Heat equation problem so confusing

[Chestermiller]

For getting b1, b2,b3 and b4 do I just multiply the f(0) with sin(nπx) and integrate 4 times? with n being 1,2,3 and 4 consecutively?

Yes. This is one way to do it. Or you could just treat n as an algebraic parameter, and do it only once.

My question is for integration of Bn the general formula is 2/L* integration of f(0)*sin(nπx) to give it the oddness. So as the sin function is already there why did we still multiply with another sin(mπx)?

These questions indicate to me that you have not yet mastered the basics of Fourier series. You need to go back and review your notes and your textbook. Also, another good reference is section 9.2 of Advanced Engineering Mathematics by Kreizig. You are not going to be able to complete the solution to this problem until you better understand how to apply Fourier Series. Unfortunately, Physics Forums is not an appropriate venue for a complete tutorial on Fourier Series.

Chet

 

[JI567]

Yes. This is one way to do it. Or you could just treat n as an algebraic parameter, and do it only once.

These questions indicate to me that you have not yet mastered the basics of Fourier series. You need to go back and review your notes and your textbook. Also, another good reference is section 9.2 of Advanced Engineering Mathematics by Kreizig. You are not going to be able to complete the solution to this problem until you better understand how to apply Fourier Series. Unfortunately, Physics Forums is not an appropriate venue for a complete tutorial on Fourier Series.

Chet

Okay I will check it out but however for the continuation of this question after doing the three integrations you told me to do we find Bm = something. After that do we just write U(x,t) as Bm*e^-4n^2*π^2t*sin(mπx) ? For Bm subbing the value that we obtained from integrations. Is that it? Or there are more things needed to do to find Bn or Bm?

 

[Chestermiller]

Okay I will check it out but however for the continuation of this question after doing the three integrations you told me to do we find Bm = something. After that do we just write U(x,t) as Bm*e^-4n^2*π^2t*sin(mπx) ? For Bm subbing the value that we obtained from integrations. Is that it? Or there are more things needed to do to find Bn or Bm?

It's a summation, but the counting index in the summation should be either m or n; both should not appear simultaneously in the terms of the summation.

Chet

 

[JI567]

Okay so the final solution to the Heat equation will be T(x,t) = U(x,t) - T(x,infinity) right? With U(x,t) being summation of Bn*e^-4n^2*π^2t*sin(mπx) when n = 1 below the summation sign and with an infinity symbol at top of the summation sign. Is that correct?

 

[JI567]

Okay so the final solution to the Heat equation will be T(x,t) = U(x,t) - T(x,infinity) right? With U(x,t) being summation of Bn*e^-4n^2*π^2t*sin(mπx) when n = 1 below the summation sign and with an infinity symbol at top of the summation sign. Is that correct?

Ignore the n and m terms. I am just going to write it down as n everywhere.

 

[Chestermiller]

Okay so the final solution to the Heat equation will be T(x,t) = U(x,t) - T(x,infinity) right? With U(x,t) being summation of Bn*e^-4n^2*π^2t*sin(mπx) when n = 1 below the summation sign and with an infinity symbol at top of the summation sign. Is that correct?

No, that m should be an n, and that should be a plus sign in front of T(x,infinity).

 

[JI567]

No, that m should be an n, and that should be a plus sign in front of T(x,infinity).

Alright cheers. I will get back to you with the final working out steps and final answer to the heat equation. Please check if that's correct

 

[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:
$$\int_{-1}^{+1}{\frac{(cos((m-n)\pi x)-cos((m+n)\pi x))}{2}dx}$$
$$\int_{-1}^{0}{\frac{(sin((m+2)\pi x)+sin((m-2)\pi x))}{2}dx}+\int_{0}^{+1}{\left(-\frac{(sin((m+2)\pi x)+sin((m-2)\pi x))}{2}\right)dx}$$
$$\int_{-1}^{+1}{\frac{(cos((m-1)\pi x)-cos((m+1)\pi x))}{2}dx}$$
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

No, that m should be an n, and that should be a plus sign in front of T(x,infinity).

For the second integration, isn't it just going to be 0? I was doing the m = 2 and then i saw it can be written as -sin(4πx) + sin(0πx) which can be written as 0 for both, is that correct? Also when m ≠2 even there its going to be 0 right as for any sin(π)*(m-n) or (m+n) its always going to be 0, so entire second integral is 0, is that correct?

 

[JI567]

So I am just left with Bn = -1/π^2. That's all right? I don't need to determine anything else right? any other coefficient ...

 

[JI567]

Nicely done. For the heat transfer problem, you do the exact same thing you did on this problem to get the Fourier Series. You can even use the exact same half wave equation to get Bn.

Chet

Okay I did it here. So I got Bn by

2 $\int_0^1 (cos(2πx) - \frac {1} {π^2} \$sin(πx)) * sin(nπx)

L was just 1.

We know integration of cos(mπx)*sin(nπx) is always 0 so just ignore that. Then integration of sin(πx)*sin(nπx) is 0 for when n ≠1 and when n =1 the final integration value is -$\frac {1} {π^2} \$.

So at the end for Bn we have that when n = 1, Bn = -$\frac {2} {π^2} \$. And if n otherwise then Bn = 0.

Can you please confirm if I am correct so far? Thanks.

 

[Chestermiller]

Okay I did it here. So I got Bn by

2 $\int_0^1 (cos(2πx) - \frac {1} {π^2} \$sin(πx)) * sin(nπx)

L was just 1.

We know integration of cos(mπx)*sin(nπx) is always 0 so just ignore that. Then integration of sin(πx)*sin(nπx) is 0 for when n ≠1 and when n =1 the final integration value is -$\frac {1} {π^2} \$.

So at the end for Bn we have that when n = 1, Bn = -$\frac {2} {π^2} \$. And if n otherwise then Bn = 0.

Can you please confirm if I am correct so far? Thanks.

The integral of cos(2πx)sin(nπx) is not equal to zero for any odd value of n for your integration limits.

Chet

 

[JI567]

Cos(2πx)sin(nπx) = $\frac {(sin(n+2)πx)+ (sin(n-2)πx)} {2} \$

If m is let's say a even of 4 then it becomes $\frac {1} {2} \$((sin(6πx) + sin(2πx)) which = 0 and if m is let's say an odd of 3 then it comes $\frac {1} {2} \$( sin(5πx) + sin(πx)) which = 0 as well. Am I missing something here cause I can't see how it can not equal to 0. Can you please tell...

 

[Chestermiller]

Cos(2πx)sin(nπx) = $\frac {(sin(n+2)πx)+ (sin(n-2)πx)} {2} \$

If m is let's say a even of 4 then it becomes $\frac {1} {2} \$((sin(6πx) + sin(2πx)) which = 0 and if m is let's say an odd of 3 then it comes $\frac {1} {2} \$( sin(5πx) + sin(πx)) which = 0 as well. Am I missing something here cause I can't see how it can not equal to 0. Can you please tell...

When were you planning on performing the integration?

Chet

 

[JI567]

When were you planning on performing the integration?

Chet

I thought I could just cancel the sin terms of as they equal to 0 on their own but well I will integrate and see now. I should just integrate for the limits 0 to 1 right?

 

[Chestermiller]

I thought I could just cancel the sin terms of as they equal to 0 on their own but well I will integrate and see now. I should just integrate for the limits 0 to 1 right?

Sure.

 

[JI567]

Okay so for the odd n integration of cos(2πx)*sin(nπx) after putting the limits I get a relationship of 2( $\frac {1} {2((n+2)π)} \$ + $\frac {1} {2((n-2)π)} \$). Is this correct?

 

[Chestermiller]

Okay so for the odd n integration of cos(2πx)*sin(nπx) after putting the limits I get a relationship of 2( $\frac {1} {2((n+2)π)} \$ + $\frac {1} {2((n-2)π)} \$). Is this correct?

I'm have to check it later, but it certainly looks close. But at least reduce to common denominators and factor.

Chet

 

[JI567]

okay after reducing it becomes

$\frac {1} {π} \$ ( $\frac {1} {n+2} \$+ $\frac {1} {n-2} \$)

Let me know whenever you have checked it...

 

[Chestermiller]

okay after reducing it becomes

$\frac {1} {π} \$ ( $\frac {1} {n+2} \$+ $\frac {1} {n-2} \$)

Let me know whenever you have checked it...

Reduce to least common denominator? Otherwise, OK.

 

[JI567]

$\frac {1} {π} \$ ( $\frac {2n} {n^2-4} \$)...please don't say I need to reduce more... so only when n = 1 my Bn will be the result of 2 $\int_0^1 (cos(2πx) - \frac {1} {π^2} \$sin(πx)) * sin(nπx). When n = any other odd numbers except for 1 then Bn will be the result of 2 $\int_0^1 (cos(2πx) \$ sin(nπx). For n otherwise Bn = 0. Is this correct...?

 

[Chestermiller]

$\frac {1} {π} \$ ( $\frac {2n} {n^2-4} \$)

Yes. Nicely done.

so only when n = 1 my Bn will be the result of 2 $\int_0^1 (cos(2πx) - \frac {1} {π^2} \$sin(πx)) * sin(nπx). When n = any other odd numbers except for 1 then Bn will be the result of 2 $\int_0^1 (cos(2πx) \$ sin(nπx). For n otherwise Bn = 0. Is this correct...?

So,

$B_1= -\frac{4}{3\pi} - \frac {1} {π^2} \$

$B_n=\frac {4} {π} \left(\frac {n} {n^2-4}\right)$ for all odd n≥3

$B_n=0$ for all even n.

Chet

 

[JI567]

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))

Yes. Nicely done.So,

$B_1= -\frac{4}{3\pi} - \frac {1} {π^2} \$

$B_n=\frac {4} {π} \left(\frac {n} {n^2-4}\right)$ for all odd n≥3

$B_n=0$ for all even n.

Chet

So for this heat equation, can the final solution be written like this

when n = 1
T(x,t) = (-$\frac {4} {3π} \$ - $\frac {1} {π^2} \$) $\ e^{-4π^2t} \$ sin(πx) + $\frac {1} {π^2} \$ sin(πx)

and for any odd n greater than or equal to 3

T(x,t) = (($\frac {4} {π} \$ ( $\frac {n} {n^2-4} \$)) $\ e^{-4n^2π^2t} \$ sin(nπx) + $\frac {1} {π^2} \$ sin(πx)

and for all even n

T(x,t) = $\frac {1} {π^2} \$ sin(πx)

 

[Chestermiller]

Not even close. I'll get back to you later.

Chet

 

[Chestermiller]

Here is the results you obtained rendered into a solution:

$$T(x,t)=\frac{sinπx}{π^2}-(\frac{4}{3\pi} + \frac {1} {π^2})e^{-4π^2t}sinπx+\frac{4}{π}\frac{3}{5}e^{-36π^2t}sin(3πx)+\frac{4}{π}\frac{5}{21}e^{-100π^2t}sin(5πx)+...$$
Rearranging this in a little different way gives:
$$T(x,t)=\frac{1}{π^2}(1-e^{-4π^2t})sinπx+\frac{4}{π}\sum_{k=1}^∞\frac{(2k-1)e^{-4(2k-1)^2π^2t}}{(2k-1)^2-4}sin((2k-1)πx)$$
Can you see how that works?

Chet

 

[JI567]

Here is the results you obtained rendered into a solution:

$$T(x,t)=\frac{sinπx}{π^2}-(\frac{4}{3\pi} + \frac {1} {π^2})e^{-4π^2t}sinπx+\frac{4}{π}\frac{3}{5}e^{-36π^2t}sin(3πx)+\frac{4}{π}\frac{5}{21}e^{-100π^2t}sin(5πx)+...$$
Rearranging this in a little different way gives:
$$T(x,t)=\frac{1}{π^2}(1-e^{-4π^2t})sinπx+\frac{4}{π}\sum_{k=1}^∞\frac{(2k-1)e^{-4(2k-1)^2π^2t}}{(2k-1)^2-4}sin((2k-1)πx)$$
Can you see how that works?

Chet

Oh I see...I added up each U(x,t) to the T(x) separately that's why and totally forgot about the summation thing for odd n. That makes sense. Cheers! However shouldn't the last equation be

$$T(x,t)=\frac{1}{π^2}(1-e^{-4π^2t}-\frac {4π} {3} e^{-4π^2t})sinπx+\frac{4}{π}\sum_{k=1}^∞\frac{(2k-1)e^{-4(2k-1)^2π^2t}}{(2k-1)^2-4}sin((2k-1)πx)$$

 

[Chestermiller]

Oh I see...I added up each U(x,t) to the T(x) separately that's why and totally forgot about the summation thing for odd n. That makes sense. Cheers! However shouldn't the last equation be

$$T(x,t)=\frac{1}{π^2}(1-e^{-4π^2t}-\frac {4π} {3} e^{-4π^2t})sinπx+\frac{4}{π}\sum_{k=1}^∞\frac{(2k-1)e^{-4(2k-1)^2π^2t}}{(2k-1)^2-4}sin((2k-1)πx)$$

No. Then the summation would have to be from k = 2. I purposely wrote it in this form so that the term outside the summation would represent the separate effect of the heat generation, and the summation would represent the separate effect of the initial condition.

Chet

 

[JI567]

No. Then the summation would have to be from k = 2. I purposely wrote it in this form so that the term outside the summation would represent the separate effect of the heat generation, and the summation would represent the separate effect of the initial condition.

Chet

Okay cheers. One question outside solving problems. How in actual engineering does wave equation and heat equation help? How do I use these lengthy solution of equations for like making machines or something.

 

[JI567]

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) = sin (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 wrote it like U(x,0)= - 1/π^2 * sin(πx) + sin(2πx) = B1 sin(πx) + B2sin(2πx) so B1 = -1/π^2 and B2 = 1 and Bn =0 for n otherwise

Can I then write the final solution for this heat equation as

T(x,t) = U(x,t) + T(x)
T(x,t) = $\frac {1} {π^2} \$ (1- $\ e^{-4π^2t} \$) sin(πx)+ $\ e^{-16π^2t} sin(2πx) \$

is this correct? please check...

 

[Chestermiller]

No way. What happened to all the other terms in the summation? Plus, there should be no terms containing sin(2πx). Just write out the terms in the result I gave in post #94.

Chet

 

[JI567]

No way. What happened to all the other terms in the summation? Plus, there should be no terms containing sin(2πx). Just write out the terms in the result I gave in post #94.

Chet

The one is post 94 was different initial condition which was cos(2πx). This is exactly the same question as of post 94 but the initial conditions in the previous post if you see is now sin(2πx). That's why the solution is different and thus the term with sin(2πx). Do you still think its wrong?

 

[Chestermiller]

The one is post 94 was different initial condition which was cos(2πx). This is exactly the same question as of post 94 but the initial conditions in the previous post if you see is now sin(2πx). That's why the solution is different and thus the term with sin(2πx). Do you still think its wrong?

Are you saying that this is for a different problem that we haven't discussed before?

Chet

 

[JI567]

Are you saying that this is for a different problem that we haven't discussed before?

Chet

Yes different problem. But only thing different is the initial condition...

 

[Chestermiller]

Yes different problem. But only thing different is the initial condition...

Then, the solution looks OK, but you didn't need to use Fourier Series to solve this.

Chet

 

[JI567]

Then, the solution looks OK, but you didn't need to use Fourier Series to solve this.

Chet

This is the only way I was taught to solve inhomogeneous questions like these... like finding the Bn and then solving for U(x,t)...is the solution really correct though?

 

[Chestermiller]

This is the only way I was taught to solve inhomogeneous questions like these... like finding the Bn and then solving for U(x,t)...is the solution really correct though?

As best I can tell, yes.

Chet