Use the method of separation of variables to solve non homogeneous initial-Neumann problem

Use the method of separation of variables to solve the following non homogeneous initial-Neumann problem: $$ \begin{cases} u_t − u_{xx} = tx& 0 < x < \pi, t >0\\ u (x, 0) = 1& 0 ≤ x ≤ \pi\\ u_x (0, t) = u_x (\pi, t) = 0 & t > 0. \end{cases} $$ We have homogeneous b.c., so using separation of variables, we can find the solution for $u$ in the form: $$u(x,t) = \sum_n u_n(t)X_n(x),$$ where u$_k$ are the eigenfunctions of the eigenvalue problem associated with the homogeneous equation.

To find the eigenfunctions $X_n$, substitute a trial solution $u=T(t)X(x)$ into the homogeneous part of the partial differential equation, $u_t = u_{xx} + xt$. Remember: ignore the inhomogeneous part $tx$ when finding the eigenfunctions. $$\frac{T'}{T} = \frac{X''}{X} = \mbox{constant} = \lambda.$$

We then get the following Sturm-Liouville problem for any eigenfunctions $X(x):$ $$\begin{equation}X''=\lambda X\label{x}\end{equation}$$ $$X(0)=X(L)=0.$$ Quickly go through three cases:

1. $\lambda = 0:$ $$X''=0\Rightarrow X(x)=Ax+B\Rightarrow A=B=0.$$
2. $\lambda = \mu^2 = > 0:$ $$X''=\mu^2X\Rightarrow X=Ae^{\mu x}+Be^{-\mu x}\Rightarrow A=B=0.$$
3. $\lambda = -\mu^2 = < 0:$ $$X''=-\mu^2X\Rightarrow X=A\cos{\mu x}+B\sin{\mu x}.$$ From b.c. $A=0$ and either $B = 0$ or $\sin{\mu x} = 0.$ The first case isn't interesting for us, because we would get no eigenvalues from it. The second case requires $\mu_n \pi = n\pi,$ so $\mu_n = n,$ that yields: $$X_n = A_n\sin{n x}.$$

Now write the partial differential equation $u_t = u_{xx} + tx$ using the Fourier series: $$\sum_{n=0}^{\infty}\dot{u}_n(t)X_n(x)=\sum_{n=0}^{\infty}\lambda u_n(t)X_n(x) + tx,$$ here we've used $\ref{x}.$

To solve the equation we need to represent $x$ in terms of $X_n(x),$ which is as Fourier series on $[0,\pi]$ and then equate the coefficients on the right hand and the left hand sides.

References

5.6 Inhomogeneous boundary conditions