I am asked to show that a problem of Laplace's Equation is well-posed in terms of stability with respect to small changes in the initial data. The PDE is:
$$u_{tt}+u_{xx} =0,\>\>\>\>\>\> u(x,0) = 0,\>\>\>\> u_t(x,0)=\frac{\sin(nx)}{n}$$
We are given the solution: $$u_n(x,t) = \frac{\sinh(nt)\sin(nx)}{n^2}$$ Thus we have to find out what happens to the solution at a later fixed time t. Since the solution grows exponentially, I figured that the solution wouldn't change much when $n$ is low, but as $n$ became large, there would be a lot of volatility with respect to small changes in the initial data. But how can I show this?
The idea is to look at the homogeneous version for your problem, say it's solution is $v(x, t)$. Then you should be able to say something about $|v_{t}(x,0) - u_{t}(x,0)|$ and $|v(x, t) - u(x, t)|$. That is, the difference in the initial conditions should result in large differences in the solution when $n$ is large.
EDIT:
Following from the comments,
$\frac{\sinh{nt}\sin{nx}}{n^2} = \frac{(e^{nt}-e^{-nt})\sin{nx}}{2n^2} = \frac{(e^{nt}\sin{nx}-e^{-nt}\sin{nx})}{2n^2}$
For large n, the $e^{nt}$ term dominates the entire expression, so it will diverge. The second term goes to zero, and even dividing by $2n^2$ cannot outweigh how fast the exponential term grows.