Thread: 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) +F₁(x) Where F₁(x) is a constant in terms of x.

u_{x =} 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

Hi, princessmath.

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

.

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!

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?

Hi again. I'm glad you're sticking with it on this tricky little problem!

Let's start from



We want to integrate both sides with respect to now. But we have a problem because we don't know what and are; all we know is that they are functions of .

I think if we use a concrete function for what I'm getting at might be a little more clear.

Note: I'm ignoring for now because I'm trying to demonstrate why after integrating we don't have

For example, pretend Then we have



When you take an antiderivative of the right hand side stays , it does NOT become

Does that help demonstrate what we're getting at?

Again, the main issue is that we don't know what and are specifically; they can be any functions with 's in it. So, at best, after we take an antiderivative we can write



where is a function of only, and so is constant with respect to

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

Good luck!

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!

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?

If you want to simplify the notation a bit and write



and



that's no problem at all

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