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Let $u$ be a solution of $$u' - \Delta u = 0 \quad\text{on $\Omega$}$$ $$\partial_\nu u = 0\quad\text{in $\partial \Omega$}$$ $$u(t=0)=u_0\quad\text{on $\Omega$}$$ where $\Omega$ is a bounded Lipschitz domain.

Here $u_0 \in H^1(\Omega)$ can be as regular as needed, but it must satisfy $$u_0 \geq 0 \text{ a.e., and } \int_\Omega u_0 > 0.$$

Is it at all possible to choose $u_0$ satisfying the above conditions such that we can obtain a bound on $\lVert u \rVert_{L^2((0,\infty);L^2(\Omega)}$?

The infinite time interval is the problem -- Gronwall doesn't appear to help. One can bound the gradient of $u$ though in the desired space, and also one can bound $u$ in $L^\infty((0,T);L^2(\Omega))$.

Is this even possible? Could it be achieved if there were a non-positive source term?

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    $\begingroup$ Weclome to MathOverflow! The solution $u(t)$ converges in $L^2(\Omega)$ to $\frac{1}{|\Omega|} \int_\Omega u_0(x) \,dx \, 1$ as $t \to \infty$. So how could one expect it to be square-integrable in time, unless the spatial integral of $u_0$ vanishes? $\endgroup$
    – Jochen Glueck
    Commented Jul 14, 2023 at 22:44

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