1
$\begingroup$

I completed the separation of variables step but I am very confused on how to apply the initial any boundary conditions to solve the problem. Please give me some advice or help to go about solving this problem, thank you!

Image.

Problem 1: Solution to homogeneous PDEs by separation of variables
a. Apply separation of variables to derive the general solution $u(x,t)$ to the following homogeneous PDE over the interval $0\leq x\leq 1$, with boundary conditions $u(0,t) = u(1,t) = 0$: $$u = c\frac{\partial^3u}{\partial t\,\partial x^2}$$ b. Find the particular solution with initial conditions $u(x,0)=1$ for $0<x<1$.

$\endgroup$
2
  • 1
    $\begingroup$ The boundary conditions arise in selecting the eigenfunctions. The initial conditions determine the "Fourier coefficients" of the eigenfunctions in the expansion. $\endgroup$
    – Ian
    Commented Sep 30, 2016 at 21:47
  • $\begingroup$ Formatting tips here. $\endgroup$
    – Em.
    Commented Sep 30, 2016 at 23:32

2 Answers 2

0
$\begingroup$

Since you did the separation of variables, you have something like $u(x,t) = a(x)b(t)$. The condition that $u(0,t) = 0$ means that $a(0)b(t) =0$. So you must have $a(0)=0$, because if you set $b(t) = 0$ it forces $u(x,t) =0$ and we have no use for the trivial solution. Likewise the second condition gives us $a(1) =0$. Similarly the initial condition says $b(0) =1$.

$\endgroup$
0
$\begingroup$

a. Substituting $u(x,t)=f(x)g(t)$ into the PDE, we obtain $$ f(x)g(t)=cf''(x)g'(t) \implies \frac{f''(x)}{f(x)}=\frac{g(t)}{cg'(t)}=\lambda. \tag{1} $$ Depending on the sign of $\lambda$, the solution to $f''(x)=\lambda f(x)$ is $$ f(x)=\begin{cases} a\cos(kx)+b\sin(kx)&\text{if $\lambda=-k^2<0$}, \\ a+bx&\text{if $\lambda=0$}, \\ a\cosh(\kappa x)+b\sinh(\kappa x)&\text{if $\lambda=\kappa^2>0$}. \end{cases} \tag{2} $$ The boundary conditions $f(0)=f(1)=0$ generally imply $a=b=0$, except if $\lambda=-n^2\pi^2\, (n\in\mathbb{N}^*)$, in which case $f(x)=b\sin(n\pi x)$. The corresponding solution to $g'(t)=\frac{1}{\lambda c}g(t)$ is $g(t)=g(0)e^{-t/(n^2\pi^2c)}$. Since the PDE is linear, the general solution satisfying the given boundary conditions is the linear combination $$ u(x,t)=\sum_{n=1}^{\infty}a_ne^{-t/(n^2\pi^2c)}\sin(n\pi x), \tag{3} $$ provided the series converges.

b. The particular solution satisfying the initial condition $u(x,0)=1$ for $0<x<1$ can be obtained from $(3)$ using the orthogonality relation $$ \int_0^{1}\sin(m\pi x)\sin(n\pi x)\,dx=\frac{1}{2}\delta_{m,n}\quad(m,n\in\mathbb{N}^*). \tag{4} $$ Thus, \begin{align} u(x,0)=1 &\implies \sum_{n=1}^{\infty}a_n\sin(n\pi x)\,dx=1 \\ &\implies \frac{1}{2}a_m=\int_0^1\sin(m\pi x)\,dx=\frac{1-(-1)^m}{m\pi}, \tag{5} \end{align} hence, $$ u(x,t)=\sum_{n=1}^{\infty}\frac{2[1-(-1)^n]}{n\pi}e^{-t/(n^2\pi^2c)}\sin(n\pi x). \tag{6} $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .