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

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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.