I am trying to study the following equation:
$$ F(t):=\|u_t-u^*\|_{L^2(\Omega)}^2 $$ where $u^*\in H^1(\Omega)\cap L^\infty(\Omega)$ is fixed and $\Omega\subset \mathbb R^2$ is open bounded with smooth boundary. Here $u_t$ is the unique solution for PDE $$ u_t+t\Delta u_t=u_0 \tag 1 $$ for each $t>0$, where $u_0\in H^1(\Omega)\cap L^\infty(\Omega)$ is fixed.
I am trying to find $t_0$ so that $$ F(t_0) = \min_{t>0} F(t) $$ and study some relative properties. We have $$ \frac{d}{dt}F(t) = \frac{d}{dt}\|u_t-u^*\|_{L^2(\Omega)}^2 = \frac{d}{dt}(u^\ast-(I+t\Delta)^{-1}u_0,u^\ast-(I+t\Delta)^{-1}u_0)_{L^2}=0.\tag2 $$ Note from $(1)$ we have $u_0=(I+t\Delta)u_t$ and hence $u_t=(I+t\Delta)^{-1}u_0$.
You may assume that $t=0$ is not a minimizer and $u_t\neq u^\ast$.
From the hint below I am able to obtain that $$ F'(t)=\frac{d}{dt}F(t)=2\int_\Omega[(I+t\Delta)^{-1}\Delta(I+t\Delta)^{-1}u_0][u^\ast-(I+t\Delta)^{-1}u_0]dx $$ As assumed, $u^\ast\neq (I+t\Delta)^{-1}u_0=u_t$ for any $t>0$.
I am wishing to prove the following argument:
Suppose the minimizer $t_0>0$ does exist. Then do we have
$$ \frac{d}{dt}F(t)\leq 0\text{ for }t<t_0\text{, and }\frac{d}{dt}F(t)\geq 0\text{ for }t>t_0. $$
Moreover, if I can obtain an equation or bound for $t_0$ in term of $u_0$ and $u^\ast$ without involving integration, it would be great.
Thank you!
Hint: the derivative of $(I+t\Delta)^{-1}$ w.r.t. $t$ is $-(I+t\Delta)^{-1}\Delta(I+t\Delta)^{-1}$.