I am working on a problem where I have successfully reduced a version of Black Scholes to the Heat Equation and then shown the solution to be:


I now need to show that if $f(x)$ is continuous then $$\lim_{t\rightarrow 0+}u(x,t)=f(x)$$

Further, there is a tip that a change of variables of $p=\frac{(\xi-x)}{2\sqrt{t}}$ may help.

I think that I need to do some integration by parts, show that some part of the integration go to zero as $t\rightarrow0+$ which will cancel out and then by linearity I can say that as u(x,t) is a solution and then $f(x)$ must also be a solution. I am just missing the first step and hoped someone can give me a nudge!

Thanks for any help!

  • $\begingroup$ This is just to give some intuition as to why the result holds: $\frac{1}{2\sqrt{t\pi}} e^{-\frac{(x-\xi)^2}{4t}}\to\delta(x-\xi)$ as $t\to0^+$ because the Gaussian gets infinitely thin yet the area under this curve is constant. The result would then immediately follow. As for the mathematically more rigorous solution, as I see it, after the substitution of $p$ you take the limit and you're done (the $p$ you wrote in your updated question has a typo: there should be no power of two). $\endgroup$
    – user8040
    Aug 8, 2014 at 18:26
  • $\begingroup$ I have been working through this and so far have the following. If $p=\frac{(\xi-x)}{2\sqrt{t}}$ then $\xi = p2\sqrt{t} + x$ and $d\xi = 2\sqrt{t} dp$. Now, if I substitute all that in, I can simplify to f(x) when t approaches zero and take f out of the integrand and cancelling out the 1/2root(t). I'm not sure if I'm completely off track or not here. As I am left with the integral of $e^{-p^2}$ which doesn't cancel itself out as far as I can see. Have I gone completely wrong? $\endgroup$ Aug 10, 2014 at 22:18
  • $\begingroup$ That'd be a Gaussian integral: en.wikipedia.org/wiki/Gaussian_integral $\endgroup$
    – user8040
    Aug 10, 2014 at 23:52
  • $\begingroup$ Ahhh thank you, so that will cancel the $\frac{1}{\sqrt{\pi}}$ and leave me with $f(x)$! That was a bit easier than I had expected it to be... I think I was just over complicating things and should have known. Maths always seems to come up with convenient answers! $\endgroup$ Aug 11, 2014 at 6:29

1 Answer 1


What you need to do is to first make a variable change such as $u = \frac{x-\xi}{2\sqrt{t}}$. Then change the order of the limit and the integral.

  • $\begingroup$ Thanks, I suppose if I had mentioned that there is a tip to make the change of variable, $p=\frac{(\xi-x)}{2\sqrt{t}}$ it would have made sense! which shows that your suggested route is the correct solution! $\endgroup$ Jul 29, 2014 at 18:18

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