigg\{ \bigg(\overline{\theta}_0+3\overline{M}\widetilde{\delta}^2 +\frac{2\widehat{M}_0 \widetilde{\delta}}{\underline{\theta}} +3\widehat{M}_0\overline{M}\widetilde{\delta}\bigg)\cdot \overline{K}\widehat{M}_0\widetilde{\delta}^2 +\widetilde{\delta}+\widehat{M}_0 \bigg\}\overline{M}\xi\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ &\leq \frac{4}{\underline{\theta}^2}(1+\widehat{M}_0)\overline{M}\xi\bigg(\frac{2}{3}\bigg)^{k-1}\leq 2\overline{K}\widehat{M}_0\overline{M}\xi\bigg(\frac{2}{3}\bigg)^{k-1}, \end{align} $$

by the choice of $\widetilde {\delta }$ in (3.31). Therefore, we can use Lemmas 3.1 and 3.2, the induction assumptions, (3.75), and (3.80) to estimate the terms $T_{5,6,7,8}^{(k)}$ . In detail, for the term $T_{5}^{(k)}$ , we have

(3.81) $$ \begin{align} &T_{5}^{(k)}\leq\big|A^{(k)}-A^{(k-1)}\big|\cdot\frac{(\widetilde{R}^{(k)}-\widetilde{S}^{(k)})^2}{4\tau} +|A^{(k-1)}| \nonumber \\ &\times\frac{\big|(\widetilde{R}^{(k)}-\widetilde{S}^{(k)})^2(\tau, y_{-}^{(k)}(\tau))- (\widetilde{R}^{(k-1)}-\widetilde{S}^{(k-1)})^2(\tau, y_{-}^{(k-1)}(\tau))\big|}{4\tau} \nonumber \\ &\leq 2\overline{K}\widehat{M}_0\overline{M}\xi\bigg(\frac{2}{3}\bigg)^{k-1}\cdot \frac{(3\overline{M}\tau)^2}{4\tau} +\frac{\overline{M}}{16}\cdot\frac{1}{4\tau}\cdot6\overline{M}\tau^2\cdot2\tau T_{4}^{(k)} \nonumber \\ &\leq 5\overline{K}\widehat{M}_0\overline{M}^3\widetilde{\delta}\tau^3\bigg(\frac{2}{3}\bigg)^{k-1} +\frac{3}{16}(1+3\overline{K}\widehat{M}_0\overline{M}\widetilde{\delta}^3) \overline{M}^3\tau^3\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ &\leq \bigg(5\overline{K}\widehat{M}_0\widetilde{\delta} +\frac{3}{16} +\frac{9}{16}\overline{K}\widehat{M}_0\overline{M}\widetilde{\delta}^3\bigg) \overline{M}^3\tau^3\bigg(\frac{2}{3}\bigg)^{k-1}\leq \overline{M}^3\tau^3\bigg(\frac{2}{3}\bigg)^{k-1}. \end{align} $$

For the term $T_{6}^{(k)}$ , one obtains

(3.82) $$ \begin{align} &T_{6}^{(k)}\leq\big|A^{(k)}-A^{(k-1)}\big|\cdot\big|\widetilde{R}^{(k)}-\widetilde{S}^{(k)}\big| +\big|A^{(k-1)}\big|\cdot 2\tau T_{4}^{(k)} \nonumber \\ &\leq 2\overline{K}\widehat{M}_0\overline{M}\xi\bigg(\frac{2}{3}\bigg)^{k-1}\cdot3\overline{M}\tau^2 +\frac{\overline{M}^2}{8}\cdot\tau^2(1+3\overline{K}\widehat{M}_0\overline{M}\widetilde{\delta}^3) \bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ & \leq \bigg(6\overline{K}\widehat{M}_0\widetilde{\delta} +\frac{1}{8} +\frac{3}{16}\overline{K}\widehat{M}_0\overline{M}\widetilde{\delta}^3\bigg)\overline{M}^2\tau^2 \bigg(\frac{2}{3}\bigg)^{k-1} \leq\overline{M}^2\tau^2 \bigg(\frac{2}{3}\bigg)^{k-1}. \end{align} $$

For the terms $T_{7}^{(k)}$ and $T_{8}^{(k)}$ , we also acquire

(3.83) $$ \begin{align} &T_{7}^{(k)}\leq\big|A^{(k)}-A^{(k-1)}\big|\cdot\big|\theta_{0}'\big| +\big|A^{(k-1)}\big|\cdot\big|\theta_{0}'(y_{-}^{(k)}(\tau))-\theta_{0}'(y_{-}^{(k-1)}(\tau))\big| \nonumber \\ &\leq \frac{4}{\underline{\theta}^2}(1+\widehat{M}_0)\overline{M}\widetilde{\delta}\bigg(\frac{2}{3}\bigg)^{k-1}\cdot \overline{\theta}_0 +\frac{\overline{M}}{16}\cdot \overline{\theta}_0\cdot d_{-}^{(k)} \nonumber \\ &\leq 2\overline{K}\widehat{M}_0\overline{M}\widetilde{\delta}\bigg(\frac{2}{3}\bigg)^{k-1} +\frac{\overline{M}}{16}\cdot \overline{\theta}_0\cdot \overline{K}\widehat{M}_0\overline{M}\widetilde{\delta}^3\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ = & \bigg(2\overline{K}\widehat{M}_0\widetilde{\delta} +\frac{(\overline{M}\widetilde{\delta})\cdot(\overline{\theta}_0\widetilde{\delta})}{16}\cdot (\overline{K}\widehat{M}_0\widetilde{\delta})\bigg)\overline{M}\bigg(\frac{2}{3}\bigg)^{k-1}\leq \frac{13}{96}\overline{M}\bigg(\frac{2}{3}\bigg)^{k-1} \end{align} $$

and

(3.84) $$ \begin{align} &T_{8}^{(k)}\leq\big|A^{(k)}-A^{(k-1)}\big|\cdot\big|w(t^{(k)},y_{-}^{(k)}(\tau))\big| \nonumber \\ &\quad \ \ \quad +\big|A^{(k-1)}\big|\cdot\big|w(t^{(k)},y_{-}^{(k)}(\tau))-w(t^{(k-1)}, y_{-}^{(k-1)}(\tau))\big| \nonumber \\ &\leq 2\overline{K}\widehat{M}_0\overline{M}\widetilde{\delta}\bigg(\frac{2}{3}\bigg)^{k-1}\cdot \widehat{M}_0t^{(k)}(\tau,y_{-}^{(k)}(\tau)) \nonumber \\ &\ +\overline{K}\cdot\bigg\{\widehat{M}_0\big|t^{(k)}(\tau,y_{-}^{(k)}(\tau)) -t^{(k-1)}(\tau,y_{-}^{(k-1)}(\tau))\big| +\widehat{M}_0d_{-}^{(k)}\bigg\} \nonumber \\ &\leq 2\overline{K}\widehat{M}_0\overline{M}\widetilde{\delta}\bigg(\frac{2}{3}\bigg)^{k-1}\cdot \widehat{M}_0\cdot\frac{2\tau}{\underline{\theta}} +\overline{K}\widehat{M}_0\bigg\{(|t_{y}^{(k)}|+1)d_{-}^{(k)} +\overline{M}\tau\bigg(\frac{2}{3}\bigg)^{k-1}\bigg\} \nonumber \\ &\leq \bigg\{2(\overline{K}\widehat{M}_{0}\widetilde{\delta})^2 +2(\overline{K}\widehat{M}_0\widetilde{\delta})^2\widetilde{\delta} +\overline{K}\widehat{M}_{0}\widetilde{\delta}\bigg\}\overline{M}\bigg(\frac{2}{3}\bigg)^{k-1} \leq\frac{7}{96}\overline{M}\bigg(\frac{2}{3}\bigg)^{k-1}. \end{align} $$

Here, the facts $\widehat {M}_0\geq 1$ , $16\overline {M}\widetilde {\delta }\leq 1$ , and $16\overline {K}\overline {M}_0\widetilde {\delta }\leq 1$ are used in the above estimation processes.

We now insert (3.79) and (3.81)–(3.84) into (3.77) to gain

(3.85) $$ \begin{align} &\big|\widetilde{R}^{(k+1)}(\xi,\eta)-\widetilde{R}^{(k)}(\xi,\eta)\big| \leq \ \int_{0}^\xi\bigg\{ \frac{97}{96}\overline{M}\tau\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ &\quad +\overline{M}^3\tau^3\bigg(\frac{2}{3}\bigg)^{k-1} +\overline{M}^2\tau^2\bigg(\frac{2}{3}\bigg)^{k-1} +\frac{5}{24}\overline{M}\tau\bigg(\frac{2}{3}\bigg)^{k-1}\bigg\}\ \mathrm{d}\tau \nonumber \\ &\leq \ \bigg\{ \frac{97}{192} +\frac{1}{4}(\overline{M}\widetilde{\delta})^2 +\frac{1}{3}(\overline{M}\widetilde{\delta}) +\frac{5}{48}\bigg\}\overline{M}\xi^2\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ &\leq \ \bigg\{ \frac{97}{192} +\frac{1}{4\times16^2} +\frac{1}{48} +\frac{5}{48}\bigg\}\overline{M}\xi^2\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ <&\frac{122}{192}\overline{M}\xi^2\bigg(\frac{2}{3}\bigg)^{k-1}<\overline{M}\xi^2\bigg(\frac{2}{3}\bigg)^{k}. \end{align} $$

It is easily checked that the estimate (3.85) is also valid for the term $|\widetilde {S}^{(k+1)}(\xi ,\eta )-\widetilde {S}^{(k)}(\xi ,\eta )|$ . The proof of the lemma is finished.

3.4 The existence and uniqueness of solutions

According to Lemma 3.3, it is known that the sequences $(\widetilde {R}^{(k)}, \widetilde {S}^{(k)}, t^{(k)})(\tau , y)$ are uniformly convergent. We denote the limit functions by $(\widetilde {R}, \widetilde {S}, t)(\tau , y)$ which are obviously continuous. Furthermore, by means of Lemma 3.1, the functions $(\widetilde {R}, \widetilde {S}, t)(\tau , y)$ satisfy

(3.86) $$ \begin{align} \begin{array}{l} \displaystyle \big|\widetilde{R}(\tau,y)\big|, \big|\widetilde{S}(\tau,y)\big|\leq 3\overline{M}\tau^2=\widetilde{M}\tau^2,\ \ \frac{\tau}{2\overline{\theta}}\leq t(\tau,y)\leq \frac{2\tau}{\underline{\theta}}, \\ \big|\widetilde{R}(\tau,y)-\widetilde{S}(\tau,y)\big| \leq 3\overline{M}\tau^2=\widetilde{M}\tau^2, \end{array} \end{align} $$

for any $(\tau , y)\in [0,\widetilde {\delta }]\times \mathbb {R}$ . In addition, it is clear that the functions $(\widetilde {R}, \widetilde {S}, t)(\tau , y)$ satisfy the integral system (3.19) and the boundary conditions

(3.87) $$ \begin{align} \widetilde{R}(0, y)=\widetilde{S}(0, y)=t(0, y)=0. \end{align} $$

Thus, we have $(\widetilde {R}, \widetilde {S}, t)\in \widetilde {\Sigma }(\widetilde {\delta })$ .

Next, we show that the functions $(\widetilde {R}, \widetilde {S}, t)(\tau , y)$ possess first-order continuous derivatives with respect to y. To move forward, one differentiates (3.19) with respect to $\eta $ to deduce the following linear system of integral equations:

(EQ.a1) $$ \begin{align} \begin{array}{l} \displaystyle \widetilde{R}_{\eta}(\xi,\eta) =\int_{0}^\xi\bigg\{\frac{\widetilde{R}_{y}-\widetilde{S}_{y}} {2\tau} + A\cdot\frac{(\widetilde{R}-\widetilde{S})(\widetilde{R}_{y}-\widetilde{S}_{y})}{2\tau} \\[8pt] \displaystyle \quad \qquad \qquad -A(\widetilde{R}_{y}-\widetilde{S}_{y})+A_{11}A_{\widetilde{S}}\widetilde{S}_{y} +[A_{11}A_{t}-Aw_t\tau]t_{y} \\[6pt] \displaystyle \quad \qquad \qquad + A_{11}A_{y} +A(\theta_{0}"-w_y)\tau \bigg\}\frac{\partial y_{-}(\tau)}{\partial \eta}(\tau, y_{-}(\tau))\ \mathrm{d}\tau, \end{array} \end{align} $$

(EQ.b1) $$ \begin{align} \begin{array}{l} \displaystyle \widetilde{S}_{\eta}(\xi,\eta) =\int_{0}^\xi\bigg\{\frac{\widetilde{S}_{y}-\widetilde{R}_{y}} {2\tau} + B\cdot\frac{(\widetilde{R}-\widetilde{S})(\widetilde{R}_{y}-\widetilde{S}_{y})}{2\tau} \\[8pt] \displaystyle \quad \qquad \qquad +B(\widetilde{R}_{y}-\widetilde{S}_{y}) +B_{11}B_{\widetilde{R}}\widetilde{R}_{y}+[B_{11}A_{t}+Bw_t\tau]t_{y} \\[6pt] \displaystyle \quad \qquad \qquad + B_{11}B_{y} -B(\theta_{0}"-w_y)\tau \bigg\}\frac{\partial y_{+}(\tau)}{\partial \eta}(\tau, y_{+}(\tau))\ \mathrm{d}\tau, \end{array} \end{align} $$

(EQ.c1) $$ \begin{align} t_{\eta}(\xi,\eta)=\int_{0}^\xi \bigg\{ C_{\widetilde{R}}\widetilde{R}_{y} + C_{\widetilde{S}}\widetilde{S}_{y} +C_{t}t_{y} + C_{y} \bigg\}(\tau, \eta)\ \mathrm{d}\tau. \end{align} $$

Here, the coefficient functions in (EQ.a1)–(EQ.c1) are given in (EQ.a)–(EQ.c) but with the limit functions $(\widetilde {R}, \widetilde {S}, t)$ replacing $(\widetilde {R}^{(k)}, \widetilde {S}^{(k)}, t^{(k)})$ . Set

(3.88) $$ \begin{align} (\widetilde{R}_{\eta}^{(0)}, \widetilde{S}_{\eta}^{(0)}, t_{\eta}^{(0)})(\xi,\eta)=\textbf{0}, \end{align} $$

and construct the iterative sequences $(\widetilde {R}_{\eta }^{(k)}, \widetilde {S}_{\eta }^{(k)}, t_{\eta }^{(k)})(\xi ,\eta )$ by the integral system (EQ.a1)–(EQ.c1) as follows:

(EQ.a2) $$ \begin{align} &\widetilde{R}_{\eta}^{(k+1)}(\xi,\eta)=\int_{0}^\xi\bigg\{\frac{\widetilde{R}_{y}^{(k)}-\widetilde{S}_{y}^{(k)}} {2\tau} + A\frac{(\widetilde{R}-\widetilde{S})(\widetilde{R}_{y}^{(k)}-\widetilde{S}_{y}^{(k)})}{2\tau} \nonumber \\ &\qquad -A(\widetilde{R}_{y}^{(k)}-\widetilde{S}_{y}^{(k)}) +A_{11}A_{\widetilde{S}}\widetilde{S}_{y}^{(k)} +[A_{11}A_{t}-Aw_t\tau]t_{y}^{(k)} \nonumber \\ & \qquad + A_{11}A_{y} +A(\theta_{0}"-w_y)\tau \bigg\}\frac{\partial y_{-}^{(k)}(\tau)}{\partial \eta}(\tau, y_{-}(\tau))\ \mathrm{d}\tau, \end{align} $$

(EQ.b2) $$ \begin{align} &\widetilde{S}_{\eta}^{(k+1)}(\xi,\eta) =\int_{0}^\xi\bigg\{\frac{\widetilde{S}_{y}^{(k)}-\widetilde{R}_{y}^{(k)}} {2\tau} + B\frac{(\widetilde{R}-\widetilde{S})(\widetilde{R}_{y}^{(k)}-\widetilde{S}_{y}^{(k)})}{2\tau}\nonumber \\ &\qquad +B(\widetilde{R}_{y}^{(k)}-\widetilde{S}_{y}^{(k)}) +B_{11}B_{\widetilde{R}}\widetilde{R}_{y}^{(k)} +[B_{11}A_{t}+Bw_t\tau]t_{y}^{(k)} \nonumber \\ & \qquad + B_{11}B_{y} -B(\theta_{0}"-w_y)\tau \bigg\}\frac{\partial y_{+}^{(k)}(\tau)}{\partial \eta}(\tau, y_{+}(\tau))\ \mathrm{d}\tau, \end{align} $$

(EQ.c2) $$ \begin{align} t_{\eta}^{(k+1)}(\xi,\eta)=\int_{0}^\xi \bigg\{ C_{\widetilde{R}}\widetilde{R}_{y}^{(k)} + C_{\widetilde{S}}\widetilde{S}_{y}^{(k)} +C_{t}t_{y}^{(k)} + C_{y} \bigg\}(\tau, \eta)\ \mathrm{d}\tau, \end{align} $$

where

(3.89) $$ \begin{align} \frac{\partial y^{(k)}_\pm}{\partial \eta}(\tau)=\exp\bigg\{\int_{\xi}^\tau\frac{\partial \lambda_{\pm}^{(k)}}{\partial y}(s,y_{\pm}(s))\ \mathrm{d}s\bigg\}, \end{align} $$

and

(3.90) $$ \begin{align} \begin{array}{l} \displaystyle \frac{\partial \lambda_{-}^{(k)}}{\partial y}=-s\big[A_{\widetilde{S}}\widetilde{S}_{y}^{(k)} +A_{t}t_{y}^{(k)} + A_{y}\big], \\[8pt] \displaystyle \frac{\partial \lambda_{+}^{(k)}}{\partial y}=s\big[B_{\widetilde{R}}\widetilde{R}_{y}^{(k)} +B_{t}t_{y}^{(k)} + B_{y}\big]. \end{array} \end{align} $$

Since the limit functions $(\widetilde {R}, \widetilde {S}, t)(\tau , y)$ are in the space $\widetilde {\Sigma }(\widetilde {\delta })$ , we see that the coefficients of system (EQ.a2) and (EQ.c2) still satisfy the estimates in (3.50)–(3.53). Then we have the following.

Lemma 3.4 The sequences $(\widetilde {R}_{\eta }^{(k)}, \widetilde {S}_{\eta }^{(k)}, t_{\eta }^{(k)})$ defined by system (EQ.a2) and (EQ.c2) satisfy the following inequalities:

(3.91) $$ \begin{align} \begin{array}{l} \big|\widetilde{R}_{\eta}^{(k)}(\xi,\eta)\big|, \big|\widetilde{S}_{\eta}^{(k)}(\xi,\eta)\big|\leq \overline{M}\xi^2\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j,\ \big|t_{\eta}^{(k)}(\xi,\eta)\big|\leq \overline{M}\xi\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j, \\ \big|\widetilde{R}_{\eta}^{(k)}(\xi,\eta)-\widetilde{S}_{\eta}^{(k)}(\xi,\eta)\big| \leq \overline{M}\xi^2\displaystyle\sum^{k}_{j=0} \bigg(\frac{2}{3}\bigg)^j, \end{array} \end{align} $$

for all $k\geq 1$ and any $(\xi ,\eta )\in [0,\widetilde {\delta }]\times \mathbb {R}$ .

It concludes by (3.91) that

(3.92) $$ \begin{align} \big|\widetilde{R}_{\eta}^{(k)}(\xi,\eta)\big|, \big|\widetilde{S}_{\eta}^{(k)}(\xi,\eta)\big|\leq 3\overline{M}\xi^2, \quad \big|t_{\eta}^{(k)}(\xi,\eta)\big|\leq 3\overline{M}\xi, \end{align} $$

which together with (3.89) and (3.90) and the estimates in (3.50)–(3.53) give

(3.93) $$ \begin{align} \begin{array}{l} \displaystyle \bigg|\frac{\partial \lambda_{\pm}^{(k)}}{\partial y}\bigg|\leq s\bigg(\frac{\overline{M}}{16}\cdot3\overline{M}s^2 +\frac{1}{16\widetilde{\delta}}\cdot3\overline{M}s +\frac{\overline{M}}{8}\bigg)\leq\overline{M}s, \\[10pt] \displaystyle \bigg|\frac{\partial y^{(k)}_\pm}{\partial \eta}\bigg|\leq\exp\bigg\{\int_{0}^\xi\bigg|\frac{\partial \lambda_{\pm}^{(k)}}{\partial y}(s,y_{\pm}(s))\bigg|\ \mathrm{d}s\bigg\}\leq e^{\overline{M}\xi^2}, \end{array} \end{align} $$

and then

(3.94) $$ \begin{align} & \bigg|\frac{\partial y^{(k)}_\pm}{\partial \eta}(\tau) -\frac{\partial y^{(k-1)}_\pm}{\partial \eta}(\tau)\bigg|\leq e^{\overline{M}\xi^2}\cdot\int_{0}^\xi \bigg|\frac{\partial \lambda_{\pm}^{(k)}}{\partial y}-\frac{\partial \lambda_{\pm}^{(k-1)}}{\partial y}\bigg|\ \mathrm{d}s \nonumber \\ &\leq e^{\overline{M}\widetilde{\delta}^2}\int_{0}^\xi s\bigg\{\frac{\overline{M}}{16}T_{9}^{(k)} + \frac{1}{16\widetilde{\delta}}|t_{y}^{(k)}-t_{y}^{(k-1)}|\bigg\}\ \mathrm{d}s, \end{align} $$

where

$$ \begin{align*}T_{9}^{(k)}=|\widetilde{R}_{y}^{(k)}-\widetilde{R}_{y}^{(k-1)}| +|\widetilde{S}_{y}^{(k)}-\widetilde{S}_{y}^{(k-1)}|. \end{align*} $$

By utilizing Lemma 3.4 and (3.93) and (3.94), we have the following lemma.

Lemma 3.5 Let the functions $(\widetilde {R}_{\eta }^{(k)}, \widetilde {S}_{\eta }^{(k)}, t_{\eta }^{(k)})$ be defined by the iterative system (EQ.a2) and (EQ.c2). Then, for all $k\geq 0$ , the following inequalities

(3.95) $$ \begin{align} \begin{array}{r} \displaystyle \big|\widetilde{R}_{\eta}^{(k+1)}(\xi,\eta)-\widetilde{R}_{\eta}^{(k)}(\xi,\eta)\big| \leq \overline{M}\xi^2\displaystyle\bigg(\frac{2}{3}\bigg)^k, \\[8pt] \displaystyle \big|\widetilde{S}_{\eta}^{(k+1)}(\xi,\eta)-\widetilde{S}_{\eta}^{(k)}(\xi,\eta)\big| \leq \overline{M}\xi^2\displaystyle\bigg(\frac{2}{3}\bigg)^k, \\[8pt] \displaystyle \big|t_{\eta}^{(k+1)}(\xi,\eta)-t_{\eta}^{(k)}(\xi,\eta)\big| \leq \overline{M}\xi\displaystyle\bigg(\frac{2}{3}\bigg)^k \end{array} \end{align} $$

hold in $[0,\widetilde {\delta }]\times \mathbb {R}$ .

Proof The proof of the lemma is still based on the inductive method. It is obvious by Lemma 3.4 that all inequalities in (3.95) are true for $k=0$ . We assume that each inequality in (3.95) holds for $n=k-1$ and then verify all of them valid for $n=k$ .

For the term $|t_{\eta }^{(k+1)}(\xi ,\eta )-t_{\eta }^{(k)}(\xi ,\eta )|$ , one obtains by (EQ.c2), (3.52), and the induction assumptions

(3.96) $$ \begin{align} &\big|t_{\eta}^{(k+1)}(\xi,\eta)-t_{\eta}^{(k)}(\xi,\eta)\big| \leq \int_{0}^\xi \bigg\{ \big|C_{\widetilde{R}}\big|\cdot\big|\widetilde{R}_{y}^{(k)}-\widetilde{R}_{y}^{(k-1)}\big| \nonumber \\ &\qquad + \big|C_{\widetilde{S}}\big|\cdot\big|\widetilde{S}_{y}^{(k)}-\widetilde{S}_{y}^{(k-1)}\big| +\big|C_{t}\big|\cdot\big|t_{y}^{(k)}-t_{y}^{(k-1)}\big| \bigg\} \ \mathrm{d}\tau \nonumber \\ &\leq \int_{0}^\xi \bigg\{ \frac{\overline{M}}{16}\cdot \overline{M}\tau^2\bigg(\frac{2}{3}\bigg)^{k-1} + \frac{\overline{M}}{16}\cdot \overline{M}\tau^2\bigg(\frac{2}{3}\bigg)^{k-1} +\overline{K}\widehat{M}_0\cdot \overline{M}\tau \bigg(\frac{2}{3}\bigg)^{k-1} \bigg\} \ \mathrm{d}\tau \nonumber \\ &\leq \bigg(\frac{\overline{M}\widetilde{\delta}^2}{24} +\frac{\overline{K}\widehat{M}_0\widetilde{\delta}}{2}\bigg)\overline{M}\xi \bigg(\frac{2}{3}\bigg)^{k-1}\leq \overline{M}\xi \bigg(\frac{2}{3}\bigg)^{k}. \end{align} $$

For the term $|\widetilde {R}_{\eta }^{(k+1)}(\xi ,\eta )-\widetilde {R}_{\eta }^{(k)}(\xi ,\eta )|$ , one gets by (EQ.a2)

(3.97) $$ \begin{align} &\big|\widetilde{R}_{\eta}^{(k+1)}(\xi,\eta)-\widetilde{R}_{\eta}^{(k)}(\xi,\eta)\big| \nonumber \\ &\leq \int_{0}^\xi\bigg\{ T_{10}^{(k)}\bigg|\frac{\partial y_{-}^{(k)}(\tau)}{\partial \eta}\bigg| +T_{11}^{(k)}\bigg|\frac{\partial y_{-}^{(k)}(\tau)}{\partial \eta} -\frac{\partial y_{-}^{(k-1)}(\tau)}{\partial \eta}\bigg|\bigg\}\ \mathrm{d}\tau, \end{align} $$

where

(3.98) $$ \begin{align} T_{10}^{(k)}=&\frac{T_{9}^{(k)}} {2\tau} + |A|\frac{|\widetilde{R}-\widetilde{S}|\cdot T_{9}^{(k)}}{2\tau} +|A|T_{9}^{(k)} \nonumber \\ &\ +|A_{11}A_{\widetilde{S}}|\cdot|\widetilde{S}_{y}^{(k)} -\widetilde{S}_{y}^{(k-1)}| +|A_{11}A_{t}-Aw_t\tau|\cdot|t_{y}^{(k)} -t_{y}^{(k-1)}|, \end{align} $$

(3.99) $$ \begin{align} T_{11}^{(k)}=& \frac{|\widetilde{R}_{y}^{(k-1)}-\widetilde{S}_{y}^{(k-1)}|} {2\tau} + |A|\frac{|\widetilde{R}-\widetilde{S}|\cdot|\widetilde{R}_{y}^{(k-1)}-\widetilde{S}_{y}^{(k-1)}|}{2\tau} \nonumber \\ &\ +|A|\cdot|\widetilde{R}_{y}^{(k-1)}-\widetilde{S}_{y}^{(k-1)}| +|A_{11}A_{\widetilde{S}}|\cdot|\widetilde{S}_{y}^{(k-1)}| \nonumber \\ &\ +|A_{11}A_{t}-Aw_t\tau|\cdot|t_{y}^{(k-1)}|+ |A_{11}A_{y}| +|A(\theta_{0}"-w_y)|\tau. \end{align} $$

Recalling the estimates in (3.50) yields

(3.100) $$ \begin{align} \begin{array}{l} \displaystyle |A_{11}A_{\widetilde{S}}|\leq \frac{\overline{M}^2}{4}\tau^2+\frac{\overline{M}}{8}\tau,\quad |A_{11}A_t-Aw_t\tau|\leq\frac{1}{4}, \\[8pt] \displaystyle |A_{11}A_y|\leq \frac{\overline{M}}{4}\tau,\qquad \qquad \quad |A(\theta_{0}"-w_y)|\leq\frac{\overline{M}}{8}, \end{array} \end{align} $$

which, together with (3.86), (3.91), (3.98)–(3.99), and the induction assumptions, arrive at

(3.101) $$ \begin{align} T_{10}^{(k)}&\leq \frac{1}{2\tau}\cdot 2\overline{M}\tau^2\bigg(\frac{2}{3}\bigg)^{k-1} +\frac{\overline{M}}{16}\cdot\frac{3\overline{M}\tau^2}{2\tau}\cdot2\overline{M}\tau^2\bigg(\frac{2}{3}\bigg)^{k-1} +\frac{\overline{M}^2}{8}\tau^2\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ &\ +\bigg(\frac{\overline{M}^2}{4}\tau^2+\frac{\overline{M}}{8}\tau\bigg)\cdot \overline{M}\tau^2\bigg(\frac{2}{3}\bigg)^{k-1} +\frac{1}{4}\cdot \overline{M}\tau\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ &\leq \bigg(\frac{5}{4}+\frac{1}{2}(\overline{M}\widetilde{\delta})^2 +\frac{1}{4}(\overline{M}\widetilde{\delta})\bigg)\overline{M}\tau\bigg(\frac{2}{3}\bigg)^{k-1} <\frac{41}{32}\overline{M}\tau\bigg(\frac{2}{3}\bigg)^{k-1} \end{align} $$

and

(3.102) $$ \begin{align} T_{11}^{(k)}&\leq \frac{3\overline{M}\tau^2}{2\tau} +\frac{\overline{M}}{16}\cdot\frac{3\overline{M}\tau^2\cdot 3\overline{M}\tau^2}{2\tau} +\frac{\overline{M}}{16}\cdot 3\overline{M}\tau^2 \nonumber \\ &\ +\bigg(\frac{\overline{M}^2}{4}\tau^2+\frac{\overline{M}}{8}\tau\bigg)\cdot3\overline{M}\tau^2 +\frac{1}{4}\cdot 3\overline{M}\tau +\frac{\overline{M}}{4}\tau +\frac{\overline{M}}{8}\tau \nonumber \\ & \leq \bigg(\frac{21}{8} + \frac{17}{32}(\overline{M}\widetilde{\delta})^2 +\frac{7}{32}(\overline{M}\widetilde{\delta}) \bigg)\overline{M}\tau\leq 3\overline{M}\tau. \end{align} $$

Moreover, it suggests by (3.94) and the induction assumptions that

(3.103) $$ \begin{align} &\bigg|\frac{\partial y^{(k)}_-(\tau)}{\partial \eta} -\frac{\partial y^{(k-1)}_-(\tau)}{\partial \eta}\bigg| \nonumber \\ & \leq e^{\overline{M}\widetilde{\delta}^2}\int_{0}^\xi s\bigg\{\frac{\overline{M}}{16}\cdot2\overline{M}s^2\bigg(\frac{2}{3}\bigg)^{k-1} + \frac{1}{16\widetilde{\delta}}\cdot \overline{M}s\bigg(\frac{2}{3}\bigg)^{k-1}\bigg\}\ \mathrm{d}s \nonumber \\ & \leq e^{\overline{M}\widetilde{\delta}^2}\overline{M}\xi^2\bigg(\frac{2}{3}\bigg)^{k-1}. \end{align} $$

Putting (3.101)–(3.103) into (3.97) and using (3.93) leads to

(3.104) $$ \begin{align} &\big|\widetilde{R}_{\eta}^{(k+1)}(\xi,\eta)-\widetilde{R}_{\eta}^{(k)}(\xi,\eta)\big| \nonumber \\ &\leq \int_{0}^\xi\bigg\{ \frac{41}{32}\overline{M}\tau\bigg(\frac{2}{3}\bigg)^{k-1}\cdot e^{\overline{M}\widetilde{\delta}^2} +3\overline{M}\tau\cdot e^{\overline{M}\widetilde{\delta}^2}\overline{M}\xi^2\bigg(\frac{2}{3}\bigg)^{k-1} \bigg\}\ \mathrm{d}\tau \nonumber \\ &\leq \bigg(\frac{41}{64} + \frac{3}{2}\overline{M}\widetilde{\delta}^2\bigg) e^{\overline{M}\widetilde{\delta}^2}\overline{M}\xi^2\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ &\leq \frac{21}{32} e^{1/64}\overline{M}\xi^2\bigg(\frac{2}{3}\bigg)^{k-1}<\overline{M}\xi^2\bigg(\frac{2}{3}\bigg)^{k}, \end{align} $$

by the fact $e^{1/64}<64/63$ . The estimate of the term $|\widetilde {S}_{\eta }^{(k+1)}(\xi ,\eta )-\widetilde {S}_{\eta }^{(k)}(\xi ,\eta )|$ can be derived similar to (3.104). The proof of the lemma is complete.

According to Lemmas 3.4 and 3.5, we know that the sequences $(\widetilde {R}_{\eta }^{(k)}, \widetilde {S}_{\eta }^{(k)}, t_{\eta }^{(k)})(\xi ,\eta )$ are uniformly convergent, which means that the functions $(\widetilde {R}_{\eta }, \widetilde {S}_{\eta }, t_{\eta })(\xi ,\eta )$ are continuous in $[0,\widetilde {\delta }]\times \mathbb {R}$ . Furthermore, one also has by (3.91)

(3.105) $$ \begin{align} \begin{array}{l} \big|\widetilde{R}_{\eta}(\xi,\eta)\big|, \big|\widetilde{S}_{\eta}(\xi,\eta)\big|\leq 3\overline{M}\xi^2,\quad \big|t_{\eta}(\xi,\eta)\big|\leq 3\overline{M}\xi, \\ \big|\widetilde{R}_{\eta}(\xi,\eta)-\widetilde{S}_{\eta}(\xi,\eta)\big| \leq 3\overline{M}\xi^2. \end{array} \end{align} $$

In addition, we differentiate equations (3.19) with respect to $\xi $ to gain

(3.106) $$ \begin{align} &\widetilde{R}_\xi(\xi,\eta)=\frac{\widetilde{R}-\widetilde{S}}{2\xi} +A\cdot\frac{(\widetilde{R}-\widetilde{S})^2}{4\xi} -A(\widetilde{R}-\widetilde{S}) +A\theta_{0}'\xi-Aw(t,y)\xi \nonumber \\ &\ +\int_{0}^\xi\bigg\{\frac{\widetilde{R}_{y}-\widetilde{S}_{y}} {2\tau} + A\frac{(\widetilde{R}-\widetilde{S})(\widetilde{R}_{y}-\widetilde{S}_{y})}{2\tau} -A(\widetilde{R}_{y}-\widetilde{S}_{y}) +A_{11}A_{\widetilde{S}}\widetilde{S}_{y} \nonumber \\ &\ +[A_{11}A_{t}-Aw_t\tau]t_{y} + A_{11}A_{y} +A(\theta_{0}"-w_y)\tau \bigg\} \frac{\partial y_-(\tau)}{\partial \xi}(\tau, y_-(\tau))\ \mathrm{d}\tau, \end{align} $$

(3.107) $$ \begin{align} &\widetilde{S}_\xi(\xi,\eta)=\frac{\widetilde{S}-\widetilde{R}}{2\xi} +B\cdot\frac{(\widetilde{R}-\widetilde{S})^2}{4\xi} +B(\widetilde{R}-\widetilde{S}) -B\theta_{0}'\xi+Bw(t,y)\xi \nonumber \\ &\ +\int_{0}^\xi\bigg\{\frac{\widetilde{S}_{y}-\widetilde{R}_{y}} {2\tau} + B\frac{(\widetilde{R}-\widetilde{S})(\widetilde{R}_{y}-\widetilde{S}_{y})}{2\tau} +B(\widetilde{R}_{y}-\widetilde{S}_{y}) +B_{11}B_{\widetilde{R}}\widetilde{R}_{y} \nonumber \\ &\ +[B_{11}B_{t}+Bw_t\tau]t_{y} + B_{11}B_{y} -B(\theta_{0}"-w_y)\tau \bigg\} \frac{\partial y_+(\tau)}{\partial \xi}(\tau, y_+(\tau))\ \mathrm{d}\tau, \end{align} $$

and

(3.108) $$ \begin{align} t_\xi(\xi,\eta)=C(\xi,\eta). \end{align} $$

The terms $\partial _\xi y_\pm (\tau )$ in (3.106) and (3.107) are given by

(3.109) $$ \begin{align} \frac{\partial y_\pm(\tau)}{\partial \xi}=-\lambda_\pm\frac{\partial y_\pm(\tau)}{\partial \eta}=-\lambda_\pm\exp\bigg\{\int_{\xi}^\tau\frac{\partial \lambda_{\pm}}{\partial y}(s,y_{\pm}(s))\ \mathrm{ d}s\bigg\}. \end{align} $$

In view of the expressions of $\widetilde {R}_\xi $ and $\widetilde {S}_\xi $ in (3.106) and (3.107), we employ the estimates in (3.86) and (3.105) to find that the functions $(\widetilde {R}_{\tau }, \widetilde {S}_{\tau }, t_\tau )(\tau , y)$ are continuous in $[0,\widetilde {\delta }]\times \mathbb {R}$ and satisfy

(3.110) $$ \begin{align} \widetilde{R}_{\tau}(0,y)=\widetilde{S}_{\tau}(0,y)=0. \end{align} $$

All in all, the functions $(\widetilde {R}, \widetilde {S}, t)(\tau , y)$ satisfy the integral equations (3.19) and the homogeneous initial conditions (3.87) and (3.110), and they also own the required differentiability properties. Therefore, they are a smooth solution to the singular problem (3.7), (3.8).

In order to verify the uniqueness, we suppose that $(\widetilde {R}_a, \widetilde {S}_a, t_a)(\tau , y)$ and $(\widetilde {R}_b, \widetilde {S}_b, t_b)(\tau , y)$ are two smooth solutions of the problem (3.7), (3.8) and consider their difference. Denote $X=\widetilde {R}_a-\widetilde {R}_b$ , $Y=\widetilde {S}_a-\widetilde {S}_b$ , and $T=t_a-t_b$ . By (3.7), one derives the equations for $(X,Y,T)$ as follows:

(3.111) $$ \begin{align} & X_\tau-A_a\tau X_y=(A_a-A_b)\tau \widetilde{R}_{by}+\frac{X-Y}{2\tau} \nonumber \\ &\quad +\bigg\{\frac{(\widetilde{R}_a-\widetilde{S}_a)^2}{4\tau} -(\widetilde{R}_a-\widetilde{S}_a)+\theta_{0}'\tau-w(t_a,y)\bigg\}(A_a-A_b) \nonumber \\ &\quad +A_b\bigg\{\frac{(R_a-S_a+R_b-S_b)}{4\tau}(X-Y)-(X-Y)-[w(t_a,y)-w(t_b,y)]\bigg\}, \end{align} $$

(3.112) $$ \begin{align} &Y_\tau+B_a\tau Y_y=-(B_a-B_b)\tau \widetilde{S}_{by} + \frac{Y-X}{2\tau} \nonumber \\ &\quad +\bigg\{\frac{(\widetilde{R}_a-\widetilde{S}_a)^2}{4\tau} +(\widetilde{R}_a-\widetilde{S}_a)-\theta_{0}'\tau+w(t_a,y)\bigg\}(B_a-B_b) \nonumber \\ &\quad +B_b\bigg\{\frac{(R_a-S_a+R_b-S_b)}{4\tau}(X-Y)+(X-Y)+[w(t_a,y)-w(t_b,y)]\bigg\}, \end{align} $$

and

(3.113) $$ \begin{align} T_\tau=C_a-C_b, \end{align} $$

where

$$ \begin{align*} A_i=\frac{1}{f_i+\widetilde{S}_i},\quad B_i=\frac{1}{f_i+\widetilde{R}_i},\quad C_i=\frac{1}{\frac{\widetilde{R}_i+\widetilde{S}_i}{2}-\tau-\widetilde{v}(t_i,y)+\theta_0(y)}, \end{align*} $$

and $f_i=\theta _0(y)-\widetilde {v}(t_i,y)-\tau $ for $i=a,b$ . Performing a direct calculation gets

$$ \begin{align*} A_a-A_b=&-A_aA_b\{[\widetilde{v}(t_a,y)-\widetilde{v}(t_b,y)]+Y\},\\ B_a-B_b=&-B_aB_b\{[\widetilde{v}(t_a,y)-\widetilde{v}(t_b,y)]+X\},\\ C_a-C_b=&-\frac{1}{2}C_aC_b\{X+Y-2[\widetilde{v}(t_a,y)-\widetilde{v}(t_b,y)]\}. \end{align*} $$

Integrating (3.111)–(3.113) from $0$ to $\xi $ and utilizing the estimates (3.86) and (3.105) and the mean value theorem, we can find that the functions $(X,Y,T)$ satisfy a homogeneous integral inequality system as the following form

(3.114) $$ \begin{align} \left\{ \begin{array}{l} \displaystyle \big|X(\xi,\eta)\big|\leq \int_{0}^\xi\bigg\{\frac{|X-Y|}{2\tau} +M^*\big(|X|+|Y|+|T|\big)\bigg\}\ \mathrm{d}\tau, \\[8pt] \displaystyle \big|Y(\xi,\eta)\big|\leq \int_{0}^\xi\bigg\{\frac{|X-Y|}{2\tau} +M^*\big(|X|+|Y|+|T|\big)\bigg\}\ \mathrm{d}\tau, \\[8pt] \displaystyle \big|T(\xi,\eta)\big|\leq \int_{0}^\xi M^*\big(|X|+|Y|+|T|\big) \ \mathrm{d}\tau, \end{array} \right. \end{align} $$

for some positive constant $M^*$ . One repeats the insertion of the right side of (3.114) to see that the functions $(X,Y,T)$ must satisfy

$$ \begin{align*} \big|X\big|, \big|Y\big|, \big|T\big|\leq \overline{M}^{*}\bigg(\frac{2}{3}\bigg)^\ell, \end{align*} $$

for arbitrary integer $\ell \geq 1$ and some positive constant $\overline {M}^*$ . This implies that $X(\tau ,y)=Y(\tau ,y)=T(\tau ,y)\equiv 0$ . Hence, the smooth solution of the singular initial value problem (3.7), (3.8) is unique.

3.5 The problem in the original plane

Based on the unique smooth solution $(\widetilde {R}, \widetilde {S}, t)(\tau , y)$ of problem (3.7), (3.8) obtained in Section 3.4, we establish the existence and uniqueness of smooth solutions for the initial value problem (3.2), (3.3) in this subsection.

By virtue of (3.86), we know that the functions $(\widetilde {R}, \widetilde {S}, t)(\tau , y)$ are in the space $\widetilde {\Sigma }(\widetilde {\delta })$ , from which and (3.16) one acquires

(3.115) $$ \begin{align} 0<\frac{1}{2}\underline{\theta}\leq\frac{\widetilde{R}(\tau,y)+\widetilde{S}(\tau,y)}{2}-\tau -\widetilde{v}(t(\tau,y),y)+\theta_0(y) \leq2\overline{\theta}, \end{align} $$

for $(\tau ,y)\in [0,\widetilde {\delta }]\times \mathbb {R}$ . Then we can use (3.5) and (3.7) to construct the functions $(t,x)$ as follows:

(3.116) $$ \begin{align} x=y,\quad t=t(\tau,y)=\int_{0}^\tau\frac{1}{\frac{\widetilde{R}(s,y)+\widetilde{S}(s,y)}{2}-s -\widetilde{v}(t(s,y),y)+\theta_0(y)}\ \mathrm{d}s, \end{align} $$

for any $(\tau ,y)\in [0,\widetilde {\delta }]\times \mathbb {R}$ . Obviously, by (3.115), the mapping $(\tau ,y)\mapsto (t,x)$ is global one-to-one in the region $[0,\widetilde {\delta }]\times \mathbb {R}$ . Now, set

(3.117) $$ \begin{align} \overline{\delta}=\frac{\widetilde{\delta}}{2\overline{\theta}}. \end{align} $$

Then, for any point $(t^*,x^*)\in [0,\overline {\delta }]\times \mathbb {R}$ , we see that there exists a unique corresponding point $(\tau ^*,y^*)\in [0,\widetilde {\delta }]\times \mathbb {R}$ such that

(3.118) $$ \begin{align} x^*=y^*,\quad t^*=t(\tau^*, y^*). \end{align} $$

Thus, one can define the functions $(\theta , \widetilde {R}, \widetilde {S})(t,x)$ as follows:

(3.119) $$ \begin{align} \theta(t^*,x^*)=\tau^*,\quad \widetilde{R}(t^*,x^*)=\widetilde{R}(\tau^*,y^*),\quad \widetilde{S}(t^*,x^*)=\widetilde{S}(\tau^*,y^*). \end{align} $$

In sum, we have obtained the functions $(\theta , \widetilde {R}, \widetilde {S})(t, x)$ in the region $[0,\overline {\delta }]\times \mathbb {R}$ .

We next verify that the functions defined in (3.119) satisfy the initial value conditions (3.3) and equations (3.2). By means of (3.86) and (3.119), one concludes

(3.120) $$ \begin{align} \displaystyle \big|\widetilde{R}(t,x)\big|, \big|\widetilde{S}(t,x)\big|\leq 3\overline{M}\tau^2=3\overline{M}\theta^2,\ \ \frac{1}{2}\underline{\theta}t\leq\theta(t,x)\leq 2\overline{\theta}t, \end{align} $$

from which we get $\theta (0,x)=\widetilde {R}(0,x)=\widetilde {S}(0,x)=0$ . Moreover, we calculate

(3.121) $$ \begin{align} \begin{array}{l} \displaystyle \widetilde{R}_t(t,x)=\widetilde{R}_\tau(\tau,y)\tau_t \\ =\widetilde{R}_\tau(\tau,y) \bigg(\frac{\widetilde{R}(\tau,y)+\widetilde{S}(\tau,y)}{2}-\tau -\widetilde{v}(t(\tau,y),y)+\theta_0(y)\bigg), \\[10pt] \displaystyle \widetilde{S}_t(t,x)=\widetilde{S}_\tau(\tau,y)\tau_t \\ =\widetilde{S}_\tau(\tau,y) \bigg(\frac{\widetilde{R}(\tau,y)+\widetilde{S}(\tau,y)}{2}-\tau -\widetilde{v}(t(\tau,y),y)+\theta_0(y)\bigg), \end{array} \end{align} $$

which along with (3.105) and (3.115) achieve $\widetilde {R}_t(0,x)=\widetilde {S}_t(0,x)=0$ . Thus, the functions $(\theta , \widetilde {R}, \widetilde {S})(t, x)$ satisfy the initial value conditions (3.3). Furthermore, we apply the fact $\tau _x=(\widetilde {R}-\widetilde {S})/2\tau $ by (3.6) to arrive at

$$ \begin{align*}\widetilde{R}_x=\frac{\widetilde{R}-\widetilde{S}}{2\tau}\widetilde{R}_\tau+\widetilde{R}_y, \end{align*} $$

which combined with (3.121) and (3.7) leads to

(3.122) $$ \begin{align} &\widetilde{R}_t-\theta\widetilde{R}_x= \bigg(\frac{\widetilde{R}(\tau,y)+\widetilde{S}(\tau,y)}{2}-\tau -\widetilde{v}(t(\tau,y),y)+\theta_0(y)\bigg)\widetilde{R}_\tau \nonumber \\ &\quad -\tau\cdot \bigg( \frac{\widetilde{R}-\widetilde{S}}{2\tau}\widetilde{R}_\tau+\widetilde{R}_y\bigg) \nonumber \\ =& \big(\widetilde{S}-\tau-\widetilde{v}+\theta_0\big)\widetilde{R}_\tau-\tau \widetilde{R}_y =(\widetilde{S}+f)\bigg(\widetilde{R}_\tau-\frac{\tau}{\widetilde{S}+f} \widetilde{R}_y\bigg) \nonumber \\ =& (\widetilde{S}+f)\bigg(\frac{\widetilde{R}-\widetilde{S}}{2\tau} +\frac{1}{f+\widetilde{S}}\cdot\frac{(\widetilde{R}-\widetilde{S})^2}{4\tau} -\frac{\widetilde{R}-\widetilde{S}}{f+\widetilde{S}} +\frac{\theta_{0}'}{f+\widetilde{S}}\tau-\frac{w(t,y)}{f+\widetilde{S}}\tau\bigg) \nonumber \\ =& \frac{(\widetilde{S}+f)(\widetilde{R}-\widetilde{S})}{2\tau} +\frac{(\widetilde{R}-\widetilde{S})^2}{4\tau} -(\widetilde{R}-\widetilde{S}) +\theta_{0}'\tau-w(t,y)\tau \nonumber \\ =& \frac{2(\widetilde{S}+f)(\widetilde{R}-\widetilde{S})+(\widetilde{R}-\widetilde{S})^2}{4\theta} -(\widetilde{R}-\widetilde{S}) +\theta_{0}'\theta-w(t,x)\theta \nonumber \\ =& \frac{\widetilde{R}+\widetilde{S}-2\widetilde{v} +2\theta_0}{4\theta}(\widetilde{R}-\widetilde{S})-(\widetilde{R}-\widetilde{S}) +\theta\theta_{0}'-\theta w, \end{align} $$

which is the desired equation for $\widetilde {R}$ in (3.2). One can check the other equations in (3.2) in a similar way. The inequalities in (3.4) come from (3.120). Therefore, the proof of Theorem 3.1 is completed.

In addition, we check that the following relation holds:

(3.123) $$ \begin{align} \theta_x=\frac{\widetilde{R}-\widetilde{S}}{2\theta}. \end{align} $$

To show (3.123), we set $\Phi =\widetilde {R}-\widetilde {S}-2\theta \theta _x$ and apply (3.2) to compute

(3.124) $$ \begin{align} &\partial _t\Phi=\widetilde{R}_t-\widetilde{S}_t-2\theta_t\theta_x-2\theta\theta_{tx} \nonumber \\ =&[\widetilde{R}_t-\theta\widetilde{R}_x]-[\widetilde{S}_t+\theta\widetilde{S}_x] -2\bigg(\frac{\widetilde{R}+\widetilde{S}}{2}-\theta-\widetilde{v}+\theta_0\bigg)\theta_x \nonumber \\ &\quad -2\theta(-\theta_x-\widetilde{v}_x +\theta_{0}') \nonumber \\ =&\frac{\widetilde{R}+\widetilde{S}-2\widetilde{v} +2\theta_0}{2\theta}(\widetilde{R}-\widetilde{S})-(\widetilde{R}-\widetilde{S}) -[\widetilde{R}+\widetilde{S}-2\theta-2\widetilde{v}+2\theta_0]\theta_x +2\theta\theta_x \nonumber \\ =&\frac{[\widetilde{R}+\widetilde{S}-2\widetilde{v} +2\theta_0-4\theta]t}{2\theta}\cdot\frac{\Phi}{t}. \end{align} $$

Recalling (3.17) gives

$$ \begin{align*}\bigg|\frac{[\widetilde{R}+\widetilde{S}-2\widetilde{v} +2\theta_0-4\theta]t}{2\theta}\bigg|\leq \frac{24\overline{M}\overline{\theta} t^2 +\widehat{M}_0t+2\overline{\theta}+8\overline{\theta} t}{\underline{\theta}}<\infty, \end{align*} $$

which, along with (3.124) and the initial condition $(\Phi /t)|_{t=0}=0$ , arrives at $\Phi \equiv 0$ . Hence, the relation (3.123) holds.

Finally, for later applications, we summarize some properties for the functions $(\theta , \widetilde {R}, \widetilde {S})(t, x)$ in the region $[0,\overline {\delta }]\times \mathbb {R}$

(3.125) $$ \begin{align} \begin{array}{l} \displaystyle \big|\widetilde{R}(t,x)\big|, \big|\widetilde{S}(t,x)\big|\leq 3\overline{M}\theta^2\leq 12\overline{M}\overline{\theta}^2 t^2,\ \ \frac{1}{2}\underline{\theta}t\leq\theta(t,x)\leq 2\overline{\theta}t, \\ \big|\widetilde{R}(t,x)-\widetilde{S}(t,x)\big|\leq 3\overline{M}\theta^2\leq 12\overline{M}\overline{\theta}^2 t^2, \\ \big|\widetilde{R}_t(t,x)\big|,\big|\widetilde{S}_t(t,x)\big|\leq 8\overline{M}\overline{\theta}^2 t, \quad |\theta_t(t,x)|\leq 2\overline{\theta}, \\ \big|\widetilde{R}_x(t,x)\big|,\big|\widetilde{S}_x(t,x)\big|\leq 24\overline{M}^2\overline{\theta}^2 t^2, \quad |\theta_x(t,x)|\leq 3\overline{M}\overline{\theta}t,\\ \displaystyle |\theta_{xx}(t,x)|\leq \frac{3}{2}\overline{M}\theta +\frac{9}{4}\overline{M}^2\theta\leq 3\overline{M}^2\theta \leq 6\overline{M}^2\overline{\theta} t. \end{array} \end{align} $$

Here, we used the following estimates:

$$ \begin{align*} \begin{array}{l} \displaystyle \tau=\theta(t,x)\leq 2\overline{\theta}t,\quad |\theta_x(t,x)|=\bigg|\frac{\widetilde{R}-\widetilde{S}}{2\tau}\bigg|\leq \frac{3}{2}\overline{M}\tau, \\[6pt] \displaystyle |\theta_t(t,x)|=\bigg|\frac{\widetilde{R}+\widetilde{S}}{2} -\tau -\widetilde{v} +\theta_0\bigg|\leq 3\overline{M}\tau^2 +\tau +\widehat{M}_0\cdot\frac{2}{\underline{\theta}}\tau +\overline{\theta}\leq 2\overline{\theta}, \\[6pt] \displaystyle \big|\widetilde{R}_t(t,x)\big|\leq |\widetilde{R}_\tau|\cdot|\theta_t|,\quad \big|\widetilde{S}_t(t,x)\big|\leq |\widetilde{S}_\tau|\cdot|\theta_t|, \\[6pt] \displaystyle \big|\widetilde{R}_x(t,x)\big|\leq \frac{3}{2}\overline{M}\tau\cdot \big|\widetilde{R}_\tau\big| +\big|\widetilde{R}_y\big|,\quad \big|\widetilde{S}_x(t,x)\big|\leq \frac{3}{2}\overline{M}\tau\cdot \big|\widetilde{S}_\tau\big| +\big|\widetilde{S}_y\big|, \end{array} \end{align*} $$

and

$$ \begin{align*} \big|\widetilde{R}_\tau\big|&\leq |A|\tau|\widetilde{R}_y| +\frac{|\widetilde{R}-\widetilde{S}|}{2\tau} +|A|\frac{|\widetilde{R}-\widetilde{S}|^2}{4\tau} +\frac{|A|}{2}|\widetilde{R}-\widetilde{S}| +|A\theta_{0}'|\tau +|Aw|\tau \\ &\leq \frac{\overline{M}}{16}\tau\cdot 3\overline{M}\tau^2 +\frac{3\overline{M}\tau}{2} +\frac{\overline{M}}{16}\cdot\frac{9\overline{M}^2\tau^3}{4}+\frac{\overline{M}}{32}\cdot3\overline{M}\tau^2 +\frac{\overline{M}}{16}\cdot\tau +\frac{\overline{M}}{16}\cdot\tau \\ &\leq \overline{M}\tau\bigg\{\frac{3\overline{M}\widetilde{\delta}^2}{16} +\frac{3}{2} + \frac{9(\overline{M}\widetilde{\delta})^2}{64}+\frac{\overline{M}\widetilde{\delta}}{32}+ \frac{1}{8}\bigg\}\leq 2\overline{M}\tau. \end{align*} $$

4.1 Preliminary results for heat equation

We first introduce some known results for the heat equation. These results can be found in the many excellent texts (see, e.g., [Reference Evans14]). Let $b(t,x)\in C^1$ be a function defined in the region $[0,\overline {\delta }]\times \mathbb {R}$ . Here, the positive number $\overline {\delta }$ is given in Theorem 3.1. Consider the following initial value problem for the heat equation:

(4.1) $$ \begin{align} \left\{ \begin{array}{l} \displaystyle \frac{\partial \widetilde{v}}{\partial t}-\frac{\partial ^2 \widetilde{v}}{\partial x^2}=b(t,x), \\[3pt] \widetilde{v}(t,x)|_{t=0}=0. \end{array} \right. \end{align} $$

We know that the fundamental solution of the heat equation is

(4.2) $$ \begin{align} G(t,x)=\frac{1}{\sqrt{4\pi t}}\exp\bigg\{-\frac{x^2}{4t}\bigg\}, \end{align} $$

that is, $G(t,x)$ satisfies

(4.3) $$ \begin{align} \displaystyle \frac{\partial G}{\partial t}-\frac{\partial ^2 G}{\partial x^2}=0, \ \mathrm{with}\ G(0,x)=\delta_0(x), \end{align} $$

where $\delta _0(x)$ is the Dirac function at $x=0$ . Thus, the smooth solution of problem (4.1) can be expressed by the fundamental solution

(4.4) $$ \begin{align} \begin{array}{l} \displaystyle \widetilde{v}(t,x)=\int_{0}^t\int_{\mathbb{R}}G(t-\varsigma,x-z)b(\varsigma,z)\ \mathrm{d}z\mathrm{ d}\varsigma \\[8pt] \displaystyle \qquad \quad =\int_{0}^t\int_{\mathbb{R}}G(t-\varsigma,z)b(\varsigma,x-z)\ \mathrm{d}z\mathrm{ d}\varsigma. \end{array} \end{align} $$

Furthermore, the functions $\widetilde {v}_x, \widetilde {v}_t, \widetilde {v}_{xx}$ , and $\widetilde {v}_{xt}$ can also be stated as follows:

(4.5) $$ \begin{align} \begin{array}{l} \displaystyle \widetilde{v}_x(t,x)=\int_{0}^t\int_{\mathbb{R}}\frac{\partial G(t-\varsigma,x-z)}{\partial x}b(\varsigma,z)\ \mathrm{d}z\mathrm{d}\varsigma \\[8pt] \ \displaystyle \qquad \quad =\int_{0}^t\int_{\mathbb{R}}\frac{\partial G(\varsigma,z)}{\partial z}b(t-\varsigma,x-z)\ \mathrm{d}z\mathrm{d}\varsigma \\[8pt] \ \displaystyle \qquad \quad =\int_{0}^t\int_{\mathbb{R}}\frac{\partial G(t-\varsigma,z)}{\partial z}b(\varsigma,x-z)\ \mathrm{d}z\mathrm{d}\varsigma \\[8pt] \ \displaystyle \qquad \quad =\int_{0}^t\int_{\mathbb{R}}G(t-\varsigma,z)b_x(\varsigma,x-z)\ \mathrm{d}z\mathrm{ d}\varsigma, \end{array} \end{align} $$

(4.6) $$ \begin{align} \begin{array}{l} \displaystyle \widetilde{v}_t(t,x)=\int_{0}^t\int_{\mathbb{R}}G(t-\varsigma,z)b_\varsigma(\varsigma,x-z)\ \mathrm{ d}z\mathrm{d}\varsigma +\int_{\mathbb{R}}G(t,z)b(0,x-z)\ \mathrm{d}z, \end{array} \end{align} $$

(4.7) $$ \begin{align} \begin{array}{l} \displaystyle \widetilde{v}_{xx}(t,x)=\int_{0}^t\int_{\mathbb{R}}\frac{\partial G(t-\varsigma,z)}{\partial z}b_x(\varsigma,x-z)\ \mathrm{d}z\mathrm{d}\varsigma, \\[8pt] \ \ \displaystyle \qquad \quad =\int_{0}^t\int_{\mathbb{R}}\frac{\partial ^2 G(t-\varsigma,x-z)}{\partial x^2}[b(\varsigma,z) -b(\varsigma,x)]\ \mathrm{d}z\mathrm{d}\varsigma, \end{array} \end{align} $$

and

(4.8) $$ \begin{align} \displaystyle \widetilde{v}_{xt}(t,x)=\int_{0}^t\int_{\mathbb{R}}\frac{\partial G(t-\varsigma,z)}{\partial z}b_\varsigma(\varsigma,x-z)\ \mathrm{d}z\mathrm{d}\varsigma +\int_{\mathbb{R}}G(t,z)b_x(0,x-z)\ \mathrm{d}z. \end{align} $$

We set

(4.9) $$ \begin{align} \displaystyle H_\sigma(t,x)=\frac{1}{\sqrt{4\pi}}t^{-\frac{\sigma}{2}}\exp\bigg\{-\frac{x^2}{16t}\bigg\}. \end{align} $$

Then one has

(4.10) $$ \begin{align} \int_{0}^t\int_{\mathbb{R}}H_\sigma(t-\varsigma,z)\ \mathrm{d}z\mathrm{ d}\varsigma=\frac{4}{3-\sigma}t^{\frac{3-\sigma}{2}}\ \ \mathrm{for}\ \sigma<3, \end{align} $$

and

(4.11) $$ \begin{align} \begin{array}{l} \displaystyle \quad \int_{t-\gamma}^t\int_{\mathbb{R}}H_\sigma(t-\varsigma,z)\ \mathrm{d}z\mathrm{ d}\varsigma=\frac{4}{3-\sigma}\gamma^{\frac{3-\sigma}{2}}\ \ \mathrm{for}\ \sigma<3\ \mathrm{and}\ 0<\gamma\leq t, \\[8pt] \displaystyle \int_{0}^{t_1-\gamma}\int_{\mathbb{R}}H_\sigma(t-\varsigma,z)\ \mathrm{ d}z\mathrm{d}\varsigma\leq\frac{4}{\sigma-3}\gamma^{\frac{3-\sigma}{2}}\ \ \mathrm{for}\ \sigma>3\ \mathrm{and}\ 0<\gamma\leq t_1\leq t. \end{array} \end{align} $$

The results in (4.9)–(4.11) can be found in [Reference Li, Yu and Shen31]. Thus, we obtain for $\sigma =1,2$ ,

(4.12) $$ \begin{align} \int_{0}^t\int_{\mathbb{R}}H_1(t-\varsigma,z)\ \mathrm{d}z\mathrm{d}\varsigma=2t, \quad \int_{0}^t\int_{\mathbb{R}}H_2(t-\varsigma,z)\ \mathrm{d}z\mathrm{d}\varsigma=4\sqrt{t}. \end{align} $$

In addition, it is clear that the following inequality holds

(4.13) $$ \begin{align} z^\beta\exp\bigg\{-\frac{z}{4}\bigg\}\leq C_\beta\exp\bigg\{-\frac{z}{16}\bigg\}, \end{align} $$

for $z\geq 0$ , where $\beta \geq 0$ is an arbitrary number and $C_\beta $ is a positive constant depending only on $\beta $ . Then, by (4.2), (4.9), and (4.13), one acquires

(4.14) $$ \begin{align} G(t-\varsigma,z)\leq H_1(t-\varsigma,z), \quad \big|\partial _zG(t-\varsigma,z)\big|\leq H_2(t-\varsigma,z), \end{align} $$

and

(4.15) $$ \begin{align} \bigg||z|^\beta \frac{\partial ^jG(t-\varsigma,z)}{\partial z^j}\bigg|\leq C_{j,\beta}H_{j+1-\beta}(t-\varsigma,z), \end{align} $$

where $C_{j,\beta }(j\geq 1)$ are positive constants depending only on $\beta $ . The inequalities in (4.15) can also be found in [Reference Li, Yu and Shen31]. Making use of (4.12) and (4.14), we gain

(4.16) $$ \begin{align} \int_{0}^t\int_{\mathbb{R}}\big|G(t-\varsigma,z)\big|\ \mathrm{d}z\mathrm{d}\varsigma\leq2t, \quad \int_{0}^t\int_{\mathbb{R}}\bigg|\frac{\partial G(t-\varsigma,z)}{\partial z}\bigg|\ \mathrm{d}z\mathrm{ d}\varsigma\leq4\sqrt{t}. \end{align} $$

4.2 The iterative sequence for the heat equation

For any $(t,x)\in [0,\overline {\delta }]\times \mathbb {R}$ , we set

$$ \begin{align*}\widetilde{v}^{(0)}(t,x)\equiv0, \end{align*} $$

which obviously satisfies $\widetilde {v}^{(0)}\in \Sigma (\overline {\delta })$ . Here, the space $\Sigma (\overline {\delta })$ is defined in (3.1) but with the number $\overline {\delta }$ replacing $\delta _0$ . In view of the results in Section 3, one concludes the functions $(\widetilde {R}^{(0)},\widetilde {S}^{(0)},\theta ^{(0)})(t,x)$ for $(t,x)\in [0,\overline {\delta }]\times \mathbb {R}$ . According to (4.4), we define the function $\widetilde {v}^{(1)}(t,x)$ in the region $[0,\overline {\delta }]\times \mathbb {R}$ as

(4.17) $$ \begin{align} \widetilde{v}^{(1)}(t,x)=&\int_{0}^t\int_{\mathbb{R}}G(t-\varsigma,x-z)b^{(0)}(\varsigma,z)\ \mathrm{d}z\mathrm{ d}\varsigma \nonumber \\ =&\int_{0}^t\int_{\mathbb{R}}G(t-\varsigma,z)b^{(0)}(\varsigma,x-z)\ \mathrm{d}z\mathrm{d}\varsigma, \end{align} $$

where

$$ \begin{align*}b^{(0)}(t,x)=\frac{\widetilde{R}^{(0)}(t,x)+\widetilde{S}^{(0)}(t,x)}{2}-\theta^{(0)}(t,x) -\widetilde{v}^{(0)}(t,x)+\theta_0(x)+v_{0}"(x). \end{align*} $$

After determining the function $\widetilde {v}^{(k)}(t,x)$ in $[0,\overline {\delta }]\times \mathbb {R}$ , one can achieve the functions $(\widetilde {R}^{(k)},\widetilde {S}^{(k)},\theta ^{(k)})(t,x)$ based on the results in Section 3. Then we define the function $\widetilde {v}^{(k+1)}(t,x)$ in the region $[0,\overline {\delta }]\times \mathbb {R}$ by the following relation:

(4.18) $$ \begin{align} \widetilde{v}^{(k+1)}(t,x)=&\int_{0}^t\int_{\mathbb{R}}G(t-\varsigma,x-z)b^{(k)}(\varsigma,z)\ \mathrm{d}z\mathrm{d}\varsigma \nonumber \\ =&\int_{0}^t\int_{\mathbb{R}}G(t-\varsigma,z)b^{(k)}(\varsigma,x-z)\ \mathrm{d}z\mathrm{d}\varsigma, \end{align} $$

where

$$ \begin{align*}b^{(k)}(t,x)=\frac{\widetilde{R}^{(k)}(t,x)+\widetilde{S}^{(k)}(t,x)}{2}-\theta^{(k)}(t,x) -\widetilde{v}^{(k)}(t,x)+\theta_0(x)+v_{0}"(x). \end{align*} $$

Therefore, we have constructed an iterative sequence $\{\widetilde {v}^{(k)}(t,x)\}$ .

Denote

(4.19) $$ \begin{align} \widehat{K}=\max\bigg\{&1, \max_{x\in\mathbb{R}}|\theta_0(x)+v_{0}"(x)|,\ \max_{x\in\mathbb{R}}|\theta_{0}'(x)+v_{0}"'(x)|, \nonumber \\ &\max_{x\in\mathbb{R}}|\theta_{0}"(x)+v_{0}""(x)| \bigg\}, \end{align} $$

which originates from the last two terms in the expression of $b^{(k)}$ . Set

(4.20) $$ \begin{align} \begin{array}{l} \displaystyle \widehat{M}_0=\max\bigg\{32\widehat{K},\ 12\overline{M}^2(1+\overline{\theta})^2, 32\overline{\theta}\big(C_{2,1}+2C_{3,1/2}+C_{3,0}\big)\bigg\},\\[8pt] \displaystyle \delta=\min\bigg\{\overline{\delta}, \frac{1}{\widehat{M}_0}, \frac{\underline{\theta}^2}{16\widehat{M}_0\overline{\theta}} \bigg\}. \end{array} \end{align} $$

Here, the constant $\overline {M}\geq 16$ is given in (3.30), which depends only on $\underline {\theta }$ and the $C^3$ norm of $\theta _0(y)$ , $\overline {\delta }$ is defined in (3.117), and $C_{j,\beta }$ are presented in (4.15). Then we see by (3.125) that if $\widetilde {v}(t,x)\in \Sigma (\delta )$ ,

(4.21) $$ \begin{align} \begin{array}{l} \displaystyle \big|\widetilde{R}(t,x)\big|, \big|\widetilde{S}(t,x)\big|\leq 3\overline{M}\theta^2 \leq \widehat{M}_0 t^2,\ \ \frac{1}{2}\underline{\theta}t\leq\theta(t,x)\leq 2\overline{\theta}t\leq \frac{1}{32}\widehat{M}_0t, \\ \displaystyle \big|\widetilde{R}(t,x)-\widetilde{S}(t,x)\big|\leq 3\overline{M}\theta^2 \leq \widehat{M}_0 t^2, \quad |\theta_t(t,x)|\leq \frac{1}{32}\widehat{M}_0, \\ \displaystyle \big|\widetilde{R}_t(t,x)\big|,\big|\widetilde{S}_t(t,x)\big|\leq \widehat{M}_0 t, \quad \big|\widetilde{R}_x(t,x)\big|,\big|\widetilde{S}_x(t,x)\big|\leq 6\overline{M}^2\theta^2 \leq \widehat{M}_{0} t^2, \\ \displaystyle |\theta_x(t,x)|\leq \frac{3}{2}\overline{M}\theta \leq \frac{1}{4}\widehat{M}_0t,\quad |\theta_{xx}(t,x)|\leq 3\overline{M}^2\theta\leq \frac{1}{4}\widehat{M}_0t. \end{array} \end{align} $$

For the sequence $\{\widetilde {v}^{(k)}(t,x)\}$ , we have the following lemma.

Lemma 4.1 For all $k\geq 0$ , there hold $\widetilde {v}^{(k)}(t,x)\in \Sigma (\delta )$ , that is, the functions $\widetilde {v}^{(k)}(t,x)(k=1,2,\ldots )$ satisfy $\widetilde {v}^{(k)}(t,x)\in C^1,\ \widetilde {v}_{x}^{(k)}(t,x)\in C^1$ , and $\forall \ (t,x)\in [0,\delta ]\times \mathbb {R}$

(4.22) $$ \begin{align} \begin{array}{l} \big|\widetilde{v}^{(k)}(t,x)\big|, \big|\widetilde{v}_{x}^{(k)}(t,x)\big|\leq \widehat{M}_0t,\quad \big|\widetilde{v}_{t}^{(k)}(t,x)\big|\leq \widehat{M}_0,\\ \big|\widetilde{v}_{xt}^{(k)}(t,x)\big|\leq \widehat{M}_0,\qquad \qquad \qquad \big|\widetilde{v}_{xx}^{(k)}(t,x)\big|\leq \widehat{M}_0\sqrt{t}. \end{array} \end{align} $$

Proof We mainly verify that all the inequalities in (4.22) are true for any $k\geq 0$ . The proof is also based on the argument of induction.

Due to $\widetilde {v}^{(0)}\equiv 0\in \Sigma (\delta )$ , we see by (4.21) that the functions $(\widetilde {R}^{(0)},\widetilde {S}^{(0)},\theta ^{(0)})(t,x)$ satisfy

(4.23) $$ \begin{align} \begin{array}{l} \displaystyle \big|\widetilde{R}^{(0)}(t,x)\big|, \big|\widetilde{S}^{(0)}(t,x)\big|\leq \widehat{M}_0 t^2,\ \ \frac{1}{2}\underline{\theta}t\leq\theta^{(0)}(t,x)\leq \frac{1}{32}\widehat{M}_0t, \\ \displaystyle \big|\widetilde{R}^{(0)}(t,x)-\widetilde{S}^{(0)}(t,x)\big|\leq \widehat{M}_0 t^2, \quad |\theta_{t}^{(0)}(t,x)|\leq \frac{1}{32}\widehat{M}_0, \quad |\theta_{x}^{(0)}(t,x)|\leq \frac{1}{4}\widehat{M}_0t, \\ \displaystyle \big|\widetilde{R}_{t}^{(0)}(t,x)\big|,\big|\widetilde{S}_{t}^{(0)}(t,x)\big|\leq \widehat{M}_0 t, \quad \big|\widetilde{R}_{x}^{(0)}(t,x)\big|,\big|\widetilde{S}_{x}^{(0)}(t,x)\big|\leq \widehat{M}_{0} t^2, \end{array} \end{align} $$

from which we find that

(4.24) $$ \begin{align} &\big|b^{(0)}(t,x)\big| \nonumber \\ &\leq \frac{|\widetilde{R}^{(0)}(t,x)|+|\widetilde{S}^{(0)}(t,x)|}{2}+|\theta^{(0)}(t,x)| + |\widetilde{v}^{(0)}(t,x)|+|\theta_0(x)+v_{0}"(x)| \nonumber \\ &\leq \widehat{M}_0t^2 +\frac{1}{32}\widehat{M}_0t +0 +\widehat{K} \leq \widehat{M}_0\delta^2 +\frac{1}{32}\widehat{M}_0\delta +\frac{1}{16}\widehat{M}_0\leq \frac{1}{4}\widehat{M}_0 \end{align} $$

and

(4.25) $$ \begin{align} &\big|b^{(0)}_{x}(t,x)\big| \nonumber \\ &\leq \frac{|\widetilde{R}_{x}^{(0)}(t,x)|+|\widetilde{S}_{x}^{(0)}(t,x)|}{2}+|\theta_{x}^{(0)}(t,x)| + |\widetilde{v}_{x}^{(0)}(t,x)|+|\theta_{0}'(x)+v_{0}"'(x)| \nonumber \\ &\leq \frac{1}{2}\widehat{M}_{0}\delta^2 +\frac{1}{4}\widehat{M}_{0}\delta +0 +\widehat{K} \leq \frac{1}{4}\widehat{M}_0, \end{align} $$

(4.26) $$ \begin{align} &\big|b^{(0)}_{t}(t,x)\big|\leq \frac{|\widetilde{R}_{t}^{(0)}(t,x)|+|\widetilde{S}_{t}^{(0)}(t,x)|}{2}+|\theta_{t}^{(0)}(t,x)| + |\widetilde{v}_{t}^{(0)}(t,x)| \nonumber \\ &\leq \widehat{M}_{0} \delta +2\overline{\theta} +0\leq 3\overline{\theta}\leq \frac{3}{32}\widehat{M}_{0}\leq \frac{1}{4}\widehat{M}_0. \end{align} $$

One combines (4.17) and (4.5)–(4.8) and utilizes (4.24)–(4.26) to arrive at

(4.27) $$ \begin{align} \begin{array}{l} \displaystyle \big|\widetilde{v}^{(1)}(t,x)\big| \leq \int_{0}^t\int_{\mathbb{R}}|G(t-\varsigma,z)|\cdot|b^{(0)}|\ \mathrm{d}z\mathrm{d}\varsigma \leq \frac{1}{4}\widehat{M}_0 \int_{0}^t\int_{\mathbb{R}}|G(t-\varsigma,z)| \ \mathrm{d}z\mathrm{d}\varsigma, \\[8pt] \displaystyle \big|\widetilde{v}_{x}^{(1)}(t,x)\big| \leq \int_{0}^t\int_{\mathbb{R}}|G(t-\varsigma,z)|\cdot|b_{x}^{(0)}|\ \mathrm{d}z\mathrm{d}\varsigma \leq \frac{1}{4}\widehat{M}_0 \int_{0}^t\int_{\mathbb{R}}|G(t-\varsigma,z)| \ \mathrm{d}z\mathrm{d}\varsigma, \\[8pt] \displaystyle \big|\widetilde{v}_{t}^{(1)}(t,x)\big| \leq \int_{0}^t\int_{\mathbb{R}}|G(t-\varsigma,z)|\cdot|b_{\varsigma}^{(0)}|\ \mathrm{d}z\mathrm{d}\varsigma + \int_{\mathbb{R}}|G(t,z)|\cdot|b^{(0)}(0,x-z)|\ \mathrm{d}z \\[8pt] \displaystyle \qquad \qquad \ \ \leq \frac{1}{4}\widehat{M}_0 \int_{0}^t\int_{\mathbb{R}}|G(t-\varsigma,z)|\ \mathrm{d}z\mathrm{d}\varsigma +\frac{1}{4}\widehat{M}_0\int_{\mathbb{R}}|G(t,z)|\ \mathrm{d}z, \end{array} \end{align} $$

and

(4.28) $$ \begin{align} \begin{array}{l} \displaystyle \big|\widetilde{v}_{xx}^{(1)}(t,x)\big| \leq \int_{0}^t\int_{\mathbb{R}}\bigg|\frac{\partial G(t-\varsigma,z)}{\partial z}\bigg|\cdot|b_{x}^{(0)}\bigg|\ \mathrm{d}z\mathrm{d}\varsigma \\[8pt] \displaystyle \leq \frac{1}{4}\widehat{M}_0 \int_{0}^t\int_{\mathbb{R}}\bigg|\frac{\partial G(t-\varsigma,z)}{\partial z}\bigg| \ \mathrm{d}z\mathrm{d}\varsigma, \\[8pt] \displaystyle \big|\widetilde{v}_{xt}^{(1)}(t,x)\big| \leq \int_{0}^t\int_{\mathbb{R}}\bigg|\frac{\partial G(t-\varsigma,z)}{\partial z}\bigg|\cdot|b_{\varsigma}^{(0)}|\ \mathrm{d}z\mathrm{d}\varsigma \\[8pt] \displaystyle \quad +\int_{\mathbb{R}}|G(t,z)|\cdot|b_{x}^{(0)}(0,x-z)|\ \mathrm{d}z \\[8pt] \displaystyle \leq \frac{1}{4}\widehat{M}_0 \int_{0}^t\int_{\mathbb{R}}\bigg|\frac{\partial G(t-\varsigma,z)}{\partial z}\bigg|\ \mathrm{d}z\mathrm{d}\varsigma +\frac{1}{4}\widehat{M}_0\int_{\mathbb{R}}|G(t,z)|\ \mathrm{d}z. \end{array} \end{align} $$

Making use of (4.16), (4.27), and (4.28), we get

(4.29) $$ \begin{align} \begin{array}{l} \displaystyle \big|\widetilde{v}^{(1)}(t,x)\big|, \big|\widetilde{v}_{x}^{(1)}(t,x)\big| \leq \frac{1}{4}\widehat{M}_0\cdot 2t=\frac{1}{2}\widehat{M}_0t\leq \widehat{M}_0t, \\[6pt] \displaystyle \big|\widetilde{v}_{t}^{(1)}(t,x)\big| \leq \frac{1}{4}\widehat{M}_0 \cdot 2t +\frac{1}{4}\widehat{M}_0\cdot 2\leq \frac{1}{2}(1+\delta)\widehat{M}_0\leq \widehat{M}_0, \\[6pt] \displaystyle \big|\widetilde{v}_{xx}^{(1)}(t,x)\big|\leq \frac{1}{4}\widehat{M}_0\cdot 4\sqrt{t}=\widehat{M}_0 \sqrt{t}, \\[6pt] \displaystyle \big|\widetilde{v}_{xt}^{(1)}(t,x)\big|\leq \frac{1}{4}\widehat{M}_0\cdot 4\sqrt{t} +\frac{1}{4}\widehat{M}_0 \cdot 2 =\frac{1}{2}(1+2\sqrt{\delta})\widehat{M}_0\leq \widehat{M}_0, \end{array} \end{align} $$

from which we gain that (4.22) holds for $k=1$ .

Assume that $\widetilde {v}^{(k)}(t,x)\in \Sigma (\delta )$ , that is, (4.22) is true for $n=k$ . Thanks to (4.21), the functions $(\widetilde {R}^{(k)},\widetilde {S}^{(k)},\theta ^{(k)})(t,x)$ obtained in Section 3 satisfy

(4.30) $$ \begin{align} \begin{array}{l} \displaystyle \big|\widetilde{R}^{(k)}(t,x)\big|, \big|\widetilde{S}^{(k)}(t,x)\big|\leq \widehat{M}_0 t^2,\ \ \frac{1}{2}\underline{\theta}t\leq\theta^{(k)}(t,x)\leq \frac{1}{32}\widehat{M}_0t, \\ \displaystyle |\theta_{t}^{(k)}(t,x)|\leq \frac{1}{32}\widehat{M}_0, \big|\widetilde{R}^{(k)}(t,x)-\widetilde{S}^{(k)}(t,x)\big|\leq \widehat{M}_0 t^2, \\ \displaystyle |\theta_{x}^{(k)}(t,x)|\leq \frac{1}{4}\widehat{M}_0t, \quad |\theta_{xx}^{(k)}(t,x)|\leq \frac{1}{4}\widehat{M}_0t, \\ \displaystyle \big|\widetilde{R}_{t}^{(k)}(t,x)\big|,\big|\widetilde{S}_{t}^{(k)}(t,x)\big|\leq \widehat{M}_0 t, \quad \big|\widetilde{R}_{x}^{(k)}(t,x)\big|,\big|\widetilde{S}_{x}^{(k)}(t,x)\big|\leq \widehat{M}_{0} t^2. \end{array} \end{align} $$

It suggests by (4.30) and the induction assumptions that

(4.31) $$ \begin{align} \begin{array}{l} \displaystyle \big|b^{(k)}(t,x)\big| \leq \widehat{M}_{0}\delta^2 +\frac{1}{32}\widehat{M}_{0}\delta +\widehat{M}_0\delta +\widehat{K}\leq \frac{1}{4}\widehat{M}_0, \\[6pt] \displaystyle \big|b_{x}^{(k)}(t,x)\big| \leq \frac{1}{2} \widehat{M}_{0}^2\delta^2 +\frac{1}{4}\widehat{M}_{0}\delta +\widehat{M}_0\delta +\widehat{K}\leq \frac{1}{4}\widehat{M}_0, \\[6pt] \displaystyle \big|b_{t}^{(k)}(t,x)\big| \leq \widehat{M}_{0}\delta +2\overline{\theta} +\widehat{M}_0 \leq 3\overline{\theta} +\widehat{M}_0 \leq \frac{5}{4}\widehat{M}_0. \end{array} \end{align} $$

Thus, by applying (4.18), (4.5)–(4.8), (4.16), and (4.31), one achieves

(4.32) $$ \begin{align} \begin{array}{l} \displaystyle \big|\widetilde{v}^{(k+1)}(t,x)\big| \leq \int_{0}^t\int_{\mathbb{R}}|G(t-\varsigma,z)|\cdot|b^{(k)}|\ \mathrm{d}z\mathrm{d}\varsigma \leq \frac{1}{4}\widehat{M}_0 \cdot 2t\leq\widehat{M}_0t, \\[8pt] \displaystyle \big|\widetilde{v}_{x}^{(k+1)}(t,x)\big| \leq \int_{0}^t\int_{\mathbb{R}}|G(t-\varsigma,z)|\cdot|b_{x}^{(k)}|\ \mathrm{d}z\mathrm{d}\varsigma \leq \frac{1}{4}\widehat{M}_0 \cdot 2t\leq\widehat{M}_0t, \\[8pt] \displaystyle \big|\widetilde{v}_{t}^{(k+1)}(t,x)\big| \\[8pt] \displaystyle \leq \int_{0}^t\int_{\mathbb{R}}|G(t-\varsigma,z)|\cdot|b_{\varsigma}^{(k)}|\ \mathrm{d}z\mathrm{d}\varsigma + \int_{\mathbb{R}}|G(t,z)|\cdot|b^{(k)}(0,x-z)|\ \mathrm{d}z \\[8pt] \displaystyle \leq \frac{5}{4}\widehat{M}_0 \cdot2t +\frac{1}{4}\widehat{M}_0\cdot2\leq\bigg(\frac{5}{2}\delta+\frac{1}{2}\bigg)\widehat{M}_0 \leq\bigg(\frac{5}{64}+\frac{1}{2}\bigg)\widehat{M}_0\leq \widehat{M}_0, \end{array} \end{align} $$

and

(4.33) $$ \begin{align} \begin{array}{l} \displaystyle \big|\widetilde{v}_{xx}^{(k+1)}(t,x)\big| \leq \int_{0}^t\int_{\mathbb{R}}\bigg|\frac{\partial G(t-\varsigma,z)}{\partial z}\bigg|\cdot|b_{x}^{(k)}\bigg|\ \mathrm{d}z\mathrm{d}\varsigma \leq \widehat{M}_0 \sqrt{t}, \\[8pt] \displaystyle \big|\widetilde{v}_{xt}^{(k+1)}(t,x)\big| \\[8pt] \displaystyle \leq \int_{0}^t\int_{\mathbb{R}}\bigg|\frac{\partial G(t-\varsigma,z)}{\partial z}\bigg|\cdot|b_{\varsigma}^{(k)}|\ \mathrm{d}z\mathrm{d}\varsigma +\int_{\mathbb{R}}|G(t,z)|\cdot|b_{x}^{(k)}(0,x-z)|\ \mathrm{d}z \\[8pt] \displaystyle \leq \frac{5}{4}\widehat{M}_0 \cdot 4\sqrt{t} +\frac{1}{4}\widehat{M}_0\cdot 2\leq \bigg(5\sqrt{\delta} +\frac{1}{2}\bigg)\widehat{M}_0\leq \widehat{M}_0. \end{array} \end{align} $$

From (4.32) and (4.33), we acquire $\widetilde {v}^{(k+1)}(t,x)\in \Sigma (\delta )$ and then complete the proof of the lemma.

By virtue of Lemma 4.1 and (4.21), we find that for any $k\geq 0$ and $(t,x)\in [0,\delta ]\times \mathbb {R}$ ,

(4.34) $$ \begin{align} \begin{array}{l} \displaystyle \big|\widetilde{R}^{(k)}(t,x)\big|, \big|\widetilde{S}^{(k)}(t,x)\big|\leq 3\overline{M}(\theta^{(k)})^2 \leq \widehat{M}_0 t^2,\\ \displaystyle \frac{1}{2}\underline{\theta}t\leq\theta^{(k)}(t,x)\leq 2\overline{\theta}t\leq \frac{1}{32}\widehat{M}_0t, \\ \displaystyle \big|\widetilde{R}^{(k)}(t,x)-\widetilde{S}^{(k)}(t,x)\big|\leq 3\overline{M}(\theta^{(k)})^2 \leq \widehat{M}_0 t^2, \quad |\theta_{t}^{(k)}(t,x)|\leq \frac{1}{32}\widehat{M}_0, \\ \displaystyle \big|\widetilde{R}_{t}^{(k)}(t,x)\big|,\big|\widetilde{S}_{t}^{(k)}(t,x)\big|\leq \widehat{M}_0 t, \\ \displaystyle \big|\widetilde{R}_{x}^{(k)}(t,x)\big|,\big|\widetilde{S}_{x}^{(k)}(t,x)\big|\leq 6\overline{M}^2(\theta^{(k)})^2 \leq \widehat{M}_{0} t^2, \\ \displaystyle |\theta_{x}^{(k)}(t,x)|\leq \frac{3}{2}\overline{M}\theta^{(k)} \leq \frac{1}{4}\widehat{M}_0t,\quad |\theta_{xx}^{(k)}(t,x)|\leq 3\overline{M}^2\theta^{(k)}\leq \frac{1}{4}\widehat{M}_0t. \end{array} \end{align} $$

Furthermore, the function $\widetilde {v}^{(k)}_{xx}(t,x)$ owns the following regularity.

Lemma 4.2 For any $k\geq 0$ , the function $\widetilde {v}^{(k)}_{xx}(t,x)$ is uniformly $\alpha $ -Hölder continuous with respect to x, that is, there exists a positive constant $\widehat {C}_\alpha $ independent of k such that for any two points $(t,x_1)$ and $(t,x_2)$ in $[0,\delta ]\times \mathbb {R}$ , there hold for $\alpha \in (0,1)$

(4.35) $$ \begin{align} \big|\widetilde{v}^{(k)}_{xx}(t,x_1)-\widetilde{v}^{(k)}_{xx}(t,x_2)\big|\leq \widehat{C}_\alpha|x_1-x_2|^{\alpha}. \end{align} $$

Proof The proof of the lemma is similar to that of Lemma 3.3 in [Reference Li, Yu and Shen31], but we list it here for the sake of completeness. By means of (4.7) and (4.18), we see that

(4.36) $$ \begin{align} \widetilde{v}_{xx}^{(k+1)}(t,x)=\int_{0}^t\int_{\mathbb{R}}\frac{\partial ^2 G(t-\varsigma,x-z)}{\partial x^2}[b^{(k)}(\varsigma,z) -b^{(k)}(\varsigma,x)]\ \mathrm{d}z\mathrm{d}\varsigma, \end{align} $$

and then for any $x_1<x_2$ ,

(4.37) $$ \begin{align} &\widetilde{v}_{xx}^{(k+1)}(t,x_1)-\widetilde{v}_{xx}^{(k+1)}(t,x_2) \nonumber \\ =&\int_{0}^t\int_{\mathbb{R}}\frac{\partial ^2 G(t-\varsigma,x_1-z)}{\partial x^2}[b^{(k)}(\varsigma,z) -b^{(k)}(\varsigma,x_1)]\ \mathrm{d}z\mathrm{d}\varsigma \nonumber \\ &-\int_{0}^t\int_{\mathbb{R}}\frac{\partial ^2 G(t-\varsigma,x_2-z)}{\partial x^2}[b^{(k)}(\varsigma,z) -b^{(k)}(\varsigma,x_2)]\ \mathrm{d}z\mathrm{d}\varsigma. \end{align} $$

Denote $\gamma =(x_2-x_1)^2$ . The proof is divided into two cases: $\gamma> t$ and $\gamma \leq t$ .

If $\gamma> t$ , we use (4.31), (4.15) with $\beta =1$ , and (4.12) to estimate (4.37) directly

(4.38) $$ \begin{align} &\big|\widetilde{v}_{xx}^{(k+1)}(t,x_1)-\widetilde{v}_{xx}^{(k+1)}(t,x_2) \big|\leq \int_{0}^t\int_{\mathbb{R}}\bigg|\frac{\partial ^2 G(t-\varsigma,x_1-z)}{\partial x^2}\bigg|\cdot|b_{x}^{(k)}|\cdot|z-x_1| \ \mathrm{d}z\mathrm{d}\varsigma \nonumber \\ &\quad +\int_{0}^t\int_{\mathbb{R}}\bigg|\frac{\partial ^2 G(t-\varsigma,x_2-z)}{\partial x^2}\bigg|\cdot|b_{x}^{(k)}|\cdot|z-x_2|\ \mathrm{d}z\mathrm{d}\varsigma \nonumber \\ &\leq \frac{1}{2}\widehat{M}_0\int_{0}^t\int_{\mathbb{R}}C_{2,1}H_2(t-\varsigma, z)\ \mathrm{d}z\mathrm{d}\varsigma \nonumber \\ &\leq 2\widehat{M}_0C_{2,1}\delta^{\frac{1-\alpha}{2}}t^{\frac{\alpha}{2}} \leq 2\widehat{M}_0C_{2,1}|x_2-x_1|^{\alpha}. \end{align} $$

If $\gamma \leq t$ , the relation (4.37) can be rewritten as

(4.39) $$ \begin{align} \widetilde{v}_{xx}^{(k+1)}(t,x_1)-\widetilde{v}_{xx}^{(k+1)}(t,x_2) =I_{1}^{(k)}+I_{2}^{(k)}+I_{3}^{(k)}+I_{4}^{(k)}, \end{align} $$

where

$$ \begin{align*} \begin{array}{l} \displaystyle I_{1}^{(k)}=\int_{t-\gamma}^t\int_{\mathbb{R}}\frac{\partial ^2 G(t-\varsigma,x_1-z)}{\partial x^2}[b^{(k)}(\varsigma,z) -b^{(k)}(\varsigma,x_1)]\ \mathrm{d}z\mathrm{d}\varsigma, \\[10pt] \displaystyle I_{2}^{(k)}=-\int_{t-\gamma}^t\int_{\mathbb{R}}\frac{\partial ^2 G(t-\varsigma,x_2-z)}{\partial x^2}[b^{(k)}(\varsigma,z) -b^{(k)}(\varsigma,x_2)]\ \mathrm{d}z\mathrm{d}\varsigma, \\[10pt] \displaystyle I_{3}^{(k)}=\int_{0}^{t-\gamma}\int_{\mathbb{R}}\frac{\partial ^2 G(t-\varsigma,x_2-z)}{\partial x^2}[b^{(k)}(\varsigma,x_2) -b^{(k)}(\varsigma,x_1)]\ \mathrm{d}z\mathrm{ d}\varsigma, \\[10pt] \displaystyle I_{4}^{(k)}=\int_{0}^{t-\gamma}\int_{\mathbb{R}}\bigg(\frac{\partial ^2 G(t-\varsigma,x_1-z)}{\partial x^2} -\frac{\partial ^2 G(t-\varsigma,x_2-z)}{\partial x^2}\bigg) \\[10pt] \displaystyle \qquad \qquad \qquad \times[b^{(k)}(\varsigma,z) -b^{(k)}(\varsigma,x_1)]\ \mathrm{d}z\mathrm{ d}\varsigma. \end{array} \end{align*} $$

For the term $I_{1}^{(k)}$ , it concludes by (4.31), (4.15) with $\beta =1$ , and (4.11) that

(4.40) $$ \begin{align} \big|I_{1}^{(k)}\big|&\leq \int_{t-\gamma}^t\int_{\mathbb{R}}\bigg|\frac{\partial ^2 G(t-\varsigma,x_1-z)}{\partial x^2}\bigg|\cdot|b_{x}^{(k)}|\cdot|z-x_1|\ \mathrm{d}z\mathrm{d}\varsigma \nonumber \\ &\leq \frac{1}{4}\widehat{M}_{0}\int_{t-\gamma}^t\int_{\mathbb{R}}C_{2,1}H_2(t-\varsigma,z)\ \mathrm{d}z\mathrm{ d}\varsigma =\widehat{M}_{0}C_{2,1}\gamma^{\frac{1}{2}} \nonumber \\ &\leq \widehat{M}_{0}C_{2,1}\delta^{\frac{1-\alpha}{2}}|x_1-x_2|^\alpha\leq \widehat{M}_{0}C_{2,1}|x_1-x_2|^\alpha. \end{align} $$

The estimate (4.40) also holds for the term $I_{2}^{(k)}$ . Furthermore, one acquires by the integration by parts that $I_{3}^{(k)}=0$ . For the term $I_{4}^{(k)}$ , we use (4.31), (4.15), and (4.11) again to find

(4.41) $$ \begin{align} \big|I_{4}^{(k)}\big|&=\bigg|\int_{x_1}^{x_2}\int_{0}^{t-\gamma}\int_{\mathbb{R}} \frac{\partial ^3 G(t-\varsigma,x-z)}{\partial x^3}[b^{(k)}(\varsigma,z) -b^{(k)}(\varsigma,x_1)]\ \mathrm{d}z\mathrm{ d}\varsigma\mathrm{d}x\bigg| \nonumber \\ &\leq \int_{x_1}^{x_2}\int_{0}^{t-\gamma}\int_{\mathbb{R}} \bigg|\frac{\partial ^3 G(t-\varsigma,x-z)}{\partial x^3}\bigg|\cdot \frac{1}{2}\widehat{M}_0\cdot|z-x_1|^\alpha\ \mathrm{d}z\mathrm{ d}\varsigma\mathrm{d}x \nonumber \\ &\leq \frac{1}{2}\widehat{M}_0\int_{x_1}^{x_2}\int_{0}^{t-\gamma}\int_{\mathbb{R}} \bigg|\frac{\partial ^3 G(t-\varsigma,x-z)}{\partial x^3}\bigg|\big(|z-x|^\alpha+|x-x_1|^\alpha\big)\ \mathrm{d}z\mathrm{ d}\varsigma\mathrm{d}x \nonumber \\ &\leq \frac{1}{2}\widehat{M}_0 \int_{x_1}^{x_2}\int_{0}^{t-\gamma}\int_{\mathbb{R}} \bigg\{ C_{3,\alpha}H_{4-\alpha}(t-\varsigma, z) \nonumber \\ &\quad + C_{3,0}|x-x_1|^\alpha H_{4}(t-\varsigma, z)\bigg\}\ \mathrm{d}z\mathrm{d}\varsigma\mathrm{d}x \nonumber \\ &\leq \frac{1}{2}\widehat{M}_0\bigg\{C_{3,\alpha}|x_2-x_1|\cdot\frac{4}{(4-\alpha)-3}\gamma^{\frac{3-(4-\alpha)}{2}} \nonumber \\ &\quad +C_{3,0}|x_2-x_1|^{1+\alpha}\cdot\frac{4}{4-3}\gamma^{\frac{3-4}{2}}\bigg\} \nonumber \\ &=\ 2\widehat{M}_0\bigg(\frac{C_{3,\alpha}}{1-\alpha}+C_{3,0}\bigg)|x_2-x_1|^\alpha. \end{align} $$

Substituting (4.40) and (4.41) into (4.39) yields

(4.42) $$ \begin{align} \big|\widetilde{v}_{xx}^{(k+1)}(t,x_1)-\widetilde{v}_{xx}^{(k+1)}(t,x_2)\big| \leq 2\widehat{M}_0\bigg(C_{2,1}+\frac{C_{3,\alpha}}{1-\alpha}+C_{3,0}\bigg)|x_2-x_1|^\alpha. \end{align} $$

We denote

$$ \begin{align*}\widehat{C}_\alpha=2\widehat{M}_0\bigg(C_{2,1}+\frac{C_{3,\alpha}}{1-\alpha}+C_{3,0}\bigg), \end{align*} $$

and combine (4.38) and (4.42) to achieve

(4.43) $$ \begin{align} \big|\widetilde{v}_{xx}^{(k+1)}(t,x_1)-\widetilde{v}_{xx}^{(k+1)}(t,x_2)\big| \leq \widehat{C}_\alpha|x_1-x_2|^\alpha, \end{align} $$

which is the desired inequality (4.35).

4.3 The convergence of the iterative sequence (I)

In order to establish the uniform convergence of the iterative sequence $\{\widetilde {v}^{(k)}(t,x)\}$ , we need to derive its series of properties in the space $\Sigma (\delta )$ .

By direct calculations, one applies (4.18) and (4.5) to get

(4.44) $$ \begin{align} &\ \big|\widetilde{v}^{(k+1)}(t,x)- \widetilde{v}^{(k)}(t,x)\big| \nonumber \\ &\leq \int_{0}^t\int_{\mathbb{R}}G(t-\varsigma,z)\bigg\{\frac{1}{2}(T_{12}^{(k)} +T_{13}^{(k)}) +T_{14}^{(k)} + T_{15}^{(k)} \bigg\}(\varsigma,x-z)\ \mathrm{d}z\mathrm{d}\varsigma \end{align} $$

and

(4.45) $$ \begin{align} &\ \big|\widetilde{v}_{x}^{(k+1)}(t,x)- \widetilde{v}_{x}^{(k)}(t,x)\big|\leq \int_{0}^t\int_{\mathbb{R}}\bigg|\frac{\partial G(t-\varsigma,z)}{\partial z}\bigg| \nonumber \\ &\quad \times \bigg\{\frac{1}{2}(T_{12}^{(k)} +T_{13})^{(k)} +T_{14}^{(k)} + T_{15}^{(k)} \bigg\}(\varsigma,x-z)\ \mathrm{ d}z\mathrm{d}\varsigma, \end{align} $$

where

$$ \begin{align*} T_{12}^{(k)}(t,x)=&\big|\widetilde{R}^{(k)}(t,x)-\widetilde{R}^{(k-1)}(t,x)\big|, \ T_{13}^{(k)}(t,x)=\big|\widetilde{S}^{(k)}(t,x)-\widetilde{S}^{(k-1)}(t,x)\big|, \\ T_{14}^{(k)}(t,x)=&\big|\theta^{(k)}(t,x)-\theta^{(k-1)}(t,x)\big|, \ T_{15}^{(k)}(t,x)=\big|\widetilde{v}^{(k)}(t,x)-\widetilde{v}^{(k-1)}(t,x)\big|. \end{align*} $$

To estimate $T_{12-14}^{(k)}$ , we first recall (3.2) to gain the equations for $(\widetilde {R}^{(k)}, \widetilde {S}^{(k)}, \theta ^{(k)})(t,x)$ in $[0,\delta ]\times \mathbb {R}$

(4.46) $$ \begin{align} \left\{ \begin{array}{l} \displaystyle \widetilde{R}_{t}^{(k)}-\theta^{(k)} \widetilde{R}_{x}^{(k)} =(\Psi^{(k)}-1)(\widetilde{R}^{(k)}-\widetilde{S}^{(k)}) +\theta^{(k)}\theta_{0}'-\theta^{(k)} \widetilde{v}_{x}^{(k)}=:F_{1}^{(k)}(t,x),\\ \displaystyle \widetilde{S}_{t}^{(k)}+\theta^{(k)} \widetilde{S}_{x}^{(k)}=(\Psi^{(k)}-1)(\widetilde{S}^{(k)}-\widetilde{R}^{(k)}) -\theta^{(k)}\theta_{0}'+\theta^{(k)} \widetilde{v}_{x}^{(k)}=:F_{2}^{(k)}(t,x),\\ \displaystyle \theta_{t}^{(k)}=\frac{\widetilde{R}^{(k)}+\widetilde{S}^{(k)}}{2}-\theta^{(k)}-\widetilde{v}^{(k)}+\theta_1 =:F_{3}^{(k)}(t,x), \end{array} \right. \end{align} $$

where

(4.47) $$ \begin{align} \Psi^{(k)}=\frac{\widetilde{R}^{(k)}+\widetilde{S}^{(k)}-2\widetilde{v}^{(k)} +2\theta_0}{4\theta^{(k)}}. \end{align} $$

It follows by (4.46) that

$$ \begin{align*} \left\{ \begin{array}{l} \displaystyle \big(\widetilde{R}^{(k)}-\widetilde{R}^{(k-1)}\big)_{t}-\theta^{(k)} \big(\widetilde{R}^{(k)}-\widetilde{R}^{(k-1)}\big)_x \\[6pt] \displaystyle \quad =F_{1}^{(k)}-F_{1}^{(k-1)} +(\theta^{(k)}-\theta^{(k-1)})\widetilde{R}_{x}^{(k-1)},\\[6pt] \displaystyle \big(\widetilde{S}^{(k)}-\widetilde{S}^{(k-1)}\big)_{t}+\theta^{(k)} \big(\widetilde{S}^{(k)}-\widetilde{S}^{(k-1)}\big)_x \\[6pt] \displaystyle \quad =F_{2}^{(k)}-F_{2}^{(k-1)} -(\theta^{(k)}-\theta^{(k-1)})\widetilde{S}_{x}^{(k-1)},\\[6pt] \displaystyle \big(\theta^{(k)}-\theta^{(k-1)}\big)_{t}=F_{3}^{(k)}-F_{3}^{(k-1)}, \end{array} \right. \end{align*} $$

from which one acquires for any $(\xi ,\eta )\in [0,\delta ]\times \mathbb {R}$ ,

(4.48) $$ \begin{align} \left\{ \begin{array}{l} \displaystyle T_{12}^{(k)}(\xi,\eta)\leq \int_{0}^\xi \bigg\{ \big|F_{1}^{(k)}-F_{1}^{(k-1)}\big| + |\widetilde{R}_{x}^{(k-1)}|T_{14}^{(k)} \bigg\}(t,x_{-}^{(k)}(t))\ \mathrm{d}t, \\[8pt] \displaystyle T_{13}^{(k)}(\xi,\eta)\leq \int_{0}^\xi \bigg\{ \big|F_{2}^{(k)}-F_{2}^{(k-1)}\big| + |\widetilde{S}_{x}^{(k-1)}|T_{14}^{(k)} \bigg\}(t,x_{+}^{(k)}(t))\ \mathrm{d}t, \\[8pt] \displaystyle T_{14}^{(k)}(\xi,\eta)\leq \int_{0}^\xi \big|F_{3}^{(k)}-F_{3}^{(k-1)}\big|(t,\eta)\ \mathrm{d}t, \end{array} \right. \end{align} $$

where $x_{\pm }^{(k)}(t)=x_{\pm }^{(k)}(t;\xi ,\eta )(t\in [0,\xi ])$ are provided by

$$ \begin{align*} \left\{ \begin{array}{l} \displaystyle \frac{\mathrm{d}x_{\pm}^{(k)}(t;\xi,\eta)}{\mathrm{ d}t}=\pm\theta^{(k)}(t,x_{\pm}^{(k)}(t;\xi,\eta)),\\ x_{\pm}^{(k)}(\xi;\xi,\eta)=\eta. \end{array} \right. \end{align*} $$

Performing direct calculations achieves

(4.49) $$ \begin{align} \begin{array}{l} \displaystyle \big|F_{1}^{(k)}-F_{1}^{(k-1)}\big|, \big|F_{2}^{(k)}-F_{2}^{(k-1)}\big| \\ \leq \displaystyle \big(|\Psi^{(k)}| +1\big)(T_{12}^{(k)}+T_{13}^{(k)}) +|\theta^{(k-1)}|T_{16}^{(k)} \\ \ \displaystyle + |\widetilde{R}^{(k-1)}-\widetilde{S}^{(k-1)}|T_{17}^{(k)} + \big( |\theta_{0}'|+|\widetilde{v}_{x}^{(k)}|\big)T_{14}^{(k)}, \\ \displaystyle \big|F_{3}^{(k)}-F_{3}^{(k-1)}\big|\leq \frac{1}{2}(T_{12}^{(k)}+T_{13}^{(k)}) +T_{14}^{(k)} +T_{15}^{(k)}, \end{array} \end{align} $$

where

$$ \begin{align*} T_{16}^{(k)}= \big|\widetilde{v}_{x}^{(k)}-\widetilde{v}_{x}^{(k-1)}\big|, \quad T_{17}^{(k)}=\big|\Psi^{(k)}-\Psi^{(k-1)}\big|. \end{align*} $$

Moreover, thanks to (4.34) and Lemma 4.1, we can get a more precise estimate for $\theta ^{(k)}$

$$ \begin{align*} \theta_0-\bigg(\widehat{M}_0t^2 +\frac{1}{32}\widehat{M}_0t +\widehat{M}_0t\bigg)\leq \theta_{t}^{(k)} \leq \theta_0+\bigg(\widehat{M}_0t^2 +\frac{1}{32}\widehat{M}_0t +\widehat{M}_0t\bigg), \end{align*} $$

and thus

$$ \begin{align*} \theta_0-\frac{17}{16}\widehat{M}_0t \leq \theta_{t}^{(k)} \leq \theta_0+\frac{17}{16}\widehat{M}_0t, \end{align*} $$

and then

(4.50) $$ \begin{align} \theta_0t-\frac{17}{32}\widehat{M}_0t^2\leq \theta^{(k)} \leq \theta_0t +\frac{17}{32}\widehat{M}_0t^2. \end{align} $$

It suggests by (4.34) and (4.50) that

(4.51) $$ \begin{align} \displaystyle\big|\Psi^{(k)}\big| &\leq \frac{2\theta_0+ 2\widehat{M}_{0}t^2 +2\widehat{M}_{0}t}{4t\big\{\theta_0- \frac{17}{32}\widehat{M}_0t\big\}} = \frac{1}{2t} + \frac{\frac{17}{32}\widehat{M}_0+\widehat{M}_{0}t +\widehat{M}_{0}}{2(\theta_0- \widehat{M}_0t)} \nonumber \\ &\leq \frac{1}{2t} +\frac{2\widehat{M}_0}{\underline{\theta}}. \end{align} $$

Furthermore, for the term $T_{17}^{(k)}$ , we obtain

(4.52) $$ \begin{align} T_{17}^{(k)}&\leq \frac{T_{12}^{(k)}+T_{13}^{(k)} +2T_{15}^{(k)}}{4\theta^{(k)}} +\frac{\big|\widetilde{R}^{(k-1)}+\widetilde{S}^{(k-1)}-2\widetilde{v}^{(k-1)} +2\theta_0\big|}{4\theta^{(k)}\theta^{(k-1)}}T_{14}^{(k)} \nonumber \\ &\leq \frac{1}{2\underline{\theta}t} (T_{12}^{(k)}+T_{13}^{(k)} +2T_{15}^{(k)}) +\frac{2\widehat{M}_0\delta^2 +2\widehat{M}_{0}\delta +2\overline{\theta}}{\underline{\theta}^2t^2}T_{14}^{(k)} \nonumber \\ &\leq \frac{1}{2\underline{\theta}t} (T_{12}^{(k)}+T_{13}^{(k)} +2T_{15}^{(k)}) +\frac{3\overline{\theta}}{\underline{\theta}^2t^2}T_{14}^{(k)}, \end{align} $$

by $8\widehat {M}_{0}\delta \leq \underline {\theta }<\overline {\theta }$ . Putting (4.51) and (4.52) into (4.49) and applying (4.34) again yield

(4.53) $$ \begin{align} \displaystyle & \big|F_{1}^{(k)}-F_{1}^{(k-1)}\big|+ |\widetilde{R}_{x}^{(k-1)}|T_{14}^{(k)}, \ \ \big|F_{2}^{(k)}-F_{2}^{(k-1)}\big|+ |\widetilde{S}_{x}^{(k-1)}|T_{14}^{(k)} \nonumber \\ &\leq \frac{T_{12}^{(k)}+T_{13}^{(k)}}{2t} +\bigg(\frac{2\widehat{M}_0}{\underline{\theta}} +1 +\frac{\widehat{M}_0\delta}{2\underline{\theta}}\bigg)(T_{12}^{(k)}+T_{13}^{(k)})+\frac{1}{32}\widehat{M}_0t T_{16}^{(k)} \nonumber \\ &\quad +\frac{\widehat{M}_0t}{\underline{\theta}} T_{15}^{(k)} +\bigg(\frac{3\widehat{M}_{0}\overline{\theta}}{\underline{\theta}^2} +\overline{\theta}_1+ \frac{1}{6}(\widehat{M}_0\delta)^2+ \widehat{M}_0\delta\bigg)T_{14}^{(k)} \nonumber \\ &\leq \frac{T_{12}^{(k)}+T_{13}^{(k)}}{2t} +\frac{3\widehat{M}_0}{\underline{\theta}} \big(T_{12}^{(k)}+T_{13}^{(k)}\big)+\frac{4\widehat{M}_{0}\overline{\theta}}{\underline{\theta}^2}T_{14}^{(k)} \nonumber \\ &\ +\bigg(\frac{\widehat{M}_0\delta}{32}+\frac{\widehat{M}_0\delta}{\underline{\theta}}\bigg) \big(T_{15}^{(k)} + T_{16}^{(k)}\big) \nonumber \\ &\leq \frac{T_{12}^{(k)}+T_{13}^{(k)}}{2t} +\frac{1}{4\delta} \big(T_{12}^{(k)}+T_{13}^{(k)}+ T_{14}^{(k)} \big)+\frac{1}{4} \big(T_{15}^{(k)} + T_{16}^{(k)}\big). \end{align} $$

We insert (4.53) and (4.49) into (4.48) to see that

(4.54) $$ \begin{align} \displaystyle & T_{12}^{(k)}(\xi,\eta), T_{13}^{(k)}(\xi,\eta) \nonumber \\ &\leq \int_{0}^\xi \bigg\{ \frac{T_{12}^{(k)}+T_{13}^{(k)}}{2t} +\frac{1}{4\delta} \big(T_{12}^{(k)}+T_{13}^{(k)}+ T_{14}^{(k)} \big)+\frac{1}{4} \big(T_{15}^{(k)} + T_{16}^{(k)}\big) \bigg\} \ \mathrm{d}t \end{align} $$

and

(4.55) $$ \begin{align} T_{14}^{(k)}(\xi,\eta)\leq \int_{0}^\xi \bigg\{ \big(T_{12}^{(k)}+T_{13}^{(k)} +T_{14}^{(k)}\big) +T_{15}^{(k)}\bigg\}(t,\eta)\ \mathrm{d}t. \end{align} $$

Based on (4.54) and (4.55), we can show the following lemma.

Lemma 4.3 For any $k\geq 1$ and $(\xi ,\eta )\in [0,\delta ]\times \mathbb {R}$ , if $T_{15}^{(k)}(\xi ,\eta )$ and $T_{16}^{(k)}(\xi ,\eta )$ satisfy

(4.56) $$ \begin{align} T_{15}^{(k)}(\xi,\eta),T_{16}^{(k)}(\xi,\eta)\leq \widehat{M}_0\xi\bigg(\frac{2}{3}\bigg)^{k-1}, \end{align} $$

then there hold

(4.57) $$ \begin{align} T_{12}^{(k)}(\xi,\eta),T_{13}^{(k)}(\xi,\eta),T_{14}^{(k)}(\xi,\eta)\leq 2\widehat{M}_0\xi^2\bigg(\frac{2}{3}\bigg)^{k-1}. \end{align} $$

Proof By (4.34) and the definitions of $T_{12-14}^{(k)}$ in (4.45), we first find that

(4.58) $$ \begin{align} T_{12}^{(k)}(\xi,\eta),T_{13}^{(k)}(\xi,\eta)\leq 2\widehat{M}_0\xi^2,\quad T_{14}^{(k)}(\xi,\eta)\leq \frac{1}{16}\widehat{M}_0\xi. \end{align} $$

Substituting (4.58) into (4.55) and utilizing the assumptions in (4.56) give

(4.59) $$ \begin{align} T_{14}^{(k)}(\xi,\eta)&\leq \int_{0}^\xi \bigg\{ 4\widehat{M}_0t^2 +\frac{1}{16}\widehat{M}_0t +\widehat{M}_0t\bigg(\frac{2}{3}\bigg)^{k-1}\bigg\}(t,\eta)\ \mathrm{d}t \nonumber \\ &\leq \bigg[\frac{4\delta}{3} +\frac{1}{32} +\frac{1}{2}\bigg(\frac{2}{3}\bigg)^{k-1}\bigg] \widehat{M}_0\xi^2\leq 2\widehat{M}_0\xi^2. \end{align} $$

We now put the estimates (4.58) and (4.59) into (4.54) and apply the assumptions in (4.56) to deduce

(4.60) $$ \begin{align} &\displaystyle T_{12}^{(k)}(\xi,\eta), T_{13}^{(k)}(\xi,\eta) \nonumber \\ &\leq \int_{0}^\xi \bigg\{ 2\widehat{M}_0t +\frac{1}{4\delta}\cdot 3\cdot 2\widehat{M}_0t^2 +\frac{1}{4} \cdot 2\widehat{M}_0t\bigg(\frac{2}{3}\bigg)^{k-1} \bigg\} \ \mathrm{d}t \nonumber \\ &\leq \widehat{M}_0\xi^2 +\frac{1}{2\delta}\widehat{M}_0\xi^3 + \frac{1}{4} \widehat{M}_{0}\xi^2\bigg(\frac{2}{3}\bigg)^{k-1} \leq \bigg[\frac{3}{4}\cdot 2 +\frac{1}{2}\bigg(\frac{2}{3}\bigg)^{k-1}\bigg]\widehat{M}_0\xi^2. \end{align} $$

One also has by (4.59)

(4.61) $$ \begin{align} \displaystyle T_{14}^{(k)}(\xi,\eta)&\leq \int_{0}^\xi \bigg\{ 6\widehat{M}_0t^2 +\widehat{M}_0t\bigg(\frac{2}{3}\bigg)^{k-1}\bigg\}(t,\eta)\ \mathrm{d}t \nonumber \\ &\leq \bigg[2\delta +\frac{1}{2}\bigg(\frac{2}{3}\bigg)^{k-1}\bigg] \widehat{M}_0\xi^2 \leq \bigg[\frac{3}{4}\cdot 2 +\frac{1}{2}\bigg(\frac{2}{3}\bigg)^{k-1}\bigg]\widehat{M}_0\xi^2. \end{align} $$

Inserting the estimates of $T_{12-14}^{(k)}$ in (4.60) and (4.61) into (4.54) and (4.55) and using the assumptions in (4.56) again arrive at

(4.62) $$ \begin{align} &\displaystyle T_{12}^{(k)}(\xi,\eta), T_{13}^{(k)}(\xi,\eta) \nonumber \\ &\leq \int_{0}^\xi \bigg\{ \bigg[\frac{3}{4}\cdot 2 +\frac{1}{2}\bigg(\frac{2}{3}\bigg)^{k-1}\bigg]\widehat{M}_0t +\frac{3}{4\delta}\bigg[\frac{3}{4}\cdot 2 +\frac{1}{2}\bigg(\frac{2}{3}\bigg)^{k-1}\bigg]\widehat{M}_0t^2 \nonumber \\ &\qquad + \frac{1}{2}\widehat{M}_{0}t\bigg(\frac{2}{3}\bigg)^{k-1} \bigg\} \ \mathrm{d}t \nonumber \\ &\leq \bigg[\frac{3}{4}\cdot 2 +\frac{1}{2}\bigg(\frac{2}{3}\bigg)^{k-1}\bigg]\cdot\frac{1}{2}\widehat{M}_0\xi^2 +\bigg\{\frac{3}{8} +\frac{1}{8}\bigg(\frac{2}{3}\bigg)^{k-1}\bigg\}\widehat{M}_0\xi^2 + \frac{1}{4}\widehat{M}_0\xi^2\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ &\leq \bigg[\frac{3}{4}\cdot 2 +\frac{1}{2}\bigg(\frac{2}{3}\bigg)^{k-1}\bigg]\cdot\frac{3}{4}\widehat{M}_0\xi^2 +\frac{1}{2}\widehat{M}_0\xi^2\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ = &\bigg[\bigg(\frac{3}{4}\bigg)^2\cdot 2 +\frac{1}{2}\displaystyle\sum_{j=0}^{1}\bigg(\frac{3}{4}\bigg)^j\bigg(\frac{2}{3}\bigg)^{k-1}\bigg] \widehat{M}_0\xi^2 \end{align} $$

and

(4.63) $$ \begin{align} &\displaystyle T_{14}^{(k)}(\xi,\eta)\leq \int_{0}^\xi \bigg\{ 3\cdot \bigg[\frac{3}{4}\cdot 2 +\frac{1}{2}\bigg(\frac{2}{3}\bigg)^{k-1}\bigg]\widehat{M}_0t^2 + \widehat{M}_0t\bigg(\frac{2}{3}\bigg)^{k-1}\bigg\}\ \mathrm{d}t \nonumber \\ = &\bigg[\frac{3}{4}\cdot 2 +\frac{1}{2}\bigg(\frac{2}{3}\bigg)^{k-1}\bigg]\widehat{M}_0\xi^3 +\frac{1}{2}\widehat{M}_0\xi^2\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ &\leq \bigg[\bigg(\frac{3}{4}\bigg)^2\cdot 2 +\frac{1}{2}\displaystyle\sum_{j=0}^{1}\bigg(\frac{3}{4}\bigg)^j\bigg(\frac{2}{3}\bigg)^{k-1}\bigg] \widehat{M}_0\xi^2. \end{align} $$

We next repeatedly substitute the new obtained estimates of $T_{12-14}^{(k)}$ into (4.54) and (4.55) to get for arbitrary integer $\ell \geq 1$ ,

(4.64) $$ \begin{align} \displaystyle &T_{12}^{(k)}(\xi,\eta), T_{13}^{(k)}(\xi,\eta), T_{14}^{(k)}(\xi,\eta) \nonumber \\ &\leq \bigg[\bigg(\frac{3}{4}\bigg)^\ell\cdot 2 +\frac{1}{2}\displaystyle\sum_{j=0}^{\ell}\bigg(\frac{3}{4}\bigg)^j\bigg(\frac{2}{3}\bigg)^{k-1}\bigg] \widehat{M}_0\xi^2. \end{align} $$

It follows by the arbitrariness of $\ell $ that

(4.65) $$ \begin{align} \displaystyle &T_{12}^{(k)}(\xi,\eta), T_{13}^{(k)}(\xi,\eta), T_{14}^{(k)}(\xi,\eta) \nonumber \\ &\leq \frac{1}{2}\displaystyle\sum_{j=0}^{\infty}\bigg(\frac{3}{4}\bigg)^j\bigg(\frac{2}{3}\bigg)^{k-1}\widehat{M}_0\xi^2\leq 2\widehat{M}_0\xi^2\bigg(\frac{2}{3}\bigg)^{k-1}, \end{align} $$

which finishes the proof of the lemma.

By virtue of Lemma 4.3 and (4.44) and (4.45), we can achieve the following lemma.

Lemma 4.4 Let the iterative sequence $\{\widetilde {v}^{(k)}\}$ be defined by (4.18). Then, for all $k\geq 0$ , the following inequalities

(4.66) $$ \begin{align} \big|\widetilde{v}^{(k+1)}(t,x)- \widetilde{v}^{(k)}(t,x)\big|, \big|\widetilde{v}_{x}^{(k+1)}(t,x)- \widetilde{v}_{x}^{(k)}(t,x)\big|\leq \widehat{M}_0t\bigg(\frac{2}{3}\bigg)^k \end{align} $$

hold for $(t,x)\in [0,\delta ]\times \mathbb {R}$ .

Proof We show the lemma by using the argument of induction again. Obviously, the results of Lemma 4.1 and the fact $\widetilde {v}^{(0)}(t,x)\equiv 0$ indicate that (4.66) is valid for $k=0$ .

Now, suppose that all inequalities in (4.66) are true for $n=k$ . Recalling the definitions of $T_{15}^{(k)}$ and $T_{16}^{(k)}$ , the induction assumptions mean that

(4.67) $$ \begin{align} \begin{array}{l} \displaystyle T_{15}^{(k)}(t,x)=\big|\widetilde{v}^{(k)}(t,x)- \widetilde{v}^{(k-1)}(t,x)\big|\leq \widehat{M}_0t\bigg(\frac{2}{3}\bigg)^{k-1}, \\[8pt] \displaystyle T_{16}^{(k)}(t,x)=\big|\widetilde{v}_{x}^{(k)}(t,x)- \widetilde{v}_{x}^{(k-1)}(t,x)\big|\leq \widehat{M}_0t\bigg(\frac{2}{3}\bigg)^{k-1}, \end{array} \end{align} $$

from which and Lemma 4.3 one has

(4.68) $$ \begin{align} T_{12}^{(k)}(t,x),T_{13}^{(k)}(t,x),T_{14}^{(k)}(t,x)\leq 2\widehat{M}_0t^2\bigg(\frac{2}{3}\bigg)^{k-1}, \end{align} $$

for any $(t,x)\in [0,\delta ]\times \mathbb {R}$ . Inserting (4.67) and (4.68) into (4.44) and (4.45) and making use of (4.16) give

(4.69) $$ \begin{align} &\ \big|\widetilde{v}^{(k+1)}(t,x)- \widetilde{v}^{(k)}(t,x)\big| \nonumber \\ &\leq \int_{0}^t\int_{\mathbb{R}}G(t-\varsigma,z)\bigg\{4\widehat{M}_0\varsigma^2\bigg(\frac{2}{3}\bigg)^{k-1} + \widehat{M}_0\varsigma\bigg(\frac{2}{3}\bigg)^{k-1} \bigg\}(\varsigma,x-z)\ \mathrm{d}z\mathrm{d}\varsigma \nonumber \\ &\leq (4t^2+t)\widehat{M}_0\bigg(\frac{2}{3}\bigg)^{k-1}\int_{0}^t\int_{\mathbb{R}}G(t-\varsigma,z)\ \mathrm{ d}z\mathrm{d}\varsigma \nonumber \\ &\leq (4t^2+t)\widehat{M}_0\bigg(\frac{2}{3}\bigg)^{k-1}\cdot 2t\leq (8\delta^2+2\delta)\widehat{M}_0t\bigg(\frac{2}{3}\bigg)^{k-1} \leq \widehat{M}_0t\bigg(\frac{2}{3}\bigg)^{k} \end{align} $$

and

(4.70) $$ \begin{align} &\ \big|\widetilde{v}_{x}^{(k+1)}(t,x)- \widetilde{v}_{x}^{(k)}(t,x)\big| \nonumber \\ &\leq \int_{0}^t\int_{\mathbb{R}}\bigg|\frac{\partial G(t-\varsigma,z)}{\partial z}\bigg|\bigg\{4\widehat{M}_0\varsigma^2\bigg(\frac{2}{3}\bigg)^{k-1} + \widehat{M}_0\varsigma\bigg(\frac{2}{3}\bigg)^{k-1} \bigg\}(\varsigma,x-z)\ \mathrm{d}z\mathrm{d}\varsigma \nonumber \\ &\leq (4t^2+t)\widehat{M}_0\bigg(\frac{2}{3}\bigg)^{k-1}\int_{0}^t\int_{\mathbb{R}}\bigg|\frac{\partial G(t-\varsigma,z)}{\partial z}\bigg|\ \mathrm{d}z\mathrm{d}\varsigma \nonumber \\ &\leq (4t^2+t)\widehat{M}_0\bigg(\frac{2}{3}\bigg)^{k-1}\cdot4\sqrt{t} \leq (16\delta\sqrt{\delta}+4\sqrt{\delta})\widehat{M}_0t\bigg(\frac{2}{3}\bigg)^{k-1} \nonumber \\ &\leq \widehat{M}_0t\bigg(\frac{2}{3}\bigg)^{k}, \end{align} $$

by the choice of $\delta $ in (4.20). We combine (4.69) and (4.70) to complete the proof of the lemma.

4.4 The improved regularity of the iterative sequence

In this subsection, we improve the regularity of the function $\widetilde {v}_{xx}^{(k)}(t,x)$ with respect to x, which will be used to establish the uniform convergence of sequences $\{(\widetilde {v}_{t}^{(k)}, \widetilde {v}_{xt}^{(k)}, \widetilde {v}_{xx}^{(k)})(t,x)\}$ in the space $\Sigma (\delta )$ .

Integrating system (4.46) along the characteristic curves $x_{i}^{(k)}(t)(i=\pm ,0)$ and differentiating the resulting with respect to $\eta $ , we deduce for any $(\xi ,\eta )\in [0,\delta ]\times \mathbb {R}$ ,

(4.71) $$ \begin{align} \left\{ \begin{array}{l} \displaystyle \widetilde{R}_{\eta}^{(k)}(\xi,\eta)=\int_{0}^{\xi}\frac{\partial F_{1}^{(k)}}{\partial x}\cdot\frac{\partial x_{-}^{(k)}}{\partial \eta}(t, x_{-}^{(k)}(t))\ \mathrm{d}t, \\[8pt] \displaystyle \widetilde{S}_{\eta}^{(k)}(\xi,\eta)=\int_{0}^{\xi}\frac{\partial F_{2}^{(k)}}{\partial x}\cdot\frac{\partial x_{+}^{(k)}}{\partial \eta}(t, x_{+}^{(k)}(t))\ \mathrm{d}t, \\[8pt] \displaystyle \theta_{\eta}^{(k)}(\xi,\eta)=\int_{0}^{\xi}\frac{\partial F_{3}^{(k)}}{\partial \eta}(t, \eta)\ \mathrm{d}t, \end{array} \right. \end{align} $$

where $x_{\pm }^{(k)}(t)=x_{\pm }^{(k)}(t;\xi ,\eta )$ are determined by (4.48), and

(4.72) $$ \begin{align} \frac{\partial x_{\pm}^{(k)}}{\partial \eta}=\exp\bigg\{ \int_{\xi}^t\theta_{x}^{(k)}(s,x_{\pm}^{(k)}(s))\ \mathrm{d}s\bigg\}. \end{align} $$

From (4.71), one obtains for any $(\xi ,\eta _1), (\xi ,\eta _2)\in [0,\delta ]\times \mathbb {R}$ ,

(4.73) $$ \begin{align} \begin{array}{l} \displaystyle I_{5}^{(k)}:=\big|\widetilde{R}_{\eta}^{(k)}(\xi,\eta_1)-\widetilde{R}_{\eta}^{(k)}(\xi,\eta_2)\big| \leq\int_{0}^{\xi}I_{7}^{(k)}\bigg|\frac{\partial x_{-}^{(k)}}{\partial \eta}\bigg| +I_{8}^{(k)}\bigg|\frac{\partial F_{1}^{(k)}}{\partial x}\bigg|\ \mathrm{d}t, \\[8pt] \displaystyle I_{6}^{(k)}:=\big|\widetilde{S}_{\eta}^{(k)}(\xi,\eta_1)-\widetilde{S}_{\eta}^{(k)}(\xi,\eta_2)\big| \leq\int_{0}^{\xi}I_{9}^{(k)}\bigg|\frac{\partial x_{+}^{(k)}}{\partial \eta}\bigg| +I_{10}^{(k)}\bigg|\frac{\partial F_{2}^{(k)}}{\partial x}\bigg|\ \mathrm{d}t, \end{array} \end{align} $$

where

$$ \begin{align*} I_{7}^{(k)}&=\bigg|\frac{\partial F_{1}^{(k)}}{\partial x}(t, x_{-}^{(k)}(t;\xi,\eta_1)) -\frac{\partial F_{1}^{(k)}}{\partial x}(t, x_{-}^{(k)}(t;\xi,\eta_2))\bigg|, \\ I_{8}^{(k)}&=\bigg|\frac{\partial x_{-}^{(k)}}{\partial \eta}(t, x_{-}^{(k)}(t;\xi,\eta_1)) -\frac{\partial x_{-}^{(k)}}{\partial \eta}(t, x_{-}^{(k)}(t;\xi,\eta_2))\bigg|, \\ I_{9}^{(k)}&=\bigg|\frac{\partial F_{2}^{(k)}}{\partial x}(t, x_{+}^{(k)}(t;\xi,\eta_1)) -\frac{\partial F_{2}^{(k)}}{\partial x}(t, x_{+}^{(k)}(t;\xi,\eta_2))\bigg|, \\ I_{10}^{(k)}&=\bigg|\frac{\partial x_{+}^{(k)}}{\partial \eta}(t, x_{+}^{(k)}(t;\xi,\eta_1)) -\frac{\partial x_{+}^{(k)}}{\partial \eta}(t, x_{+}^{(k)}(t;\xi,\eta_2))\bigg|. \end{align*} $$

We point out by the estimate $\theta _{\eta \eta }^{(k)}$ in (4.34) that there is no need to consider the difference between $\theta _{\eta }^{(k)}(\xi ,\eta _1)$ and $\theta _{\eta }^{(k)}(\xi ,\eta _2)$ . One performs direct calculations and uses (4.34) and (4.72) to achieve

(4.74) $$ \begin{align} \displaystyle \bigg|\frac{\partial x_{\pm}^{(k)}}{\partial \eta}\bigg|\leq \exp\bigg(\int_{0}^\xi \frac{1}{4}\widehat{M}_0s\ \mathrm{d}s\bigg)\leq e^{\widehat{M}_0\delta^2} \end{align} $$

and

(4.75) $$ \begin{align} &\bigg|\frac{\partial x_{\pm}^{(k)}}{\partial \eta}(t, x_{\pm}^{(k)}(t;\xi,\eta_1)) -\frac{\partial x_{\pm}^{(k)}}{\partial \eta}(t, x_{\pm}^{(k)}(t;\xi,\eta_2))\bigg| \nonumber \\ &\leq \exp\bigg(\int_{0}^\xi \frac{1}{4}\widehat{M}_0s\ \mathrm{ d}s\bigg)\int_{t}^\xi|\theta_{xx}^{(k)}|\cdot\big|x_{\pm}^{(k)}(s;\xi,\eta_1) -x_{\pm}^{(k)}(s;\xi,\eta_2)\big| \ \mathrm{d}s \nonumber \\ &\leq \frac{1}{4}e^{\widehat{M}_0\delta^2}\overline{M}\widehat{M}_0\xi^2D_{\pm}^{(k)}, \end{align} $$

where

$$ \begin{align*}D_{\pm}^{(k)}=\max_{t\in[0,\xi]}\big|x_{\pm}^{(k)}(t;\xi,\eta_1) -x_{\pm}^{(k)}(t;\xi,\eta_2)\big|. \end{align*} $$

Recalling the definitions of $x_{\pm }^{(k)}(t;\xi ,\eta )$ in (4.48) and utilizing (4.34) get

$$ \begin{align*} &\big|x_{\pm}^{(k)}(t;\xi,\eta_1) -x_{\pm}^{(k)}(t;\xi,\eta_2)\big| \\ &\leq |\eta_1-\eta_2| +\int_{t}^\xi\big|\theta^{(k)}(s,x_{\pm}^{(k)}(s;\xi,\eta_1)) -\theta^{(k)}(s,x_{\pm}^{(k)}(s;\xi,\eta_2))\big| \ \mathrm{d}s \\ &\leq |\eta_1-\eta_2| +\int_{t}^\xi|\theta_{x}^{(k)}|\cdot\big|x_{\pm}^{(k)}(s;\xi,\eta_1) - x_{\pm}^{(k)}(s;\xi,\eta_2)\big| \ \mathrm{d}s \\ &\leq |\eta_1-\eta_2| +\frac{1}{8}\widehat{M}_0\xi^2D_{\pm}^{(k)}, \end{align*} $$

for any $t\in [0,\xi ]$ , from which one finds that

(4.76) $$ \begin{align} D_{\pm}^{(k)}\leq 2|\eta_1-\eta_2|. \end{align} $$

We put (4.76) into (4.75) to gain

(4.77) $$ \begin{align} I_{8}^{(k)}, I_{10}^{(k)} \leq \frac{1}{2}e^{\widehat{M}_0\delta^2}\overline{M}\widehat{M}_0\xi^2|\eta_1-\eta_2|\leq \frac{1}{12}\widehat{M}_{0}^2\xi^2|\eta_1-\eta_2|. \end{align} $$