Given the problem $$u_t-\Delta u=0$$ in $D \times (0,T)$ $$u=0$$ in $\partial D \times [0,T]$ $$u=g$$ in $ D \times \{t=0\}$ Where $D$ is some open and bounded subset of $\mathbb{R}^n$, $T>0$, $\partial D$ is $C^1$ and $u$ is $C^2$.
Given the energy $$E(t)=\int_{D} u^2 (r,t) dr$$ Prove that $h(t)=\ln E$ is convex.
My attempt is the following, I take the second derivative of $h$ to get $$h''=\frac{E''E-(E')^2 }{E^2}$$
Then take the derivative of the energy to get $$E'(t)=2\int_D u u_t dr$$ since $u_t=\Delta u$ I get $$E'(t)=2\int_D u \Delta u dr$$ and after using Green's identity I get $$E'(t)=2\int_{\partial D} u \frac{\partial u}{\partial n} -2\int_{D} | \nabla u |^2 dr $$ since $u=0$ on the boundary the I am left with $$E'(t)=-2\int_{D} | \nabla u |^2 dr $$
Now if I take the second derivative I get $$E''(t)=-4\int_D \nabla u \nabla u_t dr $$ Using Green's identity I get $$E''=-4\int_{\partial D} u_t \frac{\partial u }{\partial n}+4\int_D u_t \Delta u dr$$ now since $u=0$ on the boundary then $u_t=0$ on the boundary and I am left with $$E''(t)=4 \int_D u_t \Delta u dr$$ Since $\Delta u=u_t$ I get $$E''(t)=4 \int_D u_t^2 dr$$ If I plug those in $h''$ I get $$h''(t)=\frac{4\int_D \int_D u_t^2(r,t) u^2 (r',t) drdr'-(E')^2 }{E^2} $$ $E^2$ and $(E') $ are always non-negative, but this doesn't give me anything useful.
For a function to be convex, it's second derivative must be non-negative and here for $h''$ I can't see how this is true, can you help?