Answer to Question #250183 in Differential Equations for Wajid Ali

Question

uxx= c^2ut by variable separation method

Answer 1

We will solve the problem in next full description

1) Differential equation

$u_t=c^2u_{xx}$ $0<x<l.\space t>0$

2) Boundary conditions (zero conditions for example)

u(0,t)=0, u(l,t)=0;

3) Initial condition

u(x,0)=$\phi(x),0<x<l$

Solution:

Let u(x,y)=X(x)T(t) or have separeted variables.

Inserting this form in differential equation we have:

$X(x)T'(t)=c^2T(t)X''(x)$

Dividing both paths by $c^2X(x)T(t)$ we have

$\frac{1}{c^2}\cdot \frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}$

In this equation left part depend on t bur right part depend on x therefore both part are equal some constant which may be denoted as -$\lambda$

Thus we have the system:

$x = \begin{cases} T'+\lambda c^2T(t)=0\\ X''(x)+\lambda\cdot X(x)=0 \end{cases}$

Let we consider second eqution

$X''(x)+\lambda\cdot X(x)=0$

Boundary conditions at x=0 and x=l take the form:

X(0)=X(l)=0.

1) $\lambda<0$

General solution of ode is $X(x)=C_1e^{-\sqrt{-\lambda}x}+C_2e^{\sqrt{\lambda }x}$

Then $X(0)=C_1+C_2=0$ ,$C_2=-C_1$

Therefore $X(x)=C\cdot (e^{-\sqrt{-\lambda}x}-e^{\sqrt{-\lambda}x})$

Condition X(l)=0 implies

$C\cdot (e^{-\sqrt{-\lambda}l}-e^{\sqrt{-\lambda}l})=0$ and hence C=0, therefore X(x)$\equiv0$

but we need nontrival solution hence case $\lambda<0$ is impossible, case $\lambda=0$ is impossible too with similar argues, typical case is $\lambda>0$

In this case $X(x)=C_1sin(\sqrt \lambda x)+C_2cos(\sqrt \lambda x)$

X(0)= $C_2=0=>$ X(x)=$Csin(\sqrt \lambda x)$

X(l)=0=>$sin(\sqrt\lambda l)=0=>\sqrt \lambda l=\pi\cdot n.n>0$

$\lambda_n=\left(\frac{\pi\cdot n}{l} \right)^2$ and $X_n(x)=sin\left(\frac{\pi\cdot n\cdot x}{l}\right)$

For this $\lambda_n$ an equation $T'+\lambda c^2T(t)=0$ has solution

$T_n(t)=e^{-\lambda_n\cdot c^2\cdot t}$ and $u+n(x,t)=T_n(t)X_n(x)=e^{-\lambda_nc^2t}sin\left( \frac{\pi nx}{l}\right)$

After that we search solution in the form

$\sum_{n=1}^{\infty}C_ne^{-\lambda_nc^2t}sin\left( \frac{\pi nx}{l}\right)$ for initial condition to satisfy.

Insertinf t=0 we have

$\sum_{n=1}^{\infty}C_nsin\left( \frac{\pi nx}{l}\right)=\phi(x),0<x<l$

Then $C_n=\frac{2}{l}\int_0^l\phi(x)sin\frac{\pi n z}{l}dz$ -Fourier coefficients. With such coefficients solution of the problem is:

$\sum_{n=1}^{\infty}C_ne^{-\lambda_nc^2t}sin\left( \frac{\pi nx}{l}\right)$