# Uniqueness of Solution for Boundary Valued Problems

Is it true that all boundary problems (ODE's n-order, PDE's, etc.) in any dimension always have an unique solution (if a solution exists)? If not, what are some counterexamples to this? Furthermore, is there a general set of restrictions that will allow the uniqueness of solution (ODE, PDE, etc.). For example, forcing both the derivative and the position function to have certain values at the boundary, or required the solution to be infinitely smooth and the space compact. Maybe it has to do with weak derivative, etc.

My intuition is that the answer should be no. Suppose two non-identical solutions for the BVP exist. Then I can "follow" both equations "out" (away) from the Boundary to a point where they differ. This means the vector field points in two directions at this point, which is not allowed to happen as learned in elementary differential equation.

Thank you.

• Short answer: no. It can happen that a BVP has no solution at all, or infinitely many solutions. The latter is particularly of interest. See characteristic function. – Vim Mar 22 '16 at 9:57
• I am only interested in the case that at least one solution exists. I do feel like the answer should be no. But suppose I have two functions both solve the differential equation and satisfy the boundary. Then I "will" be able to find a point where the vector field points in two different directions due to the 2 equations not being the same. – Paichu Mar 24 '16 at 6:20

Your question is equivalent to the question which boundary value problems (ODEs or PDEs) are well posed, in the sense of Hadamard. To this question alone, a major part of PDE research has been devoted for the last century or so, especially in the case of boundary conditions which are considered to be physically meaningful. To quote Wikipedia's lemma on boundary value problems:

Much theoretical work in the field of partial differential equations is devoted to proving that boundary value problems arising from scientific and engineering applications are in fact well-posed.

A famous example is the question when the Navier-Stokes equations possess a global, smooth solution; it is one of the one million dollar Clay Institute problems. For more thoughts on the issue of uniqueness for this particular problem, see here. For some recent well-posedness results for the Navier-Stokes equations, see here.

Even in the ODE setting, the question of uniqueness related to boundary conditions is a difficult one. Quoting Scholarpedia's lemma on boundary value problems, which focuses exclusively on the ODE context:

Questions of existence and uniqueness for BVPs are much more difficult than for IVPs. Indeed, there is no general theory. However, there is a vast literature on individual cases [...]

I hope this helps!

• Thank you. But rather than focusing on certain special problems. I was wondering if there is a general requirement such that if the solution satisfy it, then the solution must be unique. – Paichu Apr 5 '16 at 8:06
• There is not, I'm afraid. It's not so simple. – Frits Veerman Apr 5 '16 at 10:01

Counterexample to all BVP with existing solutions being unique

Take $y'' = -16y$ with $y(0) = 0$ and $y(\pi) = 0$ then:

$r^2 = -16 \implies r = \pm 4i \implies y(x) = c_1\cos(4x) + c_2 \sin(4x)$

Apply BC:

$y(0) = c_1 = 0$ and $y(\pi) = 0 = c_2 \sin(0)$

Note the second BC cannot uniquely determine $c_2$, thus there are infinitely many solutions:

$y(x) = c_2\sin(4x)$ for $c_2 \in \mathbb{R}$.

Theorem about Uniqueness and Existence of BVP Solutions

Given a problem of the form: \begin{cases} y'' =f(y',y,x) \\ y(a) = \alpha, y(b) = \beta \end{cases}

If $f$ and $\frac{df}{dy}$ and $\frac{df}{dy'}$ continuous on $x \in [a,b]$, $y,y' \in \mathbb{R}$ then if the both of the following are satisfied:

i) $\left| \frac{df}{dy} \right| > 0$

ii) $\frac{df}{dy'}$ satisfies a Lipschitz condition

then the problem given has a unique solution.

Note that the above counterexample works as it fails the (i) requirement of the theorem.

• Would you care to provide a reference for your uniqueness theorem? It might be helpful for curious readers (like me). – Vim Mar 24 '16 at 8:48