Heat flow equation via Fourier Series

I know how to solve heat equations and wave equations defined on $\mathbb{R}^n\times(0,\infty)$ using Fourier transform. But I am having trouble solving similar equations defined on finite intervals using Fourier series. To be precise, how do I start on the equation $$u_x = u_{tt}$$ with initial condition $$u(x,0) = f(x),$$ and $u$ is defined on $(a, b)\times(0, \infty)$?

The boundary conditions are $u(a,t) = u(b ,t) = 0$ for every $t>0$.

• What are the boundary conditions? – velut luna Mar 11 '16 at 12:50
• I just added them. – dezdichado Mar 11 '16 at 12:54
• The functions $X_n(x)=\sin\left(\frac{x-a}{b-a}n\pi\right)$ for $n=1,2,3,\cdots$ form an orthogonal basis of $L^2[a,b]$. – DisintegratingByParts Mar 11 '16 at 15:01

There is something contradictory in the wording of the question. The written equation is : $$u_x=u_{tt}$$ which, in fact, is not the heat equation : $$u_t=u_{xx}$$ as raised in the title and claimed the wording.
Supposing that this is a pypo, the next answer is the solving of the heat equation $u_t=u_{xx}$ with the specified boundary conditions ( and via Fourier series as requested ).
Your endpoint condition is not sufficient. You'll need two conditions. For example, $$-f''=\lambda f \\ f(a)=f(b) \\ f'(a)=f'(b).$$ is a well-posed eigenfunction problem with eigenfunctions $\{ \exp(2\pi n\frac{x-a}{b-a})\}_{n=-\infty}^{\infty}$, and eigenvalues $\lambda=\left(\frac{2\pi n}{b-a}\right)^2$
And the following problem is also a well-posed eigenfunction problem: $$-f''-\lambda f \\ f(a)=f(b) \\ f'(a)=-f'(b).$$ Solutions include $\{\sin((2n+1)\pi\frac{x-a}{b-a})\}_{n=1}^{\infty}$