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The parabolic heat equation is a partial differential equation given by $\frac{du}{dt}=\nabla^2u+f$. If i impose an initial condition u(x,0) and pure homogeneous neumann boundary conditions that satisfy the compatibility conditions with respect to the source term f(x), does this result in a well posed problem. That is, does a unique solution exist?

I know that if we consider the steady state heat equation $-\nabla^2u=f$, pure neumann conditions imply that a solution exists, but the solution is not unique without imposing additional constraints. Of course, the steady state heat equation is an elliptic pde, so i'm not sure if the same can be said about the parabolic heat equation pde as well.

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Yes, it is well-posed (provided $f$ and the initial data are suitable, at least). There is a nice physical interpretation of the difference between the two situations.

Under a pure Neumann condition with a time-constant source, an equilibrium only exists if the net heat input from the source being zero. (More generally, given an inhomogeneous pure Neumann condition, the existence of an equilibrium distribution depends on the net heat input from the source minus the loss from the boundary being zero.) When the heat input from the source is zero, which equilibrium you get depends on the initial condition. One reason is because the total heat at the start is the same as the total heat at the end.

But given a suitable initial condition and source, the temperature will still evolve, even if it doesn't equilibrate.

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I assume you're working on a nice bounded domain with piecewise smooth boundary. For the elliptic problem $-\nabla^{2}u = f$, the difference of two solutions $u_1,u_2$ would have to be a constant because $w = u_1-u_2$ would satisfy $\nabla^{2}w =0$ and $\frac{\partial w}{\partial n}$, which gives $$ 0=\int_{\Omega}w\nabla^{2}u dV = \int_{\Omega}\nabla\cdot(w\nabla w)-|\nabla w|^{2}dV =-\int_{\Omega}|\nabla w|^{2}dV. $$

For the parabolic problem, the difference $w$ of two solutions would satisfy $$ \frac{d}{dt}\frac{1}{2}\int_{\Omega}w^{2}dV=\int_{\Omega}w\nabla^{2}wdV=-\int_{\Omega}|\nabla w|^{2}dV \le 0 \\ \implies 0\le\int_{\Omega}w^{2}(x,t)dV \le \int_{\Omega}w^{2}(x,0)dV=0. $$ Interesting difference.

The earlier post of Ian deals with existence of solutions.

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