# heat equation on bounded and unbounded domain

I have been reading about the heat equation and I am confused about uniqueness in the case when the domain is bounded and when it is not. In the book I am following, it is common to write the heat equation over [0,1], with zero values on the boundaries and shows that a series solves that equation. I can follow that, and the claim that the solution is unique.

Then, the same book can give an example on unbounded domain and write the solution as an integral of the heat kernel and again it is a solution and it is unique.

So, do boundary conditions affect uniqueness? Looks like not as I can write a unique solution in both cases. What is the effect then?

## 1 Answer

This question is more related with mathematics than Scientific Computing. But I will mention some things that come to my mind. I am not trying to be rigorous here, though.

You are mentioning more than one thing here:

1. You are changing your domain. In one case I suppose that you have $x \in [0,1]$, but for the other one $x \in [0, \infty)$.
2. The effect of having different boundary conditions.
3. Uniqueness dependance on boundary conditions.

Regarding 1, you need to consider that the functions that you allow in your solution are different when you change the domain. For one you would have something like

$$u = A e^{-kx} + B e^{kx},\quad x \in [0,1],$$

for the second you will enforce $B = 0$, to satisfy $\lim_{x\rightarrow \infty} u < \infty$, thus

$$u = A e^{-kx} \quad x \in [0,\infty),$$

assuming that $k$ is positive.

For 2, I can mention that the boundary conditions change the values that your constants $A$ and $B$ would take, assuming that the solution is unique.

About 3, for some boundary conditions the solutions is not unique. Your problem can have infinite solutions, e.g., $\frac{d^2 u}{dx^2} = 0$, with Neumann conditions at both sides. Your problem can have zero solutions if your boundary conditions cannot be satisfied. And, I suppose, that you can have multiple solutions for non-linear boundary conditions.

• I think you are mainly considering Laplace's equation not the heat equation. Is this true? – fred Aug 1 '16 at 13:33
• @fred, more or less, when the system evolves it should converge to a solution that depends on the boundary conditions and not in the initial conditions. – nicoguaro Aug 1 '16 at 13:40