The problem is from a previous Qual exam. Consider the following initial boundary value problem: find $u(x,t)$ such that $$ \begin{aligned} \frac{\partial}{\partial t}(u-\Delta u)-\mu \Delta u &=f, x \in \Omega, T \geq t>0 \\ u(x, t) &=0, x \in \partial \Omega, T \geq t>0 \\ u(x, 0) &=u_{0}(x), x \in \Omega \end{aligned} $$ where $\mu>0$ is a given constant.
Question: Consider the forward Euler approximation for the derivative in $t$ and the finite element method in space with continuous piecewise linear functions. Find the Courant condition for stability of the resulting method.
Define $a(u,v)=\int_\Omega\nabla u\nabla vdx$. Then the weak form of the BVP is: find $u(t)\in H_0^1(\Omega)$ such that $$ \left(u_{t}, v\right)+a\left(u_{t}, v\right)+\mu a(u, v)=(f, v) $$ for all $v\in H_0^1(\Omega)$.
Let $V_h$ be space of continuous piecewise linear functions. Let $t_n=nk$ for $1\leq n\leq N$ and $k>0$ such that $T=kN$. Let $u_h^n:=u_h(x,t_h)$ and $f^n=f(x,t_n)$. Then the fully discrete problem becomes: given $u_h^n\in V_h$, find $u_h^{n+1}\in V_h$ such that $$ \left(\frac{u_{h}^{n+1}-u_{h}^{n}}{k}, v_{h}\right)+a\left(\frac{u_{h}^{n+1}-u_{h}^{n}}{k}, v_{h}\right)+\mu a\left(u_{h}^{n}, v_{h}\right)=\left(f^{n}, v_{h}\right). $$ Take $v_h=ku^{n+1}_h$. Then we have $$ \Vert u_h^{n+1}\Vert_{H^1}^2\leq k\Vert f^n\Vert_{L^2}\Vert u_h^{n+1}\Vert_{L^2}+\Vert u_h^n\Vert_{H^1}\Vert u_h^{n+1}\Vert_{H^1}+ k\mu \vert u_h^n\vert_{H^1}\vert u_h^{n+1}\vert_{H^1} $$ And thus $$ \Vert u_h^{n+1}\Vert_{H^1}\leq k\Vert f^n\Vert_{L^2}+(k\mu+1)\Vert u_h^n\Vert_{H^1}. $$
Let $a_ n=\Vert u_h^{n}\Vert_{H^1}$. Then we have $$ a_{n+1}-a_{n}\leq k\Vert f^n\Vert_{L^2}+k\mu a_n. $$ Take sum of above inequality over $n$, $$ a_{n+1}\leq a_0+k\sum_{i=0}^n \Vert f^i\Vert_{L^2}+\sum_{i =0}^n k\mu a_n $$ Using the discrete gronwall inequality (see this for example) $$ a_{n+1}\leq (a_ 0+k\sum_{i=0}^N \Vert f^i\Vert_{L^2}) exp(k\mu(n+1)) $$ Note that $T=kN$. Thus, $$ a_{n+1}\leq (a_0+\int_0^T\Vert f\Vert_{L^2}dt )exp (T\mu) $$ It seems that it’s unconditionally stable?
Stability would mean that, for $f=0$, the norm is bounded independent of $T$. You show that the norm is bounded for all $T$ but with a bound that grows exponentially with respect to $T$. Refining your argument will show a bound independent of $T$, but only if the timestep is small enough. The fact that for some stepsizes there is actually exponential growth is called instability.