# Diffusion Problem (Conduction)

1. Jul 4, 2008

### aeroguy2008

1. The problem statement, all variables and given/known data

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)

3. 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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2. Jul 5, 2008

### aeroguy2008

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. Jul 5, 2008

### HallsofIvy

Staff Emeritus
You separated variables and determined that the "parts" of the solution are $X(x)= A cos(\lambda x)+ B sin(\lambda x)$ 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 $\lambda$ can have (so that $sin(2\lambda)= 0$) as well as the fact that the coefficients of $cos(\lamda x)$ 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
$$\frac{a}{b\pi^2}sin(\frac{5\pi x}{2}e^(-(c/d)\pi^2t} 4. Jul 5, 2008 ### Defennder Woah I'm surprised HallsOfIvy could view the attachment even before it has been approved. Just something available only to mentors I suppose. 5. Jul 5, 2008 ### Hootenanny Staff Emeritus 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. Jul 5, 2008 ### Borek ### Staff: Mentor What makes me wonder is that while HOI have already seen the attachement it is still pending approval EOT, we will get banned for OT posts 7. Jul 5, 2008 ### aeroguy2008 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 $\lambda$ can have (so that $sin(2\lambda)= 0$) as well as the fact that the coefficients of $cos(\lamda x)$ 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. Jul 5, 2008 ### aeroguy2008 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. Jul 6, 2008 ### aeroguy2008 Can anyone help me with this problem please? I'm stuck... 10. Jul 9, 2008 ### aeroguy2008 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. Aug 17, 2008 ### aerospace08 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. Aug 17, 2008 ### Astronuc ### Staff: Mentor 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 $\lambda$. One will note that sin (n$\pi$) = 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}$$

$$X(x) = A cos {\lambda x} + B sin {\lambda x} = B sin {\lambda x}$$

$$T = C e^{-\lambda^2t}$$

then

$$X(x) T(t) = C'_n sin\,{\lambda x}\,e^{-\lambda^2{t}}$$ and compare to the above expression

Last edited: Aug 17, 2008