Approximaths

A nonlinear ODE (Part I)

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In this post we would like to solve the following non-linear first-order equation with corresponding initial condition:

\displaystyle y' + y + y^2 = 0, y(0) = 1

We rewrite the equation (inserting \epsilon):

\displaystyle y' + y + \epsilon y^2 = 0

and y(x)

\displaystyle y_{\epsilon}(x) = \sum_{n=0}^{\infty} a_n(x) \epsilon^n

The differential equation becomes:

\displaystyle y' + y + \epsilon y^2 = 0
\displaystyle \sum_{n=0}^{\infty} a_{n}'(x)\epsilon^{n} + \sum_{n=0}^{\infty} a_{n}(x)\epsilon^{n} + \epsilon\biggl(\sum_{k=0}^{\infty}\sum_{l=0}^{k}a_{l}(x) a_{k-l}(x)\epsilon^{k}\biggr) = 0

Selecting the n-th term we have:

\displaystyle a_n'(x) + a_n(x) + \sum_{k=1}^{n} a_{k-1} a_{n-k} = 0

for n=0 we obtain:

\displaystyle a_{0}'(x) + a_{0}(x) = 0
\displaystyle a_{0}'(x) = - a_{0}(x)
\displaystyle a_{0}(x) = e^{-x}

for n=1 we obtain:

\displaystyle a_{1}'(x) + a_{1}(x) + a_{0}^2(x) = 0
\displaystyle a_{1}'(x) + a_{1}(x) + e^{-2x} = 0
\displaystyle a_{1}'(x) = - a_{1}(x) - e^{-2x}

which implies:

\displaystyle a_{1}(x) = -e^{-x} + e^{-2x}

Finally we have:

\displaystyle y_{\epsilon}(x) = e^{-x} + (-e^{-x} + e^{-2x})\epsilon + ...

This equation is related to the logistic equation, a classical model used in population dynamics to describe growth and decline. Here, the negative terms indicate that y(x) decreases over time and tends toward zero. The quadratic term adds a nonlinear effect to this decay. Similar equations can also be found in chemical kinetics and in other models involving nonlinear growth or decay.

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