Partial Differential Equations result -- How to simplify trig series?

[AnotherParadox]

Homework Statement

Solve the Boundary value problem
$u_{t}=u_{xx}$



$u(0, t) = u(\pi ,t)=0$



$u(x, 0) = f(x)$

$$f(x)=\left\{\begin{matrix} x; 0 < x < \frac{\pi}{2}\\ \pi-x; \frac{\pi}{2} < x < \pi \end{matrix}\right.$$

Relevant Equations

$$u(x, t) = \sum_{n=1}^{\infty} A_{n} \exp(- \lambda t)\sin(\sqrt{\lambda}\cdot x))$$
$$A_{n}=\frac{2}{L}\int_{0}^{L}f(x)sin(\sqrt{\lambda}\cdot x)dx$$

Solve the boundary value problem
Given
$u_{t}=u_{xx}$

$u(0, t) = u(\pi ,t)=0$

$u(x, 0) = f(x)$
$$f(x)=\left\{\begin{matrix} x; 0 < x < \frac{\pi}{2}\\ \pi-x; \frac{\pi}{2} < x < \pi \end{matrix}\right.$$

L is π - 0=π

λ = α² since 0 and -α lead to trivial solutions
Let
$u = XT$

$X{T}'={X}''T$

$\frac{{X}''}{X}=\frac{{T}'}{T}=-\lambda$

${X}''+\lambda =0$

$X(0)=X(\pi)=0$

${T}'+\lambda T=0$

$X(x)=C_{1}\cos (\sqrt{\lambda }\cdot x)+C_{2}\sin(\sqrt{\lambda }\cdot x)$

$\sqrt{\lambda } \cdot \pi = n\pi$

$\lambda =n ^{2}; n=1,2,3,..$

${T}'+n^{2}T=0$

(I'm not entirely sure how t got into this next step, I know T is actually T(t) but still not sure what it's doing in the next step with the exponential)

$T(t)=C_{3}\exp (-n ^{2}t)$

$u_{n}(x,t)=C_{2}C_{3} \exp(-n^{2}t)\sin(nx)=A_{n} \exp(-n^{2}t)\sin(nx)); n = 1, 2, 3, ..$
$$u(x, t) = \sum_{n=1}^{\infty} A_{n} \exp(-n^{2}t)\sin(nx))$$
$$u(x,0)=f(x)$$
$$A_{n}=\frac{2}{\pi}\left [\int_{0}^{\pi/2}x\sin(nx)dx + \int_{\pi/2}^{\pi}(\pi-x)\sin(nx)dx \right ]$$
$$=\frac{2}{\pi}\left [\left [ \frac{x(-\cos(nx))}{n} \right ]_{0}^{\pi/2} + \int_{0}^{\pi/2} \frac{\cos(nx)}{n}dx + \left [\frac{(\pi-x)(-\cos(nx))}{n} \right ] _{\pi/2}^{\pi} - \int_{\pi}^{\pi/2}\frac{\cos(nx)}{n}dx \right ]$$
$$=\frac{-1}{n}\cos (\frac{n \pi}{2}) + \frac{2}{\pi n^{2}}\sin(\frac{n \pi}{2}) + \frac{1}{n}\cos (\frac{n \pi}{2}) + \frac{2}{\pi n^{2}}\sin(\frac{n \pi}{2})$$
$$=\frac{4}{\pi n^{2}}\sin(\frac{n \pi}{2})$$
$$u(x, t) = \sum_{n=1}^{\infty} \frac{4}{\pi n^{2}}\sin(\frac{n \pi}{2}) \exp(-n^{2}t)\sin(nx))$$

Ok so far so good. A_n was found and put back into the general form of the solution.

Here's where it gets tricky
Apparently there is further simplification of this, now these next steps were provided to me by my instructor (as were the previous too) during lecture and I cannot make sense of them. I see what is supposed to happen, cos and sin follow a pattern on intervals of π or nπ where they evaluate to predictable values like 1,0,-1 etc. and we can ignore some values because they lead to all 0s or the trivial solution.

Apparently he continues by the following steps

$$u(x, t) = \sum_{k=0}^{\infty} \frac{4}{\pi (4k+1)^{2}}\exp(-(4k+1)^{2}t)\sin((4k+1)x)) -\sum_{k=0}^{\infty} \frac{4}{\pi (4k+3)^{2}}\exp(-(4k+3)^{2}t)\sin((4k+3)x))$$

Separated into two series and replaced n with 4k+1 and 4k+3, eliminated the first trig function with some version of previously stated pattern recognition and started the series at 0 instead of 1

Now I can't even verify if this is equivalent to the other series let alone figure out the pattern or logic to this to figure it out for other scenarios.

And someone stated that it can be reduced further into

$$u(x, t) = \sum_{k=0}^{\infty} (-1)^{k}\exp(-(2k+1)^{2}t)\sin((2k+1)x))$$
If someone can please break it down and explain how this is done I will be very grateful as the exam is soon and I cannot seem to figure out how to follow the process to come up with these series reductions

Thank you!

Also it would be nice to know how t ends up in this step${T}'+n^{2}T=0$
$T(t)=C_{3}\exp (-n ^{2}t)$

but not nearly as important.

Thanks again

 

[haruspex]

What values can $\sin(n\frac\pi 2)$ take?
For each such value, what values of n produce that result?

 

[Orodruin]

Also it would be nice to know how t ends up in this stepT′+n2T=0T′+n2T=0{T}'+n^{2}T=0
T(t)=C3exp(−n2t)T(t)=C3exp⁡(−n2t)T(t)=C_{3}\exp (-n ^{2}t)

but not nearly as important.

This is just solving a first order ODE in $t$. Clearly the result must depend on $t$ unless $T’ = 0$ or you have a trivial solution.

Having taught and authored a book on this subject, I would say that if you have an exam soon then understanding this would be far more important than being able to simplify the resulting expressions.

Edit: Rereading this I see an ambiguity in my phrasing so just for clarity: I did not teach the book, I taught students ...

 

[AnotherParadox]

What values can $\sin(n\frac\pi 2)$ take?
For each such value, what values of n produce that result?

Thanks for the response

Here's my thought process

It can't take 0 since sin(0*pi/2) and sin(0*x) is 0 and that leads to the trivial solution.
It can take 1 since sin(1*pi/2) does not equal 0 it equals 1
It can't take 2 since sin(2*pi/2) = 0 and the trivial solution
It can take 3 since sin(3pi/2) = -1 and not zero
It can't take 4 since sin(4pi/4) = 0 and the trivial solution

So I know it can't take any even number since they all lead to 0 and thus the trivial solution

Odd numbers i.e. n = 1, 3, 5, 7.. lead to the pattern 1, -1, 1, -1

Which is a similar pattern as (-1)^n which is for n = 0, 1, 2, 3,: 1, -1, 1, -1

So this can be (-1)^(n) which is for n = 0, 1, 2, 3: To match that part of the expression

Evaluating the rest of the series 2k+1 is needed instead of n is needed so that it's 1,3,5,7 for the 0,1,3,4.. I see how this logic was applied to come up with the 4k+1 and 4k+3 so that checks out too now

So far so good however I think he may of made a mistake and I would like confirmation of this

So he changes

$$u(x, t) = \sum_{n=1}^{\infty} \frac{4}{\pi n^{2}}\sin(\frac{n \pi}{2}) \exp(-n^{2}t)\sin(nx)$$

to
$$u(x, t) = \sum_{k=0}^{\infty} \frac{4}{\pi (4k+1)^{2}}\exp(-(4k+1)^{2}t)\sin((4k+1)x) -\sum_{k=0}^{\infty} \frac{4}{\pi (4k+3)^{2}}\exp(-(4k+3)^{2}t)\sin((4k+3)x)$$

I think this checks out and the series match however on the next reduction I think a few things are left out because he shows
$$u(x, t) = \sum_{k=0}^{\infty} (-1)^{k}\exp(-(2k+1)^{2}t)\sin((2k+1)x)$$

Which looks even better but I believe it's missing the front part and it should be$$u(x, t) = \sum_{k=0}^{\infty} \frac{4}{\pi (2k+1)^2} (-1)^{k}\exp(-(2k+1)^{2}t)\sin((2k+1)x)$$

Or even more

$$u(x, t) = \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{1}{ (2k+1)^2} (-1)^{k}\exp(-(2k+1)^{2}t)\sin((2k+1)x)$$

If anyone can confirm this and/or tell me about other short-cuts/quick verifying tricks/ and ways of thinking that would be much appreciated.

This is just solving a first order ODE in $t$. Clearly the result must depend on $t$ unless $T’ = 0$ or you have a trivial solution.

Having taught and authored a book on this subject, I would say that if you have an exam soon then understanding this would be far more important than being able to simplify the resulting expressions.

Edit: Rereading this I see an ambiguity in my phrasing so just for clarity: I did not teach the book, I taught students ...

Np, I understood what you meant, thanks for the response.

Here's my work

$${T}'+n^2T=0$$
$${T}'=-n^2T$$
$$\frac{{T}'}{T}=-n^2$$
$$\ln\left ( \frac{{T}'}{T} \right )=-n^2$$
$$\int \left ( \frac{{T}'}{T} \right )=\int-n^2dt$$

I'll stop here since I figured it out. the -n^2 has to be integrated with respect to t which multiplies t in. Though it makes me wonder why not with respect to n. I suppose for the boundary conditions it needs to be t? I kind of see what is being said with T'=/=0

I agree it's more important to understand this ODE but I can easily write in the correct step and receive full credit but he counts off significantly for not reducing the series if I understand his grading correctly.

 

[haruspex]

however on the next reduction I think a few things are left out because he shows
$$u(x, t) = \sum_{k=0}^{\infty} (-1)^{k}\exp(-(2k+1)^{2}t)\sin((2k+1)x)$$

Which looks even better but I believe it's missing the front part and it should be$$u(x, t) = \sum_{k=0}^{\infty} \frac{4}{\pi (2k+1)^2} (-1)^{k}\exp(-(2k+1)^{2}t)\sin((2k+1)x)$$

I agree, he seems to have dropped the factor $\frac 4{\pi}$.

 

[AnotherParadox]

I agree, he seems to have dropped the factor $\frac 4{\pi}$.

Also the term for 1/(2k+1)^2, right?

 

[haruspex]

Also the term for 1/(2k+1)^2, right?

Ah, yes... and that's worse.

 

[Orodruin]

Though it makes me wonder why not with respect to n.

Because the derivative is with respect to t while n is a constant.

I agree it's more important to understand this ODE but I can easily write in the correct step and receive full credit but he counts off significantly for not reducing the series if I understand his grading correctly.

This is really bad pedagogy if it is correct.

 

[AnotherParadox]

Because the derivative is with respect to t while n is a constant.

I like this reasoning

Here's some work I thought up with considering this more

$${T}'+n^2T=0$$
$$\frac{{dT}}{dt}+n^2T=0$$
$$\frac{{dT}}{dt}=-n^2T$$
$$\frac{\frac{{dT}}{dt}}{T}=-n^2$$
$$\int \left ( \frac{dT}{T} \right )=\int-n^2dt$$

from there it's ln -> exp

Let me know if this checks out

This is really bad pedagogy if it is correct.

I agree. I wasn't aware of how bad it was though. To be fair it is not a math major course, it's an engineering elective (still important imho). I could very well leave out a lot of these steps entirely and not be counted off however the reduction portion could be anywhere to 10%-30% of the question worth. He often teaches from memory (impressively) and makes mistakes that don't get corrected until I look over the class notes on my own which can be awfully difficult at times (wish he would check /confirm his answers from a record). Still a decent teacher, just a bit aged and dated in his ways. Also different standards for non math majors I suppose.

 

[Orodruin]

Let me know if this checks out

Yes. The more standard approach is to assume $T = C e^{rt}$ and inserting into the differential equation to solve for $r$, but what you did is certainly fine and does not depend on an ansatz.