Is u=0 the only solution for the PDE on the unit disc?

[somethingstra]

Homework Statement

We the domain be the unit disc D:
D=\left \{(x,y):x^{2}+y^{2}<1 \right \}

let u(x,y) solve:
-\triangle u+(u_{x}+2u_{y})u^{4}=0 on D

boundary:
u=0 on \partial D

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:
-\triangle u_{1}+(u_{x1}+2u_{y1})u_{1}^{4}=0
and
-\triangle u_{2}+(u_{x2}+2u_{y2})u_{2}^{4}=0

subtract the two to get:
-\triangle (u_{2}-u_{1})+(u_{x1}+2u_{y1})u_{1}^{4})-(u_{x2}+2u_{y2})u_{2}^{4}=0

let: v=u_{2}-u_{1}

Multiply by v and integrate. After employing the divergence theorem and the fact that v=0 on the boundary, we get:

\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

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

\int_{\Omega }^{}|\triangledown (u')|^{2}=-\int_{\Omega }^{}u'^{5}(u'_{x}+2u'_{y}))

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?

 

[somethingstra]

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