I am on the domain $(t,x) \in [0,T] \times [0,1]$ and I consider the following PDE:
$$\partial_t \theta(t,x) + L\theta(t,x) = f(x)\theta(t,x)^2$$
with $L$ a non-degenerate elliptic operator (think $a(x)\partial_{xx}+ b(x) \partial_x$ with $a$ and $b$ smooth and $a > 0$) and $f$ smooth and positive.
I consider smooth Dirichlet conditions like $\theta(t,x) = 1$ if $x=0$, $x=1$, or $t=T$.
How to prove existence and uniqueness of a $C^{1,2}$ solution to the PDE?
Clearly the assumption $f>0$ should play a role but I do not know how to proceed.
This is a partial answer to the uniqueness. Since $\theta(x,T) = 1$, there is a time $t_1$ such that $\theta(x,t) > 0$ when $t_1 \le t \le T$. We prove such solutions are unique in the case that $c = a_x-b$ is constant.
Suppose there are two, and set $u = \theta_1-\theta_2$ and $$ p(x,t) = f(x)\big(\theta_1(x,t)+\theta_2(x,t)\big). $$ We have $$ p \ge 0 $$ and $$ \frac{d}{dt}\int_0^1 u^2 dx = \int_0^1 2u(-au_{xx}-bu_x+pu)dx. $$ Integrate by parts, using $u(0,t) = u(1,t) = 0$. The previous line becomes $$ = \int_0^1 \big(2a u_x^2+2(a_x-b)uu_x +pu^2\big)dx \ge \int_0^1 2(a_x-b)uu_x dx, $$ and with our assumption this is $$ = cu^2\big|_0^1 = 0. $$ Thus we have that $$ \int_0^1 u^2 dx $$ is nonnegative, nondecreasing with time, and finally 0 when $t = T$. So it is zero for $t_1 \le t \le T$. So $u = 0$ there, and $\theta_1 = \theta_2$.
[Edit:] Integrating again by parts we get $$ \int_0^1 2(a_x-b)uu_x dx = -\int_0^1 (a_x-b)_x u^2 dx. $$ So uniqueness also holds if $a_x-b$ is nonincreasing.