Solution of PDE (heat equation) unique? I am new to PDEs and need some help. In fact, I am not studying PDEs but I need the following for another result not related to this topic.
Consider the initial value problem:
$\frac{d}{dt}w(t,x) - \frac{1}{2}\Delta_xw(t,x) -c(x)w(t,x)=0 \\w(t,x) \text{ is continous and } w(0,x)=f(x)$
where $c(x)$ is bounded and continous. Under which further conditions (on $f$) is the solution of this PDE unique and why? Is there a simple argument which is often used?
Thanks in advance!
 A: More or less fleshing out the comment of Winther to the question itself, and supplying some necessary initial/boundary conditions, we have:
If we have two solutions $w_1(t, x)$, $w_2(t, x)$ both satisfying
$w_t = \dfrac{1}{2} \nabla^2w + cw, \tag 1$
with identical initial conditions 
$w_i(0, x) = f(x), \; i = 1, 2,  \tag{1.1}$
on some region $\Omega$ with a sufficiently smooth boundary $\partial \Omega$ on which each also satisfies
$w_i(t, x) = 0, \; x \in \partial \Omega, \tag{1.2}$
then by linearity their difference $w(t, x)$ also obeys (1), and we may multiply through by $w$ to obtain
$ww_t = \dfrac{1}{2} w\nabla^2w + cw^2; \tag 2$
we have
$\nabla \cdot (w \nabla w) = \langle \nabla w, \nabla w\rangle + w\nabla^2w, \tag 3$
from which
$\dfrac{1}{2} w\nabla^2 w = \dfrac{1}{2} \nabla \cdot (w \nabla w) - \dfrac{1}{2} \langle \nabla w, \nabla w\rangle; \tag 4$
we integrate over $\Omega$, and apply the divergence theorem to the first integral on the right, which then vanishes in accord with (1.2):
$\dfrac{1}{2} \displaystyle \int_\Omega w\nabla^2 w\; dV$
$= \displaystyle \dfrac{1}{2}\int_\Omega \nabla \cdot (w \nabla w)\; dV - \dfrac{1}{2} 
\int_\Omega \langle \nabla w, \nabla w\rangle\; dV$
$= \displaystyle \dfrac{1}{2}\int_{\partial \Omega} w \nabla w\ \cdot \vec {dS} - \dfrac{1}{2} 
\int_\Omega \langle \nabla w, \nabla w\rangle\; dV$
$= -\dfrac{1}{2} 
\displaystyle \int_\Omega \langle \nabla w, \nabla w\rangle\; dV  \tag 5$
from (2),
$\dfrac{1}{2}(w^2)_t = \dfrac{1}{2} w\nabla^2w + cw^2; \tag 6$
$\displaystyle \dfrac{1}{2} \int_\Omega (w^2)_t \; dV = \dfrac{1}{2} \int_\Omega w\nabla^2w \; dV + \int_\Omega cw^2 \; dV; \tag 7$
using (5),
$\displaystyle \dfrac{1}{2} \int_\Omega (w^2)_t \; dV = -\dfrac{1}{2} 
\displaystyle \int_\Omega \langle \nabla w, \nabla w\rangle\; dV  + \int_\Omega cw^2 \; dV; \tag 7$
also,
$\displaystyle \dfrac{1}{2} \int_\Omega (w^2)_t \; dV =  \dfrac{1}{2} \dfrac{d}{dt} \int_\Omega w^2 \; dV; \tag 8$
thus, clearing the factors of $1/2$,
$\dfrac{d}{dt} \displaystyle \int_\Omega w^2 \; dV$
$= -\displaystyle \int_\Omega \langle \nabla w, \nabla w\rangle\; dV  + 2 \int_\Omega cw^2 \; dV < 0, \tag 9$
whenever $w(t, x) \ne 0$ provided the hypothesis $c(t, x) < 0$ binds.
Applying this inequality in the case $w(0, x) = 0$, we conclude that the left-hand integral remains $0$ for all $t$, and hence so $w(t, x)$ itself.  Thus,
$w_1(t, x) = w_2(t, x), \; \forall t, x, \tag{10}$
and the solution to (1) is unique.  $OE\Delta$.
