# General Solution of a PDE

• Aug 17th 2012, 05:27 PM
princessmath
General Solution of a PDE
Find the general solution of the following PDE.

uxxyy = 0

I know I have to go from right to left i.e integrate with respect to y, then y then x and x.

So here it goes :)

uxxy = F(x) Where F(x) is a constant in terms of x.

uxx =yF(x) +F1(x) Where F1(x) is a constant in terms of x.

ux = yxF(x) + xF1(x) + F2(y) Where F2(y) is a constant in terms of y.

u = u(x,y) = 1/2 (yx2F(x) ) + 1/2 (x2F1(x) ) + xF2(y) + F3(y) Where F3(y) is a constant in terms of y.

Is that correct?

Thanks
• Aug 17th 2012, 09:23 PM
GJA
Re: General Solution of a PDE
Hi, princessmath.

Nice work so far. Something did catch my eye. You have the line

Quote:

$\displaystyle u_{xx}=yF(x) + F_{1}(x)$
.

We need to be careful at this point, because the next step is to take an antiderivative with respect to x. Since there are functions of x on the right hand side we cannot do this by putting x's in front of them like we were able to put y in front of F(x) in the previous integration.

Does this make sense? Let me know if there are any questions.

Good luck!
• Aug 17th 2012, 10:27 PM
princessmath
Re: General Solution of a PDE
Hey GJA,

Do you mean I have to change F(x) to F(x^2/2) if we were to integrate wrt x?

The y in front is fine because the LHS has no y.

is that what you mean?
• Aug 18th 2012, 07:14 AM
GJA
Re: General Solution of a PDE
Hi again. I'm glad you're sticking with it on this tricky little problem!

Let's start from

$\displaystyle u_{xx}=yF(x)+F_{1}(x).$

We want to integrate both sides with respect to $\displaystyle x$ now. But we have a problem because we don't know what $\displaystyle F(x)$ and $\displaystyle F_{1}(x)$ are; all we know is that they are functions of $\displaystyle x$.

I think if we use a concrete function for $\displaystyle F_{1}(x)$ what I'm getting at might be a little more clear.

Note: I'm ignoring $\displaystyle F(x)$ for now because I'm trying to demonstrate why after integrating we don't have $\displaystyle xF_{1}(x).$

For example, pretend $\displaystyle F_{1}(x)=e^{x}.$ Then we have

$\displaystyle u_{xx}=yF(x)+e^{x}.$

When you take an antiderivative of the right hand side $\displaystyle e^{x}$ stays $\displaystyle e^{x}$, it does NOT become $\displaystyle xe^{x}=xF_{1}(x).$

Does that help demonstrate what we're getting at?

Again, the main issue is that we don't know what $\displaystyle F(x)$ and $\displaystyle F_{1}(x)$ are specifically; they can be any functions with $\displaystyle x$'s in it. So, at best, after we take an antiderivative we can write

$\displaystyle u_{x}=y\int F(x)dx+\int F_{1}(x)dx+F_{2}(y),$

where $\displaystyle F_{2}(y)$ is a function of $\displaystyle y$ only, and so is constant with respect to $\displaystyle x.$

Again, good work. Keep working hard and asking questions and it will make sense!

Good luck!
• Aug 18th 2012, 11:06 PM
princessmath
Re: General Solution of a PDE
Oh I see what you mean. Yeah I never thought about it like that. So it's a function in terms of x. Hmm let me think about it and see if I can produce a final solution.

Thanks a lot!
• Aug 18th 2012, 11:17 PM
princessmath
Re: General Solution of a PDE
Can you generalize it to say Integral of F(x) is equal to I(x) where I(x) = S F(x) dx and then will continue the same process? Do you think that is valid?
• Aug 19th 2012, 07:31 AM
GJA
Re: General Solution of a PDE
If you want to simplify the notation a bit and write

$\displaystyle I(x)=\int F(x)dx$

and

$\displaystyle I_{1}(x)=\int F_{1}(x)dx,$

that's no problem at all :)
• Aug 19th 2012, 06:09 PM
princessmath
Re: General Solution of a PDE
I got it!

So basically I get this general solution where I have something that looks like this.

U(x,y) = yM(x) + M1(x) + xF2(y) +F3(y)

Where M(x) = S I(x) and where I(x) = S F(x)
sames goes for M1 = S I1(x) where I1(x) = S F1 (x).

Is there a simpler way of doing it?

Thanks