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How to deal with this Neumann boundary conditions?

  1. Aug 21, 2010 #1
    hi all,

    I am trying to solve this PDE by separation of variables, it goes like this:


    [tex]\frac{\partial u}{\partial t} = \alpha\frac{\partial ^2 u}{\partial z^2} [/tex] for [tex]0\leq z\leq infty[/tex]

    the initial condition I have is: t=0; u = uo.
    the boundary condtions:


    z=0; [tex]\frac{\partial u}{\partial z}\right\rfloor_{z=0} = k\left(u-u_{b}\right) ... ...(1)[/tex]

    z= [tex]\infty[/tex]; [tex]\frac{\partial u}{\partial z}\right\rfloor_{z=\infty} = 0......(2)[/tex]

    where, uo, [tex]k[/tex],[tex]u_{b}[/tex], and h are constants.

    I write:

    u(z,t) = F(z)G(t),
    with the subsitition I got:

    [tex]\frac{1}{G}\frac{\partial G}{\partial t} = \frac{\alpha}{F}\frac{\partial ^2 F}{\partial Z^2}[/tex]

    Setting this to some constant: [tex]omega^2[/tex], I can have the two ODEs.

    The problem is when I try to use the first boundary condition, bcos of the second term in the parenthesis of the RHS, I seem to have [tex]\frac{u_{b}}{G}[/tex], which I have no clue how to deal with.
    pls could some one point to me how i can go about this?

    thanks in advance.
     
    Last edited: Aug 21, 2010
  2. jcsd
  3. Aug 21, 2010 #2
    I'm not sure if I understand your first boundary condition:
    [tex]
    \frac{\partial u}{\partial z}\right\rfloor_{z=0} = k\left(u-u_{b}\right) ... ...(1)
    [/tex]

    Is that u = u(0,t)?

    Maybe its my fault, for not understanding what that u is. If u is a constant also (like ub), then your approach is right. You need to find the eigenfunction of z, F(Z). Once you do that, use the Boundary conditions to find omega^2. Once you do this, you can solve for G(t).

    If its not constant, then you need to define a new function v(z,t)= u(z-t)- h(z), such that v has a PDE of the same form as u. However, v will have homogeneous BC's.

    Hence, I personally need to have a better understanding of the first BC to help you. Good Luck!
     
    Last edited: Aug 21, 2010
  4. Aug 21, 2010 #3
    Hi aq1q,

    Pardon me for the messy representation of the equations,
    the bcs should read:

    1) [tex]\ z = 0[/tex]; [tex]\left[\frac{\partial u}{\partial z}\right]_{z=0}= k\left(u - u_{b}\right)[/tex]


    2) z = [tex]\infty[/tex]; [tex]\left[\frac{\partial u}{\partial z}\right]_{\infty} = 0[/tex]

    where k, and ub are constants
     
  5. Aug 21, 2010 #4
    what about u (on the left side of the parenthesis)? is that constant? Is it a function of t? Is that u(z,t)? what does that stand for?
     
  6. Aug 21, 2010 #5
    Hi aq1q,

    In the RHS of bc 1, u is not a constant but rather the dependant variable(the u in the derivative). so I guess your second suggestion will be the option for me, but could u pls tell me more abt this step:

    'v(z,t)= u(z-t)- h(z)'

    do you mean u(z,t) as in the orginal problem?
     
  7. Aug 21, 2010 #6
    So which variable is it dependent of? t?
     
  8. Aug 21, 2010 #7
    Hi aq1q,

    I apologised, the first condition is actually convective boundary.

    so the u on the left is actually the same u in the derivative: u(z,t).
     
  9. Aug 22, 2010 #8
    ahh, no, no its not your fault. It's mine. I was a bit confused because I thought it was a form of neuman boundary.

    In that case, I'm lost. I'll try to figure it out. Sorry I couldn't be of more help. It doesn't make sense how Uz(0,t)= K(U(z,t)- Ub) on the boundary (excuse my capitalization of u). Anyhow, I doubt thats the case. Would you mind me asking if this is a problem from a textbook? I looked through my PDE textbook and couldn't find anything on convective boundary conditions.

    If it is u(0,t), then I can probably help
     
  10. Aug 22, 2010 #9
    Hi aq1q,


    the problem is getting me stirred up as well that I cant figure out this error, yes certainly you are right it should read u(0,t) because it is the temperature at z = 0 that we are dealing with at any point in time as we implement the bc. So any suggestions in that case?

    thanks.
     
  11. Aug 22, 2010 #10
    Hello aq1q,

    No the problem is not from a text book, I set it up myself. But I can asure is no home work as well :)

    thanks.
     
  12. Aug 22, 2010 #11
    ahh ok. So, i'll just tell you up front. I'm still a second year undergrad student. Obviously, I'm NOTHING compared to some of the people here. So please excuse me if I have bit of a difficulty with it. Nonetheless, this looks like an interesting problem! Let's attempt it together.

    Ok, going back to what I was saying earlier. Let u(0,t)= [tex]\phi[/tex](t). Hence, we have to define a h(z,t) such that v(z,t)= u(z,t) - h(z,t). This is important because we want V to have homogeneous boundary conditions; hence, when we solve V using separation of variables we won't get stuck on the boundary conditions.

    So lets try solving for this h(z,t). I'll try it also

    I think this approach will work. I could be wrong, of course. Also, we might need to make some assumptions about Ub (we will get to that later)
     
  13. Aug 22, 2010 #12
    well yes, i didn't think this was a homework. Normally for non-trivial homeworks you would be given a hint. You set this yourself?! Then you should be able to answer my questions about "u"? So, obviously the "u" can't be u(z,t) for the first boundary condition... you understand that right?
     
  14. Aug 22, 2010 #13
    hi aq1q,

    OK, that very encouraging, we do it together as you said. I am sure you are very capable to guide me through.
    thanks alot.
     
  15. Aug 22, 2010 #14
    Understand that when you make up your own problem (what is this for btw? for a research? or just for fun?) you have to check for uniqueness. We can't assume uniqueness for u, this would have to be proven (and I highly doubt its unique). Hence, h(z,t) is not unique either, and you might have infinite solutions (this is a possibility).

    Anyway, I got something like this for h(z,t).

    let u(0,t)= [tex]\phi(t)[/tex]

    h(z,t)= [tex] e^{-k(\phi(t)-u_b)z} [/tex]

    do you see why this works? (assume [tex]\phi(t) >= u_b [/tex]

    we set v(z,t)= u(z,t) - h(z,t). Find Vz(0,t) and Vz(inf, t). what do you get?
     
    Last edited: Aug 22, 2010
  16. Aug 22, 2010 #15
    Hi aq1q,

    yes to experment research possibilities, so I need to have some kind of solution for the heat equation (which is standard) , but imposing these bcs.

    But you are faster than I thought though, cos I you got the h(z,t) even b4 I figure out how to go about it.
     
  17. Aug 22, 2010 #16
    Ignore what i got for h(z,t), it doesn't work for the given PDE. Vt does not equal to (alpha)*Vzz. However, its still important to realize u(z,t) is probably not unique, and uniqueness is important.

    Hm its about 2:15 am in my time. I'm really tired and sleepy, but I have a feeling you will need to modify your PDE or your boundary conditions. Otherwise, it might not be solvable. I'll promise to help tomorrow, but I really need to sleep now.. I can't think straight haha. Good Luck!
     
    Last edited: Aug 22, 2010
  18. Aug 22, 2010 #17
    hi aq1q,

    My line of thinking is that I am to use the soln for u(z,t) which is for the case at hand wll be of the form:

    Aexp(wt) + Bexpt(-wt).

    where A and B are to be determined.

    so if we introduce h(z,t)

    V(z,t) = Aexpt(wt) + Bexpt(-wt) - expt(-k(phi -ub)z)


    right?
     
  19. Aug 22, 2010 #18
    Hi aq1q,
    Thank you very much, you have been absolutely amazing for your kind assistance.
    I guess you should go and rest, I shall meet you up tomorrow hopefully with some leeway.

    Thank you alot!
     
  20. Aug 22, 2010 #19
    hey daz71,

    Have you made any progress? So, obviously I'm getting stuck on the bounds you imposed. I'm guessing that you reached those bounds through experimentation, which is why I am not questioning it. Moreover, I don't have much experience with experimentation. How sure are you of the second boundary condition for z= infinity?

    Here's what I think: your PDE might need some modification. How sure are you that you are dealing with a homogeneous diffusion equation. It is a possibility that you have a source function. Example: [tex] u_t - \alpha u_{zz} = f(z,t) [/tex]. If you don't think this is the case, let me know.

    Sorry, I know I am not actually helping with these questions but your PDE and the boundary values must agree with each other.
     
    Last edited: Aug 22, 2010
  21. Aug 23, 2010 #20
    Hello aq1q,

    thank you for your reply, I am afraid I did not make any break through. Apart from a paper I have seen that have a similar problem formulation, the only diffrence is that the boundary condition they have at z = infinity is a Dritchlet condition i.e, u(infinity, t) = uc, where uc is a constant temperature. I did not see their general solution, but they reported an expression for the u(0,t) as:

    [tex] u(0,t) = u_{b} + (u0 - u_{b})exp(k^2\alpha t)erf(k\sqrt{\alpha t}[/tex])

    can you help figure ouit how they arrive at this? I have no clue at all!

    thanks in advance.
     
  22. Aug 23, 2010 #21
    It is very likely that they used wolfram alpha, Mathematica, or MATLAB to solve for the two ODE's: F(z) and G(t). You can let u(0,t) = h(t). Then once u(z,t) is solved for, its not hard to find something like that. But again, I would need to see what their PDE was, and the boundary conditions. I am starting to think that your the PDE :

    [tex]
    u_{tt} = \alpha u_{zz} [/tex] is NOT homogeneous... [tex] \Rightarrow
    \frac{\partial u}{\partial t} \neq \alpha\frac{\partial ^2 u}{\partial z^2} [/tex]
     
    Last edited: Aug 23, 2010
  23. Aug 24, 2010 #22
    hi aq1q,

    As I have said, their PDE is exactly like mine, their boundary condition at z = 0 is also same as mine, the only difference is their bc at z = infinity, in which they have:
    u(infnity, t) = uc (a constant), and I have du/dz = 0 instaed. But let me remark that they did not claim to have used separtion of variables to solve the equation in the paper, so we are free to think of other methods as well.

    thanks
     
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