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Partial Diff qn

by fredrick08
Tags: diff, partial
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fredrick08
#1
May21-09, 08:52 AM
P: 376
1. The problem statement, all variables and given/known data
Does the following differential equation for u(x,y) have solutions which have the form of a product of functions of each independent variable:

[tex]\partial[/tex]2u/[tex]\partial[/tex]x[tex]\partial[/tex]y=u


[b]3. The attempt at a solution[/]

=>[tex]\int[/tex][tex]\int[/tex]u dx dy =uxy=f(x)f(y) ???
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fredrick08
#2
May21-09, 08:57 AM
P: 376
im not sure if that even makes sense, but if u(x,y)=f(x)f(y) and d*f(x)/dx=u/constant and d*f(y)/dy=constant... i think, i don't know im confused..
Dick
#3
May21-09, 09:14 AM
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Set u(x,y)=f(x)*g(y). What is your partial derivative in terms of f(x), f'(x), g(y), and g'(y)?

fredrick08
#4
May21-09, 10:00 AM
P: 376
Partial Diff qn

so [tex]\partial[/tex]2u/[tex]\partial[/tex]x[tex]\partial[/tex]y=f(x)g(y)??
so in terms of f'(x) and g'(y) im not sure..
Cyosis
#5
May21-09, 11:31 AM
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P: 1,495
You have an equation that equates the derivative of u to u itself. You can already guess what u will be, but lets not guess.

Setting u(x,y)=f(x)g(y), then the equation becomes [tex]\frac{\partial^2 f(x)g(y)}{\partial x \partial y}=f(x)g(y)[/tex]. Now what Dick asks you to do is to calculate [tex]\frac{\partial^2 f(x)g(y)}{\partial x \partial y}[/tex].

Hint:

[tex]
\frac{d}{dx}f(x)= ?
[/tex]

In terms of f'(x)
Dick
#6
May21-09, 01:32 PM
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I actually meant u(x,y)=f(x)g(y). There's no reason why the function of x would have to be the same as the function of y. And yes, evaluate the derivative.
Cyosis
#7
May21-09, 01:35 PM
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Sorry, that was sloppy of me I actually meant f(x)g(x). I will fix it.
fredrick08
#8
May21-09, 06:00 PM
P: 376
hmm d/dx of f(x).. isn't that just f'(x)??.... is it something like the integral of f'(x)g'(y)??
fredrick08
#9
May21-09, 06:02 PM
P: 376
so... [tex]
\frac{\partial^2 f(x)g(y)}{\partial x \partial y}
[/tex] =[tex]\int[/tex][tex]\int[/tex]f'(x)g'(y) dx dy??
Cyosis
#10
May21-09, 06:12 PM
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What you've written down now is the equivalent to [itex]\frac{d}{dx}f(x)=\int f'(x) dx[/itex]. Which is wrong. Lets test it, f(x)=x then [itex]\frac{d}{dx}x=1 \neq \int 1 dx=x+c[/itex].

hmm d/dx of f(x).. isn't that just f'(x)??
This is right, so if you evaluate [tex]\frac{\partial^2 f(x)g(y)}{\partial x \partial y}[/tex] you get....?
fredrick08
#11
May21-09, 06:20 PM
P: 376
u get f'(x)g'(y)?? Im confused becasue when u take partial of f(x), u get d/dy(d/dx*f(x))=0??
Cyosis
#12
May21-09, 06:25 PM
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P: 1,495
Yes that's correct, but you're not taking the partial of f(x), you're talking the partial derivative of f(x)g(y).

[tex]
\frac{\partial^2 f(x)g(y)} {\partial x \partial y}=\frac{\partial}{\partial x} f(x) \frac{\partial}{\partial y}g(y)=f'(x)g'(y)
[/tex]
fredrick08
#13
May21-09, 06:26 PM
P: 376
ok so since f(x)g(y)=f'(x)g(y)=u(x,y)=??
Cyosis
#14
May21-09, 06:30 PM
HW Helper
P: 1,495
You forgot a prime it should be [itex]f'(x)g'(y)=f(x)g(y)=u(x,y)[/itex]. You can instantly see which function satisfies the equation, but you can also separate the variables.
fredrick08
#15
May21-09, 06:32 PM
P: 376
e^x+y?? so f(x)=e^x and g(y)=e^y same as the derivativeS?
Cyosis
#16
May21-09, 06:35 PM
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P: 1,495
That would be a solution indeed, you're on the right track. it is not the general solution however. The general solution will involve integration constants and a constant in the argument of the exponential. Can you see how to integrate the differential equation?
fredrick08
#17
May21-09, 06:38 PM
P: 376
e^xy?? so f(x)=e^x and g(y)=e^y same as the derrivativeS?

srry this was an accident
Cyosis
#18
May21-09, 06:41 PM
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Don't randomly guess, [itex]f'(x)g'(y)=y e^{xy}xe^{xy} \neq e^{xy}[/itex]. Separate the variables first, that is all the x dependent stuff on one side and all the y dependent stuff on the other side.


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