General Solution of a PDE

Find the general solution of the following PDE.

u_{xxyy} = 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 :)

u_{xxy = }F(x) Where F(x) is a constant in terms of x.

u_{xx =}yF(x) +F_{1}(x) Where F_{1}(x) is a constant in terms of x.

u_{x = }yxF(x) + xF_{1}(x) + F_{2}(y) Where F_{2}(y) is a constant in terms of y.

u = u(x,y) = 1/2 (yx^{2}F(x) ) + 1/2 (x^{2}F_{1}(x) ) + xF_{2}(y) + F_{3}(y) Where F_{3}(y) is a constant in terms of y.

Is that correct?

Thanks

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!

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?

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!

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!

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?

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 :)

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) + M_{1}(x) + xF_{2}(y) +F_{3}(y)

Where M(x) = S I(x) and where I(x) = S F(x)

sames goes for M_{1} = S I_{1}(x) where I_{1}(x) = S F_{1} (x).

Is there a simpler way of doing it?

Thanks