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Critiquing separation of variables method for PDE.

  1. Nov 17, 2013 #1
    "Critiquing" separation of variables method for PDE.

    I am currently taking a course in PDE's and it has been very "applied" and not so much theory based. I can say its been separate this separate that separate this separate that… Enough! We are always "separating variables" and it always "works". I get the feeling that much more is going on with the select few equations that we are solving with this method given that I can easily come up with a linear PDE where separation of variables falls.

    I have looked at a few books and at various information on the web, and it all seems vague and up straight with the process of this method.

    I'll base the rest of my post on a boundary value problem regarding the heat equation in it's simplest form. One dimensional, homogeneous boundary conditions, and an initial distribution f(x). The text in bold can be thought of as my comments/questions if someone was showing me how to solve this problem. Here we go:

    We let the solution in its most general form be denoted by u = u(x, t).

    The PDE:
    [tex]u_t=au_{xx}, a > 0[/tex]
    The boundary conditions:
    [tex]u(0,t) = u(l,t) = 0, l > 0[/tex]
    The initial condition:
    [tex]u(x,0) = f(x)[/tex]
    Assume f(x) is well defined in the region [0,l].

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

    Okay, I realize we are reducing our function space to those separable in the independent variables, but if we can arrive at a solution in the end with all conditions met and can show that this solution is unique, it must be the solution. Does this boundary value problem have a unique solution?

    Simplifying notation by letting X(x) = X, T(t) = T, it must then follow that:
    [tex]XT' = aX''T[/tex]

    Dividing this by aXT, we get:
    [tex]\frac{T'}{aT} = \frac{X''}{X}[/tex]

    Okay, but this equation is not equivalent to the previous. What about division by zero? We are not assuming X(x)T(t) ≠ 0 for all x, t in (0,l), (0, ∞) are we? Given we are not, do we just let any countable pairs of (x0, t0) where U(x0, t0) = 0 belong to the solution so as to keep the solution continuous (rather than having holes)?

    Given the left and right sides are independent, they must both equal the same constant. Therefore, in defining a constant λ, we have:
    [tex]\frac{T'}{aT} = \frac{X''}{X} = \lambda[/tex]
    Or:
    [tex]T' = a \lambda T \\
    X'' = \lambda X[/tex]
    Now, if u(0,t) = 0, X(0)T(t) = 0. So either X(0) = 0 or T(t) = 0. We don't want T(t) identically zero, so X(0) = 0. Likewise, if u(l,t) = 0, then X(l) = 0.

    The ODE in X with these boundary conditions has the following non-trivial solutions (eigenfunctions):
    [tex]X_n = \sin(\frac{n\pi x}{l})[/tex]
    with corresponding eigenvalues:
    [tex]\lambda = - \frac{n^2\pi^2}{l^2}, n = 1,2,...[/tex]
    These eigenfunctions form a complete orthogonal family.

    That seems awfully convenient. Why should they?

    The ODE in T then becomes a corresponding family of ODE's Tn:
    [tex]T_n' = -\frac{an^2\pi^2}{l^2}T[/tex]
    from which has the general solutions:
    [tex]T = c_n e^{\frac{an^2\pi^2}{l^2}t}[/tex]

    We can then construct a general Fourier series solution by:
    [tex]\sum_{n=1}^{\infty} c_n e^{\frac{an^2\pi^2}{l^2}t} \sin(\frac{n\pi x}{l}) [/tex]
    From the initial condition, the coefficients can be found by orthogonality:
    [tex]c_n = \frac{2}{l} \int_0^l f(x)\sin(\frac{n\pi x}{l}) dx[/tex]
    where this is for t = 0.

    So we have found a solution to the boundary value problem where u(x, 0) = f(x), at least in the L2 sense. That being said, ut(x, 0) is not necessarily f''(x) as the PDE would imply. What we did was find solutions to the PDE satisfying the boundary conditions individually and used the homogeneity/linearity of the problem to construct a general solution as the sum of these. Therefore, on a term by term basis, we know that (un)t = (un)xx. For many f(x) were there is L2 convergence, it is not so that g'(x) = f(x) where g(x) is the expansion of f(x). In fact, f(x) must have continuous periodic extension. So indeed, if u(0, t) = f(x), it is not necessary that ut(x,0) = uxx(x,0) = f''(x) because our solution satisfies the PDE on a term by term basis. So what is it about this method that allows it to actually work? It seems as if there are some stability properties to the PDE.


    Sorry for the long post. I have moved much further in my PDE course as far as the various problems I am dealing with, but it is these same questions that trouble me. Any answers or direction pointing is appreciated.
     
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  3. Nov 17, 2013 #2

    joshmccraney

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    to answer your uniqueness question, the question you pose is actually one of clay institute's seven most important. if you can prove uniqueness for non-linear pde's in a general context, congrats, because now youre a millionaire (literally, they'll give you a million dollars). to further up what im saying, consider the navier-stokes equations (as it seems you're already looking at the simplified ones).
     
  4. Nov 17, 2013 #3
    So as it stands, attempt to find a solution; if it tends to describe a physical problem well, take it for what it is and disregard other possible solutions? Like the method here, is it used because it is a practical way of finding solutions that seem to model heat flow well?
     
  5. Nov 18, 2013 #4

    joshmccraney

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    for many of these problems that would seem the case. separation is a powerful idea, but if you can't get rid of those pesky partials, it doesn't seem anyone to date can generally prove existence or uniqueness.
     
  6. Nov 18, 2013 #5
    If you're saying that in general solutions would form a complete basis, then that's obviously not the case. However, in many particular situations it does turn out to be true.

    Recall and/or check out Sturm-Liouville theory of ODEs: A well-known result is that eigenfunctions of an S-L operator in a well-posed S-L problem form a complete orthogonal set of functions in L2(a,b) restricted to the functions satisfying the given boundary conditions. As it happens, the PDEs you're presented with often reduce to such problems after the separation of variables.
     
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