Heat equation problem so confusing

[JI567]

I tend to agree with Ray. But, maybe you can give us an example of a specific problem you are struggling with (from start to finish, not just the final equation), and we can help you through it. I don't think that the equation you presented in the initial post of this thread is a correct relationship for any heat equation problem that I have ever seen (and I've seen lots of them).

Chet

Okay like you suggested I have given the whole problem.

Tt = 4Txx + 4 sin(πx)

with 0 < x < 1, x is actually less than and equal too 0, greater than and equal too 1

Boundary is T(0,t) = T(1,t) = 0
Initial condition is T (x,0) = cos (2πx)

So I attempted to find the U(x,t) by U(x,0) = T(x,0) - T(x)

I found T(x) by boundary condition and got -1/π^2 * (sin(πx))

So then I wanted to find U(x,t) by using the summation of An * cos(nπx) + summation of Bn* sin (nπx). So now to find An and Bn there are separate formulas that involves integration but its going to be complex. So i tried comparing coefficients.

So I had cos(2πx) - 1/π^2 * sin(πx) = summation of An*cos(nπx) + summation of Bn * sin(nπx). So my An = 1 and Bn = -1/π^2. Is this correct then?

The idea of examples is that they illustrate a method; it is then up to you to apply the method in situations that are NOT covered in the textbook.

You can find relevant articles yourself; just Google "Fourier Series", and you will find dozens of web pages at various levels of sophistication. YOU choose the ones you want. I have no way of knowing which of the many articles out there are the ones that will "speak to you".

As for you sir, there is absolutely nothing on the web about a sin function and cosine function for the same equation. I have checked everywhere, even youtube. So please cut me some slack as I don't have experience of solving heat equations for 5 or 10 years. I need a solid example similar to this to understand as I just had lecture for 2 weeks on heat equation. If reading was helpful I wouldn't be asking for help on this forum. Cheers.

 

[LCKurtz]

Okay like you suggested I have given the whole problem.

Tt = 4Txx + 4 sin(πx)

with 0 < x < 1, x is actually less than and equal too 0, greater than and equal too 1

Boundary is T(0,t) = T(1,t) = 0
Initial condition is T (x,0) = cos (2πx)

So I attempted to find the U(x,t) by U(x,0) = T(x,0) - T(x)

I found T(x) by boundary condition and got -1/π^2 * (sin(πx))

I get $-\frac {4}{\pi^2}\sin(\pi x)$

So then I wanted to find U(x,t) by using the summation of An * cos(nπx) + summation of Bn* sin (nπx).

Where did you get those summations? Did you separate variables and solve the $X$ eigenvalue problem? I don't get any cosine eigenfunctions, so please show your work.

So now to find An and Bn there are separate formulas that involves integration but its going to be complex. So i tried comparing coefficients.
So I had cos(2πx) - 1/π^2 * sin(πx) = summation of An*cos(nπx) + summation of Bn * sin(nπx). So my An = 1 and Bn = -1/π^2. Is this correct then?

No, it isn't. I thought we had cleared this up in post #21. Equating coefficients would give $A_2=1$, not $A_n=1$ and similarly for the B. But that's assuming your summation was correct, which I don't think it is. Show us your separation of variables steps.

 

[Chestermiller]

I get $-\frac {4}{\pi^2}\sin(\pi x)$

That's not what I get. For the solution at long times, it get
T(x)=+\frac {1}{\pi^2}\sin(\pi x)

Chet

 

[LCKurtz]

That's not what I get. For the solution at long times, it get
T(x)=+\frac {1}{\pi^2}\sin(\pi x)

Chet

Yeah, I think my substitution was slightly different than the OP's, so what I am calling T(x) is slightly different. That's not the major issue for the OP though.

 

[Chestermiller]

Okay like you suggested I have given the whole problem.

OK. I can see what you were doing, and you kinda had the right idea.

Tt = 4Txx + 4 sin(πx)

with 0 < x < 1, x is actually less than and equal too 0, greater than and equal too 1

Boundary is T(0,t) = T(1,t) = 0
Initial condition is T (x,0) = cos (2πx)

So I attempted to find the U(x,t) by U(x,0) = T(x,0) - T(x)

This last equation is incorrect. It should be U(x,t)=T(x,t) - T(x,∞)
I assume you are using T(x) to represent T(x,∞)

I found T(x) by boundary condition and got -1/π^2 * (sin(πx))

I get T(x)=+1/π^2 * (sin(πx))
So, $U(0,t)=cos(2πx)-\frac{1}{π^2}sin(πx)$

Are we together so far?

Chet

 

[JI567]

OK. I can see what you were doing, and you kinda had the right idea.

This last equation is incorrect. It should be U(x,t)=T(x,t) - T(x,∞)
I assume you are using T(x) to represent T(x,∞)

I get T(x)=+1/π^2 * (sin(πx))
So, $U(0,t)=cos(2πx)-\frac{1}{π^2}sin(πx)$

Are we together so far?

Chet

Yeah I am with you. What are the next steps, i am confused there...

 

[Ray Vickson]

Okay like you suggested I have given the whole problem.

Tt = 4Txx + 4 sin(πx)

with 0 < x < 1, x is actually less than and equal too 0, greater than and equal too 1

Boundary is T(0,t) = T(1,t) = 0
Initial condition is T (x,0) = cos (2πx)

So I attempted to find the U(x,t) by U(x,0) = T(x,0) - T(x)

I found T(x) by boundary condition and got -1/π^2 * (sin(πx))

So then I wanted to find U(x,t) by using the summation of An * cos(nπx) + summation of Bn* sin (nπx). So now to find An and Bn there are separate formulas that involves integration but its going to be complex. So i tried comparing coefficients.

So I had cos(2πx) - 1/π^2 * sin(πx) = summation of An*cos(nπx) + summation of Bn * sin(nπx). So my An = 1 and Bn = -1/π^2. Is this correct then?As for you sir, there is absolutely nothing on the web about a sin function and cosine function for the same equation. I have checked everywhere, even youtube. So please cut me some slack as I don't have experience of solving heat equations for 5 or 10 years. I need a solid example similar to this to understand as I just had lecture for 2 weeks on heat equation. If reading was helpful I wouldn't be asking for help on this forum. Cheers.

Well, when I Googled "nonhomogeneous heat equation" I got dozens of hits that all go through examples exactly like yours (but sometimes with a general function $f(x)$ or $F(x,t)$ in place of your specific forcing term $4 \sin(\pi x)$). So, OK, they use $f(x)$ instead of a sin function, but they give detailed formulas that can be applied to any function, and they work out examples from start to finish. Different web pages work out different examples.

 

[JI567]

Well, when I Googled "nonhomogeneous heat equation" I got dozens of hits that all go through examples exactly like yours (but sometimes with a general function $f(x)$ or $F(x,t)$ in place of your specific forcing term $4 \sin(\pi x)$). So, OK, they use $f(x)$ instead of a sin function, but they give detailed formulas that can be applied to any function, and they work out examples from start to finish. Different web pages work out different examples.

Integrating with f(x) as a function to get Bn or An is way more simpler then actually putting sin function or cos function and trying to integrate. Specially in this one where its cos and sin both combined. I just don't get it and I probably won't unless I actually see it. I know when its cos its an even function so we use just An and when its odd its sin and we use Bn. For this problem both of them are there so I think we need to add An and Bn but I don't know how to integrate or do we even need to integrate, My head is going to burst, I just don't get it.

 

[Chestermiller]

In post #12, you were close to having the correct summation, but not quite. I got:
U(x,t)=\sum_{n=1}^∞ B_ne^{-4π^ 2 n^2t}sin(n\pi x)
So, at time t= 0, you need to satisfy:
U(x,0)=\sum_{n=1}^∞ B_nsin(n\pi x)=cos(2πx)-\frac{1}{π^2}sin(πx)
We need to find the infinite set of numbers B_n (n = 1,...,π) such that this equation is satisfied.

Are we still together?

Chet

 

[JI567]

In post #12, you were close to having the correct summation, but not quite. I got:
U(x,t)=\sum_{n=1}^∞ B_ne^{-4π^ 2 n^2t}sin(n\pi x)
So, at time t= 0, you need to satisfy:
U(x,0)=\sum_{n=1}^∞ B_nsin(n\pi x)=cos(2πx)-\frac{1}{π^2}sin(πx)
We need to find the infinite set of numbers B_n (n = 1,...,π) such that this equation is satisfied.

Are we still together?

Chet

How did you get the 4 in the exponential term?

And yes I am together with the Bn part, please continue.

 

[Chestermiller]

How did you get the 4 in the exponential term?

First, substitute into the differential equation and see that each term satisfies the differential equation. The 4 comes from the coefficient of Txx.

And yes I am together with the Bn part, please continue.

Now, the next step to determine the B_ns is a little new to you and tricky, so you are going to have to bear with me. This is where the "half wave" feature comes in. Even though the problem is only defined over the interval 0<x<1, we are going to make use of the entire interval from -1<x<+1. We are going to extend U(x,0) to the interval -1<x<0 in the following way:

$U(x,0) = -cos(2πx)-\frac{1}{π^2}sin(πx)$ (-1<x<0)

$U(x,0) = +cos(2πx)-\frac{1}{π^2}sin(πx)$ (0<x<1)

Note that, when we define U(x,0) in this way, U(x,0) is odd function of x, such that U(-x,0)=-U(x,0). As such, it is in perfect mathematical form for representation as a sin(nπx) series, which is also an odd function of x.

Still together?

Chet

 

[JI567]

First, substitute into the differential equation and see that each term satisfies the differential equation. The 4 comes from the coefficient of Txx.

Now, the next step to determine the B_ns is a little new to you and tricky, so you are going to have to bear with me. This is where the "half wave" feature comes in. Even though the problem is only defined over the interval 0<x<1, we are going to make use of the entire interval from -1<x<+1. We are going to extend U(x,0) to the interval -1<x<0 in the following way:

$U(x,0) = -cos(2πx)-\frac{1}{π^2}sin(πx)$ (-1<x<0)

$U(x,0) = +cos(2πx)-\frac{1}{π^2}sin(πx)$ (0<x<1)

Note that, when we define U(x,0) in this way, U(x,0) is odd function of x, such that U(-x,0)=-U(x,0). As such, it is in perfect mathematical form for representation as a sin(nπx) series, which is also an odd function of x.

Still together?

Chet

I get the entire odd function approach as I am familiar with it from the wave equation questions. But however for the negative range of x why did you change the sign of cos function only and not the sin function? For the wave equation, I know the negative part in D'alembert is always (x-ct) so you just take it as -f(-x) there but not sure how it works here...

 

[Chestermiller]

I get the entire odd function approach as I am familiar with it from the wave equation questions. But however for the negative range of x why did you change the sign of cos function only and not the sin function?

The sine term is already odd, so I don't need to do anything with it. But cosine is an even function, so, to use the sin(nπx) series to represent it, we have to give it "oddness."

Note that all we really care about is accurately representing U(x,0) over the region from x = 0 to x = 1. By doing what we have done, we are able to get an accurate representation with the sin(nπx) series.

Chet

 

[JI567]

The sine term is already odd, so I don't need to do anything with it. But cosine is an even function, so, to use the sin(nπx) series to represent it, we have to give it "oddness."

Note that all we really care about is accurately representing U(x,0) over the region from x = 0 to x = 1. By doing what we have done, we are able to get an accurate representation with the sin(nπx) series.

Chet

Yeah alright makes sense. Continue please, I am with you. next step

 

[Chestermiller]

OK. Now we are going to take the equation $U(x,0)=\sum_{n=1}^∞ B_nsin(n\pi x)$ (including the part from x = -1 to x = 0), multiply both sides of it by sin(mπx), and integrate both sides from x = -1 to x = +1. This will enable you to determine each B_m individually, since, for each value of m (where m is a positive integer), only one term in the summation on the right hand side will survive (i.e., be non-zero). Do you feel comfortable doing these integrations? (You need to integrate the cosine term from -1 to 0, and then switch the sign and continue the integration from 0 to +1).

Chet

 

[JI567]

OK. Now we are going to take the equation $U(x,0)=\sum_{n=1}^∞ B_nsin(n\pi x)$ (including the part from x = -1 to x = 0), multiply both sides of it by sin(mπx), and integrate both sides from x = -1 to x = +1. This will enable you to determine each B_m individually, since, for each value of m (where m is a positive integer), only one term in the summation on the right hand side will survive (i.e., be non-zero). Do you feel comfortable doing these integrations? (You need to integrate the cosine term from -1 to 0, and then switch the sign and continue the integration from 0 to +1).

Chet

How do you integrate cos(2πx)*sin(mπx)?

I will have this right

-cos(2πx)*sin(mπx)-1/π^2 * sin(πx)* sin(mπx) = summation of Bn*sin(nπx)*sin(mπx)?

 

[Chestermiller]

How do you integrate cos(2πx)*sin(mπx)?

Consider this:
sin(A+B)=sin(A)cos(B)+cos(A)sin(B)
sin(A-B)=sin(A)cos(B)-cos(A)sin(B)
What happens if you add these two equations together?

I will have this right

-cos(2πx)*sin(mπx)-1/π^2 * sin(πx)* sin(mπx) = summation of Bn*sin(nπx)*sin(mπx)?

Yes. That will be what you will be integrating from -1 to 0. You will complete the integral over the full interval by integrating
cos(2πx)*sin(mπx)-1/π^2 * sin(πx)* sin(mπx) = summation of Bn*sin(nπx)*sin(mπx) from 0 to 1.

Chet

 

[JI567]

Consider this:
sin(A+B)=sin(A)cos(B)+cos(A)sin(B)
sin(A-B)=sin(A)cos(B)-cos(A)sin(B)
What happens if you add these two equations together?

Yes. That will be what you will be integrating from -1 to 0. You will complete the integral over the full interval by integrating
cos(2πx)*sin(mπx)-1/π^2 * sin(πx)* sin(mπx) = summation of Bn*sin(nπx)*sin(mπx) from 0 to 1.

Chet

Then it just become 2 sin a cos b. But that's not an integral...or do we do 1/2 * integration of sin (a+b) + integration of sin (a-b). But then m is our b here. How can we add or subtract m from 2. I am confused

 

[Ray Vickson]

Then it just become 2 sin a cos b. But that's not an integral...or do we do 1/2 * integration of sin (a+b) + integration of sin (a-b). But then m is our b here. How can we add or subtract m from 2. I am confused

Everything you need is in here: http://mathworld.wolfram.com/FourierSeries.html That is what makes Fourier analysis possible

 

[JI567]

Everything you need is in here: http://mathworld.wolfram.com/FourierSeries.html That is what makes Fourier analysis possible

What's this kronecker delta? How do I convert it into actual numbers or simple things such as π because this delta thing is not present in the answers.

 

[Chestermiller]

Then it just become 2 sin a cos b. But that's not an integral...or do we do 1/2 * integration of sin (a+b) + integration of sin (a-b). But then m is our b here. How can we add or subtract m from 2. I am confused

Yes. This is correct.
cos(2πx)sin(mπx)=\frac{sin((m+2)πx)+sin((m-2)πx)}{2}
So you can integrate that easily.

Please note that m and n are just dummy indices, and that, in the end, we will be able to change m back to n (including on the B's). So please, just bear with it.

You can use the same kind of trigonometic identities on the B summation terms, except there, they will involve cos(m+n)πx and cos(m-n)πx. See if you can do that.

Chet

 

[JI567]

Yes. This is correct.
cos(2πx)sin(mπx)=\frac{sin((m+2)πx)+sin((m-2)πx)}{2}
So you can integrate that easily.

Please note that m and n are just dummy indices, and that, in the end, we will be able to change m back to n (including on the B's). So please, just bear with it.

You can use the same kind of trigonometic identities on the B summation terms, except there, they will involve cos(m+n)πx and cos(m-n)πx. See if you can do that.

Chet

I am confused.

I thought we just integrate the -cos(2πx)*sin(mπx)-1/π^2 * sin(πx)* sin(mπx) and then compare the coeffecients on the other side with Bn's m. Do I have to integrate summation of Bn side as well?

And to check, integration of ( sin(m+2)πx)/2 equals to (-cos(m+2)πx)/(2(m+2)π)

Is that correct? x values I can use -1 to 0 for this one but what about the m values? Do I just leave it as m?

 

[Chestermiller]

I am confused.

I thought we just integrate the -cos(2πx)*sin(mπx)-1/π^2 * sin(πx)* sin(mπx) and then compare the coeffecients on the other side with Bn's m. Do I have to integrate summation of Bn side as well?

Yes. But you will find that, for any value of m, all the integrated terms are zero except when m = n.

And to check, integration of ( sin(m+2)πx)/2 equals to (-cos(m+2)πx)/(2(m+2)π)

Yes, but don't forget to apply the integration limits.

Is that correct? x values I can use -1 to 0 for this one but what about the m values? Do I just leave it as m?

Yes.

You need to perform the following integrations:
\int_{-1}^{+1}{\frac{(cos((m-n)\pi x)-cos((m+n)\pi x))}{2}dx}
\int_{-1}^{0}{\frac{(sin((m+2)\pi x)+sin((m-2)\pi x))}{2}dx}+\int_{0}^{+1}{\left(-\frac{(sin((m+2)\pi x)+sin((m-2)\pi x))}{2}\right)dx}
\int_{-1}^{+1}{\frac{(cos((m-1)\pi x)-cos((m+1)\pi x))}{2}dx}
Don't forget that, when m = n, $cos((m-n)\pi x)=1$ and $sin((m-n)\pi x)=0$
For the first integral, you need to consider the two cases m≠n and m = n.
For the second integral, you need to consider the two cases, m≠2 and m = 2.
For the third integral, you need to consider the two cases, m≠1 and m = 1.

Chet

 

[JI567]

Yes. But you will find that, for any value of m, all the integrated terms are zero except when m = n.

Yes, but don't forget to apply the integration limits.

Yes.

You need to perform the following integrations:
\int_{-1}^{+1}{\frac{(cos((m-n)\pi x)-cos((m+n)\pi x))}{2}dx}
\int_{-1}^{0}{\left(-\frac{(sin((m+2)\pi x)+sin((m-2)\pi x))}{2}\right)dx}+\int_{0}^{+1}{\left(\frac{(sin((m+2)\pi x)+sin((m-2)\pi x))}{2}\right)dx}
\int_{-1}^{+1}{\frac{(cos((m-1)\pi x)-cos((m+1)\pi x))}{2}dx}
Don't forget that, when m = n, $cos((m-n)\pi x)=1$ and $sin((m-n)\pi x)=0$
For the first integral, you need to consider the two cases m≠n and m = n.
For the second integral, you need to consider the two cases, m≠2 and m = 2.
For the third integral, you need to consider the two cases, m≠1 and m = 1.

Chet

Only the second integral makes complete sense. Is the first integral representing the integration of the Bn summation side of sin(nπx)*sin(mπx)? And the third integral the integration of sin(mπx)*-1/π^2 *(sin(πx))?

Then, why are we splitting it up to two parts for the second integral? Shouldn't the 0 to 1 range of integral for the second one be for the positive sin function, because remember we were doing positive cos for the 0 to 1 range and when we convert cos by angle identity we can't get negative sin. So I am really confused there.

And about these "cases", so let's say for the first integral. Do I need to integrate with respect to x then put limits -1 to +1? And after doing that I will end up with an equation with no x terms. Only then do I put m equal to n and put m as m and n as n to get two different equations? Should I add up the two equations or what do I do...

Explain please...

 

[Chestermiller]

Only the second integral makes complete sense. Is the first integral representing the integration of the Bn summation side of sin(nπx)*sin(mπx)?

Yes.

And the third integral the integration of sin(mπx)*-1/π^2 *(sin(πx))?

Yes.

Then, why are we splitting it up to two parts for the second integral?

Because of what we did with the cos(2πx) to make U(x,0) an odd function.

Shouldn't the 0 to 1 range of integral for the second one be for the positive sin function, because remember we were doing positive cos for the 0 to 1 range and when we convert cos by angle identity we can't get negative sin.

Yes. I made a mistake here. I should have had a negative sign on the integral from -1 to 0, and + sign from 0 to 1. I was hoping you wouldn't catch that error until I had a chance to change it. But your understanding is so good now that you were able to spot the error. After I finish this response, I'm going to go back and make the correction.

And about these "cases", so let's say for the first integral. Do I need to integrate with respect to x then put limits -1 to +1? And after doing that I will end up with an equation with no x terms. Only then do I put m equal to n and put m as m and n as n to get two different equations?

You can do it before you integrate and apply the integration limits or afterwards. Doing it before is easier, and doing it after requires you to take a limit.

Should I add up the two equations or what do I do...
Explain please...

Just do the integrations first and see what you get, OK.

Chet

 

[Chestermiller]

Regarding the first integration, see if you can show that:

\int_{-1}^{+1}{\left(\sum_{n=1}^∞{B_nsin(n\pi x)}\right)sin(m\pi x)dx}=B_m

Note that the right hand side has no n's in it. The integration has picked out and isolated the coefficient of the mth term of the summation B_m.

Chet

 

[JI567]

Regarding the first integration, see if you can show that:

\int_{-1}^{+1}{\left(\sum_{n=1}^∞{B_nsin(n\pi x)}\right)sin(m\pi x)dx}=B_m

Note that the right hand side has no n's in it. The integration has picked out and isolated the coefficient of the mth term of the summation B_m.

Chet

I have just done the first integration. let me know if its correct...

Keeping the Bn term out as its a constant so just integrating sin terms

So integration of sin(nπx)*sinπ(mπx) when m≠n after applying limits 1 and -1 is (sin((m-n)π))/((m-n)π) - (sin((m+n)π))/((m+n)π)

and integration of the same thing when m = n after applying limits 1 and -1 is 2- (sin(2nπ))/(nπ)

For the above one I first equated m = n when it was in cos form and then integrated to apply limits.

Should I add them up now? I have no idea how does this gives Bm...

 

[Chestermiller]

I have just done the first integration. let me know if its correct...

Keeping the Bn term out as its a constant so just integrating sin terms

So integration of sin(nπx)*sinπ(mπx) when m≠n after applying limits 1 and -1 is (sin((m-n)π))/((m-n)π) - (sin((m+n)π))/((m+n)π)

and integration of the same thing when m = n after applying limits 1 and -1 is 2- (sin(2nπ))/(nπ)

For the above one I first equated m = n when it was in cos form and then integrated to apply limits.

Should I add them up now? I have no idea how does this gives Bm...

This is not what I get for the integral. I get integral=1 for n = m and I get integral=0 for n ≠ m. Don't forget when you do the integration that sin(kπ) = 0 for any integer value of k.

Chet

 

[JI567]

This is not what I get for the integral. I get integral=1 for n = m and I get integral=0 for n ≠ m. Don't forget when you do the integration that sin(kπ) = 0 for any integer value of k.

Chet

Oh yeah, my integration is fine then. Even I get the same as yours cause I got all sin terms. So for m=n i get 1. How do I change it to Bm now?

 

[Chestermiller]

Oh yeah, my integration is fine then. Even I get the same as yours cause I got all sin terms. So for m=n i get 1. How do I change it to Bm now?

Well, if every term in the sum integrates to zero except the m'th term, and the integral of the sines in the m'th term integrate to 1, then all you are left with is 1 times the coefficient of the m'th term, which is Bm.

Now that you know all that, you should immediately be able to determine the value of the third integral. What is it when m =1 ? What is it when m ≠ 1?

Chet