Fourier series of complex diff eq

Question

Can I just use Euler's identity to construct the Fourier Series since it is complex? I was personally thinking I could, but I wanted to be doubly sure.

Answer 1

Those roots directly give you a homogeneous solution of $c_1 e^{(-1+i)t} + c_2 e^{(-1-i)t}$, where $c_1$ and $c_2$ are in general complex constants. Through judicious choice of $c_1,c_2$ using Euler's formula, this can be recast into the form $b_1 e^{-t} \cos(t) + b_2 e^{-t} \sin(t)$, so that $b_1,b_2$ can be chosen to be real (provided the initial condition is real). Which one is convenient depends on how you plan to deal with the inhomogeneity.

Answer 2

To find the particular solution, we expand $y(x)$ in its complex Fourier Series as

$$y(x)=\sum_{n=-\infty}^{\infty} c_n\,e^{inx}$$

Now, formally differentiating term by term reveals that

$$y''(x)+2y'(x)+2y(x)=\sum_{n=-\infty}^{\infty} c_n\,\left(-n^2+i2n+2\right)\,e^{inx}=|x|$$

Let's find the complex Fourier Series for $|x|$. Write,

$$|x|=\sum_{n=-\infty}^{\infty} d_n\,e^{inx}$$

with

$$\begin{align} d_n&=\frac{1}{2\pi}\int_{-\pi}^{\pi}|x|e^{-inx}dx\\\\ &=\frac{1}{\pi}\int_{0}^{\pi}x\cos(nx)dx\\\\ &=-\frac{1-(-1)^n}{\pi n^2}+(\pi/2) \delta_{n0} \end{align}$$

where $\delta_{n0}$ is the Kronecker Delta and is equal to $1$ when $n=0$ and equal to $0$ otherwise.

Thus, $$c_n=-\left(\frac{1-(-1)^n}{\pi n^2}\right)\,\left(\frac{1}{-n^2+2in+2}\right)+(\pi/4) \delta_{n0}$$.

Note that the complex Fourier Series for $y(x)$ can be converted to a Fourier Series by use of Euler's identity. Further note that the resulting series will have both cosine and sine series, and that only the odd indexed terms (other than $c_0$) are present.