Analytical solution to a nonlinear system of coupled ODEs

Question

I think I'm being rather silly but I'm trying to see if the system of nonlinear coupled ODEs given by

$$x'=-xy$$ $$y'=-xy$$

has any non-trivial solutions. After some googling I have found that a solution to this problem is given by $$ y(t)=c_{1}-\frac{c_{1}e^{c_{1}c_{2}}}{e^{c_{1}c_{2}}-e^{c_{1}t}}$$ $$ x(t)=-\frac{c_{1}e^{c_{1}c_{2}}}{e^{c_{1}c_{2}}-e^{c_{1}t}}$$

but I have no idea how this solution was obtained? Any insight into the problem or this solution would be a great help.

This all started as I was trying to see if the system given by

$$x'=-k_{1}xy-k_{2}xz$$ $$y'=-k_{1}xy$$ $$z'=-k_{2}xz$$

where $k_{1}$ and $k_{2}$ are constants, has an analytical solution. I simplified the problem and obtained what I presented at the start of this post.

Anyway, any insight into any of this is appreciated.

Answer 1

Since $y' = x'$, $y = x - A$ for some constant $A$, reducing the system to a single separable o.d.e., $$x' = x (A - x) .$$

The stationary solutions are $(x, y) = (0, A)$ and $(x, y) = (A, 0)$, $A \in \Bbb R$.

If a solution is not stationary, then separating gives $$\frac{dx}{x (A - x)} = dt ,$$ and integrating, solving for $x$, and relabeling constants gives $$x(t) = \frac{A B}{1 + B e^{A t}} ,$$ which is equivalent to your solution. On the other hand, if $A = 0$ (i.e., if $y = x$), then integrating gives $$x(t) = \frac{1}{B - t} .$$

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We can handle your original system, \begin{align*} x' &= -k_1 x y - k_2 x z \\ y' &= -k_1 x y \\ z' &= -k_2 x z , \end{align*}

similarly, though there isn't a nice closed form solution for generic values of $(k_1, k_2)$. First, observe that $$x' = y' + z',$$ so we must have $x = y + z + A$ for some constant $A$. On the other hand, dividing the second and third equations gives that when $x \neq 0$, $y \neq 0$, $z \neq 0$ (and when $k_1 \neq 0$), $$\frac{z'}{z} = \lambda \frac{y'}{y} ,$$ where $\lambda := \frac{k_2}{k_1}$, hence $z = B y^\lambda$ for some constant $B$. Substituting now yields a single, separable equation in $y$ alone: $$y' = -k_1 y \left(A + y + B y^\lambda\right) ,$$ so $$t = -\frac1{k_1} \int^y \frac{d\upsilon}{\upsilon \left(A + \upsilon + B \upsilon^\lambda\right)} .$$ For generic values of $\lambda$ this integral does not have a closed-form antiderivative in terms of elementary functions.

If $\lambda \in \{-1, 0, +1\}$, the equation has the form $$y' = \alpha y^2 + \beta y + \gamma,$$ and we can solve for $y(t)$ in closed form in terms of elementary functions.

• If $\lambda \in \{0, +1\}$ (i.e., $k_2 = 0$ or $k_2 = k_1$, respectively), then the quadratic on the right-hand side has two real roots $y$, so the behavior is essentially the same as in the first example in the question.
• If $\lambda = -1$ (i.e., $k_2 = -k_1$), then the behavior of the solutions is governed by the discriminant $\Delta := \beta^2 - 4 \alpha \gamma = k_1^2 (A^2 - 4 B)$ of the quadratic. If $\Delta \geq 0$ the roots of the quadratic are real, and the previous comment applies. If $\Delta < 0$, the roots are complex conjugates, and the solution is $$y(t) = -\frac12 \sqrt{4 B - A^2} \tan \left(\frac{k_1}2 \sqrt{4 B - A^2} (t + C)\right) - \frac{A}2 .$$

For $\lambda \in \left\{\frac12,2\right\}$ we can express $t$ in terms of $y$ in closed form, but not the other way around.

The remaining special cases, when one of $x, y, z$ has a zero, are easier to resolve.