2 dimensional PDE and a higher dimension problem.

[r.a.c.]

Homework Statement

The 2d PDE

Assume $$f\in S(\mathbb{R}^2)$$ (Schwartz space)

Then solve

$$u_{xx}(x,y) + 2u_{yy}(x,y) + 3u_{x}(x,y) -4u(x,y) = f(x,y) ; (x,y) \in\mathbb{R}^2$$

$$u_{xxxx}(x,y) - u_{yy}(x,y) + 2u(x,y) = f(x,y) ; (x,y) \in\mathbb{R}^2$$

Homework Equations

The relevant equations are a whole course. None specifically come to mind except the Fourier transform which is $$f^h = \int_{\mathbb{R}^2} f(x) e^-^2^\pi ^i^\theta^.^x dx$$

The Attempt at a Solution

I, by all honesty have no idea how to start I've been thinking but can't. In fact that's the one thing I need help in, I think once I pick up some momentum I'll be fine.

Homework Statement

Higher d problem.

Assume $$f\in S(\mathbb{R}^8)$$ (Schwartz space)

Prove That

$$\int_{\mathbb{R}^5} | \int_{\mathbb{R}^8} e^-^2^\pi ^i^(^\theta^,^0^,^0^,^0^).^u f(u) du |^2 d\theta = \int_{\mathbb{R}^5} | \int_{\mathbb{R}^3} f(x,y) dy |^2 dx$$

Homework Equations

Plancherel's Theorem: $$\int |f(x)|^2 dx = \int |f^h(\theta )|^2 dc$$

The Attempt at a Solution

Well its easy to see that inner integral on the left hand side becomes $$f^h(\theta ,0,0,0) which is = f^h(\theta)$$. From there we can use plancherel to take the h off. But from there I get a bit confused because I'm not exactly sure what's allowed or not.

 

[lippyka]

Can I just pull out the f as a constant ? Can I flip the inner integral and absolute value ? I'm thinking that the trick is to write the right hand side in terms of Fourier transform and then apply Plancherel's theorem. But I'm not sure how to do it. Any help would be appreciated.