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Steady State Solution of Forward Kolmogorov Equation

  1. Feb 11, 2013 #1
    Essentially, this looks like a differential equation problem but being rusty on differential equations I am a little stuck.

    1. The problem statement, all variables and given/known data
    Consider the following SDE
    [tex]
    d\sigma = a(\sigma,t)dt + b(\sigma,t)dW
    [/tex]
    The Forward Equation (FKE) is given by
    [tex]
    \frac{\partial p}{\partial t} = \frac{1}{2}\frac{\partial^2}{\partial\sigma^2}(b^2p) - \frac{\partial}{\partial\sigma}(ap)
    [/tex]

    The steady state solution is given by setting ##\frac{\partial p}{\partial t}=0##. Considering the boundary conditions that as ##\sigma\rightarrow\infty##, ##p\rightarrow 0## and ##\frac{\partial p}{\partial \sigma}\rightarrow 0##, show that the steady state solution is given by

    [tex]
    p(\sigma) = \frac{A}{b^2}e^{\int^\sigma\frac{2a}{b^2}d\sigma '}
    [/tex]


    2. Relevant equations



    3. The attempt at a solution
    So, as per the question, the steady state solution is:
    [tex]
    \frac{1}{2}\frac{d^2}{d\sigma^2}(b^2p) - \frac{d}{d\sigma}(ap) = 0
    [/tex]

    Integrating to convert to a first order differential equation?:

    [tex]
    \frac{1}{2}\frac{d}{d\sigma}(b^2p) - ap = A
    [/tex]

    Given the boundary conditions in the question, ##A = 0##??

    [tex]
    \frac{1}{2}\frac{d}{d\sigma}(b^2p) = ap
    [/tex]

    Re-arranging we have a variable separable D.E

    [tex]
    \frac{1}{p} dp = \frac{2a}{b^2} d\sigma
    [/tex]

    Integrating:

    [tex]
    \log p = \int \frac{2a}{b^2} d\sigma + D
    [/tex]

    [tex]
    p = e^{\int \frac{2a}{b^2} d\sigma + D}
    [/tex]

    ....and this is where I'm stuck. It looks a *little* like the expected solution but I don't know where the extra ##\frac{A}{b^2}## coefficient comes from and why there is a ##d\sigma '## in the expected solution.

    Any help appreciated. Please bear in mind that my differential equation and integration skills are very poor so any hint as to where I've gone wrong would be great, thanks!
     
  2. jcsd
  3. Feb 12, 2013 #2
    Can anybody help? Still stuck on this one. It's really a differential equation problem....
     
  4. Feb 12, 2013 #3

    Ray Vickson

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    You should avoid asking the same symbol to have two different meanings in the same expression, so writing ##p(\sigma)## in terms of an integral having ##d\sigma## in it is a no-no. Instead, you need to distinguish between the limit of integration, ##\sigma##, and the dummy integration variable. You could, for example, write
    [tex] \int^\sigma\frac{2a(x)}{b^2(x)}\, dx, [/tex] although the posted solution chooses the dummy variable to be named ##\sigma'##. That's OK: ##\sigma'## is different from ##\sigma##.

    Anyway, the problem just asks you to verify a formula, not to figure out where the formula comes from; that's what the "show that" instruction means. 'Verification' is a whole lot easier than 'discovery'.
     
  5. Feb 13, 2013 #4
    Hi Ray, thanks for the help again.

    The dummy variable point makes sense.

    On the topic of "verification" vs. "discovery" are you saying that the problem isn't to solve the differential equation? (which I have now done, more or less - see below). If not, what exactly is it that I'm supposed to do here then?!

    Even if it's not what the question is asking, somebody else pointed out that there was a mistake in my working:

    ##
    \begin{align} \frac{1}{2} \frac{d}{d\sigma} (b^2 \cdot p) &= a \cdot p \\ \Rightarrow b \cdot \frac{d b}{d \sigma} \cdot p + \frac{1}{2} b^2 \cdot \frac{dp}{d \sigma} &= a \cdot p \\ \Rightarrow \frac{1}{p} \, dp &= 2 \frac{ \left(a- b \cdot \frac{db}{d\sigma}\right) }{b^2} \, d\sigma = \frac{2a}{b^2} \, d\sigma - 2 \frac{1}{b} \cdot \frac{db}{d\sigma} \, d\sigma \end{align}
    ##

    And from there, integrating and taking exponential arrives at the correct equation from the question.
     
  6. Feb 13, 2013 #5

    Ray Vickson

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    The problem is absolutely clear: it says "show that the solution is ..." How do you show that some given formula satisfies a differential equation? You plug it into the DE to see if it works!
     
  7. Feb 26, 2013 #6
    OK, I'm back on this one. What you're suggesting seems to be more of a challenge than the derivation I computed. I'm having a complete mental block.

    Can somebody get me started?

    ##
    \frac{∂}{∂\sigma}(ap) = \frac{∂}{∂\sigma} (a \frac{A}{b^2}e^{\int\frac{2a}{b^2}d\sigma})
    ##

    But the question says that p tends to 0. So don't both terms disappear? I just don't get it. Need a pointer please.
     
    Last edited: Feb 26, 2013
  8. Feb 26, 2013 #7

    haruspex

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    No. You don't know that ##\frac d{d\sigma}\left(b^2p\right) → 0##.
    Given the form of the desired answer, I would substitute ##p = \frac q{b^2}##
     
  9. Feb 26, 2013 #8
    Hi, apparently that whole line of attack is not what the question is asking. I'm supposed to show that the equation given, satisfies the PDE.
     
  10. Feb 26, 2013 #9

    haruspex

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    Doesn't stop you doing a simplifying substitution.
     
  11. Feb 27, 2013 #10
    OK, but how does that help me show that the given equation for p satisfies the PDE? I'm honestly more lost now than I was at the beginning of the thread.
     
  12. Feb 27, 2013 #11
    So,

    $$
    \frac{\partial}{\partial\sigma} (ap) = a.\frac{\partial p}{\partial\sigma} + \frac{\partial a}{\partial\sigma}.p
    $$

    Given the boundary conditions in the question ##p → 0## and ##\frac{\partial p}{\partial\sigma} → 0##, does this mean that:

    $$
    \frac{\partial}{\partial\sigma} (ap) = a.0 + \frac{\partial a}{\partial\sigma}.0
    $$

    $$
    \frac{\partial}{\partial\sigma} (ap) = 0??
    $$

    Am I on the right track here? Do I have to do the same for the ##b^2## term?
     
    Last edited: Feb 27, 2013
  13. Feb 27, 2013 #12
    Doing the same for the second order term, I get:

    $$
    \frac{d^2}{d\sigma^2} (b^2.p) = 2b.p.\frac{d^2b}{d\sigma^2} + 4b.\frac{db}{d\sigma}.\frac{dp}{d\sigma} + 2p.(\frac{db}{d\sigma})^2 + b^2.\frac{d^2p}{d\sigma^2}
    $$

    Which doesn't seem to help me show that the equation for ##p## satisfies the differential equation.

    Using the boundary conditions ##p → 0## and ##\frac{dp}{d\sigma} → 0##, all of the terms dissappear but I'm left with:

    $$
    b^2.\frac{d^2p}{d\sigma^2}
    $$

    Where am I going wrong? Getting desperate now.
     
  14. Feb 27, 2013 #13

    haruspex

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    I posted earlier that you could not do that because you don't know that ##\frac{d}{d\sigma}(b^2p)→0##. I.e. although it expands to ##2bp\frac{d}{d\sigma}b+b^2\frac{d}{d\sigma}p##, you don't know that either of those terms tends to 0. However, you are going to have to assume that to solve the problem, so maybe you can justify it on some physical grounds constraining the magnitude of b.
    That last step doesn't look valid. It becomes more obvious with the substitution I proposed:
    ## \frac{dq}{d\sigma} = \frac{2aq}{b^2} ##
    ## \frac{dq}q = \frac{2a}{b^2} d\sigma##
     
  15. Feb 28, 2013 #14
    Hi haruspex,

    I really appreciate the time and effort you are putting in to help me but I think we're having trouble communicating.

    I *don't* need to find a solution to the differential equation - this was my mistake so you can ignore my first post and attempted solution which you seem to be focusing on.

    I need to show that the equation for p already given *satisfies* the differential equation. So I need to evaluate the differentials accordingly and plug it into the equation. This is the part I'm having trouble with.

    Looks like i'm about to fail this coursework, thanks anyway.
     
  16. Feb 28, 2013 #15

    Ray Vickson

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    To answer specifically where you went wrong, it was in going from
    [tex] \frac{1}{2}\frac{d}{d\sigma}(b^2p) = ap [/tex]
    to
    [tex]
    \frac{1}{p} dp = \frac{2a}{b^2} d\sigma
    [/tex]

    Using ##x## instead of ##\sigma## and setting ##u = b^2 p##, ##r = a/b^2## and ##D = \frac{d}{dx}## we have
    [tex] D^2 u = D(r u) \: \Longrightarrow \; D u = r u + c,[/tex]
    where ##c## is a constant of integration. One of the solutions (obtained by fixing c in some way) will be exactly what you are asked to show.

    However, there is a much more serious issue: the system may not have an equilibrium distribution at all; whether or not it has one depends on behavior of a(x) and b(x). For example, if a and b > 0 are constants there is no equilibrium distribution. The differential equation's written solution still exists, but cannot satisfy the boundary conditions on p(x) for large |x|. One can offer some sufficient conditions, such as requiring that
    [tex] R(x) = \int_0^x r(y) \, dy [/tex]
    be < 0 for large |x| and that either |R(x)| → ∞ as |x| → ∞ and b(x) remaining bounded, or that R(x) (> 0 or < 0) remain bounded and |b(x)| → ∞ as |x| → ∞, or some other conditions. It is difficult to give a set of general conditions on a(x) and b(x), but it is crucial to recognize that *some* conditions are absolutely vital.
     
  17. Feb 28, 2013 #16

    haruspex

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    The OP says
    To me, that says to deduce it from the set-up in the usual way. If it had said "show that a steady state solution is given by" then you could take the easier route of plugging it into the D.E. and see that it satisfies it. The second is never harder than the first.
    For steady state you had the ODE:
    ##\frac{d^2}{d \sigma^2}\left(b^2p\right) = 2 \frac{d}{d \sigma}\left(ap\right)##
    Substituting ##p = \frac{A}{b^2}e^{\int^{\sigma}\frac{2a}{b^2}d \sigma '}##:
    ##\frac{d}{d \sigma}\left(b^2p\right) = \frac{d}{d \sigma}Ae^{\int^{\sigma}\frac{2a}{b^2}d \sigma '} = Ae^{\int^{\sigma}\frac{2a}{b^2}d\sigma '} \frac{d}{d \sigma}\int^{\sigma}\frac{2a}{b^2}d \sigma ' = b^2p\frac{2a}{b^2} = 2ap##
    Differentiating both sides wrt sigma produces the ODE.
    It would remain to show that the boundary conditions are met. There you run into a snag working this way round. I don't believe it is possible to show those boundary conditions follow from the posited solution. All you can hope to do is show that there is no other solution given those conditions. At this point you are forced back into the deductive approach.
    But as Ray and I agree, there is a flaw in the problem statement. You need to know that ##bp\frac{d b}{d\sigma}## and ##b^2\frac{d p}{d\sigma}## tend to 0.
     
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