Particular solution for a periodic Fourier forcing term

Key concept: When a linear differential equation is driven by a 2π-periodic forcing function, each Fourier mode can be handled separately and then added back together by superposition.

Step by step

Core idea

For a linear differential equation, the response to a sum of forcing terms is the sum of the responses to each term. If the forcing function is

$f(t)=a_0+\sum_{n=1}^{\infty}a_n\cos(nt)+b_n\sin(nt),$

then the particular solution is written as the corresponding sum of the particular solutions for the constant term and each sine-cosine mode.

Mode-by-mode particular solution

For the nonresonant modes, the usual trial form produces a response of the same frequency. In the common case where the differential operator has the form

$y''+9y=f(t),$

a constant forcing term gives

$y_{p,0}(t)=\frac{a_0}{9},$

and for each $n\ge 1$ with $n\ne 3$,

$y_{p,n}(t)=\frac{a_n}{9-n^2}\cos(nt)+\frac{b_n}{9-n^2}\sin(nt).$

Adding all of these contributions gives

$y_p(t)=\frac{a_0}{9}+\sum_{n=1}^{\infty}\left(\frac{a_n}{9-n^2}\cos(nt)+\frac{b_n}{9-n^2}\sin(nt)\right),$

provided none of the terms hits the resonant frequency.

Why this works

This formula comes from linearity and the fact that sine and cosine are eigenfunctions of constant-coefficient differential operators. Each Fourier mode is transformed independently, so the output has the same frequency unless the frequency matches a natural frequency of the homogeneous equation.

That resonance issue is the key limitation. If the operator has natural frequency $3$, then the $n=3$ term must be treated separately with a modified trial function, usually multiplied by $t$.

Important resonance check

The decomposition above is valid only for terms with $n\ne 3$. The $n=3$ mode is special because it overlaps with the homogeneous solution and cannot be handled by the ordinary undetermined-coefficients guess.

Pitfall alert

A common mistake is to apply the same denominator to every Fourier mode without checking resonance. If the differential equation has homogeneous solutions involving $\cos(3t)$ and $\sin(3t)$, then the $n=3$ term is not safe to divide by $9-3^2=0$. Another error is forgetting the constant term: the $a_0$ mode does not disappear and must be handled as its own forcing component. Students also sometimes write a single trial solution for the whole series instead of using superposition term by term.

Try different conditions

If the forcing function were truncated to a finite Fourier sum such as

$f(t)=2+5\cos(2t)-\sin(4t),$

the same idea would give a finite particular solution by adding the responses to each term. For example, under $y''+9y=f(t)$, one would write

$y_p(t)=\frac{2}{9}+\frac{5}{5}\cos(2t)-\frac{1}{5}\sin(4t).$

If the forcing instead included a resonant term like $7\cos(3t)$, the trial form would need to be changed to something like $t\big(A\sin(3t)+B\cos(3t)\big)$, because the standard nonresonant formula breaks down.

Further reading

Fourier series forcing, method of undetermined coefficients, resonance in differential equations