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Math Help - Tricky Partial Differential problem

  1. #16
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    I can - I got the same.
    Thanks!
    I should get it now. Cheers
    Bugatti79
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  2. #17
    Super Member
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    Another thing you could have done, to avoid so many calculations was to note that v(x,t)= \frac{1}{2\sqrt{\pi kt}} e^{\frac{-x^2}{4kt}} solves the heat equation v_t=kv_{xx} on \mathbb{R} \times (0,\infty ) and noting that u(x,t)=u^t \ast f where the u^t means simply that the convolution is with respect to the first variable. Now properties of this operation give you the result based on that for v. The fact that it satisfies the initial (or boundary, depending on how you view it) condition on t=0 is more delicate since you can't just plug that value.
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