How to solve $\partial_a \partial_b F(a,b) = F(a,b)$?

Question

If I had a linear PDE like $\partial_a F(a,b) + \partial_b F(a,b) = 0,$ I would employ the method of characteristics to solve it, or, just google the solution and find that, depending on some initial condition/function h, the solution is $h(a+b).$

However, I'm trying to understand a different type of PDE, namely

$\partial_a \partial_b F(a,b) = F(a,b).$

What techniques can be relied upon to solve this, which I can then study along with your solution to understand the process?

The reason I'm asking is because, for instance, if I take an initial condition $F(a,0) = f(a),$ and I know that $e^{a+b}$ is one solution, this one solution doesn't depend on the initial condition and is certainly not all possible solutions. So obviously I'm missing a lot of ground work that provides a robust means of finding all solutions and knowing that the set you've found is unique.

Answer 1

The ansatz in my comment doesn't appear to be the most general, but it shows a continuum of solutions, rather than the single one you found.

Separating variables, so they must separately equal a constant $k$:

\begin{align} F(a,b) &= f(a)g(b) \\ f(a)g(b) &= f'(a)g'(b) \\ f(a)/f'(a) &= g'(b)/g(b) \\ g'(b) &= k g(b) \\ f'(a) &= f(a) / k \\ \end{align}

We then have $f(a) = A_0 \exp(a/k)$, and $g(b) = B_0 \exp(kb)$.

This lets you get exponential/evanescent plane waves in any direction in either the first or third quadrant.

In general, any solution can be squeezed (or reflected!) in one direction while being stretched the same amount in the other direction thanks to the effect this has on the derivatives.

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The other obvious thing to do is a power series expansion: i.e. take $F(a, b) = c_{i,j} a^i b^j$. The restriction is then $c_{i,j} = (i+1)(j+1) c_{i+1,j+1}$. The product of exponentials solution above is then $c_{i,j} = A_0 B_0 k^{(j - i)}/(i! j!)$. But I think you can take each $c_{i,j}$ with either $i$ or $j$ 0 to vary independently, fully specifying the terms on each diagonal, and I think that these all converge. I.e. let $c_{i,j} = f(i - j)/(i! j!)$ for any $f$ that doesn't get big too rapidly. This gets a countable basis for a subset of solutions, but I believe it must miss some because we have a continuum of solutions seen above.

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This is a weird little PDE. Neither Fourier transforms nor two-sided Laplace transforms directly work. In either case let $x$ and $y$ be the conjugate variable to $a$ and $b$. You end up with $F\cdot(1 \pm x\cdot y) = 0$. Changing to one-sided Laplace (on either or both axes) gives what looks like a place to stick boundary-values (perhaps too many for the double case -- or perhaps not given the expansion above, which admits nearly independent variation along $a$ and $b$ axes), but I can't actually compute the inverse to solve it (What's the inverse of $\exp(c/s)$?).

However, we can view these solutions as a clue to freely expand solutions in such a basis so long as it's only non-zero on surfaces where $x \cdot y = \pm 1$, and so long as it converges. The comment about formal solutions, and the other answer mostly captures this, but I think fails to emphasize that $x$ and $y$ (referred to as $k$, $\pm 1/k$ or $\tau$ and $1/\tau$, can actually be full complex numbers, so long as things converge. (Note that unlike the comment, we can restrict to real Fourier solutions: $f_{k,\theta}(a, b) = \sin(ka - b/k + \theta)$, or sin/cos pairs).

All of the Laplace and Fourier solutions still seem like they're undercounting solutions though, as specifying on the line $f(a, 0)$ forces an exact solution on $f(0, b)$, even though the power series expansion seems to indicate that they can be specified separately (except at the intersection, of course).

Answer 2

$$\frac{\partial^2 F}{\partial x\partial y} = F(x,y)$$ First we look for particular solutions on the form $F(x,y)=X(x)Y(y)$ as proposed in the question. $$\frac{X'}{X}\frac{Y'}{Y}=1\quad\implies\quad X'(x)=e^{cx}\quad Y'(y)=e^{y/c}$$ $$F_c(x,y)=C\;\exp\left(c\,x+\frac{1}{c}y \right)$$ $C$ and $c$ are arbitrary constants.

Then any linear combination of particular solutions is a more general solution.

On discret form with arbitrary constants $c_n$ and $C_n$ : $$F(x,y)=\sum_{n}C_n\exp\left(c_n\,x+\frac{1}{c_n}y \right)$$

On more general form with arbitrary function $\Phi$ : $$F(x,y)=\int \Phi(\tau)\exp\left(\tau\,x+\frac{1}{\tau}y \right) d\tau$$ $\Phi(\tau)$ can be any continuous or discret function and/or a mixture of both insofar the integral is convergent on the considered range.