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

Answer:

Step-by-step 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

$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial^2 x}$.

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:

$\frac{\partial u}{\partial t} (x,t) = -a^2\pi^2\cos(a\pi x)e^{-a^2\pi^2t}$.

In the other hand

$\frac{\partial u}{\partial x} (x,t) =-a\pi\sin(a\pi x)e^{-a^2\pi^2t}$,

then

$\frac{\partial^2 u}{\partial^2 x} (x,t) =-a^2\pi^2\cos(a\pi x)e^{-a^2\pi^2t}$ .

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].