Solving Diff. Eqns with Constants: a & b

[KrayzBlu]

Hello,

I've been looking for the form of the solution(s) to following differential equations:

$\frac{\partial^2}{\partial x \partial y}f(x,y) = a \cdot g(x,y)$
$\frac{\partial^2}{\partial x \partial y}g(x,y) = b \cdot f(x,y)$

Where a and b are unrelated constants, and f,g are of the same general form. I've tried exponentials and different categories of Bessel functions, with no luck. Does anyone recognize what form the solution(s) might take? Or have an idea of how I could proceed to solve this?

Thanks in advance :)

 

[the_wolfman]

What are your boundary conditions?

If you have a rectangular domain take $\frac{\partial^2}{\partial x \partial y}$ of your first equation and then use $\frac{\partial^2}{\partial x \partial y}g =b f$ to get a single forth order equation $\frac{\partial^4}{\partial x^2 \partial y^2}f =ab f$.

You can use separation of variables and let $f\left(x,y\right) =X\left(x\right) Y\left(y\right)$.

This will give you two second order equations, one for $X\left(x\right)$ and one for $Y\left(y\right)$. Both of which are solvable.

 

[pasmith]

Hello,

I've been looking for the form of the solution(s) to following differential equations:

$\frac{\partial^2}{\partial x \partial y}f(x,y) = a \cdot g(x,y)$
$\frac{\partial^2}{\partial x \partial y}g(x,y) = b \cdot f(x,y)$

Where a and b are unrelated constants, and f,g are of the same general form. I've tried exponentials and different categories of Bessel functions, with no luck. Does anyone recognize what form the solution(s) might take? Or have an idea of how I could proceed to solve this?

Thanks in advance :)

Differentiating the first with respect to x and y and eliminating $\frac{\partial^2 g}{\partial x\,\partial y}$ using the second gives
$$\frac{\partial^4 f}{\partial x^2 \,\partial y^2} = ab f$$

Setting $f(x,y) = X(x)Y(y)$ then gives
$$X'' Y'' = ab XY$$
so that
$$\frac{X''}{X} \frac{Y''}{Y} = ab$$
Thus if $X'' = CX$ then $Y'' = (ab/C)Y$. The corresponding eigenfunction for $g$ is then obtained from the first equation as
$$\frac1a \frac{\partial^2 f}{\partial x\,\partial y} = \frac 1a X'(x)Y'(y)$$

 

[KrayzBlu]

Thanks for your replies.

the_wolfman, thanks for your suggestion, I wish I had some specific boundary conditions - I'll think about that and let you know, but I would be happy with anyone of the equation form(s) that would solve this.

pasmith, thank you as well. I'm not sure how in your final equation, the left hand side is any different from the right hand side, and how that solution helps.

Unfortunately, I'm still at a loss as to what functions would satisfy these equations, since exponentials or their relatives don't seem to work.

 

[KrayzBlu]

Some progress

My engineering brain wasn't used to this sort of math, so I may have been to quick to say that exponentials don't work - my apologies

I suppose that if $f(x,y) = e^{h(x,y)}$, and $g(x,y) = e^{i(x,y)}$, then we can say that $h(x,y) = a\int{\frac{g(x,y)}{f(x,y)}}$ and $i(x,y) = b\int{\frac{f(x,y)}{g(x,y)}}$. From this, we can infer that $g(x,y) = e^{ab\int{\frac{\partial^2}{\partial x \partial y}\frac{1}{h(x,y)}}}$, such that all we need is a form for h(x,y).

Is it impossible to guess at the form of h(x,y) without any boundary conditions?