 # Solving Coupled Differential Equations. I have the following differential equations, for modeling predator-prey relationships: dx/dt=Ax-Bxy curukksm 2022-09-16 Answered
Solving Coupled Differential Equations
I have the following differential equations, for modeling predator-prey relationships:
$\frac{dx}{dt}=Ax-Bxy$
$\frac{dy}{dt}=Cxy-Dy$
Where A, B, C, and D are constants. How could I go about solving this? I've only really worked with basic first order differential equations before, and I've found little to help me figure this out. Any help would be much appreciated.
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Step 1
Generally for this type of model one would start analysing what happens at equilibrium, i.e. whether there is a fixed point, if they are stable or not by looking at the Jacobian of the system.
Step 2
First find the null-clines setting: $0=Ax-Bxy={f}_{1}\left(x,y\right)$ and $Cxy-Dy={f}_{2}\left(x,y\right)=0$ then you get $x=0,y=A/B$ for ${f}_{1}\left(x,y\right)$ and $y=0,x=D/C$ for ${f}_{2}\left(x,y\right)$. Finding the intersection of the null-clines you get two fixed points (0,0) (both population are extinct) and (D/C,A,B) (both populations are non-zero). Then if you look at the Jacobian (matrix of partial derivatives of ${f}_{1}\left(x,y\right),{f}_{2}\left(x,y\right)$ computed in the two fixed points you can understand which is the stable one and where the system ends up.
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Step 1
While complete analytic solution is not likely, it's curious that we can reduce this system to a couple of nonlinear first order ODEs for x(t) and y(t).
Let's do this for y only, since the same procedure can be applied to x as well.
First, we write down some useful relationships:
$\stackrel{˙}{x}=x\left(A-By\right)$
$Cx-D=\frac{\stackrel{˙}{y}}{y}$
$Cxy=\stackrel{˙}{y}+Dy$
Now we differentiate the second equation w.r.t. t:
$\stackrel{¨}{y}=\left(Cx-D\right)\stackrel{˙}{y}+Cy\stackrel{˙}{x}=\frac{{\stackrel{˙}{y}}^{2}}{y}+\left(A-By\right)\left(\stackrel{˙}{y}+Dy\right)$
Since the equation doesn't containt t explicitly, we can reduce the order by the usual substitution:
$\stackrel{˙}{y}=u\left(y\right),\phantom{\rule{2em}{0ex}}\stackrel{¨}{y}=u\cdot {u}^{\prime }$
We obtain:
$u{u}^{\prime }=\frac{{u}^{2}}{y}+\left(A-By\right)\left(u+Dy\right)$
Now we introduce another function:
$u=y\cdot v\left(y\right),\phantom{\rule{2em}{0ex}}{u}^{\prime }=v+y{v}^{\prime }$
We get:
$yv\left(v+y{v}^{\prime }\right)=y{v}^{2}+\left(A-By\right)\left(v+D\right)y$
Step 2
Simplifying, we obtain:
$yv{v}^{\prime }=\left(A-By\right)\left(v+D\right)$
But this is a separable equation. So:
$\frac{vdv}{v+D}=\left(\frac{A}{y}-B\right)dy$
$v-D\mathrm{ln}\left(v+D\right)=A\mathrm{ln}y-By+{c}_{1}$
Getting back to the original function, we have:
$\frac{\stackrel{˙}{y}}{y}-D\mathrm{ln}\left(\frac{\stackrel{˙}{y}}{y}+D\right)=A\mathrm{ln}y-By+{c}_{1}$
Or:
$\begin{array}{}\text{(1)}& \stackrel{˙}{y}-Dy\mathrm{ln}\left(\stackrel{˙}{y}+Dy\right)=\left(A-D\right)y\mathrm{ln}y-B{y}^{2}+{c}_{1}y\end{array}$
This doesn't look like anything solvable, but it is indeed a 1st order ODE for y only.
We also need to determine c1 from the original system somehow, because the extra constant shouldn't be here.
Step 2
We can actually resolve (1) for the derivative using Lambert W (product logarithm) function. Transforming the equation:
$\left(\stackrel{˙}{y}+Dy{\right)}^{-Dy}{e}^{\stackrel{˙}{y}}={y}^{\left(A-D\right)y}{e}^{-B{y}^{2}+{c}_{1}y}$
$\frac{\stackrel{˙}{y}+Dy}{Dy}\mathrm{exp}\left(-\frac{\stackrel{˙}{y}+Dy}{Dy}\right)=\frac{{e}^{-1-{c}_{1}/D}}{D}{y}^{-A/D}\mathrm{exp}\left(\frac{B}{D}y\right)$
This has a solution:
$\frac{\stackrel{˙}{y}+Dy}{Dy}=-W\left(-\frac{{e}^{-1-{c}_{1}/D}}{D}{y}^{-A/D}\mathrm{exp}\left(\frac{B}{D}y\right)\right)$
$\begin{array}{}\text{(2)}& \stackrel{˙}{y}=-Dy\left(1+W\left(-\frac{{e}^{-1-{c}_{1}/D}}{D}{y}^{-A/D}\mathrm{exp}\left(\frac{B}{D}y\right)\right)\right)\end{array}$