Separating system of PDEs

Question

I have the following two coupled linear PDE's:

$$ \partial_{t} q_{0} + \partial_{x} q_{1} - 10^{-2} \partial_{x}^{2} q_{0} + 10^{-2} \partial_{x}^{2} q_{1} + \frac{1}{2} q_{0} = 0\\ \partial_{t} q_{1} + \partial_{x} q_{0} + 10^{-2} \partial_{x}^{2} q_{0} - 10^{-2} \partial_{x}^{2} q_{1} + 3 q_{1} = 0 $$

I would like to separate these two equations such that I would end up with two new equations

$$ F(q_0) = 0\\ G(q_1) = 0 $$ where $F$ is a function of $q_0$ and it's derivatives in $x$ and $t$ (possibly of higher order, as well as mixed time/space derivatives), and a similar situation for $G$, except in terms of $q_1$ and it's derivatives.

I'm not entirely sure how this can be done, or even to show if it is possible (or conversely, show it's impossible). I've tried the technique done for turning Maxwell's Equations into the two "decoupled" wave equations by applying a spatial derivative to both equations, but the resulting substitutions I could make from the other PDE don't eliminate $q_1$ from the first equation, or eliminate $q_0$ from the second equation.

Is there some general procedure for a linear of system of PDE's to separate them into alternative "decoupled" PDE equations? Alternatively, is there anyway to prove that such a separation is impossible for these two equations?

Answer 1

We start with:

$$ \partial_{t} q_{0} + \partial_{x} q_{1} - a \partial_{x}^{2} (q_{0} - q_{1}) + b q_{0} = 0 \\ \partial_{t} q_{1} + \partial_{x} q_{0} + a \partial_{x}^{2} (q_{0} - q_{1}) + c q_{1} = 0 $$

First, let's add the two equations:

$$\partial_{t} (q_0+q_1)+\partial_{x} (q_0+q_1)+ b q_0+ c q_1=0$$

This is a first order PDE. Now let us subtract the two equations:

$$\partial_{t} (q_0-q_1)-\partial_{x} (q_0-q_1)- 2a \partial_{x}^{2} (q_{0} - q_{1})+ b q_0- c q_1=0$$

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Let's introduce new functions:

$$f=q_0+q_1 \\ g= q_0-q_1$$

Then:

$$q_0=\frac{1}{2} (f+g) \\ q_1=\frac{1}{2} (f-g)$$

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Now we rewrite the equations we obtained as:

$$\partial_{t} f+\partial_{x} f+ \frac{b}{2} (f+g)+ \frac{c}{2} (f-g)=0$$

$$\partial_{t} g-\partial_{x} g- 2a \partial_{x}^{2} g+ \frac{b}{2} (f+g)- \frac{c}{2} (f-g)=0$$

Or:

$$\partial_{t} f+\partial_{x} f+ \frac{b+c}{2} f+ \frac{b-c}{2} g=0$$

$$\partial_{t} g-\partial_{x} g- 2a \partial_{x}^{2} g+ \frac{b-c}{2} f+ \frac{b+c}{2} g=0$$

Renaming the parameters for convenience we have:

$$\partial_{t} f+\partial_{x} f+ \beta f+ \gamma g=0$$

$$\partial_{t} g-\partial_{x} g- \alpha \partial_{x}^{2} g+ \gamma f+ \beta g=0$$

Now it's trivial to separate the equations:

$$g= -\frac{1}{\gamma} \left(\partial_{t} f+\partial_{x} f+ \beta f \right)$$

$$\partial_{t} g=-\frac{1}{\gamma} \left(\partial^2_t f+\partial^2_{xt} f+ \beta \partial_{t} f \right)$$

$$\partial_x g=-\frac{1}{\gamma} \left(\partial^2_{xt} f+\partial^2_x f+ \beta \partial_x f \right)$$

$$\partial^2_x g=-\frac{1}{\gamma} \left(\partial^3_{xxt} f+\partial^3_x f+ \beta \partial^2_x f \right)$$

Substitute all of this to the second equation to get an equation with only $f$. Do the same the other way around for $g$.