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somethingstra
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Homework Statement
We the domain be the unit disc D:
[tex]D=\left \{(x,y):x^{2}+y^{2}<1 \right \}[/tex]
let u(x,y) solve:
[tex]-\triangle u+(u_{x}+2u_{y})u^{4}=0[/tex] on D
boundary:
u=0 on [tex]\partial D[/tex]
One solution is u=0. Is it the only solution?
Homework Equations
Divergence Theorem
"Energy Method"
The Attempt at a Solution
Assume two solutions u1 and u2. We have then:
[tex]-\triangle u_{1}+(u_{x1}+2u_{y1})u_{1}^{4}=0[/tex]
and
[tex]-\triangle u_{2}+(u_{x2}+2u_{y2})u_{2}^{4}=0[/tex]
subtract the two to get:
[tex]-\triangle (u_{2}-u_{1})+(u_{x1}+2u_{y1})u_{1}^{4})-(u_{x2}+2u_{y2})u_{2}^{4}=0[/tex]
let: [tex]v=u_{2}-u_{1}[/tex]
Multiply by v and integrate. After employing the divergence theorem and the fact that v=0 on the boundary, we get:
[tex]\int_{\Omega }^{}|\triangledown (v)|^{2}=\int_{\Omega }^{}v(u_{x1}+2u_{y1})u_{1}^{4})-\int_{\Omega }^{}v(u_{x2}+2u_{y2})u_{2}^{4}=0[/tex]
Take u1 or u2 = 0. If 0 is truly the only solution, the resulting integral should equal to zero.
Resulting integral where u' = either u1 or u2
[tex]\int_{\Omega }^{}|\triangledown (u')|^{2}=-\int_{\Omega }^{}u'^{5}(u'_{x}+2u'_{y}))[/tex]
I conclude that u does not necessarily = 0, so u=0 is not the only solution
Right? or Wrong? And I have a question, what was the purpose of defining the unit disc?