Empty result set on `pdsolve` in Maple

Question

I'm trying to solve the following:

$$ U_{xy}=\sin (x) \cos (y) $$

given boundary conditions $u_x(y=\frac{\pi}{2})=2x$ and $u(x=\pi)=2 \sin(y)$.

I'm feeding these into Maple as follows:

Unfortunately, after calling pdsolve, I get an empty result set. Can you help me figure out what's going on?

Answer 1

tl;dr: The order of the PDE must exceed the order of the initial and boundary conditions in each variable. In the supplied system, the order with respect to $x$ is the same in PDE and bc[1]. This means the highest order derivative in $x$ may be eliminated between the two equations. Maple's pdsolve() assumes you have already done this and does not do it for you.

The long version (includes solution)...

The documentation for the Maple pdsolve() function indicates that pdsolve() returns any of

• an explicit result for the indeterminate function if a general solution is obtained,
• a PDESolStruc() if a conditional solution is obtained, or
• NULL if no solution is found.

Looks like Maple found no solution for your PDE+BCs.

This is because your bc[1], $u_x(x,\frac{\pi}{2}) = 2x$, is first order in $x$ and so is the PDE. Boundary and initial conditions must, in each variable, have order less than that of the PDE. Otherwise, one could eliminate the terms of highest order between the two equations, leaving a simpler system.

For instance, we can integrate your PDE with respect to $y$, yielding $\frac{\partial u}{\partial x} = \sin(x)\sin(y) + \phi(x)$ where $\phi$ is an arbitrary function of $x$. When $y$ is $\frac{\pi}{2}$, this is $$\frac{\partial u}{\partial x} \left( x,\frac{\pi}{2} \right) = \sin(x)\sin \left( \frac{\pi}{2} \right) + \phi(x) = \sin(x) + \phi(x) = 2x,$$ where the last equation is bc[1]. So $\phi(x) = 2x - \sin(x)$.

But this means our system is $$\begin{align} \frac{\partial u}{\partial x} &= \sin(x)\sin(y) + 2x - \sin(x) \\ u(\pi,x) &= 2 \sin y, \end{align}$$ which we can automatically solve to $$ 1 - \pi^2 + x^2 + \cos(x) + \sin(y) - \cos(x)\sin(y), $$ which we can verify solves the initial system you supplied.