Heat equation problem so confusing

  • #1
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Homework Statement


The problem is f(x) = sin2πx - (1/πsquare)*sinπx

and its given Bn sin (nπx) = f(x)

Question is find Bn.

Homework Equations



Bn = 2/L ∫ (sin2πx - (1/πsquare)*sinπx) * sin(nπx/L) where L is 1

The Attempt at a Solution


I did
[/B]
∫ sin2πx * sin (nπx) - (1/πsquare)*sin square πx
then I tried changing it to cos but it doesn't make sense, for the first term ∫sin2πx * sin (nπx)
∫sin2πx * sin (nπx) = -1/2 ∫cos(2π+nπ)-cos(2π-nπ)

But how do i actually integrate the above with the n in it i am so confused.....I know cosnπ = (-1)^n but I can only use that after final integration and even if I finally somehow manage to integrate it, it will be changed to sinnπ. Please can anyone help!!!
. Is there any simple way to do this?
 

Answers and Replies

  • #2
haruspex
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The problem is f(x) = sin2πx - (1/πsquare)*sinπx

and its given Bn sin (nπx) = f(x)
That doesn't make sense. Please correct the problem statement.
 
  • #4
Ray Vickson
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Homework Statement


The problem is f(x) = sin2πx - (1/πsquare)*sinπx

and its given Bn sin (nπx) = f(x)

Question is find Bn.

Homework Equations



Bn = 2/L ∫ (sin2πx - (1/πsquare)*sinπx) * sin(nπx/L) where L is 1

The Attempt at a Solution


I did
[/B]
∫ sin2πx * sin (nπx) - (1/πsquare)*sin square πx
then I tried changing it to cos but it doesn't make sense, for the first term ∫sin2πx * sin (nπx)
∫sin2πx * sin (nπx) = -1/2 ∫cos(2π+nπ)-cos(2π-nπ)

But how do i actually integrate the above with the n in it i am so confused.....I know cosnπ = (-1)^n but I can only use that after final integration and even if I finally somehow manage to integrate it, it will be changed to sinnπ. Please can anyone help!!!
. Is there any simple way to do this?
Your question seems to be asking for the ##B_n## in
[tex] \sum_{n=1}^{\infty} B_n \sin(n \pi x) =\sin(2 \pi x) - \frac{1}{\pi^2} \sin(\pi x) [/tex]
Is that really your question?
 
  • #5
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Your question seems to be asking for the ##B_n## in
[tex] \sum_{n=1}^{\infty} B_n \sin(n \pi x) =\sin(2 \pi x) - \frac{1}{\pi^2} \sin(\pi x) [/tex]
Is that really your question?
Yesss! I didn't know how to insert the symbols like that, but yes that's exactly my question. Is there any simple way to find Bn? Many thanks
 
  • #6
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Yesss! I didn't know how to insert the symbols like that, but yes that's exactly my question. Is there any simple way to find Bn? Many thanks
Just write out the first two terms of the summation on the left hand side and see what you get.

Chet
 
  • #7
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Just write out the first two terms of the summation on the left hand side and see what you get.

Chet
Sorry how do i do that? is it just Bn Sin(πx) + Bn sin (2πx)

I just used n =1 and n =2. Is that correct?
 
  • #8
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Sorry how do i do that? is it just Bn Sin(πx) + Bn sin (2πx)

I just used n =1 and n =2. Is that correct?
No. It's B1sin(πx)+B2sin(2πx).

Now compare that with the right hand side of your equation.

Chet
 
  • #9
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No. It's B1sin(πx)+B2sin(2πx).

Now compare that with the right hand side of your equation.

Chet
So B1 = -1/π square and B2 = 0, right? So do I just take Bn as -1/π square then? Isn't Bn a general constant for all the B values in the summation? I am just confused how two different B values can have the similar Bn value.
 
  • #10
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Sorry I meant B2 = 1
 
  • #11
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So do I just take Bn as -1/π square then?
No. You take B1=-1/π2, B2 = 1, B3...B = 0
Isn't Bn a general constant for all the B values in the summation? I am just confused how two different B values can have the similar Bn value.
I don't understand this question.

Chet
 
  • #12
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No. You take B1=-1/π2, B2 = 1, B3...B = 0

I don't understand this question.

Chet
You know heat equation solution is in a format of T(x,t) = Bn*sin(nπx)*e^-(nsquare*pi square* t) - T(x). So you are saying I will have one T(x,t) equation with B1 and another T(x,t) equation with B2? Do I add up both the equations together to get a final T(x,t) solution.
 
  • #13
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You know heat equation solution is in a format of T(x,t) = Bn*sin(nπx)*e^-(nsquare*pi square* t) - T(x). So you are saying I will have one T(x,t) equation with B1 and another T(x,t) equation with B2? Do I add up both the equations together to get a final T(x,t) solution.
There should be a summation sign in your equation right in front of the Bn.

You seem very confused. This is not the proper venue to teach you the entire solution approach over again. You need to go back and review your notes and text.

Chet
 
  • #14
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O
There should be a summation sign in your equation right in front of the Bn.

You seem very confused. This is not the proper venue to teach you the entire solution approach over again. You need to go back and review your notes and text.

Chet
No I am just confused which format should I present my answer. Do I just put the summation sign infront of Bn and then write when n = 1 bn = -1/pi square. And when n = 2 Bn= 1.
Is that how you present the final answer. Please confirm
 
  • #15
Ray Vickson
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You know heat equation solution is in a format of T(x,t) = Bn*sin(nπx)*e^-(nsquare*pi square* t) - T(x). So you are saying I will have one T(x,t) equation with B1 and another T(x,t) equation with B2? Do I add up both the equations together to get a final T(x,t) solution.
When you write "nsquare" do you mean n^2? By pi square do you mean pi^2? Where is the summation in front of your nth term? (In text you can just write sum_{n} B_b... if you don't want to try using LaTeX and you don't want to use fancy symbols.
 
  • #16
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Yes exactly that's what I mean. And yes there will be the summation sign, apologies. So is that it then? I just write T(x,t) = sum_{n}B_n *sin(nπx)*e^-(n^2*pi^2* t) - T(x). Then I just mention when n =1 and 2 there will be corresponding B values. Will that be the final answer?
 
  • #17
Ray Vickson
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Yes exactly that's what I mean. And yes there will be the summation sign, apologies. So is that it then? I just write T(x,t) = sum_{n}B_n *sin(nπx)*e^-(n^2*pi^2* t) - T(x). Then I just mention when n =1 and 2 there will be corresponding B values. Will that be the final answer?
Why bother with the summation? You just have two terms, so just writing them out in detail answers the question.
 
  • #18
LCKurtz
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To the OP: Setting your heat equation problem aside, what you need to understand is that some FS are finite. For example if you want to expand ##f(x) = 5\sin(3x) + 3\sin(4x)## in a half range sine expansion on ##(0,\pi)## you are wanting to write$$
5\sin(3x) + 3\sin(4x) = \sum_{n=1}^\infty b_n \sin(nx) = b_1\sin(x)+b_2\sin(2x) + b_3\sin(3x)+b_4\sin(4x)+...$$As you noticed in your original post, solving for the ##b_n## can be very tedious and requires working the integrals for ##b_3## and ##b_4## separately. Actually, I don't think you got that far in those calculations. But looking at the above equation you can see by inspection that taking ##b_3=5,~b_4=3## and all the other ##b_n=0## makes the two sides identical. The FS for this function has finitely many nonzero terms, and in fact, the function itself is its own FS.

Similarly, in your solution, you will have only a couple of nonzero terms in your FS and consequently won't need any infinite sum in your answer. Just write out the nonzero terms.
 
  • #19
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To the OP: Setting your heat equation problem aside, what you need to understand is that some FS are finite. For example if you want to expand ##f(x) = 5\sin(3x) + 3\sin(4x)## in a half range sine expansion on ##(0,\pi)## you are wanting to write$$
5\sin(3x) + 3\sin(4x) = \sum_{n=1}^\infty b_n \sin(nx) = b_1\sin(x)+b_2\sin(2x) + b_3\sin(3x)+b_4\sin(4x)+...$$As you noticed in your original post, solving for the ##b_n## can be very tedious and requires working the integrals for ##b_3## and ##b_4## separately. Actually, I don't think you got that far in those calculations. But looking at the above equation you can see by inspection that taking ##b_3=5,~b_4=3## and all the other ##b_n=0## makes the two sides identical. The FS for this function has finitely many nonzero terms, and in fact, the function itself is its own FS.

Similarly, in your solution, you will have only a couple of nonzero terms in your FS and consequently won't need any infinite sum in your answer. Just write out the nonzero terms.
Alright so my solution will be like this :

T(x,t) = b1sin(πx)*e^-(n^2*pi^2* t)+b2sin(2πx)*e^-(n^2*pi^2* t) - T(x).

Is that correct for the final answer?
 
  • #20
LCKurtz
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Alright so my solution will be like this :

T(x,t) = b1sin(πx)*e^-(n^2*pi^2* t)+b2sin(2πx)*e^-(n^2*pi^2* t) - T(x).

Is that correct for the final answer?
No. You have undefined ##n##'s on the right side and undefined ##b_1## and ##b_2## on the left.
 
  • #21
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No. You have undefined ##n##'s on the right side and undefined ##b_1## and ##b_2## on the left.
The n's are going to be 1 and 2 respectively. B1 = -1/pie square and b2 = 1. Is that correct now?
 
  • #22
LCKurtz
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If you wrote that out I think it would finally make sense. Is it correct? Who knows? You never wrote out the heat equation or showed your work getting the solution.
 
  • #23
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∑n=1∞Bnsin(nπx)=sin(2πx)−1π2sin(πx)​
\sum_{n=1}^{\infty} B_n \sin(n \pi x) =\sin(2 \pi x) - \frac{1}{\pi^2} \sin(\pi x)
Is that really your question?
How would I go about finding Bn if the right hand side was instead cos(2πx) - 1/π^2 *(sin(πx)? Can you please tell me. Thanks.
 
  • #24
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Your question seems to be asking for the ##B_n## in
[tex] \sum_{n=1}^{\infty} B_n \sin(n \pi x) =\sin(2 \pi x) - \frac{1}{\pi^2} \sin(\pi x) [/tex]
Is that really your question?
How would I go about finding Bn if the right hand side was instead cos(2πx) - 1/π^2 *(sin(πx)? Can you please tell me. Thanks.
 
  • #25
Ray Vickson
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How would I go about finding Bn if the right hand side was instead cos(2πx) - 1/π^2 *(sin(πx)? Can you please tell me. Thanks.
You wouldn't. The formula ##f(x) \equiv \sum_{n} B_n \sin(n \pi x)## describes an odd function of ##x##---that is--a function giving ##f(-x) = - f(x)##. Your new function does not have this property, so cannot be expressed in the form you want.
 

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