Let $$F(x,t)=\frac{1}{\sqrt{4\pi kt}} e^{-\frac{x^2}{4kt}}$$ be the fundamental solution to the heat equation and observe $$\int_{-\infty}^\infty F(x-y,t) dx=1$$ for arbitary $y$.
Next, assume that $\phi(x)$ satisfies $$\vert \phi(x_1)-\phi(x_2)\vert \leq e^{\vert x_1\vert + \vert x_2 \vert}\vert x_1-x_2 \vert$$ for all $x_1,x_2 \in \mathbb{R}$. I want to show that $$u(x,t)=\int_{-\infty}^\infty F(x-y,t)\phi(y)dy$$ is a solution to the wave equation $$\begin{cases}u_t-ku_{xx}=0 \\ u(x,0)=\phi(x)\end{cases}$$
Here is my approach to show $u(x,0)=\phi(x)$:
Since $\int_{-\infty}^\infty F(x-y,t) dx=1$ for arbitary $y$, we have $$\int_{-\infty}^\infty F(x-y,t)(\phi(y)-\phi(x))dy=u(x,t)-\phi(x)$$ Let $\xi=\frac{x-y}{\sqrt{4kt}}\Leftrightarrow y=x-\xi\sqrt{4kt}$. Then $$\vert u(x,t)-\phi(x) \vert =\left\vert \int_{-\infty}^\infty \frac{1}{\sqrt{\pi}}e^{-\xi^2}(\phi(x-\xi\sqrt{4kt})-\phi(x))d\xi \right\vert$$ $$\leq \left\vert \int_{-\infty}^\infty \frac{1}{\sqrt{\pi}}e^{-\xi^2}\vert \phi(x-\xi\sqrt{4kt}-\phi(x)\vert d\xi \right\vert$$ $$=\int_{-\infty}^\infty \frac{1}{\sqrt{\pi}}e^{-\xi^2}\vert \phi(x-\xi\sqrt{4kt}-\phi(x)\vert d\xi $$ $$ \leq \int_{-\infty}^\infty \frac{1}{\sqrt{\pi}}e^{-\xi^2}e^{\vert x \vert+\vert x-\xi\sqrt{4kt}\vert}\sqrt{4kt}\vert \xi \vert d\xi$$ $$=\frac{\sqrt{4kt}}{\sqrt{\pi}}\int_{-\infty}^\infty \vert \xi \vert e^{-\xi^2}e^{\vert x \vert+\vert x-\xi\sqrt{4kt}\vert} d\xi$$ $$ \leq \frac{\sqrt{4kt}}{\sqrt{\pi}}\int_{-\infty}^\infty \vert \xi \vert e^{-\xi^2}e^{2\vert x \vert}e^{\vert\xi\sqrt{4kt}\vert} d\xi$$ $$=\frac{\sqrt{4kt}}{\sqrt{\pi}} e^{2\vert x \vert}\int_{-\infty}^\infty \vert \xi \vert e^{-\xi^2+2\vert \xi \vert \sqrt{kt}}d\xi$$ $$=\frac{\sqrt{4kt}}{\sqrt{\pi}} e^{2\vert x \vert}\int_{-\infty}^\infty \vert \xi \vert e^{-(\xi-\sqrt{kt})^2+kt}d\xi$$ $$=\frac{e^{kt}\sqrt{4kt}}{\sqrt{\pi}} e^{2\vert x \vert}\int_{-\infty}^\infty \vert \xi \vert e^{-(\xi-\sqrt{kt})^2}d\xi$$ Let $\eta=\xi-\sqrt{kt}$. Then $$\vert u(x,t)-\phi(x) \vert \leq \frac{e^{kt}\sqrt{4kt}}{\sqrt{\pi}} e^{2\vert x \vert}\int_{-\infty}^\infty \vert \eta + \sqrt{kt} \vert e^{-\eta^2}d\eta $$ $$ \leq \frac{e^{kt}\sqrt{4kt}}{\sqrt{\pi}} e^{2\vert x \vert} \left( \int_{-\infty}^\infty \vert \eta \vert e^{-\eta^2}d\eta + \int_{-\infty}^\infty \vert \sqrt{kt} \vert e^{-\eta^2}d\eta\right)$$ $$=\frac{e^{kt}\sqrt{4kt}}{\sqrt{\pi}} e^{2\vert x \vert} (1+\sqrt{\pi kt})$$ by the Gaussian integral. Therefore, $$\lim_{t \rightarrow 0+} \vert u(x,t)-\phi(x)\vert \leq \lim_{t \rightarrow 0+}\frac{e^{kt}\sqrt{4kt}}{\sqrt{\pi}} e^{2\vert x \vert} (1+\sqrt{\pi kt})=0$$
I am curious about two parts:
- Does my approach have no error? Is there any shorter proof?
- Is it okay to say $\lim_{t \rightarrow 0+}\frac{e^{kt}\sqrt{4kt}}{\sqrt{\pi}} e^{2\vert x \vert} (1+\sqrt{\pi kt})=0$ for arbitrary $x$? I think $x$ increases faster than $\sqrt{4kt}$ decreases.
Could you give me any help? Thanks a lot.