Suppose u solves $u_t-u_{xx} = 16u$ on the interval $(0, \pi),$ with the homogenous Neumann condition $u_x=0$ at $x=0,\pi$. Characterize the initial data $u_0=u(x,0)$ for which $u(x,t)$ stays bounded as $t \rightarrow\infty$.
I am not sure how to approach this PDE, thus any help/guidance is highly appreciated. Thanks!
$$\begin{cases} u_t -u_{xx} = 16u\\ u_x(0,t) = u_x(\pi,t)=0\\ u(x,0) = u_0 \end{cases}$$
Let $u(x,t) = X(x)T(t)$ and $\frac{dT(t)}{dt} = \mathring{T}(t)$
Therefore, $u_t -u_{xx} = 16u$ becomes $\frac{\mathring{T(t)}}{T(t)}-16 = \frac{X''(x)}{X(x)} = \lambda$.
Here, $\lambda$ is a constant because "[t]he left hand side is independent of t. The right hand side is independent of x. The two sides are equal. So both sides must be independent of both x and t and hence equal to some constant, say [$\lambda]"$. (I actually still don't understand this sentence... If anyone can help me, please! )
$T(t):$ $T(t) = ce^{(16+\lambda )t}$ for some constant $c$.
$X(x): $
There are three cases - $\lambda = 0 , \lambda > 0, \lambda < 0$.
i) According to my experience, it $\frac{X''(x)}{X(x)} = \lambda > 0 $ ususally yields trivial solution. aka, $X(x) = 0$. Thus I am going to skip the calculation here. I have calculated to verify and it does result in trivial solution.
ii) $\frac{X''(x)}{X(x)} = \lambda = 0 $ yields trivial solution.
iii) $\frac{X''(x)}{X(x)} = \lambda = -\mu^2 < 0 $ This is the fun part.
Solving the second order DE, $X(x) = A\cos (\mu x) + B\sin(\mu x)$. Plugging in the boundary conditions, $B=0, A = a_n $when $\mu=n=1,2,3,...$
Therefore, $$u(x,t) = \sum^\infty_{n=1}a_n \cos (n\pi) e^{(16-n^2)t}$$
Now, we can use Fourier series and initial condition to solve for $a_n$.
$$a_n = \frac{2}{\pi}\int^{\pi}_0 u_0 \cos(n\pi)dx= \frac{2u_0}{\pi}\frac{(-1)^n-1}{n}$$ $$b_{2k-1} = \frac{2u_0}{\pi}\frac{-2}{2k-1}$$
Therefore, the final solution is, $$u(x,t) = \frac{-4u_0}{\pi}\sum^\infty_{k=1}\frac{cos(2k-1)\pi e^{(16-(2k-1)^2)t}}{\pi (2k-1)}$$