I'm trying to show that the solution to a PDE is unique. I'll state the problem for reference.
Let $q \in C(\bar U)$, where $U$ is a bounded domain in $\mathbb{R}^n$ with $\partial U \in C^1$. Let $u \in C^2(U) \cup C^1(\bar U)$ be a solution of $$\Delta u + q(x) u = 0 \text{ in } U \text{ and } \frac{\partial u}{\partial \nu} = \phi \text{ on } \partial U$$
where $\nu$ is the outward unit normal of $\partial U$, and $\phi$ a given continuous function on $\partial U$. Suppose $q(x) < 0$ everywhere on $U$. Show that the above problem has no more than one solution.
My work so far: let $u$ and $v$ be two solutions, and let $w=u-v$. I want to show that $w=0$. What I'd like to say is that $$\frac{\partial u}{\partial \nu} = \phi = \frac{\partial v}{\partial \nu}$$ $$\sum_{i=1}^n \frac{\partial u}{\partial x_i} \nu_i = \sum_{i=1}^n \frac{\partial v}{\partial x_i} \nu_i$$ $$\frac{\partial u}{\partial x_i} = \frac{\partial v}{\partial x_i} $$ $$\int \frac{\partial u}{\partial x_i} dx_i= \int \frac{\partial v}{\partial x_i} dx_i $$ $$u=v+C$$
Using this, I can combine it with a separate result I found using Green's Identity: namely, $0=\int_U q(x) (u-v) dx$. Together, these two statements imply that $C=0$ (because $q<0$), and so $u=v$. But I don't feel confident that I can claim $u=v+C$ using the steps I've shown. Is there any issue with my work here? If so, can someone suggest another tactic to try?