Answer to Question #93157 in Differential Equations for Subhasis Padhy

Question

Solve the following boundary value problem.

uₙ = uₓₓ , 0 < x < L , n > 0

u(0,n) = u(L,n) = 0

u(x,0) = x(L-x) , 0 ≤ x ≤ L

Answer 1

Let's assume that the solution will take the form

$u=\varphi(x)G(t)$

Substituting this from in the equation, will lead us to

$\varphi(x)G'(t)=\varphi''(x)G(t)$

$\frac{G'(t)}{G(t)}=\frac{\varphi''(x)}{\varphi(x)}$

Since the left hand side depends only of t and the right hand side only of x, the both sides can be equal only if they are equal to some constant.

Let's denote this constant as $-\lambda$

Then

$\frac{G'(t)}{G(t)}=\frac{\varphi''(x)}{\varphi(x)}=-\lambda$

Therefore we obtain two ordinary differential equations

$G'+\lambda G=0$

$\varphi''+\lambda\varphi=0$

Using boundary condition will lead us to

$\varphi(0)=\varphi(L)=0$

The general solution of the equation is

$\varphi=c_1\cos(\sqrt{\lambda}x)+c_2\sin(\sqrt{\lambda}x)$

Applying the first boundary condition gives us

$\varphi(0)=c_1=0$

Applying the second boundary condition, and using the above result, gives us

$\varphi(L)=c_2\sin(\sqrt{\lambda}L)=0$

We will only have a non-trivial solution, if $\sin(\sqrt{\lambda}L)=0$ , $\lambda \not =0.$

Hence

$\lambda_n=\left(\frac{\pi n}{L}\right)^2,\quad n\in \Zeta, n \not =0.$

Then for G(t) we will have

$G'=-\lambda_nG$

This is a linear first order differential equation and its solution is

$G=B_ne^{-\lambda_nt }$

Hence the solution is

$u=\sum\limits_{n=1}^{\infty}B_ne^{-\lambda_nt }\sin\sqrt{\lambda}x$

The initial condition will lead us to

$u(x,0)=\sum\limits_{n=1}^{\infty}B_n\sin\sqrt{\lambda}x=x(L-x)$

which is the sine Fourier series of $x(L-x)$

It's coefficients can be obtained using the formula

$B_n= \frac{2}{L}\int\limits_{0}^{L}x(L-x)sin(\frac{\pi n}{L})xdx=-\frac{2 L^2 (-2 + 2 cos(n π) + n π sin(n π))}{n^3 π^3}=-\frac{4 L^2((-1)^n-1)}{n^3 π^3}$

Hence

$B_{2n}=0$

$B_{2n-1}=\frac{8 L^2}{(2n-1)^3 π^3}$

Therefore we obtain the final form of the function

$u=\sum\limits_{n=1}^{\infty}\frac{8 L^2}{(2n-1)^3 π^3}e^{-\lambda_{2n-1}t }\sin\sqrt{\lambda_{2n-1}}x$