Extending an ODE from one input variable to two input variables

Question

I have the following parabolic-shaped, non-linear ODE: $$\frac{dx}{dt} = kx(L-x) $$ where $L$ represents some upper limit (in my case, length, but it could also be, say, carrying capacity, whatever) and $k$ is the constant of proportionality.

This ODE has the solution $$x(t) = \frac{LA_0 e^{kLt}}{1+A_0e^{kLt}}$$ where $x(0) = A_0$ is the initial amount.

If you graph the original ODE, it looks like an upside-down parabola. I would like to extend this situation from 2-dimensions to 3-dimensions. Thus, the 3-dimensional equation would look like an upside down paraboloid.

How do I make the new differential equation with this goal in mind?

I imagine that it would be as simple as $$\frac{dz}{dt}= kx(L-x) + ky(L-y)$$ if we consider everything to be symmetric.

If this is indeed so, I assume that solving the equation would turn it into a PDE somehow, but I'm not really sure about this (I've not studied PDE's before).

Am I on the right track? Let me know if you have any questions.

Answer 1

The equation

$$ \frac{dx}{dt} = f(x) $$

has the input variable $t$ and output variable $x$. If you want to extend this to a third variable, you'll either get

• 1 input variable, 2 output variables. This is a system of 2 equations

\begin{align} \frac{dx}{dt} &= f(x,y) \\ \frac{dy}{dt} &= g(x,y) \end{align}

• 2 input variables, 1 output variable. You get a partial differential equation instead

$$ a \frac{\partial u}{\partial x} + b\frac{\partial u}{\partial y} = f(u,x,y) $$

Also

If you graph the original ODE, it looks like an upside down parabola

What you did here is graph the relationship $\frac{dx}{dt}$ vs $x$. This only works when there is no dependency on the input variable. This is different from graphing the solution as $x$ vs. $t$. Different variables.

A question for you is, why? What are you trying to accomplish by adding a third variable? It doesn't seem, to me, like you fully understand this situation.