# Properties of heat solution

For $x,y,t \in \mathbb{R}$, $t\neq 0$ let

$$K(x,y,t)= (4\pi|t|)^{-\frac{1}{2}}e^{-\frac{(x-y)^2}{4t}}.$$

I need to show that

$$K(x,0,s+t)=\int_{\mathbb{R}}K(x,y,t)K(y,0,s)dy$$

when $s>0$, $t>0$.

My idea :

I wrote $$\int_{\mathbb{R}}K(x,y,t)K(y,0,s) = \frac{1}{4\pi ts}\int_{-\infty}^{\infty}e^{\frac{-x^2s + 2xys-y^2s -y^2t}{4ts}}dy$$

and tried to make the substitution $z=\sqrt{(\frac{x^2s - 2xys+y^2s +y^2t}{4ts})}$ to use that $\int_{-\infty}^{\infty}e^{-z^2}dy = \sqrt{\pi}$, but I could not finish it. Can you help me?

• why is there an $n$ here if $x\in \mathbb{R}$? – qbert Dec 4 '17 at 1:12
• You're right. I already edited. – Mallu Oliveira Dec 4 '17 at 1:15

The name of the game is completing the square. You missed a square root on the $ts$ outside the integral but you are correct that the exponent inside the integral is $$-\frac{x^2s-2xys +(t+s)y^2}{4ts}$$ so you complete the square in $y$, giving $$-\frac{x^2s+ (t+s)\left(y -\frac{xs}{t+s}\right)^2-\frac{x^2s^2}{t+s}}{4ts}.$$ So your integral is $$\frac{1}{4\pi \sqrt{ts}} e^{-\frac{x^2}{4t}+\frac{x^2s}{4t(t+s)}}\int_{-\infty}^\infty e^{-\frac{s+t}{4ts}\left(y -\frac{xs}{t+s}\right)^2} dy.$$ Doing some algebra in the exponent outside the integral, then substituting $z = \sqrt{\frac{s+t}{4ts}}\left(y -\frac{xs}{t+s}\right)$ and using the definite integral you cite yields $$\frac{1}{4\pi \sqrt{ts}} e^{-\frac{x^2}{4(t+s)}}\sqrt{\frac{4ts}{s+t}}\sqrt{\pi} = \frac{1}{\sqrt{4\pi(t+s)}}e^{-\frac{x^2}{4(t+s)}} = K(x,0,s+t)$$