How can we extend the solution of an initial value problem using Fourier series?

[evinda]

Hello! (Wave)

The following problem shall show the way with which the Fourier series can be used for the solution of initial value problems.Find the solution of the initial value problem

$$y''+ \omega^2 y=\sin{nt}, y(0)=0, y'(0)=0$$

where $n$ is a natural number and $\omega^2 \neq n^2$. What happens if $\omega^2=n^2$ ?

How can we use Fourier series in order to solve the above problem?

Do we set $y(t)=\frac{a_0}{2}+ \sum_{m=1}^{\infty} \left( a_m \cos{\frac{m \pi t}{L}}+ b_m \sin{\frac{m \pi t}{L}}\right)$ where

$a_m=\frac{1}{L} \int_{-L}^{L} f(t) \cos{\frac{m \pi t}{L}} dt$ and $b_m=\frac{1}{L} \int_{-L}^{L} f(t) \sin{\frac{m \pi t}{L}} dt$ ? (Thinking)

 

[I like Serena]

Hey evinda! (Smile)

When $\omega^2=n^2$, I think we'll get into trouble later on when doing the Fourier transforms.
We'll see soon enough.

Those Fourier series have the restriction that the function has to be periodic with period $2L$.
Can we assume that? (Wondering)

Otherwise we should use one of the non-periodic Fourier transforms like:
$$Y(\nu)=\mathscr F[y(t)](\nu) = \int_{-\infty}^\infty y(t)e^{-i\nu t}dt \quad\leftrightarrow\quad
y(t)=\mathscr F^{-1}[Y(\nu)](t)= \frac 1{2\pi} \int_{-\infty}^\infty Y(\nu)e^{i\nu t}d\nu
$$

To solve an initial value problem like yours, we take the Fourier transform $\mathscr F$ of both sides of the equation, solve for $Y(\nu)$, and apply the inverse Fourier transform.
In particular, we can use that:
$$\mathscr F[y''(t)](\nu) = (i\nu)^2 Y(\nu)$$
which follows from the definition. (Thinking)

 

[evinda]

Those Fourier series have the restriction that the function has to be periodic with period $L$.
Can we assume that? (Wondering)

Yes, we can assume this.

In my notes, we have the following formula of the Fourier series:

$$f(x)=\frac{a_0}{2}+ \sum_{m=1}^{\infty} \left( a_m \cos{\frac{m \pi x}{L}}+ b_m \sin{\frac{m \pi x}{L}}\right)$$where $a_m=\frac{1}{L} \int_{-L}^L f(x) \cos{\frac{m \pi x}{L}} dx, m=0,1,2, \dots$

and

$b_m=\frac{1}{L} \int_{-L}^{L} f(x) \sin{\frac{m\pi x}{L}} dx, m=1,2, \dots$

and the period is then $2L$.

So how could we use these Fourier series? (Thinking)

 

[I like Serena]

Yes, we can assume this.

In my notes, we have the following formula of the Fourier series:

$$f(x)=\frac{a_0}{2}+ \sum_{m=1}^{\infty} \left( a_m \cos{\frac{m \pi x}{L}}+ b_m \sin{\frac{m \pi x}{L}}\right)$$where $a_m=\frac{1}{L} \int_{-L}^L f(x) \cos{\frac{m \pi x}{L}} dx, m=0,1,2, \dots$

and

$b_m=\frac{1}{L} \int_{-L}^{L} f(x) \sin{\frac{m\pi x}{L}} dx, m=1,2, \dots$

and the period is then $2L$.

So how could we use these Fourier series? (Thinking)

What would we get if we write $y(t)$ instead of $f(x)$? And substitute into the DE? (Wondering)

 

[evinda]

What would we get if we write $y(t)$ instead of $f(x)$? And substitute into the DE? (Wondering)

First of all, we have that $y,y'$ are piecewiese continuous since $y''$ is defined.
And because of that and the fact that we assume that $y$ is periodic with period $2L$ we know that $y$ can be written as a Fourier series. Right?

So let $y(t)=\frac{a_0}{2}+ \sum_{m=1}^{\infty} \left( a_m \cos{\frac{m \pi t}{L}}+ b_m \sin{\frac{m \pi t}{L}} \right)$.

Then we have $$y'(t)=\sum_{m=1}^{\infty} \left( -a_m \frac{m \pi}{L} \sin{\frac{m \pi t}{L}}+ b_m \frac{m \pi}{L} \cos{\frac{m \pi t}{L}} \right)$$

and

$$y''(t)=\sum_{m=1}^{\infty} \left( -a_m \left( \frac{m \pi }{L}\right)^2 \cos{\frac{m \pi t}{L}}- b_m \left( \frac{m \pi}{L}\right)^2 \sin{\frac{m \pi t}{L}}\right)$$

So we get

$$\sum_{m=1}^{\infty} \left( -a_m \left( \frac{m \pi }{L}\right)^2 \cos{\frac{m \pi t}{L}}- b_m \left( \frac{m \pi}{L}\right)^2 \sin{\frac{m \pi t}{L}}\right)+ \omega^2 \frac{a_0}{2}+ \omega^2 \sum_{m=1}^{\infty} \left( a_m \cos{\frac{m \pi t}{L}}+ b_m \sin{\frac{m \pi t}{L}} \right)=\sin{nt}$$Do we deduce from this that $a_m=0, \forall m \geq 0$ and $b_m \neq 0$ only if $\frac{m \pi}{L} =n$ ?EDIT: And we will then also get that $b_{\frac{nL}{\pi}}=\frac{1}{\omega^2-n^2}$, right?

 

[I like Serena]

First of all, we have that $y,y'$ are piecewiese continuous since $y''$ is defined.
And because of that and the fact that we assume that $y$ is periodic with period $2L$ we know that $y$ can be written as a Fourier series. Right?

I believe we need that $y$ is continuous, and that $y'$ has to be square integrable.
See for instance this page.

So let $y(t)=\frac{a_0}{2}+ \sum_{m=1}^{\infty} \left( a_m \cos{\frac{m \pi t}{L}}+ b_m \sin{\frac{m \pi t}{L}} \right)$.

Then we have $$y'(t)=\sum_{m=1}^{\infty} \left( -a_m \frac{m \pi}{L} \sin{\frac{m \pi t}{L}}+ b_m \frac{m \pi}{L} \cos{\frac{m \pi t}{L}} \right)$$

and

$$y''(t)=\sum_{m=1}^{\infty} \left( -a_m \left( \frac{m \pi }{L}\right)^2 \cos{\frac{m \pi t}{L}}- b_m \left( \frac{m \pi}{L}\right)^2 \sin{\frac{m \pi t}{L}}\right)$$

So we get

$$\sum_{m=1}^{\infty} \left( -a_m \frac{m \pi}{L} \sin{\frac{m \pi t}{L}}+ b_m \frac{m \pi}{L} \cos{\frac{m \pi t}{L}} \right)+ \omega^2 \frac{a_0}{2}+ \omega^2 \sum_{m=1}^{\infty} \left( a_m \cos{\frac{m \pi t}{L}}+ b_m \sin{m t} \right)=\sin{nt}$$Do we deduce from this that $b_m=0 \forall m>0$ and $a_m=0, \forall m$ such that $\frac{m \pi}{L} \neq n$ ?

Shouldn't that be:
$$\sum_{m=1}^{\infty} \left( -a_m (\frac{m \pi}{L})^2 \cos{\frac{m \pi t}{L}} - b_m (\frac{m \pi}{L})^2 \sin{\frac{m \pi t}{L}} \right)+ \omega^2 \frac{a_0}{2}+ \omega^2 \sum_{m=1}^{\infty} \left( a_m \cos{\frac{m \pi t}{L}}+ b_m \sin{\frac{m \pi t}{L}} \right)=\sin{nt}$$
$$\quad \Rightarrow\quad\omega^2 \frac{a_0}{2}+ \sum_{m=1}^{\infty} a_m \left(\omega^2 - (\frac{m \pi}{L})^2\right)\cos{\frac{m \pi t}{L}}+ \sum_{m=1}^{\infty} b_m \left(\omega^2 - (\frac{m \pi}{L})^2\right) \sin{\frac{m \pi t}{L}} =\sin{nt}$$
(Wondering)

If we make the assumption that the period is $2\pi$, it becomes simpler yet.
And since $\{\cos mt, \sin mt\}$ is a basis, it follows that $a_m = 0$ for every $m$, and $b_m \left(\omega^2 - m^2\right) = \delta_{mn}$.
Can we solve for $b_m$ and find $y(t)$? (Wondering)

 

[evinda]

I believe we need that $y$ is continuous, and that $y'$ has to be square integrable.
See for instance this page.
Shouldn't that be:
$$\sum_{m=1}^{\infty} \left( -a_m (\frac{m \pi}{L})^2 \cos{\frac{m \pi t}{L}} - b_m (\frac{m \pi}{L})^2 \sin{\frac{m \pi t}{L}} \right)+ \omega^2 \frac{a_0}{2}+ \omega^2 \sum_{m=1}^{\infty} \left( a_m \cos{\frac{m \pi t}{L}}+ b_m \sin{\frac{m \pi t}{L}} \right)=\sin{nt}$$
$$\quad \Rightarrow\quad\omega^2 \frac{a_0}{2}+ \sum_{m=1}^{\infty} a_m \left(\omega^2 - (\frac{m \pi}{L})^2\right)\cos{\frac{m \pi t}{L}}+ \sum_{m=1}^{\infty} b_m \left(\omega^2 - (\frac{m \pi}{L})^2\right) \sin{\frac{m \pi t}{L}} =\sin{nt}$$
(Wondering)

If we make the assumption that the period is $2\pi$, it becomes simpler yet.
And since $\{\cos mt, \sin mt\}$ is a basis, it follows that $a_m = 0$ for every $m$, and $b_m \left(\omega^2 - m^2\right) = \delta_{mn}$.
Can we solve for $b_m$ and find $y(t)$? (Wondering)

I edited my post previously. We get that $y(t)=\frac{1}{\omega^2-n^2} \sin{(nt)}$. Or am I wrong? (Thinking)

 

[evinda]

And since $\{\cos mt, \sin mt\}$ is a basis, it follows that $a_m = 0$ for every $m$, and $b_m \left(\omega^2 - m^2\right) = \delta_{mn}$.
(Wondering)

How do we deduce that $b_m \left(\omega^2 - m^2\right) = \delta_{mn}$ ? (Thinking)

 

[I like Serena]

I edited my post previously. We get that $y(t)=\frac{1}{\omega^2-n^2} \sin{(nt)}$. Or am I wrong? (Thinking)

That is what we would get yes.
Does it satisfy the boundary conditions of the initial value problem? (Wondering)

How do we deduce that $b_m \left(\omega^2 - m^2\right) = \delta_{mn}$ ? (Thinking)

If we make the assumption that the period is $2\pi$, we would get that:
$$\omega^2 \frac{a_0}{2}+ \sum_{m=1}^{\infty} a_m \left(\omega^2 - m^2\right)\cos{m t}+ \sum_{m=1}^{\infty} b_m \left(\omega^2 - m^2\right) \sin{m t} =\sin{nt}$$
And since $\{\cos mt, \sin mt\}$ is a basis, all coefficients of $\cos mt$ and $\sin mt$ must match.
That means that every coefficient must be zero, except that for $m=n$ we must have that $b_m \left(\omega^2 - m^2\right) = 1$. (Thinking)

 

[evinda]

That is what we would get yes.
Does it satisfy the boundary conditions of the initial value problem? (Wondering)

No, it doesn't... (Worried) We have that $y'(0)=\frac{n}{\omega^2-n^2}$, right?

If we make the assumption that the period is $2\pi$, we would get that:
$$\omega^2 \frac{a_0}{2}+ \sum_{m=1}^{\infty} a_m \left(\omega^2 - m^2\right)\cos{m t}+ \sum_{m=1}^{\infty} b_m \left(\omega^2 - m^2\right) \sin{m t} =\sin{nt}$$
And since $\{\cos mt, \sin mt\}$ is a basis, all coefficients of $\cos mt$ and $\sin mt$ must match.
That means that every coefficients must be zero, except that for $m=n$ we must have that $b_m \left(\omega^2 - m^2\right) = 1$. (Thinking)

Ah, I see.. (Nod)

 

[I like Serena]

No, it doesn't... (Worried) We have that $y'(0)=\frac{n}{\omega^2-n^2}$, right?

Correct.

But we missed something. (Worried)
We had that $b_m(\omega^2-n^2) = 1$ if $m=n$, and $b_m(\omega^2-m^2) = 0$ otherwise.
That means that $b_n = \frac{1}{\omega^2-n^2}$ for $m=n$.
And that $b_m = 0 \lor \omega^2-m^2 = 0$ if $n\ne m$.
So $b_m$ can be non-zero for $m\ne n$ depending on $\omega$. (Thinking)
(Btw, see why we must have that $\omega^2 \ne n^2$?)

Checking with Wolfram shows that there are also non-periodic solutions if $\omega$ is not an integer. See for instance here. (Nerd)

 

[evinda]

Correct.

But we missed something. (Worried)
We had that $b_m(\omega^2-n^2) = 1$ if $m=n$, and $b_m(\omega^2-n^2) = 0$ otherwise.

You meant that otherwise we have $b_m(\omega^2-m^2)=0$ ? Because then we have that $m \neq n$.

That means that $b_n = \frac{1}{\omega^2-n^2}$ for $m=n$.
And that $b_m = 0 \lor \omega^2-m^2 = 0$ if $n\ne m$.
So $b_m$ can be non-zero for $m\ne n$ depending on $\omega$. (Thinking)

Ah I see... So can we not find a general formula of the Fourier series? (Thinking)

(Btw, see why we must have that $\omega^2 \ne n^2$?)

For $m=n$ this must hold in order to divide by $\omega^2-m^2$. Why does it hold for $m \neq n$ ? (Thinking)

 

[I like Serena]

You meant that otherwise we have $b_m(\omega^2-m^2)=0$ ? Because then we have that $m \neq n$.

Yes.

Ah I see... So can we not find a general formula of the Fourier series? (Thinking)

Yes. The general formula, assuming a periodic function and with $\omega$ an integer, is:
$$y(t) = b_{|\omega|} \sin \omega t + \frac{1}{\omega^2 - n^2} \sin nt$$
Can we take the boundary conditions into account now? (Wondering)

For $m=n$ this must hold in order to divide by $\omega^2-m^2$. Why does it hold for $m \neq n$ ? (Thinking)

For $m=n$ we must have $\omega^2 \ne n^2$ since otherwise $b_n(\omega^2 - n^2)=1$ does not have a solution.
For $m\ne n$ we can always find a solution for $b_m(\omega^2 - m^2)=0$, if necessary by setting $b_m$ to zero. (Thinking)

 

[evinda]

Yes. The general formula, assuming a periodic function and with $\omega$ an integer, is:
$$y(t) = b_{|\omega|} \sin \omega t + \frac{1}{\omega^2 - n^2} \sin nt$$
Can we take the boundary conditions into account now? (Wondering)

I am a little bit confused right now. Either $b_m=0$ or $\omega^2-m^2=0$, will the resulting Fourier series not be the same?

For $m=n$ we must have $\omega^2 \ne n^2$ since otherwise $b_n(\omega^2 - n^2)=1$ does not have a solution.
For $m\ne n$ we can always find a solution for $b_m(\omega^2 - m^2)=0$, if necessary by setting $b_m$ to zero. (Thinking)

Yes, so for $m \neq n$ it doesn't have to hold that $\omega^2 \ne n^2$, right?

 

[I like Serena]

I am a little bit confused right now. Either $b_m=0$ or $\omega^2-m^2=0$, will the resulting Fourier series not be the same?

If $\omega^2-m^2=0$, then $b_m$ can be anything - we have a free choice - $b_m$ doesn't have to be zero.
And if $b_m \ne 0$ for that particular $m$, we have a different Fourier series. (Thinking)

Yes, so for $m \neq n$ it doesn't have to hold that $\omega^2 \ne n^2$, right?

Erm... for $m \neq n$ it doesn't have to hold that $\omega^2 \ne {\color{red}m}^2$.

 

[evinda]

Yes. The general formula, assuming a periodic function and with $\omega$ an integer, is:
$$y(t) = b_{|\omega|} \sin \omega t + \frac{1}{\omega^2 - n^2} \sin nt$$
Can we take the boundary conditions into account now? (Wondering)

You mean without assuming that the period is $2 \pi$ ?

It holds that $y(0)=0$.

We have that $y'(t)=b_{|\omega|} \omega \cos{\omega t}+ \frac{n}{\omega^2-n^2} \cos{nt}$.

So $y'(0)=\omega b_{|\omega|}+\frac{n}{\omega^2-n^2}$, right?

 

[I like Serena]

You mean without assuming that the period is $2 \pi$ ?

It holds that $y(0)=0$.

We have that $y'(t)=b_{|\omega|} \omega \cos{\omega t}+ \frac{n}{\omega^2-n^2} \cos{nt}$.

So $y'(0)=\omega b_{|\omega|}+\frac{n}{\omega^2-n^2}$, right?

We have first assumed that the function is periodic, and also that it satisfies the conditions to write it as a Fourier series (continuous and with a derivative that is square integrable).
At the end we have also assumed that it has period $2\pi$, so that we could easily find a simple solution.
It turns out that it suffices to find the solution that satisfies the boundary conditions.

And yes, that's the expression for $y'(0)$. (Nod)

 

[evinda]

We have first assumed that the function is periodic, and also that it satisfies the conditions to write it as a Fourier series (continuous and with a derivative that is square integrable).
At the end we have also assumed that it has period $2\pi$, so that we could easily find a simple solution.
It turns out that it suffices to find the solution that satisfies the boundary conditions.

And yes, that's the expression for $y'(0)$. (Nod)

I think that we can also find easily the same without assuming that the period is $2 \pi$... (Thinking)

Ah I see... (Happy)

And how can we check what happens when $\omega^2=n^2$ ? (Thinking)

 

[I like Serena]

I think that we can also find easily the same without assuming that the period is $2 \pi$... (Thinking)

How? (Wondering)

And how can we check what happens when $\omega^2=n^2$ ? (Thinking)

We can't with a Fourier series, and not with a non-periodic Fourier transform either for that matter.
We'll need a different method to solve the IVP.
For instance this wiki page gives an alternate method to solve it.

 

[evinda]

How? (Wondering)

At post #35, we got this:

$$\sum_{m=1}^{\infty} \left( -a_m \left( \frac{m \pi }{L}\right)^2 \cos{\frac{m \pi t}{L}}- b_m \left( \frac{m \pi}{L}\right)^2 \sin{\frac{m \pi t}{L}}\right)+ \omega^2 \frac{a_0}{2}+ \omega^2 \sum_{m=1}^{\infty} \left( a_m \cos{\frac{m \pi t}{L}}+ b_m \sin{\frac{m \pi t}{L}} \right)=\sin{nt}$$

So for $m \neq \frac{nL}{\pi}$ we have that $b_m \left( \omega^2- \left( \frac{m \pi}{L}\right)^2 \right)=0$.

And from the latter we deduce that either $b_m=0$ or $\omega^2- \left( \frac{m \pi}{L}\right)^2=0 \Rightarrow m=\frac{L}{\pi} |\omega|$.

We can't with a Fourier series, and not with a non-periodic Fourier transform either for that matter.
We'll need a different method to solve the IVP.
For instance this wiki page gives an alternate method to solve it.

I found the exercise at the chapter with the method of the Fourier series we were talking about previously. Is it maybe meant to find the limit? (Thinking)

 

[I like Serena]

At post #35, we got this:

$$\sum_{m=1}^{\infty} \left( -a_m \left( \frac{m \pi }{L}\right)^2 \cos{\frac{m \pi t}{L}}- b_m \left( \frac{m \pi}{L}\right)^2 \sin{\frac{m \pi t}{L}}\right)+ \omega^2 \frac{a_0}{2}+ \omega^2 \sum_{m=1}^{\infty} \left( a_m \cos{\frac{m \pi t}{L}}+ b_m \sin{\frac{m \pi t}{L}} \right)=\sin{nt}$$

So for $m \neq \frac{nL}{\pi}$ we have that $b_m \left( \omega^2- \left( \frac{m \pi}{L}\right)^2 \right)=0$.

And from the latter we deduce that either $b_m=0$ or $\omega^2- \left( \frac{m \pi}{L}\right)^2=0 \Rightarrow m=\frac{L}{\pi} |\omega|$.

I found the exercise at the chapter with the method of the Fourier series we were talking about previously. Is it maybe meant to find the limit? (Thinking)

Ah yes... (Nod)

We have for some $k$ with $\frac{k \pi}{L}=n$ that $b_k \left( \omega^2- \left( \frac{k \pi}{L}\right)^2 \right)=1$.
And for all $m \ne k$ that $b_m \left( \omega^2- \left( \frac{m \pi}{L}\right)^2 \right)=0$.
So $L=\frac k n \pi$.
Suppose $b_m \ne 0$, then $\omega^2- \left( \frac{m \pi}{L}\right)^2 = 0 \Rightarrow |\omega|=\frac{m \pi}{L}=\frac{mn}{k}$.
It means that $\omega$ has to be a rational number, say $\omega=\frac pq$.
So that we can pick $k=|q|n$ and $m=|p|$.
Then the solution for any rational $\omega$ becomes:
$$y(t) = b_m \sin \left(\frac{m\pi}{L} t\right) + \frac{1}{\omega^2 - (k\pi/L)^2}\sin \left( \frac{k \pi}{L}t\right)
=b_m\sin(|\omega| t) + \frac{1}{\omega^2 - n^2}\sin nt
=B\sin \omega t + \frac{1}{\omega^2 - n^2}\sin nt$$
with a free choice for $B$.Can we take the limit now for $\omega\to n$? And also the limit for $\omega$ to any irrational value? (Wondering)

 

[evinda]

Ah yes... (Nod)

We have for some $k$ with $\frac{k \pi}{L}=n$ that $b_k \left( \omega^2- \left( \frac{k \pi}{L}\right)^2 \right)=1$.
And for all $m \ne k$ that $b_m \left( \omega^2- \left( \frac{m \pi}{L}\right)^2 \right)=0$.
So $L=\frac k n \pi$.
Suppose $b_m \ne 0$, then $\omega^2- \left( \frac{m \pi}{L}\right)^2 = 0 \Rightarrow |\omega|=\frac{m \pi}{L}=\frac{mn}{k}$.
It means that $\omega$ has to be a rational number, say $\omega=\frac pq$.
So that we can pick $k=|q|n$ and $m=|p|$.
Then the solution for any rational $\omega$ becomes:
$$y(t) = b_m \sin \left(\frac{m\pi}{L} t\right) + \frac{1}{\omega^2 - (k\pi/L)^2}\sin \left( \frac{k \pi}{L}t\right)
=b_m\sin(|\omega| t) + \frac{1}{\omega^2 - n^2}\sin nt
=B\sin \omega t + \frac{1}{\omega^2 - n^2}\sin nt$$
with a free choice for $B$.

From the initial condition $y'(0)=0$ we get that $B=-\frac{n}{|\omega| (\omega^2-n^2)}$, don't we?

Can we take the limit now for $\omega\to n$? And also the limit for $\omega$ to any irrational value? (Wondering)

We have that $\lim_{|\omega| \to n} \frac{|\omega| \sin{nt}-n \sin{|\omega| t}}{|\omega| (\omega^2-n^2)}\overset{\frac{0}{0}}{ = } \lim_{|\omega| \to n} \frac{\sin{nt}-nt \cos{|\omega| t}}{\omega^2-n^2+|\omega| 2 \omega}=\frac{\sin{nt}-nt \cos{nt}}{2n^2}$, if $\omega$>0. So do we maybe assume the latter? (Thinking)

 

[evinda]

Isn't it as follows?

$$\lim_{|\omega| \to n} \frac{|\omega| \sin{nt}-n \sin{|\omega| t}}{|\omega| (\omega^2-n^2)}=\left\{\begin{matrix}
\frac{\sin{nt}-nt \cos{nt}}{2n^2} , \text{ if } \omega=n \\
\frac{\sin{nt}-nt \cos{nt}}{-2n^2}, \text{ if } \omega=-n &
\end{matrix}\right.$$

Or am I wrong? (Thinking)

 

[I like Serena]

From the initial condition $y'(0)=0$ we get that $B=-\frac{n}{|\omega| (\omega^2-n^2)}$, don't we?

We'd get that $b_m=-\frac{n}{|\omega| (\omega^2-n^2)}$.
I intended that $B = \operatorname{sgn}(\omega) b_m$, so that $b_m \sin(|\omega| t) = B\sin\omega t$.
Thus:
$$B=-\frac{n}{\omega(\omega^2-n^2)}$$
It rids us of the absolute function. (Whew)

So:
$$y(t) = -\frac{n}{\omega(\omega^2-n^2)} \sin\omega t + \frac{1}{\omega^2 -n^2}\sin nt
= \frac{\omega\sin nt - n\sin\omega t}{\omega(\omega^2-n^2)}$$

We have that $\lim_{|\omega| \to n} \frac{|\omega| \sin{nt}-n \sin{|\omega| t}}{|\omega| (\omega^2-n^2)}\overset{\frac{0}{0}}{ = } \lim_{|\omega| \to n} \frac{\sin{nt}-nt \cos{|\omega| t}}{\omega^2-n^2+|\omega| 2 \omega}=\frac{\sin{nt}-nt \cos{nt}}{2n^2}$, if $\omega$>0. So do we maybe assume the latter? (Thinking)

Isn't it as follows?

$$\lim_{|\omega| \to n} \frac{|\omega| \sin{nt}-n \sin{|\omega| t}}{|\omega| (\omega^2-n^2)}=\left\{\begin{matrix}
\frac{\sin{nt}-nt \cos{nt}}{2n^2} &, \text{ if } \omega=n \\
\frac{\sin{nt}-nt \cos{nt}}{-2n^2}, \text{ if } \omega=-n &
\end{matrix}\right.$$

Or am I wrong? (Thinking)

Let's not try and do everything at the same time shall we? Otherwise it makes my head hurt.
$$\lim_{\omega\to n} y(t)
= \lim_{\omega\to n} \frac{\omega\sin nt - n\sin\omega t}{\omega(\omega^2-n^2)}
\ \overset{\small\text{L'Hôpital}}{=}\ \lim_{\omega\to n} \frac{\sin nt - nt\cos\omega t}{(\omega^2-n^2) + \omega(2\omega)}
= \frac{\sin nt - nt\cos n t}{n(2n)}
$$
Check. (Nod)

$$\lim_{\omega\to -n} y(t)
= \lim_{\omega\to -n} \frac{\omega\sin nt - n\sin\omega t}{\omega(\omega^2-n^2)}
= \lim_{\omega\to -n} \frac{\sin nt - nt\cos\omega t}{(\omega^2-n^2) + \omega(2\omega)}
= \frac{\sin nt - nt\cos(-n t)}{-n(-2n)}
= \frac{\sin nt - nt\cos(n t)}{2n^2}
$$
Not check. (Shake)

Makes sense doesn't it?
Because the DE is exactly the same whether $\omega$ is positive or negative, since $\omega^2$ will be the same either way.

 

[evinda]

We'd get that $b_m=-\frac{n}{|\omega| (\omega^2-n^2)}$.

Couldn't we not also take this and then calculate two limits , when $\omega \to n$ and when $\omega \to -n$ ?

I was also wondering why the solution of the initial value problem for $\omega^2=n^2$ will be equal to the limit that we get... (Thinking)

 

[I like Serena]

Couldn't we not also take this and then calculate two limits , when $\omega \to n$ and when $\omega \to -n$ ?

Yes. (Nod)

I was also wondering why the solution of the initial value problem for $\omega^2=n^2$ will be equal to the limit that we get... (Thinking)

We can consider it an educated guess.
It also means that we have to verify that the limit does indeed satisfy the DE and the boundary conditions.
And it does. It's just that our solution method with a Fourier series does not allow us to find it.

 

[evinda]

We can consider it an educated guess.
It also means that we have to verify that the limit does indeed satisfy the DE and the boundary conditions.
And it does. It's just that our solution method with a Fourier series does not allow us to find it.

So if would want to know the solution for $\omega^2=n^2$ we would have to solve the initial value problem for this value of $\omega^2$ with an other method, to check if the limit is indeed the solution, right? (Thinking)

 

[I like Serena]

So if would want to know the solution for $\omega^2=n^2$ we would have to solve the initial value problem for this value of $\omega^2$ with an other method, to check if the limit is indeed the solution, right? (Thinking)

We just have to fill in the potential solution for $\omega^2=n^2$ in the differential equation, see that it is satisfied, and do the same thing for the boundary conditions.

Whatever method we use, we should (formally: we have to) do that always anyways, since there is always the risk that our reasoning introduces a new solution that doesn't satisfy the original problem statement. (Nerd)
In practice this is usually not really painstakingly necessary (and often neglected (Smirk)), as long as we're sharp on steps that could conceivably introduce new invalid solutions (like squaring, or multiplying by an expression that could be zero).

In this case, it's very likely that our 'limit' solution will satisfy the IVP.
That's because the solution $y(t)$ that we find, can be seen as a function of $\omega$ as well. That is, $y(t) = y(t;\omega)$. That function is continuous in $\omega$, just like the IVP is continuous in $\omega$.
The $y(t;\omega)$ we found, is just not defined for $\omega^2=n^2$ (nor for any $\omega \in \mathbb R \setminus \mathbb Q$), even though the IVP is.
So we can extend the function by defining the value at its limit, making it continuous in that point (and actually in any irrational point) as well.

 

[evinda]

We just have to fill in the potential solution for $\omega^2=n^2$ in the differential equation, see that it is satisfied, and do the same thing for the boundary conditions.

Whatever method we use, we should (formally: we have to) do that always anyways, since there is always the risk that our reasoning introduces a new solution that doesn't satisfy the original problem statement. (Nerd)
In practice this is usually not really painstakingly necessary (and often neglected (Smirk)), as long as we're sharp on steps that could conceivably introduce new invalid solutions (like squaring, or multiplying by an expression that could be zero).

In this case, it's very likely that our 'limit' solution will satisfy the IVP.
That's because the solution $y(t)$ that we find, can be seen as a function of $\omega$ as well. That is, $y(t) = y(t;\omega)$. That function is continuous in $\omega$, just like the IVP is continuous in $\omega$.
The $y(t;\omega)$ we found, is just not defined for $\omega^2=n^2$ (nor for any $\omega \in \mathbb R \setminus \mathbb Q$), even though the IVP is.
So we can extend the function by defining the value at its limit, making it continuous in that point (and actually in any irrational point) as well.

I see.. Thanks a lot! (Smile)