Doing an integral (solution of heat equation)

Question

I'm doing an exercise of EDP about computing the solution of a heat solution example, this is a convolution with the fundamental solution but I don't know how to do this integral. What is the method? The integral is the following: $$u(x,t)=\int_{\mathbb{R}}e^{-|x-y|^{2}/4t}e^{ay}dy,t>0.$$ $a$ is a constant.

Answer 1

$$e^{-|x-y|^{2}/4t}e^{ay}=e^{(-(y-x)^2+4tay)/4t}$$

$$\dfrac{-(y-x)^2+4tay}{4t}=\dfrac{-y^2+2xy-x^2+4tay}{4t}=\dfrac{-y^2+(2xy+4ta)y-x^2}{4t}=$$

$$=\dfrac{-((2x+4ta)/2-y)^2-x^2+(2x+4ta)^2/4}{4t}=$$

$$=-\left(\dfrac{2x+4ta}{4\sqrt{t}}-\dfrac{y}{2\sqrt{t}}\right)^2+\dfrac{-4x^2+(2x+4ta)^2}{16t}$$

Now

$$e^{-|x-y|^{2}/4t}e^{ay}=\exp{\left(-\left(\dfrac{2x+4ta}{4\sqrt{t}}-\dfrac{y}{2\sqrt{t}}\right)^2+\dfrac{-4x^2+(2x+4ta)^2}{16t}\right)}=$$

$$=\exp{\left(-\left(\dfrac{2x+4ta}{4\sqrt{t}}-\dfrac{y}{2\sqrt{t}}\right)^2\right)}\exp{\left(\dfrac{-4x^2+(2x+4ta)^2}{16t}\right)}$$

And the integral is

$$u(x,t)=\exp{\left(\dfrac{-4x^2+(2x+4ta)^2}{16t}\right)}\int_{\mathbb R}\exp{\left(-\left(\dfrac{x+2ta}{2\sqrt{t}}-\dfrac{y}{2\sqrt{t}}\right)^2\right)}dy$$

Answer 2

First of all $|x-y|^2=(x-y)^2$, so we have: \begin{align} u(x,t) = \int_{\mathbb{R}} \exp\left(-\frac{(x-y)^2}{4t}+ay\right) \mathrm dy ,\hspace{15pt} t>0\end{align} The idea is to get this in such form that you can do a substitution that yields the Gaussian integral. Note: \begin{align} -\frac{(x-y)^2}{4t}+ay&=-\frac{x^2}{4t}+\frac{xy}{2t}-\frac{y^2}{4t}+ay\\ &=-\frac{x^2}{4t}+ \left(\frac{x}{2t}+a\right)y-\frac{y^2}{4t} \\ &= -\frac{1}{4t}\left[y^2-\left( 2x+4at\right)y \right] - \frac{x^2}{4t}\\ &= -\frac{1}{4t} \left[( y- x-2at )^2 -(x+2at)^2 \right] -\frac{x^2}{4t}\\ &=-\left(\frac{1}{2\sqrt[]{t}} (y-x-2at)\right)^2 +xa+a^2t \\ \end{align}

Now we are ready for the substitution: $z=\frac{1}{2\sqrt[]{t}} (y-x-2at)$, so that $dz=\frac{dy}{2\sqrt[]{t}}$. The integral becomes: \begin{align} u(x,t) = \int_{\mathbb{R}} \exp(-z^2) \exp(xa+a^2t) 2\sqrt[]{t}\mathrm dz = 2\sqrt[]{\pi t}\exp(xa+a^2t) \end{align} I hope it is clear.