Positive solutions to a PDE equation

Question

In this question I am concerned with nonlinear positive harmonic solutions to the following problem $$Δu(x,y)=0, (x,y)∈(a,b)×ℝ$$ $$u(x,y)=0, (x,y)∈{a₀}×ℝ$$

where $a₀$ is a real constant in the interval $(a,b)$. i.e., find $u$ harmonic solution of that PDE such that $u>0$ or $u<0$ for all $(x,y)∈(a,a₀)×ℝ$. Can we speak on the unicity, or some additional conditions are required.

what about $u(x,y)=x-a_0$? What is a harmonic solution? Simply a solution of the pde? — Quickbeam2k1, Feb 22 '14 at 13:14
@Quickbeam2k1 : That certainly work. So does $y(x-a_0)$ and possibly a few others in that kind. — user88595, Feb 22 '14 at 13:16
@MarkMcClure: Formally, if $f$ is a harmonic function, then $f$ cannot exhibit a true local maximum within the domain of definition of $f$. — DER, Feb 22 '14 at 13:40
So probably, you checked the function $u(x,y)= (x-a_0)^2-y^2$ which fulfills the boundary data, but not the sign requirement on the cylinder $(a,a_0)\times \mathbb{R}$ , right? — Quickbeam2k1, Feb 22 '14 at 13:44

score 1 · Accepted Answer · answered Feb 22 '14 at 14:11

Define $c:=a_0-a >0$. For $n>0$ the functions $$u_n(x,y):=exp\left(\frac{y}{cn}\right) \sin\left(\frac{x-a_0}{cn}\right)$$ are harmonic solutions to your problem. In particular, there is no uniqueness. (Of course, one could add some $\pi/2$ somewhere in here)

Positive solutions to a PDE equation

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