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The doubt of Seperation of Variables

  1. Feb 26, 2012 #1
    The method of seperation of variables is used to solve the problem of partial diffrential equation. For example, when the partial differential equation is:

    [tex]
    \frac{\partial u}{\partial t}-\alpha\frac{\partial^{2}u}{\partial x^{2}}=0
    [/tex]

    We could suppose that [tex]u(x,t)[/tex] is a solution concerning both x and t. It could also be represented as the product of two functions:

    [tex]
    u(x,t) = X(x)T(t)
    [/tex]

    My question is that the method of seperation of variables is based on the assumption that our solution could be represented as the product of two functions, which are with respect to x and t repsectively. My question is that what if the solution u could not be represented as the product of two functions of x and t, like:

    [tex]
    u(x,t) = \frac{1}{xt+1}
    [/tex]

    We then could not use the method of seperation of variables.

    I know that the solution I provided above is NOT RIGHT. But is it possible that the solution is of the form of the solution I provided above? Which means that the solution could not be seperated that easy?
     
  2. jcsd
  3. Feb 26, 2012 #2

    kai_sikorski

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    Okay two things, when you first apply the method of separation of variables to a PDE, there's basically nothing a priori to tell you that it will work. In fact in practice when you're doing a real problem as opposed to something from a textbook separation of variables is almost guaranteed not to work because the equation won't separate. But I think due to uniqueness results for the PDEs your working with, you can basically employ any trick you want to find a solution even if it involves an unjustified assumption, and if in the end it leads to a solution, you know that this is the unique solution.

    The second thing is that you're not really strictly limited to functions of the type X(x)T(t). You're limited to SUMS of functions like that. And it turns out that this is in fact flexible enough to capture any solution.
     
    Last edited: Feb 26, 2012
  4. Feb 26, 2012 #3

    Matterwave

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    I remember my friend and I had a discussion about this once. We were lead to some spectral theory stuff, and in the end I sort of forgot our conclusions...
     
  5. Feb 27, 2012 #4
    Can I get the rough idea fron your reply that in fact, some solution to the partial differential equation could not be represented as X(x)T(t)? X(x)T(t) is one type of the solution?
     
  6. Feb 27, 2012 #5

    Pengwuino

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    It's highly unlikely you can find a solution to a PDE as [itex] \Psi(x,t) = X(x)T(t)[/itex], but it turns out you can find a solution to many (maybe even all? I'm not confident in saying all) PDEs as a series solution given by

    [itex]\Psi(x,t) = X_1(x)T_1(t) + X_2(x)T_2(t) + X_3(x)T_3(t) + ...[/itex]

    where each X(x) can be a unique function of x and similarly with T(t)
     
  7. Feb 27, 2012 #6

    kai_sikorski

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    Lets say we're thinking about the above problem (heat equation) on some fixed domain with some fixed boundary conditions. So any function (maybe assuming some technical regularity conditions) f(x) can be used as the initial conditions for our problem and supposedly should correspond to some solution u(x,t), such that u satisfies the pde and u(x,0)=f(x). So do we know how to find the solution for any f?

    Well if the f(x) just happens to be one of the Xλ we found while working through the separation of variables then obviously we're done. u(x,t) = Xλ (x)Tλ(t). Similarly if f(x) is some linear combination of such Xs then the situation is again pretty simple. Say
    f(x) = α Xλ1(x) + β Xλ2(x)
    then
    u(x,t) =α Xλ1(x)Tλ1(t) + β Xλ2(x) Tλ2(t)

    But the genius of Fourier (if i got the history wrong someone please correct me) was figuring out that you could actually express ANY f(x) as a linear combination of Xs, as long as you're willing to use infinitely many. So since there is a 1-to-1 mapping between solutions and fs (I'm sure you need to qualify this statement somehow with regularity conditions of f to make it actually true) and since for any f we can get a solution with separation of variables, it must be that we can express any solution u(x,t) using a linear combination of the X T products.
     
  8. Feb 27, 2012 #7

    Matterwave

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    One should note that the above prescriptions work for linear systems. Non-linear PDE's will not have this nice property where you can add solutions to form new solutions.
     
  9. Feb 27, 2012 #8

    kai_sikorski

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    Yes that's true. I was just speaking to the problem the OP mentioned, which is linear.
     
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