Using Green's First Identity

Question

Use Green's first identity to show that,

$$\Delta_2u=u^3\qquad on\qquad x^2+y^2<1$$ $$u=0\qquad on\qquad x^2+y^2=1$$

has no twice continuously differentiable solution other than $u(x,y)$ identically zero.

I really don't have any idea what I'm doing. Any help would be appreciated.

Answer 1

(I assume that the $\Delta_2$ in your DE is the Laplacian, which I'll call just $\Delta$.) Show, using the divergence theorem (or Green's Theorem), that $\nabla \cdot(u \nabla u)$ integrates to $0$ on your region (using the boundary condition). But also: $\nabla \cdot(u \nabla u)=u \Delta u + |\nabla u|^2 = u^4 + |\nabla u|^2 \geq 0$ everywhere on your region. So in fact $\nabla \cdot(u \nabla u) \equiv 0$ on your region and $u \nabla u \equiv const$, where the constant is $0$ by the boundary condition. Finally, Green's first identity with u in place of both functions and an easy to see $0$ on one side yields $\int_R u \nabla u +|\nabla u|^2 dV = 0$ and since we know $u \nabla u \equiv 0$, that means that $|\nabla u|^2$ and therefore also $\nabla u$ is constant $0$. Hence $u \equiv const.$ and by boundary condition, $u \equiv 0$.

Answer 2

And I had never seen that notation for the lapacian before either. My professor said it just means 2 variable lapacian.