# PDE: Is 0 the only solution?

1. Mar 28, 2009

### somethingstra

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Divergence Theorem
"Energy Method"

3. 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?

2. Mar 29, 2009

### somethingstra

bump...

Last edited: Mar 29, 2009