# Can we enforce two boundary conditions for a second order PDE?

Suppose we try to solve a second order PDE on a finite interval $$[a,b]$$, for concreteness, let us consider the heat equation $$\partial_t u = \partial_{xx} u$$ on $$x \in (a,b)$$ with some smooth initial condition $$u_0(x)$$. I am wondering whether the PDE is still well-posed if we impose Dirichlet as well as Neumann boundary condition simultaneously. i.e., we require $$u(a,t) = u(b,t) = 0$$ and $$\partial_x u(a,t) = \partial_x u(b,t) = 0$$ for any $$t > 0$$. Assume that the initial datum $$u_0(x)$$ is compatible with the boundary constraints as well (i.e., $$u_0(a) = u_0(b) = 0$$ and $$\partial_x u_0(a) = \partial_x u_0(b) = 0$$). May I know is it too crazy to put two different types of boundary conditions at the same time in this scenario? Does the resulting initial-boundary-value-problem still admits a unique classical solution? Thanks for any help!

Edit: Based on Rhys' comment, it seems that if we require a non-trivial solution when both type of boundary conditions are enforced, we need the original PDE problem (with only the Dirichlet BC or Neumann BC) to have multiple solutions (non-uniqueness...) Probably that's the reason why I haven't seen any sort of thing like this... (impose different types of BC for a PDE problem). I would also like to know whether someone is aware of such type of problems that is well-posed. (Obviously, heat equation here is a "bad" example).

• Suppose $u_D$ solves the problem with only Dirichlet conditions, and $u_N$ solves the Neumann version. Then, if $u_D$ and $u_N$ are unique (as I recall, they are) then you require $u_D=u_N$ for the unique solution to the problem with both. May 28, 2021 at 0:28
• Following Rhys' comment above, considering the non-trivial solutions to the Dirichlet and Neumann problems on a finite interval are known to be $u_{D} \propto \sin$ and $u_{N} \propto \cos$, then the only solution would be $u \equiv 0$, so the problem would have no solution if $u_{0} \ne 0$. May 28, 2021 at 1:53

In Rhys' answer he split the boundary into Dirichlet and Neumann components, which will make the problem well-posed. A different interpretation is imposing both Dirichlet and Neumann conditions over the whole domain. This leads to an overdetermined problem, which is not solvable in general domains. The question then becomes which domains admit solutions to the problem?

For concreteness, a classical example of this is Serrin's problem in which one asks which domains $$\Omega$$ admit a function $$u \in C^2(\Omega) \cap C^1(\bar{\Omega})$$ that satisfies $$\begin{equation*} \left \{ \begin{array}{rcll} -\Delta u &=& 1 & \text{in } \Omega \\ u&=& 0 & \text{on } \partial \Omega \\ \frac{\partial u}{\partial \nu} &=& \text{constant} & \text{on } \partial \Omega . \end{array} \right . \end{equation*}$$ Serrin proved that the only such domain was if $$\Omega$$ is a ball.

Overdetermined problems often lead to free boundary problems (i.e problems in which the boundary/ a portion of the boundary are a priori not know) which is a particularly rich and active field of current research.

• Thanks for this informative answer. It's good for me to know "Serrin's problem" anyway! May 29, 2021 at 3:47

I'll make my comment into an answer.

The Dirichlet and Neumann conditions can both be considered separately, as they are independent. In other words, you can find the sets $$\mathfrak D=\{u\mid u \text{ solves Dirichlet version} \}$$ $$\mathfrak N =\{u\mid u \text{ solves Neumann version} \}$$

and then your solutions to the combined problem are the set $$\mathfrak D \cap \mathfrak N$$

Further, it would be well-posed if both the respective parts are well posed.