Question

Show that u(x,t)=cos(a*pi*x)e^-a^2*pi^2*t is a solution of the heat equation with k=1, on any interval [0,L].

Answer 1

Explanation:

We have the function

`u(x,t) = \cos(a\pi x)e^(-a^2\pi^2t)`,

while the heat equation in one spacial dimension is

`(\partial u)/(\partial t) = (\partial^2 u)/(\partial^2 x)`.

So, to solve this exercise we only need to calculate the derivatives that appears in the equation. Let us start by the derivative with respect to t:

`(\partial u)/(\partial t) (x,t) = -a^2\pi^2\cos(a\pi x)e^(-a^2\pi^2t)`.

In the other hand

`(\partial u)/(\partial x) (x,t) =-a\pi\sin(a\pi x)e^(-a^2\pi^2t)`,

then

`(\partial^2 u)/(\partial^2 x) (x,t) =-a^2\pi^2\cos(a\pi x)e^(-a^2\pi^2t)` .

Notice that,

`-a\pi\sin(a\pi x)e^(-a^2\pi^2t)=-a\pi\sin(a\pi x)e^(-a^2\pi^2t)`

So, the function
`u` satisfies the heat equation with
`k=1` on any interval [0,L].