Heat equation problem so confusing

[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