Let $x \in D \subset R^m, t \ge 0$
I would like to prove the theorem of uniqueness of the solution
$\frac {\partial u}{\partial t} - \Delta u + F(u,x,t) = f(x, t)$$ u \Bigg|_{\partial D} = \mu(x, t); u\Bigg|_{t = 0} = \varphi(x)$
What I do.
Let there are two solutions $u_1$ and $u_2$ Then $v = u_2 - u_1$
Subtract the two equations: $$\frac {\partial v}{\partial t} - \Delta v + F(u_2, x, t) - F(u_1, x, t) = f(x, t)-f(x, t)=0; v \Bigg|_{\partial D} = 0; v\Bigg|_{t = 0} = 0$$
$$\frac {\partial v}{\partial t} = \Delta v - F(u_2, x, t) + F(u_1, x, t)$$ Multiply by v and integrate in D:
$$\int\limits_D v \frac {\partial v}{\partial t} \, dx = \int\limits_D \frac {\partial }{\partial t}( \frac {v^2} {2}) \, dx = \frac{1}{2} \frac{d}{dt} \int\limits_D v^2(x, t) \, dx = \int\limits_D v \Delta v \, dx - \int\limits_D v \{ - F(u_2, x, t) + F(u_1, x, t) \} dx =$$ By first Green's formula: $$=\int\limits_{\partial D} v \frac{\partial v}{\partial n} \, dS - \int\limits_D (\nabla v)^{2} \, dx - \int\limits_D v \{ - F(u_2, x, t) + F(u_1, x, t)\} \, dx$$
Consider separately:
$$v \Bigg|_{\partial D} = 0 \Rightarrow \int\limits_{\partial D} v \frac{\partial v}{\partial n} \, dS = 0$$
$$\int\limits_D (\nabla v)^{2} \, dx \ge 0$$
If I find conditions that $\int\limits_D v \{ - F(u_2, x, t) + F(u_1, x, t)\} \, dx \ge 0$, I can say that right part $\le 0$
$\Rightarrow$
$$\int\limits_D v^2(x, t) \, dx $$ does not increase as function depends on t $$\int\limits_D v^2(x, t) \, dx \le \int\limits_D v^2(x, 0) \, dx = 0 \Rightarrow v(x,t) \equiv 0$$
But how to do it if we don't know anything about F? And the second question, what will happen if $F=F(u, \frac {\partial u}{\partial x_{1}}, \frac {\partial u}{\partial x_{2}}, \cdots , \frac {\partial u}{\partial x_{m}}, x, t)$?