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Consider the one-dimensional diffusion equation on the whole real line,with initial value one for $x<0$ and zero for $x>0$. This is an idealized Heaviside step function profile, not a measured material or an apparatus specification. The heat kernel gives a convolution with the initial profile:The change of variables $s=(x-y)/(2\sqrt{Dt})$ turns the convolution into a Gaussian integral tail. The resulting similarity variable $\eta =x/(2\sqrt{Dt})$ shows that the transition width grows like $\sqrt{Dt}$: multiplying elapsed time by four doubles every fixed-profile distance from the original interface.

At every positive time, $u(0,t)=1/2$, while $u(x,t)\to 1$ as $x\to -\infty$ and $u(x,t)\to 0$ as $x\to +\infty$. For every fixed $x\ne 0$, the limit as $t\to 0^{+}$ returns the specified initial value. The value assigned at the single initial discontinuity does not change this convolution. The error function therefore explains smoothing without pretending that a discontinuous initial profile remains sharp.

Direct differentiation checks the partial differential equation. Writing $\eta =x/(2\sqrt{Dt})$ givesThe model assumes constant diffusion coefficient, no advection, and no finite-domain boundary effects. For an interpretation as heat conduction, the heat equation applies to a normalized temperature difference under its own material assumptions.

The local conservation laws. The first equation is the radial Euler equation: the material acceleration $Du/Dt={\partial }_{t}u+u{\partial }_{r}u$ is caused by the pressure gradient force per unit mass, $-{\partial }_{r}p/\rho$. It neglects gravity and viscosity. The second equation is mass conservation in spherical coordinates,The term $2\rho u/r$ appears because the area of a spherical surface grows as $r^2$.

To obtain the third equation, let $e=p/[(\gamma -1)\rho ]$ be the specific internal energy of an ideal gas with constant heat capacity ratio $\gamma$. The first law of thermodynamics for a smooth adiabatically moving element isDifferentiating $e$ and collecting terms givesand henceThus the entropy parameter $p/{\rho }^{\gamma }$ is conserved along each smooth pathline; for fixed composition, specific entropy differs from $c_v\ln (p/{\rho }^{\gamma })$ only by a constant. This is local isentropic flow behind the shock. A physical shock wave itself generates entropy and must be treated with jump conditions, rather than by applying this smooth equation across the discontinuity.

The front-radius scaling. In the self-similar blast wave approximation there is no scale set by ambient pressure, cooling, gravity, or the initial source radius. Write $R\propto E^a{\rho }_0^b{t}^d$. Using $[E]=\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}$ and $[{\rho }_0]=\mathsf{M}{\mathsf{L}}^{-3}$, dimensional analysis givesThus $a=1/5$, $b=-1/5$, $d=2/5$ andThe dimensionless coefficient ${\xi }_0$ depends on $\gamma$ and the resolved similarity profiles; dimensions alone do not fix it. Define the length scale $\ell (t)=(E{t}^2/{\rho }_0{)}^{1/5}$, so $\xi =r/\ell$ and the front is at the fixed coordinate $\xi ={\xi }_0$.

Kinetic energy. Substitute $\rho ={\rho }_0{f}_{\rho }(\xi )$ and $u=(r/t)f_u(\xi )$ into the volume integral. Since $r=\ell \xi$ and $dr=\ell d\xi$,But ${\rho }_0{\ell }^5/t^2=E$, so the first of the spherical blast-wave similarity energy integrals isThe conclusion requires a fixed, integrable similarity profile, as in the finite-energy blast solution. Expansion increases the swept-up mass while reducing the characteristic squared speed, leaving this product constant.

Pressure and internal energy. The derivatives of the similarity variable areWriting a prime for differentiation with respect to $\xi$, the material acceleration becomesThe radial momentum equation therefore requires a pressure gradient on the scale ${\rho }_0\ell /t^2$. Integrating over radius gives pressure similarity in a spherical blast wave:whereThe prefactor has units $\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}$, as required for pressure. Strictly, the momentum equation alone permits an additional pressure uniform in space and arbitrary in time. In the scale-free strong-blast solution, shock matching and the remaining thermodynamic equation fix that freedom with the same similarity scaling; any allowed constant of integration is included in $f_p$. An independently imposed external pressure would introduce another scale and invalidate the simple dimensional ansatz.

For constant $\gamma >1$, the internal energy is consequentlyIf $E$ is the total excess explosion energy, its normalization isThe two constant fractions need not be equal. No explicit solution for $f_p$ is needed to establish their time independence.

Let $x$ increase away from the tip along the wall and $y$ point normally into the fluid. Put $u=u_x$, $v=u_y$ for velocity components and $U(x)=-A/x^2$ for the outer tangential velocity. The local Cartesian momentum and pressure equations areFor a two-dimensional flat layer the usual continuity equation is ${\partial }_{x}u+{\partial }_{y}v=0$. For this axisymmetric cone, however, the circumference is proportional to $x$, so its leading continuity equation isThe circumference term is of the same order as ${\partial }_{x}u$ and cannot be omitted when deriving the cone's similarity equation. Conditions at the stationary wall are $u=v=0$; at the layer edge $u\to U$.

Balance $U^2/x\sim \nu |U|/{\delta }^2$. With $|U|=A/x^2$ this givesThe choice of multiplicative constant fixes the normalization of the similarity variable. To enforce axisymmetric continuity set $xu={\Psi }_y$, $xv=-{\Psi }_x$, withThenDifferentiation gives $u{\partial }_{x}u+v{\partial }_{y}u=-(A^2/x^5)[2F^{\prime 2}+F{F}^{\prime \prime }/2]$, whereas $U{U}^{\prime }=-2A^2/x^5$ and $\nu u_{yy}=-A^2{F}^{\prime \prime \prime }/x^5$. Consequently the conical sink boundary layer satisfiesThe first condition is no penetration, the second no slip, and the last matches the inward core flow. If one instead makes a strictly planar approximation and omits the circumference term, the streamfunction is $-\sqrt{A\nu }{x}^{-1/2}F$ and the equation becomes $F^{\prime \prime \prime }+F{F}^{\prime \prime }/2+2(1-F^{\prime 2})=0$. The thickness exponent is the same, but that planar equation is not the axisymmetric cone equation.

Across a cap near $R=x$, the inward flux in a layer truncated at $y=Y$ is, to leading order,To match to the core, use the finite displacement-flux deficit rather than integrate the uniform outer flow to infinite $Y$. If $I={\int }_0^{\infty }(1-F^{\prime })d\eta ={\lim }_{\eta \to \infty }(\eta -F)$, thenConservation of the fixed total extraction rate requires an extra core flux of this order. Dividing by cap area, proportional to $R^2$, provesThe correction is a relative $O(\sqrt{\nu R/A})$ contribution. Its angular structure requires outer matching; the requested radial power follows from the displacement flux alone. Its cap-average direction increases the inward core speed to compensate for the wall deficit.

Set the similarity variableThenThe factors of $t$ cancel, leaving the self-similar ordinary differential equation

Let $ϵ$ be the group parameter. The infinitesimal generator of a Lie point symmetryhas flow equationsTherefore it generates the scaling group

Here ${\xi }^t=t$, ${\xi }^x=x$, and $\eta =0$. The total-derivative formulas for the second prolongation of a Lie point symmetry giveandThusFor the wave equation expression $F=u_{tt}-u_{xx}$,It therefore vanishes whenever $F=0$, proving that $V$ generates the simultaneous spacetime scaling symmetry of the wave equation.

The independent invariant of the scaling orbits is the similarity variablewhile $u$ itself is invariant. Seek a group-invariant solution $u(t,x)=f(z)$. Direct partial differentiation givesThe wave equation reduces toHence, on any interval avoiding $z=\pm 1$,

The diffusion length suggests the similarity variableThe Neumann solution of the Stefan problem, written using the error function, isThese expressions satisfy all four thermal boundary conditions. At the interface,Since $\dot{h}=\lambda \sqrt{\kappa /t}$ and $k=\rho c_p\kappa$, the Stefan condition becomesThe second term represents heat supplied from the warm water and therefore reduces the freezing rate.

Write $\Delta T=T_0-T_B$, $\Delta C=C_0-C_B$ and assume these scales are nonzero. The two reference liquidus relations imply $\Delta T=-m\Delta C$. For the similarity variable $\eta =z/(2\sqrt{\kappa t})$ and front $h=2\lambda \sqrt{\kappa t}$,Thus a temperature profile $T=T_{\ast }+\Delta T\theta (\eta )$ has $T_t=-\Delta T\eta {\theta }^{\prime }/(2t)$ and $\kappa T_{zz}=\Delta T{\theta }^{\prime \prime }/(4t)$. Each heat equation becomes ${\theta }^{\prime \prime }+2\eta {\theta }^{\prime }=0$. The solute diffusion equation becomes ${\mathcal{C}}^{l\prime \prime }+2Le\eta {\mathcal{C}}^{l\prime }=0$, withThis Lewis number compares thermal and solutal diffusivities. Inside the mushy layer, $T^s-T_B=-m(C^s-C_B)$ gives $\boxed{{\mathcal{C}}^s={\theta }^s}$. The wall and far-field conditions become ${\theta }^s(0)=0$ and ${\theta }^l,{\mathcal{C}}^l\to 0$ as $\eta \to \infty$.

At the moving phase boundary, equality of temperatures gives ${\theta }^s={\theta }^l+(T_{\infty }-T_B)/\Delta T$. Equality of concentrations gives ${\mathcal{C}}^s={\mathcal{C}}^l+1$. Therefore the dimensionless parameters with the convention used here areNeither concentration nor temperature offsets should be dropped when enforcing continuity. In particular ${\mathcal{R}}_{\theta }$ can be negative. If $k=\rho c_p\kappa$, then $S=L/[c_p\Delta T]$: this is the inverse Stefan number, although the paper calls $S$ its Stefan number.

To check every factor of two in the flux conditions, divide the thermal Stefan condition by $k\Delta T/(2\sqrt{\kappa t})$, and the solute balance by $D\Delta C/(2\sqrt{\kappa t})$. This givesFinally $T_z^l=-m{C}_z^l$ and $-m\Delta C=\Delta T$ imply $\boxed{{\theta }^{l\prime }={\mathcal{C}}^{l\prime }}$. This is marginal equilibrium at a mush-liquid boundary. Together these calculations produce all the rescaled differential equations and all three interface balances.

The transformationsform a one-parameter Lie point symmetry of $u_{XT}=\sin u$. The product $z=XT$ is a similarity variable, and a group-invariant solution $u=F(z)$ satisfies