Analytic solution of the heat equation with a source term

Question

I have the heat equation with Dirichlet boundary conditions $$u_t(t,x)=u_{xx}(t,x)+\sin(x)$$ $$u(t,0)=u(t,2\pi)=0$$ $$u(0,x)=u_0(x)$$ Now, without the source term I could write the solution as $$u(t,x) = \sum_{n=1}^\infty B_n \sin(n\pi x)e^{-n^2\pi^2t}$$ where $$B_n=2\int_0^{2\pi}u_0(x)\sin(n\pi x)dx$$ but I'm not sure what it looks like with a source term. I had a problem set question which assumed knowledge of the solution to show it converges to it's stationary form, so I'm guessing it can be derived using the homogenous equation.

Answer 1

Notice that $f(x) = \sin x$ solves \begin{align} f_{xx}+\sin x = 0, \end{align} with the given boundary conditions, that is, $f$ is a stationary solution to the above problem.

Next, consider the function $v=u-f$ where $u$ solves the above heat equation. Then, we see that \begin{align} v_t-v_{xx} = u_t-u_{xx}+f''= \sin x+f'' = 0. \end{align}

Here, we see that $v$ measures how $u$ deviates from the stationary solution $f$. The above calculation demonstrates that the deviation satisfies a heat equation that converges to zero given the boundary conditions. In short, $u$ converges to $f$ as $t\rightarrow \infty$.

Answer 2

Hint:

the $\sin x$ source term is time independent: what happens if you add $t \sin x$ to your solution?
does it respect the PDE ? and the boundary conditions ?

Answer 3

Note that since you're on $[0,2\pi]$, the appropriate argument of the eigenfunctions is not $n \pi x$. This is not just a minor technicality since it affects how you must deal with the forcing.

Once you get that straight, consider the evolution equation for the $n$th Fourier coefficient (obtained by multiplying both sides by the $n$th eigenfunction and integrating, and pulling the time derivative out of the first term). You should find that all but one of them are just of the form $\frac{d}{dt} \hat{u}_n + \lambda_n u_n = 0$ (which is what you see in the homogeneous heat equation) while the $n=1$ equation is of the form $\frac{d}{dt} \hat{u}_1 + \lambda_1 u_1 = c$. This latter ODE is simple enough that you can just solve it.