Set $t=\log x$, so $0<t<\pi$, and write $y(x)=x^{-2}w(t)$. Direct differentiation of this Cauchy-Euler equation givesThe eigenvalue equation is therefore $-w^{\prime \prime }=(\lambda -3)w$ with $w(0)=w(\pi )=0$. For $\lambda -3\le 0$, the linear/hyperbolic solutions cannot satisfy both endpoint conditions except trivially. For $\lambda -3>0$, the sine solution vanishes at both endpoints exactly when its frequency is a positive integer. ThusEach associated eigenspace is one-dimensional.

The displayed differential operator is not self-adjoint for the unweighted inner product. Let $A=x^2{\partial }_x^2+5x{\partial }_x+1$. Integration by parts on compactly supported smooth functions gives its formal adjointwhich differs from $A$. One can see an explicit failure of symmetry without boundary terms: take a nonzero real compactly supported smooth $f$ in the interval and $g=xf$. ThenBoth test functions obey the Dirichlet endpoint conditions. Changing the sign of $A$ for the eigenproblem does not repair this failure. A different, weighted inner product with weight $x^3$ does give the Sturm-Liouville form $x^{3}Ay=(x^5{y}^{\prime }{)}^{\prime }+x^{3}y$, explaining why a real discrete spectrum is nevertheless natural.

The forcing contains $x/y$, so matching it with a single similarity variable requires $\beta =1$. For $u=y^{\alpha }f(\eta )$, $\eta =x/y$, substitution givesMatching the source $y^{-1}\eta e^{-\eta }$ selects $\boxed{\alpha =0,\beta =1}$ and gives $f^{\prime \prime }+f^{\prime }=e^{-\eta }$. Multiplying the first-order equation for $f^{\prime }$ by $e^{\eta }$ yields $f^{\prime }=(\eta +C)e^{-\eta }$, so $f=-(\eta +1+C)e^{-\eta }+D$. The limits at zero and infinity give $D=0$, $C=-1$. HenceFor fixed positive $x$, the initial limit is also zero because the exponential dominates $x/y$. The function $-\eta e^{-\eta }$ decreases to $-1/e$ at $\eta =1$ and increases thereafter. ThereforeThe axis limits are zero, but there is no joint limit at $(0,0)$: approaching along $x=y$ gives $-1/e$. This does not violate any of the printed fixed-variable limits. It is this bounded similarity solution whose minimum was calculated; the unbounded corner-source additions in (b) need not have the same minimum.

Choose the positive causal sign convention $\mathcal{L}={\partial }_y-x{\partial }_x^2$, so the forced equation is $\mathcal{L}u=-x{y}^{-2}{e}^{-x/y}$. With source $(\xi ,\tau )$, a forward Green's function satisfiesIt vanishes at $x=0$, decays at spatial infinity in the chosen bounded evolution class, and has the temporal jump $G(x,{\tau }^{+};\xi ,\tau )=\delta (x-\xi )$. Away from the source it solves the homogeneous equation. Equivalently one may reverse both the Green-function operator and its source sign to use the original printed operator.

To select the physical zero-initial-data solution one also excludes extra corner-source terms of the kind found in (b). With this causal bounded convention the solution representation isA mere pointwise statement of the three boundary limits does not by itself provide that uniqueness convention.

The stated pointwise boundary limits do not imply uniqueness. A counterexample to homogeneous uniqueness isIt is smooth for $x,y>0$. Writing $\eta =x/y$, differentiation givesFor fixed positive $y$, $w\to 0$ as $x\to 0$ and as $x\to \infty$; for fixed positive $x$, it also tends to zero as $y\downarrow 0$. Thus if $u_0$ is one solution, $u_0+Cw$ satisfies the same equation and every printed boundary limit for any constant $C$. Part (d) supplies such a particular solution, so this is a genuine family of distinct solutions, not merely a formal obstruction.

This is corner-source nonuniqueness for a degenerate parabolic problem. The peak $w(y,y)=1/(ey)$ diverges towards the missing corner $(0,0)$; pointwise initial limits at each fixed $x>0$ do not exclude it. The preceding maximum argument fails to force uniqueness because a supremum over an entire past strip need not be attained or finite.

A customary bounded-solution condition restores the intended result. If the difference $z$ of two solutions is bounded on every finite time strip, subtract $\delta (1+x+y)$. Its operator $x{\partial }_x^2-{\partial }_y$ equals $\delta >0$, excluding a positive past maximum; the linear barrier controls spatial infinity, and the homogeneous initial/boundary values control the parabolic boundary. Letting $\delta \downarrow 0$ gives $z\le 0$; applying the same argument to $-z$ gives $z=0$. This proves uniqueness in that additional bounded class, which the printed pointwise limits alone do not specify.

At such an interior backward-in-time maximum, spatial differentiation gives $u_x=0$ and $u_{xx}\le 0$. Comparison with earlier times at the same spatial point gives the one-sided time derivative $u_y\ge 0$; differentiability makes this the usual derivative. Consequently $x{u}_{xx}-u_y\le 0$, contradicting the strictly positive source $x{y}^{-2}{e}^{-x/y}$. Therefore no such maximum can occur. The temporal one-sided argument matters because the condition concerns only the past of the candidate point.

Mass preservation of a nonnegative finite-mass profile requires $\alpha +\beta =0$. Balancing ${\rho }_t$ and ${\partial }_x({\rho }^4/4)$ requires $\alpha -1=4\alpha -\beta$. ThereforeThe similarity ODE isFor a source-type profile adjoining vacuum, the integration constant is zero. The positive branch is $f={\eta }^{1/3}$ for positive $\eta$, terminated by a shock into zero. Normalize the mass:Thus a unit-mass similarity solution isAt the moving right edge, $S^{\prime }={\eta }_F{t}^{-3/4}/4$ equals ${\rho }_L^3/4$, so the Rankine-Hugoniot condition is satisfied. The characteristic speeds satisfy $0<S^{\prime }<{\rho }_L^3$, making it an entropy shock. The left edge meets vacuum continuously. As $t\downarrow 0$ the mass concentrates at the origin; this is a source solution, not a claim that every finite-mass initial profile equals it at finite time.

Before intersection, the characteristic map is $x=\xi +{\rho }_0(\xi {)}^{3}t$, with derivative $1+3{\rho }_0^2{\rho }_0^{\prime }t$. Because the initial profile decreases, higher-density characteristics behind move faster than lower-density characteristics ahead. Compression therefore forms shocks, which absorb more characteristics. The far states remain one and zero, so a shock joining their limiting traces has speed tending to $[q]/[\rho ]=1/4$.

The precise large-scale interpretation uses $\zeta =x/t$. The rescaled initial transition tends to a step, and entropy comparison for the convex flux givesOne way to see the comparison is to choose distant initial points beyond which the profile is within $\delta$ of each far state. Step profiles with those nearby states bracket it; their shock speeds tend to $1/4$ as $\delta \downarrow 0$, while their fixed initial locations disappear after division by $t$.

A fixed translation of the shock can persist. If the deviation from a reference step is integrable, its signed mass fixes a phase shift. Without such a tail assumption one should assert convergence on the $x/t$ scale and $S(t)/t\to 1/4$, rather than claim that the exact line $x=t/4$ is the unshifted asymptote for every smooth initial transition.

In smooth regions the characteristic equations are $\dot{x}=q^{\prime }(\rho )={\rho }^3$ and $\dot{\rho }=0$. If $\xi$ labels the initial point, their parametrization isFor the two constant states, characteristics from the left have speed one and those from the right speed zero; they would overlap. The entropy solution replaces the overlap by a compressive shock. Its speed isThusThe left parametrization $x=\xi +t$ is retained only for $x<t/4$, and the right parametrization $x=\xi$ only for $x>t/4$. The characteristic speeds obey $0<c<1$, so characteristics enter the shock from both sides.

The quantity inside a fixed interval changes by influx at its left endpoint minus outflux at its right endpoint:In a smooth region the fundamental theorem of calculus turns this into ${\int }_a^b[{\rho }_t+{\partial }_{x}q(\rho )]dx=0$. Since the interval is arbitrary, the local conservation law is $\boxed{{\rho }_t+{\partial }_{x}q(\rho )=0}$.

For a jump at $x=S(t)$, write the two smooth pieces using a step function. The singular part of ${\rho }_t$ is $-S^{\prime }[\rho ]\delta (x-S)$ and that of ${\partial }_{x}q$ is $[q]\delta (x-S)$. Cancellation gives the Rankine-Hugoniot conditionThe same identity follows by applying conservation to a shrinking moving control interval across the shock.

Put $y=e^{x}w$. Then $y^{\prime \prime }-2y^{\prime }+y=e^x{w}^{\prime \prime }$ and the endpoint conditions become $w(0)=w(\pi )=0$. The transformed eigenvalue equation is $-w^{\prime \prime }=\lambda w$, soZero and negative $\lambda$ yield no nonzero solution with both endpoint values zero.

For the usual unweighted $L^2(0,\pi )$ inner product, the operator $D^2-2D+1$ is not self-adjoint: its formal adjoint is $D^2+2D+1$. The mismatch is already visible on compactly supported test functions, so endpoint conditions cannot remove it. With the different weight $e^{-2x}$ one does obtain a symmetric Sturm-Liouville form, since $e^{-2x}(y^{\prime \prime }-2y^{\prime }+y)=(e^{-2x}{y}^{\prime }{)}^{\prime }+e^{-2x}y$. Thus the inner-product convention matters; the real spectrum alone does not prove self-adjointness in the unweighted space.