direct relationship between diffusion and wave equation We find direct relationship between the heat and wave equation. Let $u(x,t)$ solve the wave equation on the whole line, and suppose the second derivatives of $u$ are bounded. Let: 
$$v(x,t) = \frac{c}{\sqrt{4\pi kt}}\int_{-\infty}^{\infty}e^{-s^2c^2/4kt}u(x,s)ds$$
Show that $v(x,t)$ solves the diffusion equation $u_t -ku_{xx} = 0$
I know I have to write this $v(x,t)$ in the form of $v(x,t) = \int_{-\infty}^{\infty} H(s,t)u(x,s)ds$. But, I don't know how to differentiate this to satisfy the diffusion equation. Can someone please show me how? Thank you a lot. 
 A: Let
$$H(s,t)=\frac{1}{\sqrt{4\pi \frac{k}{c^{2}} t}}e^{\frac{-s^{2}}{\frac{4kt}{c^{2}}}}$$
Then it satisfies 
$$H_{t}=\frac{k}{c^2}H_{ss}$$
And we get
$$v(x,t)=\int_{-\infty}^{\infty}H(s,t)u(x,s)ds$$
Next we calculate $v_t(x,t)$ and $v_{xx}(x,t)$,
\begin{align*}
v_t(x,t) &= \int_{-\infty}^{\infty}H_t(s,t)u(x,s)ds \\
v_{xx}(x,t) &= \int_{-\infty}^{\infty}H(s,t)u_{xx}(x,s)ds\\
&= \int_{-\infty}^{\infty}H(s,t)u_{ss}(x,s)\frac{1}{c^2}ds\\
& \quad({\rm since}\:u(x,t)\:{\rm is\:the\:solution\:of\:wave\:equation}\: u_{ss}=c^2u_{xx})\\
&= \frac{1}{c^2}\int_{-\infty}^{\infty}H(s,t)u_{ss}(x,s)ds \\
&= \frac{1}{c^2}H(s,t)u_s(x,s)\Big|_{-\infty}^{\infty}-\frac{1}{c^2}\int_{-\infty}^{\infty}H_s(s,t)u_s(x,s)ds\\
&= -\frac{1}{c^2}\int_{-\infty}^{\infty}H_s(s,t)u_s(x,s)ds\\
&= -\frac{1}{c^2}H_s(s,t)u(x,s)\Big|_{-\infty}^{\infty}+\frac{1}{c^2}\int_{-\infty}^{\infty}H_{ss}(s,t)u(x,s)ds\\
&= \frac{1}{c^2}\int_{-\infty}^{\infty}H_{ss}(s,t)u(x,s)ds\\
\end{align*}
Hence
$$v_t-kv_{xx}=\int_{-\infty}^{\infty}\Big(H_t(s,t)-\frac{k}{c^2}H_{ss}(s,t)\Big)u(x,s)ds=0$$
It implies that $v(x,t)$ solves the diffusion equation.
