Elliptic partial differential equation?

I have to "show that in a region bounded by a closed curve C that the only solution of del^(2)u = u^(2n+1) (n a positive integer) which vanishes on C is u=0." (Del is the Laplacian operator).





Obviously the (2n+1) factor means that the exponent is only ever an odd number, but I can't see what difference this makes.








The equation provided which it is suggested I use is:





∫D (u*del^(2)u + grad^(2)u)dS = ∫C u du/dn ds


(ie integral over region D) = (integral on curve C)





Now I don't even understand this equation, so I'm stumped.





Can anyone show me what I need to do? Thanks.

Elliptic partial differential equation?
To my knowledge, there is no difference between del^2 and grad^2.


The grad of a scalar is a vector.


div of a vector is a scalar.


The Laplacian


del^2 u = div*grad(u)





Also I think it should be:


∫D (u*del^(2)u + grad^(2)u)dA = ∫C u du/dn ds


D stands for the domain bounded by C, so you speak of area. C refers to the boundary fo the domain, so you speak of arc length.





Anyway, the question specifies that u vanishes on the boundary, C ie u = 0 on C.


So ∫C u du/dn ds = 0





Thus ∫D (u*del^(2)u + grad^(2)u)dA = 0


∫D (u*u^(2n+1) + u^(2n+1))dA = 0


∫D (u+1)*u^(2n+1) dA = 0





I think what's supposed to happen, is that you recognize that the term being integrated has an even power of u, so it can never be negative. Thus the only way for the integral to be zero, is for the indivual term to be identically equal to zero.





Should the u be multiplied by the del^2 only, and not the grad^2? And exactly how is grad^2 defined?
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