Similarity solution for the heat equation

Remarks

Substituting the form gives $u_t = t^{-a-1}\big(-a f - \tfrac{\eta}{2} f'\big)$ and $\alpha u_{xx} = t^{-a-1} \tfrac14 f''$. So $u$ solves the heat equation exactly when $f'' + 2 \eta f' + 4 a f = 0$. Example with $a = 0$: the equation $f'' + 2 \eta f' = 0$ gives $f' = C e^{-\eta^2}$, hence $f = A + B \operatorname{erf}(\eta)$ with the error function. The choice $A = 1$, $B = -1$ gives $u = \operatorname{erfc}\big(x / (2\sqrt{\alpha t})\big)$, the temperature of the half-space $x > 0$ initially at $0$ with surface held at $1$. Example with $a = 1/2$: the equation is $(f' + 2 \eta f)' = 0$, solved by $f = C e^{-\eta^2}$, which gives $u = C t^{-1/2} e^{-x^2/(4 \alpha t)}$. With $C = (4 \pi \alpha)^{-1/2}$ this is the heat kernel for diffusivity $\alpha$. Non-example: $u(x, t) = e^{-\alpha t} \sin x$ solves the heat equation on $\R$ but is not a similarity solution. At each time its zeros are the points of $\pi \Z$, while a similarity solution has zero set $2\sqrt{\alpha t}\, Z$ for a fixed set $Z$, which changes with $t$ unless $Z \subseteq \set{0}$. Similarity solutions arise for problems whose initial and boundary data are compatible with the scaling $x \mapsto \lambda x$, $t \mapsto \lambda^2 t$, such as constant data on a half-line or a point source.