Diffusion Problem (Conduction)

  • #1

Homework Statement



Since the problem involved formulas I have posted it as an attachment. Please check the attachment first. (I have not scanned this problem, just wrote it in word and posted it)

The Attempt at a Solution



My approach on this problem is to start with separation of variables as I don't see any indication that tells me that I am dealing with non homogeneous conditions and can't seperate. I am thinking that in order to get u, I need to separate for sure. Can anyone verify if that seems as a reasonable approach to this problem?
 

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Answers and Replies

  • #2
I will post my initial approach as an attachment. Do I apply the boundary conditions while I have solved for u?
 

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  • #3
HallsofIvy
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You separated variables and determined that the "parts" of the solution are [itex]X(x)= A cos(\lambda x)+ B sin(\lambda x)[/itex] and [/quote]T(t)= Ce^{-\lambda^2 t}.

Now you can use the boundary conditions, that u(0)= 0 and u(2)= 0, which tell you that X(0)= 0 and X(2)= 0, to determine what values [itex]\lambda[/itex] can have (so that [itex]sin(2\lambda)= 0[/itex]) as well as the fact that the coefficients of [itex]cos(\lamda x)[/itex] are all 0 (so that X(0)= 0).

Since the exponential will be 1 at t= 0, You can write the initial conditions in terms of a Fourier sine series.
Since you are told that one of the terms in the expansion is of the form
[tex]\frac{a}{b\pi^2}sin(\frac{5\pi x}{2}e^(-(c/d)\pi^2t}
 
  • #4
Defennder
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Woah I'm surprised HallsOfIvy could view the attachment even before it has been approved. Just something available only to mentors I suppose.
 
  • #5
Hootenanny
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Woah I'm surprised HallsOfIvy could view the attachment even before it has been approved. Just something available only to mentors I suppose.
It is the mentors that review and approve (or reject) attachments, so they must be able to view all attachments before they are approved.
 
  • #6
Borek
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What makes me wonder is that while HOI have already seen the attachement it is still pending approval :smile:

EOT, we will get banned for OT posts :rolleyes:
 
  • #7
You separated variables and determined that the "parts" of the solution are [itex]X(x)= A cos(\lambda x)+ B sin(\lambda x)[/itex] and
T(t)= Ce^{-\lambda^2 t}.

Now you can use the boundary conditions, that u(0)= 0 and u(2)= 0, which tell you that X(0)= 0 and X(2)= 0, to determine what values [itex]\lambda[/itex] can have (so that [itex]sin(2\lambda)= 0[/itex]) as well as the fact that the coefficients of [itex]cos(\lamda x)[/itex] are all 0 (so that X(0)= 0).

Since the exponential will be 1 at t= 0, You can write the initial conditions in terms of a Fourier sine series.
Since you are told that one of the terms in the expansion is of the form
[tex]\frac{a}{b\pi^2}sin(\frac{5\pi x}{2}e^(-(c/d)\pi^2t}[/QUOTE]

Thank you for trying to help me. I have tried a lil bit but I got confused so I guess I wanna make sure that what I am doing is correct.

X(x)=A*cos(lambda*x)+B*sin(lambda*x)
X(x=0)=A*cos(0)+B*sin(0)=>A=0
X(x=2)=0=>B*sin(2*lambda)=0

Lambda=((n*Pi)/L)^2
so the eigenfunction EF, EF=B*sin((n*Pi)/L)^2

How will I move on from here?
 
  • #8
I know that I need to consider temporal and special separately. For the spatial do I have to consider Lambda=0, Lambda<0. Lambda>0 cases?
 
  • #9
Can anyone help me with this problem please? I'm stuck...
 
  • #10
T(t)= Ce^{-\lambda^2 t}.

Now you can use the boundary conditions, that u(0)= 0 and u(2)= 0, which tell you that X(0)= 0 and X(2)= 0, to determine what values [itex]\lambda[/itex] can have (so that [itex]sin(2\lambda)= 0[/itex]) as well as the fact that the coefficients of [itex]cos(\lamda x)[/itex] are all 0 (so that X(0)= 0).

Since the exponential will be 1 at t= 0, You can write the initial conditions in terms of a Fourier sine series.
Since you are told that one of the terms in the expansion is of the form
[tex]\frac{a}{b\pi^2}sin(\frac{5\pi x}{2}e^(-(c/d)\pi^2t}

Thank you for trying to help me. I have tried a lil bit but I got confused so I guess I wanna make sure that what I am doing is correct.

X(x)=A*cos(lambda*x)+B*sin(lambda*x)
X(x=0)=A*cos(0)+B*sin(0)=>A=0
X(x=2)=0=>B*sin(2*lambda)=0

Lambda=((n*Pi)/L)^2
so the eigenfunction EF, EF=B*sin((n*Pi)/L)^2

How will I move on from here?[/QUOTE]

HallsofIvy does what I have written here seem correct to you?
 
  • #11
There seems to be something wrong with my previous account "aeroguy2008" so I have a new account now. How come nobody replies to my questions?
 
  • #12
Astronuc
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The problem is now one month old, and I suppose people were busy or waiting for some additional work.

One's solution seems correct.

One has used the boundary (spatial) conditions to eliminate the cos function and determine the eigenvalue [itex]\lambda[/itex]. One will note that sin (n[itex]\pi[/itex]) = 0, n is an integer = 0, 1, . . . . infty.

Actually, the spatial function X(x) will be written as an infinite series of sin terms, and each will have a coefficient, Bn. Has one learned how to determine the coefficients?


From the images:

[tex]\frac{a}{b\pi^2}sin{\frac{5\pi{x}}{2}}e^{-\frac{c}{d}{\pi^2}t}[/tex]

[tex]X(x) = A cos {\lambda x} + B sin {\lambda x} = B sin {\lambda x}[/tex]

[tex]T = C e^{-\lambda^2t}[/tex]

then

[tex]X(x) T(t) = C'_n sin\,{\lambda x}\,e^{-\lambda^2{t}}[/tex] and compare to the above expression
 
Last edited:

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