6
$\begingroup$

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:

$$u(x,t)=\frac{1}{2\sqrt{t\pi}}\int_{-\infty}^\infty{f(\xi)e^{-\frac{(x-\xi)^2}{4t}}}d\xi$$

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!

Improve this question
$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$
    – UnknownUser
    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! $\endgroup$
    – UnknownUser
    Aug 11, 2014 at 6:29

1 Answer 1

Reset to default
5
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$
    – UnknownUser
    Jul 29, 2014 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.