Let D be a domain inside a simple closed curve C in R2. Consider the boundary value problem(adsbygoogle = window.adsbygoogle || []).push({});

[tex] (\Delta u)(x,y) = 0, \ (x,y) \in D, \\ \frac{\partial u}{\partial n} (x,y) = 0 , \ (x,y) \in C. [/tex]

where n is the outward unit normal on C. Use Green's Theorem to prove taht every solution u is a constant.

and yes its DELTA u not nabla u .

i know that Green's theorem is something lik this

[tex] \int \int _{D} u \nabla^2 u dx dy = \oint_{C} u \frac{\partial u}{\partial n} ds - \int \int_{D} |grad u|^2 dxdy [/tex]

so cna i do this

[tex] \int \int_{D} u \Delta u dx dy = \oint_{C} u \frac{\partial u}{\partial n} ds - \int \int_{D} |\nabla u|^2 dx dy [/tex]

i mean isnt [tex] \Delta u = \nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} [/tex] isnt that true??

if that s true then moving on

[tex] \oint_{C} u \frac{\partial u}{\partial n} ds = 0 [/tex] because du/dn = 0 and sinve it is definite integration of a constant the answer is zero

so all that we're left with is

[tex] \int \int_{D} u \Delta u dx dy = - \int \int_{D} |\nabla u|^2 dx dy [/tex]

if i differentiate on both sides it yields

[tex] u \Delta u \displaystyle{|_{D}} = |\nabla^2 u| [/tex]

i m not sure how to proceed from here...

how would one figure out whether u is constnat?

Please help! Thank you.

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# Homework Help: Boundary value problem

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