The classic Protter and Weinberger: Maximum principles in Differential Equations states the following uniqueness theorem (theorem 8):
Let $u_1$, $u_2$ be solutions of the BVP $u''(x) + g(x) u'(x) + h(x) u(x) = f(x)$ with $h(x) \le 0$, $h(x)$ not identically zero. If $u_1$, $u_2$ satisfy the boundary conditions \begin{aligned} -u'(a) \cos \theta + u(a) \sin \theta &= \gamma_1 \\ u'(b) \cos \phi + u(b) \sin \phi &= \gamma_2 \end{aligned} where $0 \le \theta \le \pi/2$ and $0 \le \phi \le \pi/2$, then $u_1(x)=u_2(x)$.
The proof is fairly straight-forward involving the maximum principle.
- Assume $f(x)=0$ and all coefficients are real. Am I correct in assuming that the only solution for the BVP with $\gamma_1 = \gamma_2 = 0$ is $u(x)=0$? After all, this is a solution and, provided the theorem holds, it is the unique solution.
- Mathematica spits out a solution for the BVP $u''(x) + i\, 2 (1+\pi^2)^{1/2} u'(x) - u(x)=0$ with $u(0)=u(1)=0$, i.e. complex $g$. Now, my complex analysis skills are severely lacking but in the case $x\in \mathbb{R}$ can we separate the ODE into $u=u_1 + i u_2$ with $u_1$, $u_2$ both real? We would then have a coupled system of two ODEs which we could (possibly) solve. As a side note, are there maximum principles for systems?
- By the argument above, an ODE with real coefficients but purely imaginary boundary conditions would have a purely imaginary solution? If the boundary conditions are mixed, we would have two uncoupled ODEs which separately need to satisfy the respective boundary conditions for the real and imaginary part?
Thanks heaps!