Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does there exist a twice differentiable periodic function $f$ such that $f''(x) + f(x) =\sin(x)$ for all $x \in [-\pi, \pi]$?

How to solve this differential equation using Fourier series? I know only basics of Fourier analysis. I don`t know any inversion formula for Fourier series.

share|cite|improve this question
I guess it is more appropriate to use the Laplace Transform here, rather than Fourier's. – Pedro Tamaroff Apr 17 '12 at 0:48

The solution would be:

$$f(x) = -\frac{1}{2}x \cos(x) + C_1\sin(x) + C_2\cos(x)$$

Where the $C_1\sin(x) + C_2\cos(x)$ part is the solution to the homogeneous equation. Using Fourier Series naively, one runs into problems due to the $n = 1$ term having no solution.

share|cite|improve this answer
This shows in particular that there is no periodic solution. It would be nice to have an a priori proof of this fact. – Christian Blatter Apr 12 '12 at 11:35
@ChristianBlatter, This is common for forced oscillator problems. In fact, some authors (page 87) explicitly tell you to use a Fourier series multiplied by x in cases like these. – nbubis Apr 12 '12 at 15:52

I'm lazy and don't feel like giving you the complete answer, but here is an approach you can take.

Take the Fourier transform of both sides: $F'' + F' = T[\sin(x)]$

Fourier transforms differentiate easily:

$$F' = ik\cdot F$$ and $$F'' = -k^2\cdot F$$

So, then solve for $F$: $$F = \frac{T[\sin(x)]} {ik-k^2}$$

Then, take the inverse transform:

$$f = T^{-1}\left[\frac{T[\sin(x)]}{ik-k^2}\right]$$

You can solve the right-hand side either by looking up the transforms, or using integration by parts.

share|cite|improve this answer
The equation is $f''(x)+f(x)=\sin(x)$, not $f''(x)+f'(x)=\sin(x)$. I would at least try this before posting on it. The Fourier coefficients for $\sin(x)$ are non-zero exactly when $1-k^2=0$. This makes this approach problematic, at best. – robjohn Apr 12 '12 at 19:52

nbubis mentions that Fourier Series might not be the best method.

Suppose that $f$ has a Fourier Series: $$ f(x)=\sum_{k=-\infty}^\infty a_ke^{ikx}\tag{1} $$ Then $$ f''(x)+f(x)=\sin(x)\tag{2} $$ implies $$ \sum_{k=-\infty}^\infty a_k(1-k^2)e^{ikx}=\sin(x)=\frac{1}{2i}\left(e^{ix}-e^{-ix}\right)\tag{3} $$ However, integrating $(3)$ against $\frac{1}{2\pi}e^{-ikx}$ to get $a_k$, gives $a_k=0$ except for $k\in\{-1,1\}$. Equation $(3)$ says that $a_{-1}\cdot0=-\frac{1}{2i}$ and $a_1\cdot0=\frac{1}{2i}$. This leads one to conclude that there is no solution to $(2)$ which has a Fourier series.

share|cite|improve this answer
Actually, the solution does have a Fourier series, you just can't arrive at it using your method. The series is: $$ \frac{1}{4}\sin(x) + \sum_{n = 2 }^{\infty}{\frac{(-1)^{n+1}}{n^2-1}\sin(nx)}$$ – nbubis Apr 13 '12 at 0:35
@nbubis: I think you meant $$ \frac{1}{4}\sin(x)+\sum_{n=2}^{\infty}{\frac{(-1)^{n+1}n}{n^2-1}\sin(nx)} $$ In any case, that is the Fourier series of the solution to $$ f''(x)+f(x)=\sin(x)\tag{2} $$ on $\mathbb{R}$, restricted to $(-\pi,\pi)$. The resulting periodic function is not continuous (much less differentiable) at odd multiples of $\pi$, so it does not satisfy $(2)$ at odd multiples of $\pi$. – robjohn Apr 13 '12 at 1:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.