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What can we say about the solution of this PDE?

by mousakas
Tags: solution
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mousakas
#1
Apr28-12, 07:21 AM
P: 5
Hello!
I would like to find some functions F(x,y) which satisfy the following equation

[itex]
\frac{F(x,y)}{\partial x}=\frac{F(y,x)}{\partial y}
[/itex]

For example this is obviously satisfied for the function
[itex]
F= exp(x+y)
[/itex]

I would like however to find the most general closed form solution.
Do you have any ideas?
Could it be that it has to be a function of x+y only for example?
I tried to get some info by taylor expantions but I was not so succesful.
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JJacquelin
#2
Apr28-12, 07:52 AM
P: 761
Consider any differentiable function F(x+y)
mousakas
#3
Apr28-12, 01:47 PM
P: 5
Does it have to be nececairly a function of x+y ?

Office_Shredder
#4
Apr28-12, 08:39 PM
Emeritus
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PF Gold
P: 4,500
What can we say about the solution of this PDE?

Let u=x+y and v=x-y. Then F(x,y) can be re-written as F(u,v) and
[tex] \frac{\partial F}{\partial x} = \frac{\partial F}{\partial u} + \frac{\partial F}{\partial v}[/tex]
[tex] \frac{\partial F}{\partial y} = \frac{\partial F}{\partial u} - \frac{\partial F}{\partial v} [/tex]

So for these to be equal we get that [tex] \frac{\partial F}{\partial u} + \frac{\partial F}{\partial v}= \frac{\partial F}{\partial u} - \frac{\partial F}{\partial v}[/tex]
which reduces to [tex] \frac{\partial F}{\partial v}=0[/tex]

so F is a function of u only (i.e. F can be written as F(x+y))

This is a fairly common technique for finding the solutions to differential equations like this - divine what the answer should be then use a change of variables to prove it
mousakas
#5
Apr29-12, 12:57 PM
P: 5
Thank you both for your answers :)

BUT
look also that the function in the r.h.s. is not F(x,y) but F(y,x)
For example if
[itex]
F(x,y)=\frac{x}{x+y}
[/itex]
then
[itex]
F(y,x)=\frac{y}{x+y}
[/itex]
That's what confuses me.
Office_Shredder
#6
May1-12, 08:34 AM
Emeritus
Sci Advisor
PF Gold
P: 4,500
Ahhh, my mistake, I misread the question. Let's define
[tex] G(x,y) = \frac{\partial F(x,y)}{\partial x}[/tex]

Then all the equation in the OP is saying is that
G(x,y)=G(y,x).

So G is any function which is symmetric in x and y. Then integrating w.r.t x says that
[tex] F(x,y) = \int G(x,y) dx + H(y)[/tex]
integrate G with respect to the x variable. The "constant of integration" in this case is is a function which is constant in x, so can be any function of y.

An example of a solution:
Pick G(x,y) = x2+y2. Then [tex]F(x,y)=\frac{x^3}{3}+y^2x+H(y)[/tex] where H(y) is any function you want.
mousakas
#7
May2-12, 07:00 AM
P: 5
Thanks for the help ;)
hunt_mat
#8
May3-12, 04:19 AM
HW Helper
P: 1,583
What about the method of characteristics? Have you tried that? that should tell you that the solution are propagated along certain curves (or in this case lines), You need to know some Cauchy data first though.


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