1
$\begingroup$

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.

  • 0
    I guess it is more appropriate to use the Laplace Transform here, rather than Fourier's.2012-04-17

3 Answers 3

1

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.

  • 0
    @ChristianBlatter, This is common for forced oscillator problems. In fact, some [authors](http://www.jirka.org/diffyqs/diffyqs.pdf) (page 87) explicitly tell you to use a Fourier series multiplied by x in cases like these.2012-04-12
1

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.

  • 2
    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.2012-04-12
1

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.

  • 1
    @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 $\pi$.2012-04-13