PDE bounded smooth

Question

Let U$\subset$ $\mathbb{R}^n$ be a bounded smooth domain. Let $\lambda \in \mathbb{R}$ and $u \in C^2(\bar{U})$ such that $\Delta u=\lambda u$ and $u|_{\partial U}=0$. Prove that $u\equiv 0$ on $U$ or $\lambda<0$. We studied out of PDE Evans but I honestly don't know how to prove this.

Answer 1

If $\lambda$ is positive, then multiplying both members by $u$ and then integrating over your domain $U$ you get $$ \int_U u \Delta u = \lambda \int_U u^2 \, . $$ Integrating by parts the LHS (taking into account that $u=0$ on $\partial U$) you obtain $$ -\int_U |\nabla u|^2 = \lambda \int_U u^2 \,. $$ Since $\lambda \geq0$, necessarily $$ \int_U |\nabla u|^2\leq 0. $$ This implies $\nabla u = 0$ on $U$. Hence the function $u$ is constant in $U$. Therefore the zero function because of the boundary condition.

Answer 2

Our colleague Kosh presents the essentials of the requisite proof in his answer (+1, endorsed), but I wanted to flesh out the details, so:

We begin with the following well-known identity:

$\nabla \cdot (u \nabla u) = \nabla u \cdot \nabla u + u \nabla \cdot \nabla u$ $= \nabla u \cdot \nabla u + u \nabla^2 u = \nabla u \cdot \nabla u + u \Delta u; \tag 1$

we are given

$\Delta u = \lambda u; \tag 2$

thus

$\nabla \cdot (u \nabla u) = \nabla u \cdot \nabla u + u \Delta u = \nabla u \cdot \nabla u + \lambda u^2; \tag 3$

we integrate over $U$:

$\displaystyle \int_U \nabla \cdot (u \nabla u) \; dV = \displaystyle \int_U \nabla u \cdot \nabla u \; dV + \displaystyle \int_U \lambda u^2 \; dV$ $= \displaystyle \int_U \nabla u \cdot \nabla u \; dV + \lambda \int_U u^2 \; dV; \tag 4$

the left hand side of this equation may be converted into a surface integral over $\partial U$ by means of the divergence theorem:

$\displaystyle \int_U \nabla \cdot (u \nabla u) \; dV = \displaystyle \int_{\partial U} u \nabla u \cdot \vec n \; dS, \tag 5$

where $\vec n$ is the outward pointing unit normal vector field on $\partial U$ and $dS$ is its area element. In light of the given boundary condition

$u \mid \partial u = 0, \tag 6$

the integral on the right of (5) vanishes and so (4) becomes

$\displaystyle \int_U \nabla u \cdot \nabla u \; dV + \lambda \int_U u^2 \; dV = 0, \tag 7$

or

$\displaystyle \int_U \nabla u \cdot \nabla u \; dV = -\lambda \int_U u^2 \; dV. \tag 8$

Now if

$u \not \equiv 0 \tag 9$

on $U$, then by virtue of (6)

$\nabla u \not \equiv 0 \tag{10}$

as well, which implies (since $\nabla u$ is continuous)

$\displaystyle \int_U \nabla u \cdot \nabla u \; dV > 0; \tag{11}$

we also have, from (9),

$\displaystyle \int_U u^2 \; dV > 0, \tag{12}$

thus (7) may be written

$\lambda = -\dfrac{\displaystyle \int_U \nabla u \cdot \nabla u \; dV}{\displaystyle \int_U u^2 \; dV} < 0. \tag{13}$