Diffusion Problem (Conduction)

[aeroguy2008]

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?

 

[aeroguy2008]

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

 

[HallsofIvy]

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

 

[Defennder]

Woah I'm surprised HallsOfIvy could view the attachment even before it has been approved. Just something available only to mentors I suppose.

 

[Hootenanny]

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.

 

[Borek]

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

 

[aeroguy2008]

You separated variables and determined that the "parts" of the solution are $X(x)= A cos(\lambda x)+ B sin(\lambda x)$ 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 $\lambda$ 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 want to 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?

 

[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?

 

[aeroguy2008]

Can anyone help me with this problem please? I'm stuck...

 

[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}

Thank you for trying to help me. I have tried a lil bit but I got confused so I guess I want to 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?

 

[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?

 

[Astronuc]

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, B_n. Has one learned how to determine the coefficients?


From the images:

$$\frac{a}{b\pi^2}sin{\frac{5\pi{x}}{2}}e^{-\frac{c}{d}{\pi^2}t}$$

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

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

then

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