Heat Kernel at t=0: Dirac Delta Intuition

[i_hate_math]

Homework Statement

Show that k(x,0)=δ(x).
Where k(x,t) is the heat kernel and δ(x) is the Dirac Delta at x=0.

Homework Equations

k(x,t) = (1/Sqrt[4*π*D*t])*Exp[-x^2/(4*D*t)]

The Attempt at a Solution

I am just clueless from the beginning. I am guessing this is got to do with convolution?
I know ∫ k(x,t) dx = 1, {x, -∞, ∞} and the same goes for Dirac Delta.

 

[Orodruin]

I know ∫ k(x,t) dx = 1, {x, -∞, ∞} and the same goes for Dirac Delta.

This is not sufficient, there are many different functions that integrate to one, you need to show that
$$
\lim_{t\to 0^+} \int k(a-x,t) f(x) dx = f(a),
$$
which is the defining property of the delta distribution.

 

[i_hate_math]

This is not sufficient, there are many different functions that integrate to one, you need to show that
$$
\lim_{t\to 0^+} \int k(a-x,t) f(x) dx = f(a),
$$
which is the defining property of the delta distribution.

Thanks for ur reply! I'm still a bit confused as to how this expression is obtained? I'm not too familiar with convolution, would u care to explain why the convolution is the same as f(a) in the limit t->0

 

[Orodruin]

Thanks for ur reply! I'm still a bit confused as to how this expression is obtained? I'm not too familiar with convolution, would u care to explain why the convolution is the same as f(a) in the limit t->0

This is the definition of the delta distribution so it is what you need to show. If you show that it is true you will have shown that $k(x,0) = \delta(x)$.