# Homework Help: Find all solutions of u_(xx) + u_(yy) = 0, form u(x,y) = f(x^2 + y^2)

1. Dec 20, 2009

### coverband

1. Find all solutions of u_(xx) + u_(yy) = 0 which have the form u(x,y) = f(x^2 + y^2).

3. General solution of laplace is u(x,y)= Eexp(ky)Coskx + Fexp(-ky)Coskx + Gexp(ky)Sinkx + Hexp(-ky)Sinkx, where E,F,G,H are arbitrary constants don't see how this can be linked to f(x^2+y^2)?

2. Dec 20, 2009

### CompuChip

Re: laplace

I suppose the idea is, that you can write $u_{xx} = \partial^2 u/\partial x^2$ and $u_{yy} = \partial^2 u/\partial y^2$ in terms of the "total" derivative $df(r)/dr$, making it an ordinary differential equation for f.

Last edited: Dec 20, 2009
3. Dec 20, 2009

### HallsofIvy

Re: laplace

Why no attempt at all? Suppose u(x,y)= f(x^2+ y^2). What are u_xx and u_yy in terms of f? If you let z= x^2+ y^2, that reduces to a fairly simple d.e. for f(z).

No, that is NOT the general solution to Laplace's equation. You could also have an infinite sum of such things- i.e. any function that has a Fourier series could be a solution.

4. Dec 20, 2009

### coverband

Re: laplace

Hi Halls

Yes I was thinking of doing that initially.

u_xx=2f'(x^2+y^2)+4(x^2)f''(x^2+y^2)
u_yy=2f'(x^2+y^2)+4(y^2)f''(x^2+y^2)

5. Dec 20, 2009

### CompuChip

Re: laplace

What is the new diff.equation you get?
Hint: write it in terms of z = x^2 + y^2 as Halls suggested.

The answer is what you get when you solve this equation.

PS Final hint for solving: consider the derivative of z f'(z)

6. Dec 20, 2009

### HallsofIvy

Re: laplace

Well, no, adding those is not "the answer"- but gives, as I said, a simple equation to solve to get the answer.
Adding those gives $u_{xx}+ u_{yy}= 4f '(x^2+ y^2)+ 4(x^2+ y^2)f"(x^2+ y^2)= 0$. Setting $z= x^2+ y^2$, as I suggested, gives the ordinary differential equation $4f '(z)+ 4z^2f"(z)= 0$. Letting $g(z)= f'(z)$ reduces that to a simple first order equation for g. After finding g, integrate to get f.

7. Dec 21, 2009

### CompuChip

Re: laplace

Except that it is $4f'(z)+ 4zf"(z)= 0$, not $4f'(z)+ 4z^2f"(z)= 0$

8. Dec 21, 2009

### HallsofIvy

Re: laplace

You are right. I guess my mind shifted to "$r= \sqrt{x^2+ y^2}$" while I was writing!

9. Dec 21, 2009

### coverband

Re: laplace

Thanks guys

The answer appears to be f(x^2+y^2)=Aexp(-(x^2+y^2)) where A is an arbitrary constant