Non linear effect in differential equation

Question

Suppose I want to study the non linear ODE

$\frac{d^2}{dx^2}f(x) + a f^2(x) + b f(x) + c=0$

I know the solution for $a=0$ and I know $a$ is a small parameter. How can I study the effect of the non linear term?

Answer 1

It seems to be quite interesting. What is written below is not 100% rigorous, and I would be happy for someone to verify my thoughts on the matter.

Nonetheless, this seems a nice way to possibly approximate and study your solution.

EDIT:

The answer as it was written and accepted (In the grey box) had an error. I am adding the updated version. I am not deleting yet the original version (since it was the one to be accepted), until it this action is okayed by at least the author of the post, or one of the "moderators" here.

New Answer

As before, let us denote $f = f(t,a)$ the solution to our equation. Let us assume you have arrived at a solution for some specific $a_0$, call this solution $f_0 = f(t,a_0)$, and now you are interested in approximating the solution $f$ at $a = a_0 + \Delta a$, where $\Delta a$ is small. In your case, $a_0 = 0$, we'll substitute this later.

Now, same as before, we denote $f_a = \left. \frac{\partial}{\partial a} \right|_{a = a_0} f(t,a)$, and expanding into Taylor series and dropping the second-order terms, we can write $f = f_0 + \Delta af_a$.

$f$ is a solution to:

$$ f'' + af^2 + bf + c = 0 $$

Substitute all written above into the equation:

$$ (f_0 + \Delta a f_a)'' + (a_0+ \Delta a)(f_0 + \Delta a f_a)^2 + b(f_0 + \Delta a f_a) + c = 0 $$

Now we open the square, and rearrange the terms. We get $$ (f_0'' + a_0 f_0^2 + b f_0 + c) + (\Delta a f_a'' + 2a_0 \Delta a f_0 f_a + a_0 \Delta a^2 f_a^2 + \Delta a f_0^2 + 2\Delta a^2 f_0 f_a + b \Delta a f_a) + O(\Delta a^3) = 0 $$

The contents of the first brackets vanish since $f_0$ is a solution for $a_0$. The rest we divide by $\Delta a$. $O(\Delta a^3)$ becomes $O(\Delta a^2)$, and we disregard it, since we have already agreed to drop second-order terms. We are left with:

$$ f_a'' + 2a_0f_0 f_a + a_0 \Delta a f_a^2 + f_0^2 + 2\Delta a f_0 f_a + bf_a = 0 $$

Now, we substitute that in our case $a_0 = 0$, and $a = a_0 + \Delta a = \Delta a$, and we have an equation for $f_a$:

$$ f_a'' + f_0^2 + 2\Delta a f_0 f_a + bf_a = 0 $$

Or

$$ f_a'' + (2\Delta a f_0 + b)f_a + f_0^2 = 0 $$

Which is a linear ODE, and $f = f_0 + a f_a$ (Up to second-order terms).

Old Answer

Let us think of $f = f(t,a)$ the solution to your equation. Then we can write, using Taylor expansion (since $a$ is small):

$$ f(t,a) = f(t,0) + a \left. \frac{\partial}{\partial a} \right|_{a=0} f(t,a) + O(a^2) $$

So if we define $f = f(t,a)$, $f_0 = f(t,0)$, and $f_a = \left. > \frac{\partial}{\partial a} \right|_{a=0} f(t,a)$, and we drop the second-order remainder, we can write:

$$ f = f_0 + af_a $$

But $f$ is a solution to your ODE. So we plug this relation inside:

$$ (f_0 + af_a)'' + a(f_0 + af_a)^2 + b(f_0 + af_a) + c = 0 $$ We open the square, and rearrange the terms: $$ (f_0'' + af_0^2 + bf_0 + c) + (af_a'' + 2a^2f_0f_a + baf_a) + O(a^3) = 0 $$

Now, $f_0$ is a solution as well, so the content of the first brackets vanishes. The rest, we divide by $a$, and disregard the $O(a^3)$, since after the division it will become $O(a^2)$, and we have already agreed to disregard teh second-order terms.

We are left with an equation for $f_a$:

$$ f_a'' + (2af_0)f_a + bf_a = 0 $$

Since $f_0$ is known, this is a linear ODE, and if you find its solution, you can affroximate $f$, as before $f = f_0 + af_a$.

Answer 2

Multiply by $f'$ to yield a integrable equation. After it depends on $f$ if you can do the last integral.

$\textbf{Edit:}$

$$ f'' f' + \left[af^2+bf + c\right]f' = f'\dfrac{d}{dx}f' + \frac{d}{dx}\left[\frac{1}{3}af^3 + \frac{1}{2}bf^2+cf\right] = 0 $$ now we can re-write the last but one equation as $$ \dfrac{d}{dx}\left[\frac{f'^2}{2} + \frac{1}{3}af^3 + \frac{1}{2}bf^2+cf\right] = 0 $$ or $$ \frac{f'^2}{2} + \frac{1}{3}af^3 + \frac{1}{2}bf^2+cf = \text{Constant} = \lambda $$ then we have a separable equation $$ \dfrac{df}{dx} = \pm \sqrt{\lambda -\left(\frac{1}{3}af^3 + \frac{1}{2}bf^2+cf \right)} $$ now to solve this (maybe) heinous equation..we require the $f(x)$ to play nice ;).