# 2 dimensional PDE and a higher dimension problem.

1. Jan 17, 2010

### r.a.c.

1. The problem statement, all variables and given/known data
The 2d PDE

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

Then solve

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

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

2. Relevant equations

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

3. 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.

1. The problem statement, all variables and given/known data

Higher d problem.

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

Prove That

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

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

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution