Show Phi_(xx)=α^2Phi_t has a solution Phi=varphi(α x/(2sqrt…

Show $\Phi_{xx}=\alpha^2\Phi_t$ has a solution $\Phi=\varphi(\alpha x/(2\sqrt t))$ and find it.

Step 1 · Reduce to an ODE

With $\xi=\alpha x/(2\sqrt t)$: $\Phi_{xx}=\frac{\alpha^2}{4t}\varphi''(\xi)$, $\Phi_t=-\frac{\xi}{2t}\varphi'(\xi)$; substituting into the PDE and cancelling $\alpha^2/(4t)$: \[ \varphi''(\xi) = -2\xi\varphi'(\xi). \]

Step 2 · Solve the ODE

With $w=\varphi'$: $w'=-2\xi w \Rightarrow w=C_1e^{-\xi^2}$, so $\varphi(\xi)=C_1\int_0^\xi e^{-s^2}\,ds+C_2$, i.e. in terms of the error function $\operatorname{erf}(\xi)=\frac{2}{\sqrt\pi}\int_0^\xi e^{-s^2}ds$: \[ \varphi(\xi) = A\operatorname{erf}(\xi)+B. \]

Verification

$\varphi''=-2\xi C_1e^{-\xi^2}=-2\xi\varphi'$. \checkmark Reverting to $x,t$: both $\Phi_{xx}$ and $\alpha^2\Phi_t$ equal $-\frac{\alpha^2\xi}{2t}C_1e^{-\xi^2}$. \checkmark

Physical interpretation

With $A=B=1/2$, $\Phi=\frac12[1+\operatorname{erf}(\alpha x/(2\sqrt t))]$ models diffusion from a step initial condition, interpolating smoothly from $0$ to $1$; the front spreads as $x\sim\sqrt t$, the hallmark of diffusive scaling.