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A group-invariant solution has a graph preserved by the chosen Lie point symmetry. For a one-parameter infinitesimal generator of a Lie point symmetrya differentiable scalar solution is invariant when $V$ is tangent to its graph, equivalentlyThis is a first-order constraint in addition to the original partial differential equation. The values of $u$ need not be constant along the projected orbits if the symmetry also changes the dependent variable.

For $V=2t{\partial }_t+x{\partial }_x-u{\partial }_u$, the invariance condition is $2t{u}_t+x{u}_x+u=0$. Its solutions on $t>0$ have the similarity ansatz $u=t^{-1/2}F(x/\sqrt{t})$. The graph is invariant under $(t,x,u)\mapsto (e^{2s}t,e^{s}x,e^{-s}u)$ even though $u$ changes along that orbit. For a symmetry fixing the dependent variable, $\eta =0$, the simpler statement that $u$ is constant along projected orbits is valid.

At a positive interior maximum, $u_x=0$, $u_{xx}\le 0$, and $u_t\ge 0$ if the point is compared with its past. The PDE instead gives $u_t\le -u^3<0$, a contradiction. Thus a positive global/past-strip maximum must come from the spatial boundary or initial-time limit. The far value is positive, so any attained global maximum is positive; the argument does not prohibit a negative local maximum in isolation.

For uniqueness put $w=u_1-u_2$. It satisfiesThe parabolic comparison principle with zero difference data gives $w\le 0$ and then $-w\le 0$, hence $w=0$. As usual on this noncompact singular-time domain, this uses an admissible comparison class in which the initial and far-field limits justify truncation; for example bounded/subquadratic differences on finite past strips with the corresponding uniform zero traces suffice. A barrier $\delta (1+x^2+2t)$ controls spatial infinity, since ${\partial }_t-{\partial }_x^2+[x/(2t)]{\partial }_x+a$ applied to it is nonnegative. The printed fixed-point limits alone should not silently be substituted for all uniform/growth hypotheses of this comparison argument.

Use drift cancellation in a parabolic similarity reduction. For the similarity ansatz, derivative scaling givesThus $\boxed{\beta =1/2}$, the drift terms cancel, and $f^{\prime \prime }=f^3-f/2$ with $f(\infty )=1/\sqrt{2}$. Use the positive-equilibrium separatrix for a cubic profile equation. Multiply by $f^{\prime }$ and use the limiting equilibrium to determine the integration constant:The branch approaching the positive far equilibrium satisfies $f^{\prime }=(1/2-f^2)/\sqrt{2}$; the constant equilibrium is included as well. Separation givesFor a finite boundary value at $x=0$ and no interior pole, take $A>-1$. Values $-1<A<0$ describe the corresponding coth-type approach from above; $A=0$ is constant, and positive $A$ gives the tanh-type branch. The other first-integral sign cannot approach the positive equilibrium nontrivially as $\eta \to \infty$.

The requested deficit integral is, after $x=\sqrt{t}\eta$,Therefore $\boxed{A=1}$, and the selected solution is $\boxed{u(x,t)=(2t{)}^{-1/2}\tanh [x/(2\sqrt{t})]}$.

Balance the time derivative, cubic dispersion and quadratic transport using the cylindrical KdV equation similarity ansatz $q(x,t)=t^{-2/3}F(\xi )$, $\xi =x{t}^{-1/3}$ on $t>0$. For $t<0$ the same real cube-root convention can be used separately. Substitution and removal of the common factor $t^{-5/3}$ giveEvery term is a total derivative: $F{F}^{\prime }=(F^2/2{)}^{\prime }$ and $\xi F^{\prime }+F=(\xi F{)}^{\prime }$. Integration therefore gives the requested second-order equationwhere $C$ is a constant determined by any additional boundary data. No condition in the question forces $C=0$.

This part uses source-free viscous evolution of an accretion disk. Put $R=R_0{x}^2$ and $\Sigma ={\Sigma }_0{x}^{-3}S$. The prescribed kinematic viscosity then becomesso $\nu \Sigma R^{1/2}={\nu }_0{\Sigma }_0{R}_0^{1/2}x{S}^{3/2}$. Since ${\partial }_R=(2R_{0}x{)}^{-1}{\partial }_x$, the Keplerian viscous diffusion equation givesCanceling the prefactor and setting $\tau =3{\nu }_{0}t/(4R_0^2)$ proves

Use the similarity ansatz $S={\tau }^{-1}f(\xi )$ with $\xi =x/\sqrt{\tau }$. Derivatives at fixed $x$ giveOn the branch $1-k\xi \ge 0$, substitute $f=(1-k\xi {)}^2$. The two expressions reduce toEquality of the constant coefficients requires $\boxed{k=1/6}$, and then the other coefficients agree as well.

The physical compact self-similar disk with square-root density viscosity uses this branch up to its outer edge and vacuum beyond it:Here $(u{)}_{+}=\max (u,0)$ is the positive part. The cutoff matters: beyond $\xi =6$, the untruncated squared expression would have $f^{3/2}=|1-\xi /6{|}^3$, and its right-hand side would no longer be the polynomial used above. At the moving edge, $S$ vanishes quadratically and $x{S}^{3/2}$ vanishes cubically. The latter and its first derivative match the vacuum values continuously, so joining to zero creates no distributional mass-flux source. The joined profile satisfies the equation, including in the weak sense at the edge.

Returning to the surface density gives the explicit resultwith $\Sigma =0$ for $R\ge 36R_0\tau$. Thus the outer edge is $\boxed{R_e=36R_0\tau }$ and moves linearly in physical time, with speed $27{\nu }_0/R_0$. The density decreases monotonically with radius, has an integrable $R^{-3/2}$ central cusp, and drops quadratically to zero at the outer edge. The formal origin boundary has zero viscous torque in an accretion disk, because $\nu \Sigma \sqrt{R}\to 0$ there, even though the surface density diverges.

To characterize the spreading, let $a=6\sqrt{\tau }$ and integrate in $x$. The disk mass and angular momentum areThe mass decreases while the angular momentum remains constant: matter accretes at the origin and the remaining disk spreads outward. More explicitly, with $u=x/(6\sqrt{\tau })$, the inward mass flux isIt is inward for $u<1/4$ and outward for $u>1/4$. The dividing radius is $9R_0\tau /4$, and the origin flux $3\pi {\nu }_0{\Sigma }_0{\tau }^{-3/2}$ equals $-d{M}_d/dt$. The outer edge carries no flux beyond itself. This family is a self-similar accreting disk with finite mass at each positive time, rather than a regular finite-mass initial profile at $\tau =0$.

Use the dimensionless equation from part (a). For the similarity ansatz $h=t^{a}f(\eta )$, $\eta =x/t^b$, its three terms have time factorsA nontrivial similarity solution retaining storage, lateral drainage and nonzero rainfall requires the exponents to agree:Unlike a source-free diffusion equation, the nonzero prescribed rainfall fixes an amplitude-versus-time scale directly, so no additional conserved quantity or boundary condition is required to determine these exponents. Degenerate spatially uniform solutions are an exception to this all-terms balance. The rain-fed aquifer similarity solution satisfiesIf the dimensional equation is used directly without scaling out coefficients, its similarity profile instead obeys $\varphi (f-\eta f^{\prime })=K(f{f}^{\prime }{)}^{\prime }+R$. The normalization above is the one in which the subsequent condition $f(\infty )=1$ applies.

A similarity solution has a fixed profile after rescaling its independent variables and, when needed, its amplitude. A typical similarity ansatz isSubstitution into a partial differential equation fixes a scale balance and produces an equation for the similarity profile $F$. In one spatial dimension this is often an ordinary differential equation; in several spatial dimensions it can remain a partial differential equation unless a further symmetry, such as radiality, is imposed.

For $u_t=\Delta u$ on ${\mathbb{R}}^n$, balance of time and space derivatives gives $\beta =1/2$. Preservation of a finite nonzero spatial integral gives $\alpha =n/2$. Writing $\xi =x/\sqrt{t}$, the profile equation issolved by $F(\xi )=C{e}^{-|\xi {|}^2/4}$. The Gaussian heat kernel is therefore a spreading profile whose total mass stays fixed. A scale-invariant equation need not have a similarity solution compatible with arbitrary prescribed data; the scaling of the data and conserved quantities is part of the construction.