# Need help on solving this PDE, stuck applying the initial and boundary conditions

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!

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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 $0<x<1$.

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

## 2 Answers

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 $b(0) =1$.

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 $$0 can be obtained from $$(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}$$