Proof that diffusion-"reaction" equations yield a non-negative solution

Question

First question: I wish to prove that there exists a unique, positive solution $\phi_g(x,t)$ to the following system of $G$ PDEs with initial and boundary conditions for $t \ge 0$, $0 \le x \le X$:

$$ \left[\frac{1}{v_g}\frac{\partial}{\partial t} - \frac{\partial}{\partial x} D_g(x) \frac{\partial}{\partial x} \right] \phi_g(x,t) + \sum_{g'=1}^G \Sigma_{r, g' \to g}(x) \phi_{g'}(x,t) = 0 $$

$$ \phi_g(x,0) = \phi_{g,0}(x) > 0$$ $$ \phi_g(0,t) - 2 D_g(0) \frac{\partial\phi_g}{\partial x}(0,t) = 0 $$ $$ \phi_g(X,t) + 2 D_g(X) \frac{\partial\phi_g}{\partial x}(X,t) = 0 $$

Here, $v_g, D_g(x) > 0$ for all $g$, $\Sigma_{r, g' \to g}(x) \ge 0$ when $g = g'$, and $\Sigma_{r, g' \to g}(x) \le 0$ when $g \ne g'$.

Basically, I want to show $\phi_g(x,t) > 0$ for all $g$ and for all $x,t$ in my domain. Proving uniqueness/existence would be nice, but positivity/non-negativity is the thing I'm most interested in. From physical intuition (and from having solved these problems for a variety of physical parameters), I'm fairly confident that there must exist a unique, positive solution. However, I have no idea how to prove it. If you have any thoughts on how I could approach this proof, or know of a good reference that might lead me to such a proof, I would be very grateful!!

In Hundsdorfer's "Numerical Solution of Time-Dependent Advection-Diffusion-Reaction Equations" book, this is touched on a little bit in Chapter 1 (sections 1 and 7), but he only explicitly proves things for problems in which there is no dependence on $x$.

(Side note: In my field of study, this is the neutron diffusion equation, but I've seen in literature that it's more commonly referred to in the chemistry setting as diffusion-reaction equations, so I'm using that name here in hopes of drawing more attention.)

EDIT: Sorry, I had a small typo describing the positivity/negativity of the physical parameters. Fixed now.

EDIT2: Made it more clear which parameters are space-dependent.

EDIT3: Altered the initial condition a bit to make it easier. We also have the following inequality:

$$ \sum_{g'=1}^G \Sigma_{r,g \to g'} \ge 0 $$

I think I have a rough proof for my first question, which I will post as an answer shortly, but now I also have a new, related question.

========================================================================== New, related question: Can we show that the corresponding BVP

$$- \frac{\partial}{\partial x} D_g(x) \frac{\partial}{\partial x} \phi_g(x) + \sum_{g'=1}^G \Sigma_{r, g' \to g}(x) \phi_{g'}(x) = q_g(x) $$ $$ \phi_g(0) - 2 D_g(0) \frac{\partial\phi_g}{\partial x}(0) = 0 $$ $$ \phi_g(X) + 2 D_g(X) \frac{\partial\phi_g}{\partial x}(X) = 0 $$

has a "positive" solution? i.e., can we show that $\phi_g(x) > 0$ for all $x \in [0,X]$? Here, $q_g(x)$ is a "positive" source. That is, $q_g(x) \ge 0$ and

$$ \sum_g \int_0^X q_g(x) dx > 0 $$

Answer 1

Here's my first attempt at a proof of the positivity (with lots of help from a friend). I think there are some "holes" in it, and I would appreciate any feedback on what those holes are and how to fill them in.

First, we note that $\phi_g(x,t)$, for any $g$, cannot have a negative minimum at $x = 0$ or $x=t$ -- the boundary conditions indicate that, if $\phi_g(0,t) < 0$ or $\phi_g(X,t) < 0$, then $\phi_g(x,t)$ is decreasing "into" the domain. Thus, we can assume that, if there are any negative components to $\phi_g(x,t)$, the global minimum would occur at an interior location (i.e., not at $x = 0$ or $x = X$).

We note that at time $t = 0$, everything is positive. Let $t_0$ denote the first time at which one (or more) of the $\phi_g(x,t)$ functions become zero-valued at some point $x_0$. Let $g_0$ be the index of that function. $x_0$ is a global minimum point of $\phi_{g_0}(x,t_0)$. Thus, at this time and at this point in space, the following must be true: $$ \phi_{g_0}(x_0,t_0) = 0 $$ $$ \phi_g(x_0,t_0) \ge 0 $$ $$ \frac{\partial}{\partial x} \phi_{g_0}(x_0,t_0) = 0 $$ $$ \frac{\partial^2}{\partial x^2} \phi_{g_0}(x_0,t_0) > 0 $$

Moreover, since it has just decreased to zero, $$ \frac{\partial}{\partial t} \phi_{g_0}(x_0,t_0) \le 0 $$

Now, we revisit the terms of the PDE. The time derivative is $\le 0$. The spatial derivative term is strictly negative due to the negative sign and the strict positivity of the second derivative. The sum over $g'$ is negative unless all of the $\phi_g(x_0,t_0)$'s are zero; in that case, it is equal to zero.

Thus, we have something negative on the left hand side equal to zero on the right hand side. This is a contradiction -- our assumption that one of the $\phi_g(x,t)$'s become zero-valued at some point is impossible.

The one case I think this proof/argument leaves out is what if all the $\phi_g(x,t)$ becomes identically zero for all $x$ at time $t = t_0$. $\phi_g(x,t) = 0$ is a steady state solution for this problem as $t \to \infty$. However, I feel like this is impossible as long as we have a positive initial condition and $t$ is finite. Not sure how to show this explicitly though, I'll have to give it some more thought.