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Solution to differential equation

  1. Feb 5, 2015 #1
    1. The problem statement, all variables and given/known data
    This is actually a problem from my physics textbook, but I think it's mostly a mathematical problem, which is why I post it here:

    Show that the Langevin equation

    1: [tex]\frac{dv}{dt}=-\gamma v+\frac{1}{m} F'(t)[/tex]

    is solved by

    2: [tex]v-v_0e^{-\gamma\tau}=\sum_{k=0}^{N-1}e^{{-\gamma\tau}(N-k)}\frac{1}{m}\int_0^\tau F'(k\tau+s)ds,[/tex]



    3. The attempt at a solution

    First of all, I'm not sure how

    [tex]v-v_0e^{-\gamma\tau}[/tex] can be a solution to a differential equation where the unknown variable is [tex]v[/tex], so I put the term [tex]v_0e^{-\gamma\tau}[/tex] on the right-hand side of equation 2 instead.

    Next, I'm trying to take the derivative of equation 2 w.r.t. t:

    [tex]-\gamma V_0e^{-\gamma t}+\sum_{k=0}^{N-1}\Big(-\gamma \frac{N-k}{N}e^{-\gamma t\frac{N-k}{N}} \frac{1}{m}\int_0^\tau F'(k\tau+s)ds+e^{-\gamma t\frac{N-k}{N}}\frac{1}{m}F'(\frac{t}{N}(k+1))\Big),[/tex]


    [tex]F'(\frac{t}{N}(k+1))=\frac{d\int_0^\frac{t}{N} F'(k\tau+s)ds}{dt}.[/tex]

    My first question is if this derivative is correct so far. If so, my second question is if I could perhaps be given some clues as how to calculate the sum.
  2. jcsd
  3. Feb 5, 2015 #2

    Simon Bridge

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    ... that's not the solution.
    You have to rearrange that line to get the solution v(t)=... which you seem to have done later, so what's the problem?

    Use the chain rule and differentiate wrt ##\tau## ...less messy.
    ... you want to be able to show that the 1st derivative you just worked out is the same as the RHS of the DE. But it looks like you have cancelled out the integral for some reason ... notice that it is a definite integral, and not wrt ##\tau##?
  4. Feb 5, 2015 #3
    Just confusion on my part. It's my textbook that claims the solution to the differential equation can be written in that form.

    Ok, so differentiating w.r.t. [itex]\tau[/itex] the chain rule gives me, first of all, that

    [tex]\frac{d\int_{0}^{\tau}F'(k\tau+s)ds}{dt} =\frac{1}{N}(k+1)F'(\tau(k+1)).[/tex]

    Is this correct? I'm not very familiar with the chain rule, actually (I had to look it up).
  5. Feb 5, 2015 #4

    Simon Bridge

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    Puzzling, the chain rule is usually taught in year 1 calculus classes, right after the product and quotient rules.
    Are you self-taught? $$\frac{dv}{dt} = \frac{d\tau}{dt}\frac{dv}{d\tau}$$

    Changing variable in the DE by the chain rule:
    $$\frac{dv}{dt} = \gamma v + \frac{1}{m}F'(t) \implies N\frac{dv}{d\tau} = \gamma v + \frac{1}{m}F'(\tau): \tau=Nt$$

    The proposed solution:
    $$v-v_0e^{-\gamma\tau} = \Sigma \cdots$$... too lazy to write out the whole sum... but taking the derivative of both sides:
    N\frac{dv}{d \tau} + v_0\gamma e^{-\gamma \tau} &= \sum_{k=0}^{N-1} \left[
    e^{\gamma \tau(N-k)}\frac{d}{d \tau}\int_0^\tau F'(k\tau+s)\;\text{d}s \;\;
    -\gamma(N-k)e^{-\gamma\tau(N-k)}\int_0^\tau F'(k\tau+s)\;\text{d}s \right] \\

    & = N\gamma v + \frac{N}{m}F'(\tau) + v_0\gamma e^{-\gamma\tau}

    ... something like that. Check my work.
    So the summation (above) has to come out to the last line.
    1st you want to figure out if you can take the derivative inside the integration.
    i.e. can you say: $$\frac{d}{d\tau}\int_0^\tau F'(k\tau+s)\;\text{d}s = \int_0^\tau \left(\frac{d}{d\tau}F'(k\tau+s)\right)\;\text{d}s$$ ... if you can, you can use the chain rule on the integrand, which will go some way to simplifying the calculation.
    Can you simplify further by, say, cancelling terms?

    The way to get rid of the summation, usually, is either to show that it converges to the part that is not the sum, or see if you can make the sum here look like the RHS of the proposed solution.
    I don't see it being nice.

    The other approach is just to solve the E yourself and show that the solution you get it the same as theirs.
  6. Feb 11, 2015 #5
    I'm not self-taught and have, now that I think about it, indeed used the chain rule before. It is, however, definitely one of the rules of differentiation that I've had the least practice on.

    Ok, so I checked your work and get

    [tex]\frac{dv}{d\tau}=-N\gamma v_0 e^{-\gamma\tau N}-\gamma\sum_{k=0}^{N-1}\Big((N-k)e^{-\gamma\tau (N-k)}\frac{1}{m}\int_0^\tau F'(k\tau+s)ds\Big)+\sum_{k=0}^{N-1}\Big(e^{-\gamma\tau (N-k)}\frac{1}{m}\frac{d}{d\tau}\int_0^\tau F'(k\tau+s)ds\Big)\\
    =-N\gamma v+\frac{N}{m}F'(\tau N)[/tex]

    If I can somehow cancel the [itex]-k[/itex] in the first sum, I will at least have [itex]-N\gamma v[/itex]. I'm thinking that maybe the derivative w.r.t. [itex]\tau[/itex] of the integral in the second sum will give a term containing a factor [itex]k[/itex] times said integral, but I simply don't understand how to apply the chain rule to this integral. Wolphram Alpha suggests that

    [tex]\frac{d}{d\tau}\int_0^\tau F'(k\tau+s)ds\Big)=\int_0^\tau k\frac{d}{d\tau}F'(k\tau+s)ds+F'(\tau(k+1)),[/tex]

    but I don't know why and it doesn't quite give me what I want anyway (the derivative of [itex]F'[/itex] messes things up).
    Last edited: Feb 11, 2015
  7. Feb 11, 2015 #6


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