# Not really Laplace's Equation?

1. Mar 10, 2007

### jhuleea

Not really Laplace's Equation??

Hi all!

I've been out of school for awhile and so, some of my engineering math is still rusty. While working out a fluids problem, I got stuck on the following PDE:

$$Y''(y)}Z(z)+Y(y)Z''(z)=-1$$
$$\frac{Y''(y)}{Y(y)}+\frac{Z''(z)}{Z(z)}=-\frac{1}{Y(y)Z(z)}$$

I know that my equation looks similar to Laplace's Equation:
$$Y''(y)}Z(z)+Y(y)Z''(z)=0$$
$$\frac{Y''(y)}{Y(y)}=-\frac{Z''(z)}{Z(z)}$$

But since the right hand term in my equation is nonzero, I don't think I can solve it the same way as I would with Laplace's Equation.

Last edited: Mar 10, 2007
2. Mar 11, 2007

### Mute

No, you can't solve that with separation of variables. You might be able to solve it using some sort of integral transform like a Fourier or Laplace transform. Calculating a Green's function might also work. Otherwise, you can always try some tricks to get some sort of solution that will at least satisfy the PDE (though maybe not the boundary conditions).

For instance,

$$\phi_{xx} + \phi_{yy} = -1$$

reminds me of $\cosh^2(x) - \sinh^2(x) = 1$, so I would then guess that

$$\phi_{xx} = \sinh^2(x)$$
$$\phi_{yy} = -\cosh^2(x)$$

and solve.

If I did it correctly this time, the solution should be

$$\phi(x,y) = \frac{1}{8}\cosh(2x) + \frac{1}{4}\cosh^2(y) + \frac{1}{4}(y^2 - x^2) + C$$

with C an arbitrary constant. Because x and y are symmetric, switching x and y also gives you a valid solution. Of course, this might not satisfy your boundary conditions, and so although it solves the PDE (provided I made no mistakes), its failure to satisfy the boundary condtions probably makes it not the solution you're looking for.

Last edited: Mar 11, 2007
3. Mar 11, 2007

### Crosson

4. Mar 15, 2007

### tehno

He seems to be trying to solve:

$$\nabla^2 U(y,z) +1=0$$

Where Y(y) and Z(z) are single variable and real valued functions?
Since you have interchangable functions and symmetric derivative function forms you may want to consider playing with symmetric substitution like:

$$V(y,z)=U(y,z)+ \frac{1}{4}(y^2+z^2)$$

Then your PDE is reduced to :

$$\nabla^2 V(y,z)=0$$.

You are in the saddle,becouse you sound like you know how to deal
with Laplace equation.Only problem left is that innital & boundary conditions ,as Mute indicated,must be seriously taken care of in the process of solving (and that depends ,in turn, on the nature of your problem ).

Last edited: Mar 15, 2007