Consider the problem $$u_t = ku_{xx} \ \ \ \ \ \ (-\infty < x < \infty , 0 < t < \infty) \ \ \ \ \ \ \ \ (1)$$
$$u(x,0) = \phi(x) \ \ \ \ \ \ (2) $$
I have to Show that : $u(x,t) = \frac{1}{\sqrt{4\pi kt}} \int_{-\infty}^{\infty}e^{-(x-y)^2/4kt} \phi(y) dy $
I am studying partial differential equation in WALTER STRAUSS
Proof:
We will construct a particular solution for (1) and then construct all the other solution using this property
if S(x,t) is a solution of the given problem, then S(x-y,t) is also a solution of the given problem and also $$ v(x,t) = \int_{-\infty}^{\infty} S(x-y,t)g(y)dy$$
and it is denoted by Q(x,t) of the special form such that $$Q(x,t) = g(p) \ \ \ \ \ \ (3)$$, where p = $ \frac{x}{\sqrt{4kt}}$ and $Q(x,0) = 1 \ \ for \ \ x>0 \ \ , \ \ Q(x,0) \ \ = \ \ 0 \ \ for \ \ x<0$
why do we choose Q of the special form
Using (3), we convert (1) into an ODE for g by using chain rule
$$Q_t = -\frac{1}{2t} \frac{x}{\sqrt{4kt}} g'(p)$$
$$Q_{xx} = \frac{1}{4kt}g''(p)$$
sustitute $Q_t and \ \ Q_{xx} $ in (1) and we get g'' + 2pg' = $0$
we get g'(p) = $c_1 exp(-p^2$) and Q(x,t) = g(p) = $c_1 \int e^{-p^2} dp$ + $c_2$
and Q(x,t) satisfies $Q(x,0) = 1 \ \ for \ \ x>0 \ \ , \ \ Q(x,0) \ \ = \ \ 0 \ \ for \ \ x<0$,then find $c_1$ and $c_2$ and we get $Q(x,t) = \frac{1}{2}$ + $\frac{1}{\sqrt{\pi}}$ $\int_{0}^{x/\sqrt{4kt}} e^{-p^2} dp$
Now define S = $\frac{\partial Q}{\partial x}$.Given any function $\phi$, we also define
$$ u(x.t) = \int_{-\infty}^{\infty}S(x-y,t)\phi(y) dy \ \ \ \ \ \ \ \ t>0$$