## Partial DIfferential Equations problems

Here is one of them - i posted it in another thread and i am getting help in there http://physicsforums.com/showthread.php?t=91781

this is another of my problems
Show that if C is a piecewise continuously differentiable closed curve bounding D then the problem
$$\nabla^2 u= -F(x,y) \ in\ D$$
$$u = f \ on \ C_{1}$$
$$\frac{\partial u}{\partial n} + \alpha u = 0 \ on \ C_{2}$$
where C1 is a part of C and C2 the remainder and where alpha is a positive constant, has at most one solution.

now i know that $$\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = -F(x,y)$$

now im not quite sure how to connect the C1 part to C2 part...
would it be something liek C= C1 + C2?

but how would one go about showing that this has at most ONE solution?? I m not quite sure how to start ... Please help

another one
Show that the problem
$$\frac{\partial}{\partial x} (e^x \frac{\partial u}{\partial x} + \frac{\partial}{\partial y} (e^y \frac{\partial u}{\partial y} = 0 \ for \ x^2+y^2 < 1$$
u = x^2 for x^2 + y^2 = 1
has at most one solution
Hint Use the divergence theorem to derive an energy identity

Perhaps i dont remember a theorem i should have learnt in ap revious class... or i am not familiar with it but what would i use the divergence theorem here?
i eman i can get it down to this
$$e^x \frac{\partial}{\partial x} (u + \frac{\partial u}{\partial x}) + e^y \frac{\partial}{\partial y} (u + \frac{\partial u}{\partial y}) = 0$$
but hereafter i am stuck, please do advise!

Thank you!

 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
 ok so i can rewrite the second euqation as $$e^x \frac{\partial}{\partial x} (u + u_{x}) + e^y \frac{\partial}{\partial y} (u + u_{y}) = 0$$ also is $$u+ u_{x}$$ written as something else... how would i apply the divergence principle here?
 can anyone help me with this!

Recognitions:
Homework Help