# Show that solution to wave equation can be used to solve the Heat equation

Let $$u$$ be a regular and bounded solution of $$u_{tt}-u_{xx}=0$$ in $$\mathbb{R} \times \mathbb{R}$$ with $$u(0,x) = g(x)$$, $$u_t(0,x)=0$$ in $$\mathbb{R}$$. Define

$$w(t,x) = \frac{1}{\sqrt{4 \pi t}} \int_{\mathbb{R}} u(s,x)e^{-\frac{s^2}{4t}}ds$$

Show that $$w$$ solves the heat equation $$w_t - w_{xx} = 0$$ in $$(0,\infty) \times \mathbb{R}$$ with $$w(0,x) = g(x)$$ in $$\mathbb{R}$$.

So far I have tried this: The solution to the wave equation $$u_{tt} - u_{xx}=0$$ must be of the form found in d'Alembert's formula. So i plugged this formula into $$w$$ and then calculated $$w_t$$ and $$w_{xx}$$. But I don't see any obvious simplifications that lead to the result.

Notice that $$w(t,x)=\int_\mathbb R u(s,x) K(t,s) ds$$ where $$K$$ is the heat kernel which we know satisfies $$(\partial_t + \Delta)K=0$$ (where I use the Laplacian with a negative sign).
Now when you compute $$(\partial_t+\Delta)w$$, using your favorite theorem to justify the interchange of differentiation and integration, you can bring $$(\partial_t+\Delta)$$ inside the integral. Then, the time derivative will only interact with the heat kernel $$K$$ and the Laplacian will only act on $$u$$. With the information we have about $$u$$ and $$K$$, we arrive at an integral appearing in Green's second identity (by which I mean an integral of the form $$\int (f\Delta g - g\Delta f) ds$$) and thus show $$(\partial_t+\Delta)w=0$$.