# Partial Differential Equation with Two Dependent Variables

I've come across a PDE where I feel there should be some clever way of solving it, but I can't seem to find it. It has a form like this:

$$\frac{\partial}{\partial t}\left(y+cz\right)=\frac{\partial}{\partial x}\left(-by+acz\right)$$

Here $y$ and $z$ are the dependent variables and a, b, and c are all constants. Ideally, there would exist some other variable $u$ thats a function of $y, z$ such that we can rewrite the equation in an equivalent form like:

$$\frac{\partial}{\partial t}\left(u\right)=\frac{\partial}{\partial x}\left(u\right)$$

which could be solved with method of characteristics, though I can't seem to find such a variable. Assuming there isn't one, can you recommend any other approaches?

Edit: Here's the original two PDE's from which the above expression was obtained:

$$\frac{\partial y}{\partial t}=-a_1\frac{\partial y}{\partial x}-b_1\left(y-z\right)$$ $$\frac{\partial z}{\partial t}=a_2\frac{\partial z}{\partial x}+b_2\left(y-z\right)$$

We took a linear combination of these 2 equations to try eliminate the terms with $\left(y-z\right)$ in them and then tried to find an invariant variable to solve as I was talking about earlier. $a_1, a_2, b_1, b_2$ are all constants.

• How about Fourier transforming it? Jul 18 '18 at 16:59
• They are two unknowns $y(x,t)$ and $z(x,t)$. But there is only one equation. The problem is undetermined, or something is missing in the wording of the question, isn't it ? Jul 18 '18 at 17:07
• Yeah, this pde actually comes from the linear combination of two other pdes. One for y, and one for z. We ended up doing that combination to eliminate a term that was present in both of them. Jul 18 '18 at 17:15
• @Leif Ericson. What are the two PDEs ? Why don't you write the whole problem ? This makes doubtful what you did and what really is the key issue.. Jul 18 '18 at 17:51
• Sorry about the lateness of the reply, I edited the original post to show the PDE's from which the combined expression came from. Thanks again for your help. Jul 19 '18 at 16:21

A good way to find such solutions is too assume that $$\frac{\partial}{\partial t}\left(y+cz\right)=\frac{\partial}{\partial x}\left(-by+acz\right)=f'(t)g'(x)$$therefore $$y+cz=f(t)g'(x)+h_1(x)\\-by+acz=f'(t)g(x)+h_2(t)$$therefore$$z=\dfrac{1}{ac+bc}(bf(t)g'(x)+f'(t)g(x)+bh_1(x)+h_2(t))\\y=\dfrac{1}{a+b}(af(t)g'(x)-f'(t)g(x)+ah_1(x)-h_2(t))$$