exact solution to lotka-volterra equations

Question

I am looking for exact or perturbative solution realistic lotka-volterra (the one with logistic term in one of the equations) equations in population dynamics.

Any reference where they have done it will be useful.

Answer 1

We cannot find exact solution in a closed form, but I hope it will be useful.

So, we have the system $$ \left\{\begin{aligned} \dot x &= \alpha x - \beta xy=x(\alpha - \beta y),\\ \dot y &= -\gamma y + \delta xy = y(-\gamma + \delta x) \end{aligned}\right. $$ From first equation $$ \frac{\dot x}{x} = \alpha - \beta y\Longrightarrow \beta y = \alpha - \frac{\dot x}{x} $$ Let's substitute it into second: $$ \beta\dot y = \beta y(-\gamma + \delta x) $$ $$ \frac{d}{dt}\frac{\dot x}{x} + \Big(\alpha - \frac{\dot x}{x}\Big)(-\gamma + \delta x) = 0 $$ $$ \frac{x\ddot x - \dot x^2}{x^2} + \Big(\alpha - \frac{\dot x}{x}\Big)(-\gamma + \delta x) = 0 $$ $$ {x\ddot x - \dot x^2} + (\alpha x^2 - x\dot x)(\delta x-\gamma) = 0 $$ Let $\dot x = p$; then $\ddot x = p\frac{dp}{dx}$, $$ pxp'-p^2 + (\alpha x^2 - xp)(\delta x-\gamma) = 0 $$ Divide by $x^2$: $$ \frac{p}{x}p'-\frac{p^2}{x^2} + \Big(\alpha - \frac{p}{x}\Big)(\delta x-\gamma) = 0 $$ Let $p=xq$: $$ q(q + xq') - q^2 + (\alpha - q)(\delta x - \gamma) = 0 $$ $$ xq\frac{dq}{dx} + (\alpha - q)(\delta x - \gamma) = 0 $$ $$ \frac{qdq}{\alpha - q} + \Big(\delta - \frac{\gamma}{x}\Big)dx = 0 $$ $$ -q-\alpha\ln|q-\alpha|+\delta x-\gamma\ln x = C $$ But $q=p/x=\dot x/x=\alpha - \beta y$ ($q\le\alpha$ always) and $q_0 = \alpha - \beta y_0$ $$ -q_0-\alpha\ln|q_0-\alpha|+\delta x_0-\gamma\ln x_0 = C $$ $$ C = -q_0-\alpha\ln(\beta y_0)+\delta x_0-\gamma\ln x_0 $$ So, $$ -q-\alpha\ln(\alpha-q)+\delta x-\gamma\ln x = C $$ We may solve this equation using Lambert W function: $$ q=\alpha\left[1 + W\left(-\frac1\alpha \exp\left(-\frac1\alpha(\alpha - \delta x + \gamma\ln x + C)\right)\right)\right]. $$ But $q = \dot x/x$ and $$ \int\frac{dx/x}{1 + W\left(-\frac1\alpha \exp\left(-\frac1\alpha(\alpha - \delta x + \gamma\ln x + C)\right)\right)} = \int\alpha dt $$ I'm sure that there is no closed form for integral in LHS.