Show v_t=a^2v_(xx) has a solution v=f(x/sqrt t) , find it…

Show $v_t=a^2v_{xx}$ has a solution $v=f(x/\sqrt t)$, find it, and show $v_x$ is also a solution.

Step 1 · Reduce to an ODE

With $\xi=x/\sqrt t$: $v_t=-\frac{\xi}{2t}f'(\xi)$, $v_{xx}=\frac{1}{t}f''(\xi)$; substituting into $v_t=a^2v_{xx}$ and multiplying by $t$: \[ a^2f''(\xi)+\frac{\xi}{2}f'(\xi)=0. \]

Step 2 · Solve via the error function

With $w=f'$: $w'=-\frac{\xi}{2a^2}w \Rightarrow w=C_1e^{-\xi^2/(4a^2)}$; integrating and substituting $s=2a\tau$: \[ f(\xi) = A\operatorname{erf}\!\left(\frac{\xi}{2a}\right)+B. \]

Step 3 · $v_x$ is also a solution

Since $L=\partial_t-a^2\partial_{xx}$ has constant coefficients, $L\circ\partial_x=\partial_x\circ L$, so $L(v_x)=\partial_x(Lv)=\partial_x(0)=0$. Explicitly, \[ v_x = \frac{C_1}{\sqrt t}\exp\!\left(-\frac{x^2}{4a^2t}\right), \] which is directly verified to satisfy $v_t=a^2v_{xx}$ by substitution.

Verification

$f'(\xi)=C_1e^{-\xi^2/(4a^2)}$ satisfies $a^2f''+\frac{\xi}{2}f'=0$ directly. \checkmark Substituting $v_x$ back into the PDE confirms $\partial_tv_x=a^2\partial_{xx}v_x$ term-by-term. \checkmark

Physical interpretation

$v$ is the temperature profile from a step initial condition; $v_x$ is (up to scale) the Gaussian heat kernel $K(x,t)=\frac{1}{2a\sqrt{\pi t}}e^{-x^2/(4a^2t)}$, the fundamental point-source solution.