Partial DIfferential Equations problems

by stunner5000pt
Tags: differential, equations, partial
 P: 1,439 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!
 P: 1,439 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?