Verify that $ u(x,t) = \int_{-\infty}^{\infty} U(x-y,t)g'(y) \, dy $ is a solution of the heat equation

114 Views Asked by At

Whilst studying the free eBook 'Partial Differential Equations' by Victor Ivrii, I came across the following problem

Verify that

$$u(x,t) = \int_{-\infty}^{\infty} U(x-y,t)g'(y) \ dy$$

satisfies the heat equation

$$u_{t} = ku_{xx}$$

for $t>0, x \in \mathbb{R}$, where

$$U(x,t) = \int_{-\infty}^{x} u(s,t) \ ds$$

We also know that $U(x,t)$ satisfies the heat equation, $g$ is a smooth function, $g(-\infty) = 0$ and

$$g(x) = g(x) - g(-\infty) = \int_{-\infty}^{x} g'(y) \, dy = \int_{-\infty}^{\infty} \theta (x-y)g'(y) \ dy$$

where $\theta$ is the Heaviside function

$$\theta(x) = \begin{cases} 1, \quad \text{for } x > 0 \\ 0, \quad \text{for } x \le 0 \end{cases}$$

My first attempt was to calculate $ u_{t}, \, u_{xx} $ and then substitute in the heat equation but I got stuck. My second attempt was to prove (since $ U(x,t) $ is a solution) that the translation $ U(x-y,t) $ is also a solution and then prove that the integral $ u(x,t) = \int_{-\infty}^{\infty} U(x-y,t)g'(y) \, dy $ is also a solution but I also get stuck.