Solve by using fourier series :$y'(t)+y(t-1)=\cos^{2}\pi t$

Question

We have to solve

$y'(t)+y(t-1)=\cos^{2}\pi t$ by using fourier series ,$y(t) \displaystyle\sim \sum _{n=-\infty}^{\infty} y_ne^{i \pi nt}$

My try:I can solve it directly but unable to solve by using fourier series.Thank you.

Answer 1

The function $g(t) := \cos^2 \pi t$ is $2$-periodic. Since $p$-periodicity is preserved under differentiation and translation, it is therefore natural to seek a $2$-periodic solution to the given differential equation.

Let's suppose the solution $y$ is in $C^3[0,2]$. Then by Dirichlet's theorem $$ y(t) = \sum_{n=-\infty}^{\infty} y_n e^{i \pi n t} $$

with uniform convergence, where $y_n$ are the Fourier coefficients of $y$. Also repeated integration by parts shows $|y_n| \leq A/|n|^3$ so that Weierstrass' M-test and a standard theorem on the uniform convergence of series of functions give $$ y'(t) = \sum_{n=-\infty}^{\infty} i \pi n y_n e^{i \pi n t} $$

with uniform convergence.

Now, note that \begin{align} g(t) &= \frac{1}{2} + \frac{1}{2} \cos 2 \pi t \\ &= \sum_{n=-\infty}^{\infty} g_n e^{i \pi n t} \end{align}

with uniform convergence where the Fourier coefficients $g_n$ of $g$ are given by $$ g_n = \begin{cases} 1/2 & n = 0 \\ 1/4 & n = 2 \\ 1/4 & n = -2 \\ 0 & \text{otherwise} \end{cases} $$

Substituting in the differential equation, we get \begin{align} y'(t) + y(t-1) = \cos^2 \pi t &\iff \sum_{n=-\infty}^{\infty} i \pi n y_n e^{i \pi n t} + \sum_{n=-\infty}^{\infty} y_n e^{i \pi n (t-1)} = \sum_{n=-\infty}^{\infty} g_n e^{i \pi n t} \\ &\iff \sum_{n=-\infty}^{\infty} (i \pi n y_n + (-1)^n y_n) e^{i \pi n t} = \sum_{n=-\infty}^{\infty} g_n e^{i \pi n t} \end{align}

Multiplying by $e^{-i \pi k t}$ and integrating both sides with respect to $t$ from $0$ to $2$ shows, after interchanging the order of series and integral (which can be done because of uniform convergence), that corresponding coefficients must be equal. That is, $$ i \pi n y_n + (-1)^n y_n = g_n \quad \forall n \in \mathbb{Z} $$

One readily solves for $y_n$ and obtains \begin{align} y_0 &= g_0 = \frac{1}{2} \\ y_2 &= \frac{g_2}{2 \pi i + 1} = \frac{4}{16 + 64 \pi^2} - \frac{8 \pi}{16 + 64 \pi^2} i \\ y_{-2} &= \frac{g_{-2}}{-2 \pi i + 1} = \overline{y_2} = \frac{4}{16 + 64 \pi^2} + \frac{8 \pi}{16 + 64 \pi^2} i\\ y_n &= 0 \quad \text{otherwise} \end{align}

The resulting $y$ is $$ y(t) = \frac{1}{2} + \frac{8}{16 + 64 \pi^2} \cos 2 \pi t + \frac{16 \pi}{16 + 64 \pi^2} \sin 2 \pi t \tag{$\star$} $$

which is indeed in $C^3[0,2]$.

Conclusion: if there is a $C^3$ $2$-periodic solution to the given differential equation, then it is the one given by $(\star)$.

Finally, one easily verifies that the function given by $(\star)$ satisfies the differential equation, so that it is indeed the solution.

Note: In fact if one just wants to find a solution to the differential equation, then justifying all the steps leading to $(\star)$ isn't necessary (and is probably a loss of time in an exam).

Answer 2

$$\begin{align}y'(t)+y(t-1) &= \sum_{n=-\infty}^{\infty} y_n(in\pi + e^{-i\pi n})e^{i\pi n t}\\ &=\sum_{n=-\infty}^{\infty} y_n(in\pi + (-1)^n)e^{i\pi n t} \end{align}$$

$$\cos^2\pi t =\frac{1}{2}(\cos 2\pi t + 1) = \frac{1}{4}\left(e^{i2\pi t}+e^{-i2\pi t}+2\right)$$

So $y_n = 0$ unless $n=0,2,-2$.

Now solve for $y_0,y_2,y_{-2}$, and you are done.