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Heat Equation - Trouble Finding a General Solution

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve:

    Ut=kUxx
    U(x,0)=e^3x


    2. Relevant equations
    The Heat Equation:
    48b739a375dd1fc340c8fe456f9e165c.png

    3. The attempt at a solution

    g(y) in the heat equation for this problem is e^3y. I'm having serious trouble solving this because my professor hasn't taught us the method, and it isn't in the book. I've considered trying a change of variables by taking z=x-y, but this has led me to nowhere. I am lost, and I am not just fishing for a result.

    I actually want to know how to solve this damn thing. There's a few pages worth of **** in my notebook, and now I need some guidance.

    Thank you for your time.
     
  2. jcsd
  3. Nov 19, 2011 #2

    fzero

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    The integral that you have to do is known as a Gaussian integral. A standard result, explained, for example, at http://en.wikipedia.org/wiki/Gaussian_integral is

    [itex] \int_{-\infty}^\infty \exp\left( - A u^2\right) du = \sqrt{\frac{\pi}{A}}.~~~(1)[/itex]

    In your case, you must integrate

    [itex] \int_{-\infty}^\infty \exp\left( - \frac{(x-y)^2}{4kt} + 3y \right) dy.[/itex]

    The technique needed is known as "completing the square," namely we attempt to write the quadratic expression in [itex]y[/itex] as a sum of a square plus a constant term:

    [itex] -A y^2 + B y + C = -a (y+b)^2 + c . ~~~(2)[/itex]

    Here, [itex]a,b,c[/itex] will possibly be functions of [itex]x,t,k[/itex], but do not depend on [itex]y[/itex], so the resulting integral can be done by substituting [itex] u = y +b [/itex] and then using the formula (1).

    You should probably start with equation (2) and determine [itex]a,b,c[/itex] in terms of [itex] A, B,C[/itex].
     
  4. Nov 21, 2011 #3
    I just wanted to let you know that I haven't given up on the problem and am still working at it.

    I'll be back with results whenever I solve it.
     
  5. Nov 21, 2011 #4
    Thank you so much for your help. I reached the final solution. I'll upload the results to show you as soon as I can.

    I have a final question as to what happened with a negative sign when I did a substitution, if you don't mind.
     
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