Here is one of them - i posted it in another thread and i am getting help in there https://www.physicsforums.com/showthread.php?t=91781(adsbygoogle = window.adsbygoogle || []).push({});

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!

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# Partial DIfferential Equations problems

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