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I have an ODE of the form

$$ f”(x) + f(x) = \epsilon g(x)$$

with initial conditions

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$$ f(0) = f'(0) = 0 $$

$g(x)$ is $O(1)$ as $\epsilon \to 0$, and $g(x)$ is as smooth as necessary. Is there a way I can bound $f(x)$? In particular, I would like to be able to claim that $f(x)$ must be $O(\epsilon)$ as $\epsilon \to 0$ for all $x>0$, but I’m not sure if this is true or how to approach this. Any hints/advice would be greatly appreciated.

EDIT: I also know that $g(0) = g'(0) = 0$, if that helps with anything.

EDIT2: Upon seeing the answers, I think we can see that the values of $g$ and $g’$ at $0$ are irrelevant as long as $g(x) = O(1)$.

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You can rewrite the equation as a system of two linear differential equations. There is a closed form for the solution of such system. In particular, you have to:

- Find the solutions to the homogeneous problem and stack them in a matrix $X$
- Compute the particular solution as $X\int X^{-1} \underline{f}(x)$, where $\underline{f}$ is the right hand side in the linear system formulation.

Now, if you can compute the integral, then you have an explicit formula for the solution, and you can find a bound. Otherwise, you need to bound the integral, to find a bound for the solution.

Note: since both the initial conditions are zero, the particular solution is THE solution to your problem.

Put $f_1 = f$, $f_2 = f’$. Then $f_1′ = f_2$ and the equation gives $$f_2′ = -f_1 + \epsilon g.$$ Thus we have a system $$\frac{d}{dx} \left( \begin{matrix}f_1 \\ f_2 \end{matrix} \right) = \left( \begin{matrix}0 & 1\\ -1 & 0 \end{matrix} \right)\left( \begin{matrix}f_1 \\ f_2 \end{matrix} \right) + \epsilon \left( \begin{matrix} 0 \\ g \end{matrix} \right), \,\,\,\,\,\,\,\,\,\,\, \left( \begin{matrix}f_1(0) \\ f_2(0) \end{matrix} \right) = \left( \begin{matrix}0 \\ 0 \end{matrix} \right).$$ You can solve this systems explicitly using an integrating factor (and matrix exponentials) then look for a bound on $f_1$ since $f_1 = f$.

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