Answer to Question #160354 in Differential Equations for Saif

Question

Solve the Heat Equation du/dt=40d²u/dx² Subject to the conditions

u(0,t)=0, u(π,t)=0

u(x,0)=5sinx-2sin5x

Answer 1

Solution:

We know that the solution of general heat equation

$\begin{array}{l} \frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}} \\ u(x, 0)=f(x) \quad u(0, t)=0 \quad u(L, t)=0 \end{array}$

is as follows:

$u_{n}(x, t)=B_{n} \sin \left(\frac{n \pi x}{L}\right) \mathbf{e}^{-k\left(\frac{n \pi}{L}\right)^{2} t} \quad n=1,2,3, \ldots$ ...(i)

Now, given :

du/dt=40d²u/dx²

u(0,t)=0, u(π,t)=0

u(x,0)=5sinx-2sin5x

On comparing, we have:

$k=40,\ L=\pi,\ f(x)=5\sin x-2\sin 5x$

Choosing $n=1,\ B_n$ as 5 and -2 respectively.

Putting these values in (i), we get,

$u(x, t)=2 \sin \left(\frac{ \pi x}{\pi}\right) \mathbf{e}^{-40\left(\frac{ \pi}{\pi}\right)^{2} t} -5 \sin \left(\frac{ \pi x}{\pi}\right) \mathbf{e}^{-40\left(\frac{ \pi}{\pi}\right)^{2} t}$

$\Rightarrow u(x, t)=2 \sin x \mathbf{e}^{-40 t} -5 \sin x \mathbf{e}^{-40 t}$