
#1
Oct505, 12:06 PM

P: 1,445

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 [tex] \nabla^2 u= F(x,y) \ in\ D[/tex] [tex] u = f \ on \ C_{1} [/tex] [tex] \frac{\partial u}{\partial n} + \alpha u = 0 \ on \ C_{2} [/tex] 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 [tex] \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = F(x,y) [/tex] 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 [tex] \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 [/tex] 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 [tex] 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 [/tex] but hereafter i am stuck, please do advise! Thank you! 



#2
Oct505, 06:48 PM

P: 1,445

ok so i can rewrite the second euqation as
[tex] e^x \frac{\partial}{\partial x} (u + u_{x}) + e^y \frac{\partial}{\partial y} (u + u_{y}) = 0 [/tex] also is [tex] u+ u_{x} [/tex] written as something else... how would i apply the divergence principle here? 



#3
Oct605, 06:45 PM

P: 1,445

can anyone help me with this!




#4
Oct605, 07:56 PM

Sci Advisor
HW Helper
P: 1,322

Partial DIfferential Equations problems
For the first problem, you might begin by assuming that two solutions exist which satisfy the differential equation and boundary conditions. The difference of the two solutions satisfies a simpler set of equations, right? Maybe this is a good place to start.
For the second problem, the original equation already looks like the divergence of a vector field in 2d. Maybe you should start from this observation. 


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