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

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In summary, the conversation discusses finding solutions to the equation u_(xx) + u_(yy) = 0 in the form of u(x,y) = f(x^2 + y^2). The conversation then mentions using the variable z = x^2 + y^2 and solving the resulting ordinary differential equation for f to find the general solution. The final solution appears to be f(x^2 + y^2) = Aexp(-(x^2 + y^2)), where A is an arbitrary constant.
  • #1
coverband
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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)?
 
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  • #2


I suppose the idea is, that you can write [itex]u_{xx} = \partial^2 u/\partial x^2[/itex] and [itex]u_{yy} = \partial^2 u/\partial y^2[/itex] in terms of the "total" derivative [itex]df(r)/dr[/itex], making it an ordinary differential equation for f.
 
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  • #3


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


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)?
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


HallsofIvy said:
Suppose u(x,y)= f(x^2+ y^2). What are u_xx and u_yy in terms of f?

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)

Are you saying the addition of these is the answer!?
 
  • #5


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


coverband said:
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)

Are you saying the addition of these is the answer!?
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 [itex]u_{xx}+ u_{yy}= 4f '(x^2+ y^2)+ 4(x^2+ y^2)f"(x^2+ y^2)= 0[/itex]. Setting [itex]z= x^2+ y^2[/itex], as I suggested, gives the ordinary differential equation [itex]4f '(z)+ 4z^2f"(z)= 0[/itex]. Letting [itex]g(z)= f'(z)[/itex] reduces that to a simple first order equation for g. After finding g, integrate to get f.
 
  • #7


Except that it is [itex]
4f'(z)+ 4zf"(z)= 0
[/itex], not [itex]
4f'(z)+ 4z^2f"(z)= 0
[/itex]
 
  • #8


You are right. I guess my mind shifted to "[itex]r= \sqrt{x^2+ y^2}[/itex]" while I was writing!
 
  • #9


Thanks guys

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

Thanks for your help
 

Related to Find all solutions of u_(xx) + u_(yy) = 0, form u(x,y) = f(x^2 + y^2)

1. What is the general solution to the partial differential equation uxx + uyy = 0?

The general solution to this partial differential equation is u(x,y) = f(x2 + y2), where f is an arbitrary function.

2. How do you find all solutions to uxx + uyy = 0?

To find all solutions, we can use separation of variables. Assume u(x,y) = X(x)Y(y), then substitute this into the equation to get two ordinary differential equations. Solve these equations to get the general solution in terms of f(x2 + y2).

3. Can u(x,y) = f(x) + g(y) be a solution to uxx + uyy = 0?

No, this form of solution does not satisfy the partial differential equation. It is a linear combination of two functions, so the second derivatives would not cancel out.

4. What is the role of the arbitrary function f in the general solution?

The arbitrary function f allows for a family of solutions to the partial differential equation. Different choices of f will result in different solutions, but they will all satisfy the equation uxx + uyy = 0.

5. Can u(x,y) = f(x2 - y2) be a solution to uxx + uyy = 0?

Yes, this is a valid solution to the partial differential equation. It is a particular case of the general solution u(x,y) = f(x2 + y2), where f is an even function.

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