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2nd order Boundary Value Problem.

  1. Mar 16, 2010 #1
    I want to solve:


    With boundary condition y(0)=y(a)=0.

    First part is very easy using constant coef. which give:

    [tex]y(x)=c_1 cosh(\frac{m\pi}{a}x) + c_2 sinh(\frac{m\pi}{a}x) [/tex]

    [tex]y(0)=0 \;\Rightarrow\; c_1=0 \;\Rightarrow\; y(x) = c_2 \; sinh(\frac{m\pi}{a} x) [/tex]

    [tex] y(a)=0 \;\Rightarrow\; y(x) = c_2 sinh(\frac{m\pi}{a}(a-x)) [/tex]

    Obviously the two boundary condition give two different answer and each solution only satisfy one boundary condition and fail on the other. The book just gave:

    [tex] y_1(x)=sinh(\frac{m\pi}{a}(a-x)) \;\; and \;\; y_2(x)=sinh(\frac{m\pi}{a}x)[/tex]

    My understanding is the general solution is a combination of the two solution that satisfy both boundary conditions. But in this case, each solution meet only one of the boundary condition and not the other. The general solution is the linear combination of the two solution which will not meet either one of the boundary condition!!!

    Please help me clarify this. Thanks
    Last edited: Mar 17, 2010
  2. jcsd
  3. Mar 16, 2010 #2


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    Are you sure that your original equation has the x^2 multiplying the y(x) term? If so, this is a non-linear equation and what you gave is not a solution.
  4. Mar 16, 2010 #3
    Sorry, copy wrong. I correct the post already. Thanks for your time.
    Last edited: Mar 16, 2010
  5. Mar 16, 2010 #4
    This equation has a Wronskian which is a constant, so you can solve it to within a scale factor for the second solution and then apply to initial conditions to get an answer.

    However, like phyzguy said, I don't know if I'd trust the book's exact answer. When I did the integral using y(x) = A e^(r x), I got that the general solution was y(x) had exponents to a factor of x^2, since we take the square root of both sides which still leaves an x left over that you need to bring up. Checking this integral with Wolfram Alpha gave some parabolic cylinder function that I'm not familiar with.
  6. Mar 16, 2010 #5
    Are you refering to the post earlier on? I typed it wrong, there is no x in the coef of y(x). The associate homogeneous solution is very simple. It is the general solution that I don't understand.

    Thanks for your time.
  7. Mar 16, 2010 #6
    This is actually part of solving Poisson equation using single series expansion instead of two series by lumping y into the coef. And solve with only x as shown:

    Given . [tex] \nabla^2u(x,y)=f(x,y) \;\;,\;\;0<x<a \;\;,\;\; 0<y<b \;\;[/tex]

    We start with [tex]\phi_m(x)= \;\;,\;\; \nabla^2\phi_m(x)=-\lambda_m\phi_m(x) \;\;,\;\;\phi_m(0) =\phi_m(a) = 0 \;\;,\;\; \lambda_m=(\frac{m\pi}{a})^2 [/tex]

    [tex]u(x,y)=\sum_{m=1}^{\infty} E_m(y) \; sin(\frac{m\pi x}{a}) [/tex]

    [tex]\Rightarrow \nabla^2u(x,y)= \sum_{m=1}^{\infty} [E''_m(y) \;-\; (\frac{m\pi}{a})^2 \; E_m(y)]\;\; sin(\frac{m\pi x}{a}) \;=\; f(x,y) [/tex]

    [tex] let\;\; b_m(y) \;=\; E''_m(y) \;-\; (\frac{m\pi}{a})^2E_m(y) \;\;,\;\;using\;\; fourier \;\;expansion\;\; b_m(y)=\frac{2}{a}\int_0^a f(x,y) sin(\frac{m\pi x}{a}) dx[/tex]

    This is the part I refer to in my original post, just substitude .[tex]y(x)=Em(y)[/tex]

    [tex]E''_m(y)- (\frac{m\pi}{a})^2E_m(y) = b_m(y)[/tex] (16)

    [tex] E_m(0)=0 \;\;,\;\; E_m(b)=0 [/tex]

    The following is what I have question:

    We want to solve for [tex]E_m(y)[/tex] by solving the associate homogeneous equation:

    [tex]E''_m(y)- (\frac{m\pi}{a})^2E_m(y)=0[/tex]

    The book then gave:

    [tex] h_1(y)=sinh((\frac{m\pi}{a})(b-y)) \;\;,\;\; h_2(y)=sinh((\frac{m\pi}{a})y) [/tex]

    [tex]W(h_1,h_2)= (\frac{m\pi}{a}) sinh( (\frac{m\pi}{a}) b)[/tex]

    Where W is the Wronskian.

    I just don't follow the last two steps. I copy exactly from the book.
    Last edited: Mar 16, 2010
  8. Mar 16, 2010 #7


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    [STRIKE]Moderator's note: thread moved from "Differential Equations"[/STRIKE]

    EDIT: The OP has explained that this is not a problem "to be solved by the reader/student", but rather is an example problem worked out in a book where the author appears to be in error. For that reason, this thread has been moved back to "Differential Equations".
    Last edited: Mar 19, 2010
  9. Mar 17, 2010 #8
    Anyone please. This is not a homework. It is actually derivation shown in the Partial Differential Equation by Asmar on Poisson's equation. It does not make sense to me.
  10. Mar 18, 2010 #9


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    This has no solution for the given boundary conditions, other than the trivial y=0.

    You can't just put in the "a-x" here, since then you are no longer expressing in terms of your original function sinh(mπx/a)

    At any rate, once you have c1=0, note that sinh only crosses the x-axis once. Hence the boundary conditions are not satisfied.

    I find it easier to think in terms of the functions ebx and e-bx, where b=mπx/a. But again you would find that no combination of them will solve the given boundary conditions. (Other than y=0).
  11. Mar 19, 2010 #10
    THanks so much for your reply. That make me feel so much better. I know something is very wrong but I don't know what!!!

    As you can see on the second long post, I copy out the exact equation in the book and I think [tex]h_1 \;\;\;&\;\;\; h_2[/tex]. it is not correct

    I post the second part of the book's derivation of the particular solution of the Poisson problem in the Differential equation forum. I think it is wrong in how the book come up with the particular solution. I work out the solution in the standard way of variation of parameters and I hope someone can give me an opinion.

    It is very hard when you study a very well established book and you cannot verify the equation no matter how hard I tried. I am stuck on this page for like 4 days checking over and over and over!!!!!

    Please help.

    Many thanks

  12. Mar 31, 2010 #11
    I talked to my former ODE instructor who is a PHD for what ever it's worth. I showed him my work and he said it is ok using change of variable to get the one with (a-y)!!!

    But I asked him still the general solution [tex]y_c=c_1 y_1 + c_2 y_2 [/tex] cannot meet both the boundary condition. In fact [tex]y_c[/tex] cannot not meet any of the BC because it consists of both and one will fail in one condition and it is never equal to zero.

    Now I am totally confused!!!
  13. Apr 1, 2010 #12


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    Well, think about it. You have seen for yourself that when c_2=0 you have the function
    y = c_1 sinh(m pi x/a)​
    which is zero at x=0, thus satisfying one of the boundary conditions.
  14. Apr 1, 2010 #13
    Yes it fail the other condition where x=b. As a result, the solution has to be modify by change of variable from y to s where s=b-y.

    But after the change of variable, the new equation will not meet E_m(0)=0.

    So you end up with two solution, one can only fullfill one BC condition and none meet both BC.

    Unless I missed something totally, I still think there is no solution to this DE with the given boundary condition. Can you tell me I am correct?

    What is the difference between

    a) Solution of assoc. homogeneous DE.

    b) Solution of assoc. homogeneous DE with boundary condition?

    This is actually part of the derivation of 2D Poisson's problem using single series. I split into two question because I notice if the question is too long, nobody answer!!! Here is the remaining of it that I copied from the other thread since nobody answer that one.

    Conventional solution of [tex] \nabla^2u(x,y)=f(x,y)[/tex] involve solution [tex]u(x,y)= \sum_{n=1}^{\infty} \sum_{m=1}^{\infty}E_{mn} sin(\frac{m\pi}{a}x) sin(\frac{n\pi}{a}y)[/tex]

    This is a two series solution which is tedious to solve.

    The book PDE by Asmar suggested a method of solving Poisson problem with one series by lumping function of y into the coefficient [tex]E_{mn}[/tex] by using:

    [tex]u(x,y)= \sum_{m=1}^{\infty} E_m(y)sin(\frac{m\pi}{a}x)[/tex] .AND

    The book gave the final equation of:

    [tex]E_m(y) = (\frac{-1}{\frac{m\pi}{a} sinh(\frac{m\pi}{a}b)})[ h_1(y)\int_0^y h_2(s)b_m(s)ds \;\;+\;\; h_2(y)\int_y^b h_1(s)b_m(s)ds][/tex] (1)

    where the associate homogeneous solution is:

    [tex]h_1(y) = sinh(\frac{m\pi}{a}(b-y)) \;\;and\;\; h_2(y) = sinh(\frac{m\pi}{a}y) [/tex]

    This book claimed this is by using variation of parameters using h1 and h2 as y1 and y2 obtain from solving the associate homogeneous equation.

    I use the standard variation of parameter and cannot get the same answer. Can someone point me to a web site to verify the book? I hate to say the book is wrong but I did triple verified and fail. I have not manage to find anything on this from 4 other text book nor on the web to even talk about single series solution.

    Is it really important to use single series rather than two series because I have not problem doing in the convensional way using two series, it is very easy to understand. It is the book trying to be simple and jump steps that I don't agree with their formula all all.

    I think [tex]b_m(s)[/tex] in (1) should not be integrated like this in the definite integral. Because it really a function of y, not a function of (b-y).

    I think the solution using variation of parameters should be:

    [tex]E_m(y) = (\frac{-1}{\frac{m\pi}{a} sinh(\frac{m\pi}{a}b)})[ h_1(y)\int h_2(y)b_m(y)ds \;\;+\;\; h_2(y)\int h_1(y)b_m(y)dy][/tex]

    [tex]\Rightarrow E_m(y) = (\frac{-1}{\frac{m\pi}{a} sinh(\frac{m\pi}{a}b)})[ sinh(\frac{m\pi}{a}(b-y)) \int sinh(\frac{m\pi}{a}y) b_m(y)dy \;\;+\;\;sinh(\frac{m\pi}{a}y) \int sinh(\frac{m\pi}{a}(b-y)) b_m(y)dy][/tex]
    Last edited: Apr 2, 2010
  15. Apr 3, 2010 #14


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    Yes, that is correct, ignoring the trivial solution y=0. (I thought we had cleared this up on March 19).

    I'm unable to help with your other questions, perhaps somebody else knows...
  16. Apr 4, 2010 #15
    Thanks, I assumed that there is no non trivial solution until my ex-professor said otherwise. Two days ago, it just hit me that this has been proven in the Strum Liouville problem that one case where the solution is sinh and cosh and result in only trivial solution with the given boundary condition. I since email the page on the text book to my instructor and he since change his mind and agree with me. So this part is completely solved. I just like to see two books backing me up.

    Now, the next question is, if we don't have solution from the associate homogeneous DE, where do we get h1 and h2 for the variation of parameter in order to find the particular solution?

    Anyone can tell me this?
  17. Apr 5, 2010 #16
    After communicating with my professor again, I finally understand. For people that read this post, this is how it is:

    The given non homogeneous DE does has solution of the associate homogeneous equation h_1 and h_2. But neither one will satisfy both boundary condition. BUt you can still use it in variation of parameter to find the particular solution.

    So the general solution has the form of y = c_1 y_1 + c_2 y_2 + y_p where y_1 and y_2 are the h_1 and h_2. The difference is in this case, c_2 and c_2 are all zero and y= y_p only.
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