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Q: pde heat eqn u(x,t) history effect \int^t_0 d^2u/dx^2

  1. May 26, 2005 #1
    How do you solve this type of PDE problem:

    [tex]\int^t_0 e^{-(t-\tau)}\frac{d^2u}{dx^2} d\tau - \frac{du}{dt} = 0[/tex]
    where u(x,0) = sin x

    Any links or info on this will be appreciated.

    Last edited by a moderator: Jul 2, 2005
  2. jcsd
  3. Jul 1, 2005 #2


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    could you pleeeeease type the equation in latex format, i'm to lazy to debug this :)) i think i could solve it, i had an exam containig such PDEs a few weeks ago (heat, oscillations etc, wave equations, etc.)...
  4. Jul 2, 2005 #3


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    That's looks like an "integro-differential" equation but it isn't really!. First thing I would recommend is taking that [tex]\frac{d^2u}{dx^2}[/tex] and [tex]e^t[/tex] outside the integral since those don't depend on [tex]\tau[/tex]. The equation becomes [tex]\{\int_0^te^{-\tau}d\tau\}e^t\frac{d^2u}{dx^2}- \frac{du}{dt}= 0[/tex]. Now, just go ahead and do the integration: [tex]\int_0^t e^{-\tau}d\tau= 1- e^{-t}[/tex] so that you have a regular differential equation (with variable coefficients).
    Last edited by a moderator: Jul 2, 2005
  5. Jul 2, 2005 #4
    Then I guess you simply begin with your ansatz that

    [tex]u(x,t) = X(x) T(t)[/tex] and plug that into your PDE and begin the method of seperation of variables.
  6. Jul 2, 2005 #5


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    Might I ask how does one interpret a heat equation with history? Really, just a rod, I'm not particular. No offense Chavo, but your equation above is a bit confussing for me. For one, it should have partials but I suspect you're not use to LaTex yet and just didn't know how to include them (you can click on my equation below to see the syntax). Also in your equation, the dependent variable, u(x,t), is NOT being integrated over a history as written unless I'm missing something. Really, I think one would look something like:

    [tex]\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}+\int_0^t \phi(t-\tau)u(x,\tau)d\tau[/tex]

    But I could be wrong.
  7. Jul 3, 2005 #6


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    Yes, that's why his equation is not really an "integro-differential" equation.

    I wondered if he had copied the equation wrong.
    Last edited by a moderator: Jul 3, 2005
  8. Jul 3, 2005 #7


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    Chavo, get it right then cus' I want to see how history affects the heat equation. You know history has a lot to do with lots of things in nature. You know metal fatigue. Genetics is my favorite and is embodied in Volterra's beautiful equation for a single species:

    [tex]\frac{dy}{dt}=ay+by^2+\int_0^t K(t,x)y(s)ds[/tex]

    See! The rate that population y changes with respect to time is dependent no only on a function of the current size of the population but also on the "history" of what the population did during the time period from t=0 to the current time (adjusted by some "massaging" factor K(t,x)). Neurons too right. :smile: So, what's the equilivant deal with the heat equation?
  9. Jul 3, 2005 #8
    I'm not so sure you can pull out the u from the equation.
    Maybe I'm overcomplicating things but lacking further info
    I read this equation as

    [tex] \int_0^t e^{-(t-\tau)} \frac{\partial^2 u(x, \tau)}{\partial x^2} d \tau
    - \frac{\partial u(x, t)}{\partial t} = 0 [/tex]

    which you could simplify to
    [tex] e^{-t} \frac{d^2}{dx^2} \int_0^t e^{\tau}u(x, \tau) d \tau -
    \frac{\partial u(x, t)}{\partial t} = 0 [/tex]

    then of course i get horribly stuck.

    for what it's worth....
  10. Jul 3, 2005 #9


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    Dude, you're not over-complicating things. I mis-interpreted the equation and perhaps should have realized that the partial was being integrated with respect to [itex]\tau[/itex]. At least we're clear with the equation now:

    [tex]\text{DE:}\quad \int_0^t e^{-(t-\tau)} \frac{\partial^2 u(x, \tau)}{\partial x^2} d \tau
    - \frac{\partial u(x, t)}{\partial t} = 0 [/tex]

    [tex]\text{BC:}\quad u(0,t)=0\qquad u(\pi,t)=0[/tex]

    [tex]\text{IC:}\quad u(x,0)=Sin(x)[/tex]

    Is this realistic?

    Dang it. I'd really like to see how this is solved. I bet a doller there is someone in here, 100,000 members I believe, that can solve this. Can someone help please.
    Last edited: Jul 3, 2005
  11. Jul 3, 2005 #10


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    Ok, I believe I've come up with an interpretation of this IDE. First, consider the normal heat equation as a basis:

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

    This one is easy to interpret: it's saying that the slope in the t-direction is dependent on the curvature in the x-direction. That is, the faster the derivative is decreasing in the x-direction, the more negative the slope will be in the t-direction. That's why at the top of the Sin(x) for a rod for example, the slope in the t-direction is decreasing fastest. The first plot exhibits this. Note at the top of the sin wave, the slope in the t-direction is steepest since the derivative in the x-direction is changing quickest there.

    Now for the IDE:

    [tex]\frac{\partial u}{\partial t}=\int_0^t e^{-(t-\tau)}\frac{\partial^2 u}{\partial x^2}d\tau[/tex]

    Now, first note the kernel, [itex]e^{-(t-\tau)}[/itex] which "attunates" the historical effect exponentially. That is, the further in the past, the smaller the effect. For example, suppose we're interested in knowing what's happening at t=6. At [itex]\tau=0[/tex], six seconds ago, this factor will be [itex]e^{-6}[/tex], and so the curvature back then will contribute little to the overall sum. Compare that to [itex]\tau=6[/tex] (the present time) at which case the exponential effect will be one and the curvature at the most recent time will have maximum impact on the sum.

    Now note the partial being integrated. What we're doing is summing up ALL the curvatures since t=0 to the present time (attunated by the exponential) and then saying, that the slope at the point (x,6) in the t-direction will be equal to all these sums. You know, I bet we could come up with a numerical approach to this problem based on this interpretation.

    Qbert, Chavo, you guys good at programming or what?

    Attached Files:

    Last edited: Jul 3, 2005
  12. Jul 3, 2005 #11
    How's about:
    [tex] \int_0^t e^{-(t-\tau)} u_{xx}(x, \tau) d \tau = u_t(x, t) [/tex]
    then assuming sufficient continuity to play fast and loose
    with order of differentiation:

    [tex] e^{-t} f(x, t) = u_t(x, t) [/tex]
    [tex] f(x, t) = \frac{d^2}{dx^2} \int_0^t e^{\tau} u(x, \tau) d \tau [/tex]

    Take the derivative with respect to t giving:

    [tex] - e^{-t}f(x, t) + e^{-t} \frac{\partial f}{\partial t} = u_{tt}(x, t) [/tex]

    a quick check shows
    [tex] \frac{\partial f}{\partial t} = \frac{d}{dt} \frac{d^2}{dx^2}
    \int_0^t e^{\tau} u(x, \tau) d \tau =
    \frac{d^2}{dx^2} e^t u(x, t) =
    e^t u_{xx}(x, t) [/tex]

    and substituting for [itex] e^{-t} f(x, t) [/itex] the equation becomes:

    [tex] -u_t + e^{-t} ( e^t u_{xx} ) = u_{tt} [/tex]
    [tex] u_{tt} + u_{t} = u_{xx} [/tex]

    which doesn't look all that hard to do.
    Now I just wonder if the problem is well posed,
    but I'll leave that to some one else.
  13. Jul 3, 2005 #12


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    Jesus Qbert, you go dude! I've gone over it as best I can and, assuming continuous partial derivatives of f(x,t) to justify switching the order of differentiation, I can see nothing wrong (of course their could be, I just don't see it). That's a very elegant approach I think. :smile:

    But now:

    [tex] u_{tt} + u_{t} = u_{xx} [/tex]

    requires an extra initial condition does it not? That is:


    However, from the initial IDE:

    [tex]u_t(x,0)= f(x,0)=0 [/tex]

    Would this be correct?
    Last edited: Jul 3, 2005
  14. Jul 3, 2005 #13


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    Using Qbert's analysis, I solved the following IBVP:

    [tex]\text{DE:}\quad u_{tt}+u_t=u_{xx}[/tex]

    [tex]\text{BC:}\quad u(0,t)=0;\qquad u(\pi,t)=0[/tex]

    [tex]\text{IC:}\quad u(x,0)=Sin(x);\qquad u_t(x,0)=0[/tex]

    I got:




    The first plot shows the standard heat equation. The second plot shows u(x,t) as determined above. Note the dimple as u(x,t) goes below the x-y plane. I checked the results by back-substitution into the PDE as well as back-substitution into the IDE. That is I calculated:

    [tex]u_t(x,t)-\int_0^t e^{-(t-\tau)}u_{xx}(x,\tau)d\tau[/tex]


    The third plot shows this calculation. The results are no larger than [itex]10^{-14}[/itex]

    Really, I'd need to work on in more to obtain confidence that everything is Ok. So if anyone gets something else or sees something wrong, please say so. Chavo, I'd expect you're doing all this work too right? Just a suggestion.

    Attached Files:

  15. Jul 4, 2005 #14


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    Guys, I'm not done with this yet ok? Suppose I want a bigger dimple in the plot? Would this equation do it?

    [tex]u_t=\int_0^t e^{-a(t-\tau)}u_{xx}(x,\tau)d\tau \quad\text{for}\quad a<1[/tex]

    Also, suppose I'm not interested in "ALL" the history but rather I only want to consider "recent" history, say the last 2 seconds. Would this do it?

    [tex]u_t=\int_{t-2}^t e^{-a(t-\tau)}u_{xx}(x,\tau)d\tau [/tex]

    Or even worst, suppose my history impact is a function of t!

    [tex]u_t=\int_{t-f(t)}^t e^{-a(t-\tau)}u_{xx}(x,\tau)d\tau [/tex]

    We'd never get anywhere if I was the teacher. :yuck:
  16. Jul 4, 2005 #15
    well it looks like you solved the initial problem.
    a maple check shows that your solution satisfies
    the IDE. (and boundary conditions.)

    oh. I noticed you don't have to do any changing of
    the x and t derivatives to make it work. just differentiate
    the IDE for t. (Sometimes I overcomplimicate things.)

    the more interesting stuff are your changes.

    [tex] u_t = \int_0^t e^{-a(t-\tau)}u_{xx}(x, \tau) d \tau [/tex]
    is [itex] u_{tt} + a u_t = u_{xx} [/itex] with [itex] u_t(x, 0)=0 [/itex]
    which is easy.

    The next one is enough to scare me.
    [tex] u_t = \int_{t-2}^t e^{-a(t-\tau)}u_{xx}(x, \tau) d \tau [/tex]
    [tex] u_{tt}(x, t) + a u_t(x, t) = u_{xx}(x, t) - e^{-2a}u_{xx}(x, t-2) [/tex]
    I'm not sure what the inital condition on [itex] u_t [/itex] is.

    How does one solve an equation of this kind?
    Or even simpler, how do you solve the ode [itex] u_t(t) = u(t-k) [/itex]?

    The same procedure works for the third, but
    gives an equally confusing de.

    good questions, tho.
  17. Jul 4, 2005 #16


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    I would classify that one as a delay-PDE. One would have to supply initial conditions for the interval (0-2,0) and if I had to solve it, I'd resort to numerical means.

    Also I've come to the profound conclusion that the equation:

    [tex]u_t(x,t)=\int_0^t e^{-(t-\tau)}u_{xx}(x,\tau)d\tau[/tex]

    is not a heat equation with history but rather an integral version of a wave equation with damping factor as it is converted to such by your technique. The curve below is a cross-section of both a standard heat equation and the solution obtained for this problem at x=[itex]\pi/2[/itex]

    Note how the wave dampens as would be expected for such.

    Attached Files:

    Last edited: Jul 4, 2005
  18. Jul 4, 2005 #17


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    Wait Qbert, if indeed the equation originally posted is a wave equation in disguise and the interpretation I gave above from the perspective of a heat equation with history is valid, then can we conclude that some wave equations (with dampening anyway) embody within themselves, a history of their behavior since their origin?

    I propose the following question:

    Are all wave equations with dampening, really heat equations with history?
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