Finding a periodic solution to ODE

My main concern with this exercise is that I do not know how to verify that the solution is ##C^1## on all of ##\mathbb R##. ##g## is certainly discontinuous. I begin by computing its Fourier coefficients. They are $$c_n=\frac{1}{4\pi}\int_{-2\pi}^{2\pi}g(t)e^{-int/2}dt= \frac{1}{4\pi}\int_{-\pi}^{\pi}e^{-int/2}dt.$$ So ##c_0=\frac12## and, using WolframAlpha this time, ##c_n=\frac{\sin\left(\frac{\pi n}{2}\right)}{\pi n}## for ##n\neq 0##. If we assume ##y(t)## can be written as a sum of a Fourier series, with coefficients ##b_n##, the ODE reads $$\sum \left(-\frac{n^2}{4}+ \frac{in}{2}+1\right)b_n e^{int/2} =\sum c_n e^{int/2}.$$ From this we obtain that $$b_n=\frac{c_n}{1+\frac{in}{2}-\frac{n^2}{4}}=\frac{\sin\left(\frac{\pi n}{2}\right)}{\pi n\left(1+\frac{in}{2}-\frac{n^2}{4}\right)}.$$ But why would this function be ##C^1## on all of ##\mathbb R##?

psie said:

Homework Statement: Find a periodic solution with a continuous first derivative on ##\mathbb R## of the differential equation ##y''+y'+y=g##, where ##g## has period ##4\pi## and ##g(t)=1## for ##|t|<\pi##, ##g(t)=0## for ##\pi<|t|<2\pi##.
Relevant Equations: Complex Fourier series, complex Fourier coefficients, etc.

But why would this function be C1 on all of R?

Isn't it some sort of trivial work to prove that the function $$y(t)=\sum b_n e^{in\pi t}$$ has continuous first derivative (because it is the (infinite but countable infinite ) sum of continuous differentiable functions $$y_n(t)=b_ne^{in\pi t}$$

P.S we could have said the same thing for g(t) ..(that is continuously differentiable).

Hmm it must have something to do with that ##b_n## converges much faster to zero than ##c_n##.

Delta2 said:

Isn't it some sort of trivial work to prove that the function $$y(t)=\sum b_n e^{in\pi t}$$ has continuous first derivative (because it is the (infinite but countable infinite ) sum of continuous differentiable functions $$y_n(t)=b_ne^{in\pi t}$$

Hmm, maybe something along these lines, but for your statement to be true, we’d probably require uniform convergence of the differentiated series, which is not so trivial to verify I think.

Regarding decay of coefficients, it holds that ##c_n## decays as ##O(n^{-k})## if ##f\in C^k##, but the converse I believe is not true. One could construct an easy counter example with the indicator function.

Maybe I should just see it as an assumption on ##y## for it to have a Fourier series, because if it’s ##C^1##, its Fourier series converges. There is no reason why ##y## should inherit any properties of ##g##.

psie said:

Regarding decay of coefficients, it holds that cn decays as O(n−k) if f∈Ck, but the converse I believe is not true. One could construct an easy counter example with the indicator function.

Hmmm, maybe there is some extra condition so that the converse is also true?

Oh damn its very easy and its require only Calculus 1 to show it lol. Since we are given that the second derivative of y exists.... this means that the first derivative is continuous.

Delta2 said:

Oh damn its very easy and its require only Calculus 1 to show it lol. Since we are given that the second derivative of y exists.... this means that the first derivative is continuous.

Yeah, maybe this does it. But I’m thinking the ODE only says what happens at the points of continuity of ##g##. It doesn’t say what happens at ##\pi## for instance. We could have that ##y## is also discontinuous at ##\pi##, which makes it not ##C^1## on all of ##\mathbb R##.

psie said:

Yeah, maybe this does it. But I’m thinking the ODE only says what happens at the points of continuity of ##g##. It doesn’t say what happens at ##\pi## for instance. We could have that ##y## is also discontinuous at ##\pi##, which makes it not ##C^1## on all of ##\mathbb R##.

yes I thought of that complication too about discontinuity of y at specific points. Given that you have find the correct formula for ##c_n## and ##b_n## (something doesnt look quite right to me but lets leave that for the moment), can we prove that $$y(t)=\sum b_n e^{int/2}$$ is two times differentiable everywhere at R (for every ##t\in\mathbb{R}##?)

Seems to me we ll have to check if the series $$y''(t)=\sum \frac{n^2}{4}b_n e^{int/2}$$ converges for every t in R.

You know from general theory of linear ODEs that the solution is a linear combination of [itex]e^{\omega t}[/itex] and [itex]e^{\bar{\omega}t}[/itex] in [itex]\pi < |t| < 2\pi[/itex] and 1 plus a linear combination of those in [itex]|t| < \pi[/itex] where [itex]\omega^2 + \omega + 1 = 0[/itex] so that [itex]\omega = e^{2\pi i/3}[/itex]. The coefficients of [itex]e^{\omega t}[/itex] annd [itex]e^{\bar{\omega}t}[/itex] in the three regions [itex]-2\pi < t < -\pi[/itex], [itex]-\pi < t < \pi[/itex] and [itex]\pi < t < 2\pi[/itex] are fixed by the periodicity and continuity requirements on [itex]y[/itex] and its derivative - six conditions in all.

You could, in principle, find those coefficients, compute the fourier series of that solution and show that they are identical to those found by transforming the equation, but I would hope that there is an easier way.

Note that [tex]1 + \frac{in}2 - \frac{n^2}4 = \left(\omega - \frac{in}2\right)\left(\bar{\omega} - \frac{in}{2}\right).[/tex]

EDIT: The condition you are looking for (which is Theorem 9.4 of Körner's Fourier Analysis) is that if [tex]
\sum_{n=-\infty}^\infty |n||c_n|[/tex] converges then [itex]\sum_{n=-\infty}^\infty c_ne^{int}[/itex] is once continuously differentiable, and [itex]\sum_{n=-N}^N inc_ne^{int}[/itex] converges uniformly to its derivative. That is the case here, since [itex]|nb_n|[/itex] decreases as [itex]|n|^{-2}[/itex] for large [itex]|n|[/itex].

Delta2 said:

Isn't it some sort of trivial work to prove that the function $$y(t)=\sum b_n e^{in\pi t}$$ has continuous first derivative (because it is the (infinite but countable infinite ) sum of continuous differentiable functions $$y_n(t)=b_ne^{in\pi t}$$

Still, given an infinite sum, there may be some kinks to figure out as to what happens in the limit.

pasmith said:

EDIT: The condition you are looking for (which is Theorem 9.4 of Körner's Fourier Analysis) is that if [tex]
\sum_{n=-\infty}^\infty |n||c_n|[/tex] converges then [itex]\sum_{n=-\infty}^\infty c_ne^{int}[/itex] is once continuously differentiable, and [itex]\sum_{n=-N}^N inc_ne^{int}[/itex] converges uniformly to its derivative. That is the case here, since [itex]|nb_n|[/itex] decreases as [itex]|n|^{-2}[/itex] for large [itex]|n|[/itex].

I've been looking into Körner's book. The theorem reads:

Theorem 9.4 Let ##f:\mathbb T\to\mathbb C## be continuous. Then if ##\sum_{n=-\infty}^n |n||c_n|## converges, it follows that ##f## is once continuously differentiable and that ##\sum_{n=-N}^N in c_ne^{int}\to f'(t)## uniformly as ##N\to\infty##.

I think the continuity requirement on ##f## is crucial here. In the exercise above, we don't know if ##y## is continuous or piecewise continuous. In case of the latter, I don't think this theorem applies.

psie said:

we don't know if y is continuous or piecewise continuous.

You got to give us the full word by word statement of the problem but by the way it is already given it implies that the equation y''+y'+y=g holds everywhere, hence the first and second derivative exist everywhere in R, hence the first derivative of y (and y) are continuous.

The statement given in the book is:

Find, in the guise of a "complex" Fourier series, a periodic solution with a continuous first derivative on ##\mathbb R## of the differential equation ##y''+y'+y=g##, where ##g## has period ##4\pi## and ##g(t)=1## for ##|t|<\pi##, ##g(t)=0## for ##\pi<|t|<2\pi##.

Delta2 said:

the equation y''+y'+y=g holds everywhere

The equation holds everywhere where ##g## is defined, which is not all of ##\mathbb R##. Hence one can only speculate whether or not ##y## is continuous or piecewise continuous. In case it is piecewise continuous, I don't believe @pasmith suggestion about theorem 9.4 applies.

psie said:

The statement given in the book is:The equation holds everywhere where ##g## is defined, which is not all of ##\mathbb R##. Hence one can only speculate whether or not ##y## is continuous or piecewise continuous. In case it is piecewise continuous, I don't believe @pasmith suggestion about theorem 9.4 applies.

I think we sort of getting drowned in a spoon of water. Anyway in my opinion the theorem given by @pasmith doesn't presuppose a function f whose fourier coefficients are ##c_n##. It just presupposes that ##c_n## is a sequence of complex numbers and if that series ##\sum |n||c_n|## converges, then the fourier series with ##c_n## as coefficients is once continuously differentiable.

Delta2 said:

You got to give us the full word by word statement of the problem but by the way it is already given it implies that the equation y''+y'+y=g holds everywhere, hence the first and second derivative exist everywhere in R, hence the first derivative of y (and y) are continuous.

Still, don't know if they're considered here, but you also have weak/ distributional solutions too.https://en.m.wikipedia.org/wiki/Weak_solution

psie said:

The statement given in the book is:The equation holds everywhere where ##g## is defined, which is not all of ##\mathbb R##. Hence one can only speculate whether or not ##y## is continuous or piecewise continuous. In case it is piecewise continuous, I don't believe @pasmith suggestion about theorem 9.4 applies.

[itex]C_1[/itex] implies continuous; a function cannot be differentiable, let alone continuously so, unless it is continuous.

I'm asking myself: as far as continously differentiable goes, is the response of a second order system to a square wave fundamentally different from the response to a single step function ? (for which there are explicit solutions that can be checked more easily)

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