Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider this PDE

enter image description here

The solution to the PDE is enter image description here

So what I am having trouble is solving it using this method.

I am going to say that my $u(x,t) = \sum_{n=1}^{\infty} u_n(t) \sin(nx)$ and $x \sin(t) = \sum_{n=1}^{\infty}h_n(t)\sin(nx)$

The reason I chose sine for my inhomogeneous term is because my book recommends it. But I think it is because if I use cosine, I would get a $\frac{a_0}{2}$ term and it would be difficult.

To solve for the coefficients of $h_n(t)$, I get $h_n(t) = \frac{2}{\pi}\int_{0}^{\pi} x\sin(t) \sin(nx) dx = \frac{2\sin(t)(-1)^n}{n}$

Substituting everything into $u_{tt} = u_{xx} + x\sin(t)$ gives me

$$ \sum_{n=1}^{\infty}u''_n(t) \sin(nx) + \sum_{n=1}^{\infty}u_n(t)n^2\sin(nx) = \sum_{n=1}^{\infty}\frac{2\sin(t)(-1)^n}{n}\sin(nx)$$

Dividing out that sine, I'll get

$$ \sum_{n=1}^{\infty}u''_n(t) + \sum_{n=1}^{\infty}u_n(t)n^2 = \sum_{n=1}^{\infty}\frac{2\sin(t)(-1)^n}{n}$$

Here is where I am stuck, can someone tell me what value of n to use?

Thank you very much

share|improve this question

1 Answer 1

up vote 2 down vote accepted

I might do it this way, using Duhamel's principle. Start by looking for the solution $\phi(x,t,s)$ to the homogeneous partial differential equation (the ordinary wave equation $\dfrac{\partial^2 \phi}{\partial t^2} = \dfrac{\partial^2 \phi}{\partial x^2}$, depending on the parameter $s$, with boundary conditions $\phi(0,t,s)=\phi(\pi,t,s) = 0$ and initial conditions $\phi(x,0,s)=0$, $\dfrac{\partial \phi}{\partial t}(x,0,s) = x \sin(s)$. Then a solution of your nonhomogeneous wave equation is $u(x,t) = \int_0^t \phi(x,t-s,s)\ ds$.

EDIT: Here's another way, if you don't want to use Duhamel's principle. Let $u(x,t)$ be expanded in a Fourier series $u(x,t) = \sum_{n=1}^\infty u_n(t) \sin(n x)$ (we use sines rather than cosines here because of the boundary conditions $u(0,t) = u(\pi,t) = 0$). Correspondingly, the same type of Fourier series for the inhomogeneous term is $x \sin(t)= \sum_{n=1}^\infty a_n \sin(n x) \sin(n t)$ where $$a_n = \dfrac{2}{\pi} \int_0^\pi x \sin(n x)\ dx = \dfrac{2}{n} (-1)^{n+1}$$ Plugging these into the pde and the initial condition, for the coefficients of $\sin(n x)$ to balance we need $$ u_n''(t) = -n^2 u_n(t) + a_n \sin(t), \ u_n(0) = 0, \ u_n'(0) = 0 $$ For $n \ne 1$ the solution of this (which you can get using the method of Undetermined Coefficients) is $$u_n(t) = \dfrac{a_n}{n(n^2-1)}(n \sin(t) - \sin(nt)) = \dfrac{2 (-1)^{n+1}}{n^2 (n^2-1)} (n \sin(t) - \sin(nt))$$ For $n=1$ that solution doesn't work (it's the case of "resonance") and the solution is $$u_1(t) = \dfrac{a_1}{2} (\sin(t) - t \cos(t)) = 2 (\sin(t) - t \cos(t))$$ Thus putting it all together, $$ u(x,t) = 2 \sin(x) (\sin(t) - t \cos(t)) + \sum_{n=2}^\infty \dfrac{2 (-1)^{n+1}}{n^2 (n^2-1)} \sin(nx) (n \sin(t) - \sin(nt))$$

share|improve this answer
    
I haven't learned what Duhamel's principle. I am taking an introductory PDE class. –  sidht Jul 14 '12 at 2:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.