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Heat/Diffusion Equation on the whole real line

  1. Mar 1, 2008 #1
    1. The problem statement, all variables and given/known data

    boundary conditions:

    u(x,0) = exp[-((x-14)^2)/4

    2. Relevant equations

    u_t = u_xx

    x [tex]\in[/tex] R


    3. The attempt at a solution

    the problem says there is a way for this partial differential equation to be solved without computing any integrals. i know the general solution for the whole line is:

    u(x,t) = 1 / sqrt(4kt) [tex]\int[/tex] exp[-((x-y)^2)/4kt] * PHI(y) dy


    where PHI(y) is the boundary conditions.

    is there some kind of substitution or trick to be used here to not have to compute any integral?

    Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 2, 2008 #2
    I'm under the impression that you are mixing a few things on the notation. You use k in the resulting integral, but that is nowhere defined. I know that sometimes the partial differential equation is written as:

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

    or:

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

    In your case a² or k is equal to 1. However you still have it in the integral. There is also something wrong with a pi. I assume you mean that the solution has to be:

    [tex]u(x,t)=\frac{1}{2a\sqrt{\pi t}} \cdot \int_{-\infty}^{\infty} f(y) e^{-\frac{(x-y)^2}{4a^2t}} dy[/tex]

    or in your case:

    [tex]u(x,t)=\frac{1}{\sqrt{4 \pi t}} \cdot \int_{-\infty}^{\infty} f(y) e^{-\frac{(x-y)^2}{4t}} dy[/tex]

    Now as for the question, You can't do without the integral. However you can avoid having to calculate it explicitly by changing the argument of the exponential. This means that you need to write it as:

    [tex]u(x,t)=\frac{1}{\sqrt{4 \pi t}} \cdot \int_{-\infty}^{\infty} e^{-\frac{(y-14)^2}{4}} e^{-\frac{(x-y)^2}{4t}} dy[/tex]

    And now transform the argument of the exponential to something like:

    [tex]A+B\cdot (y-C)^2[/tex]

    The part with A can be put outside the integral and you need to use the following substitution:

    [tex]\sqrt{B}(y-C)=p[/tex]

    Giving you a final integral as:

    [tex]\int_{-\infty}^{\infty}e^{-p^2}dp[/tex]

    Which is equal to [tex]\sqrt{\pi}[/tex]. So you do not need to calculate it, just rewriting it in another form.
     
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