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Write the free surface as the zero setBecause this is a material surface, the kinematic boundary condition is $DS/Dt=0$. Thus, on $z=\eta$,Thereforeat the free surface. This is the kinematic boundary condition for a free-surface graph.

Use $[AB,C]=A[B,C]+[A,C]B$ and the canonical commutation relation. With $L_i={\epsilon }_{iab}{x}_a{p}_b$,Using these two transformation laws on $L_j={\epsilon }_{jab}{x}_a{p}_b$ givesThese are the orbital angular momentum commutation relations.

NowThe factor $[L_j,L_i]$ is antisymmetric in $j$ and its remaining product is symmetric after the two terms are combined, so the contraction vanishes:

The same commutators show that $p^2=p_j{p}_j$ and $r^2=x_j{x}_j$ are rotational scalars: their commutators with every $L_i$ vanish by contraction of the antisymmetric ${\epsilon }_{ijk}$ with a symmetric product. Therefore $U(r)$ also commutes with every $L_i$, and forwe have $[H,L_i]=0$, henceIn particular $H,L^2,L_3$ are pairwise commuting Hermitian operators. By simultaneous diagonalization, they admit a common eigenbasis, subject to the usual spectral-domain qualifications for unbounded operators. This is the rotational invariance of a central-potential Hamiltonian.

LetConvolution with the initial Dirac delta function translates the heat kernel, while part (b) propagates the impulsive source from time one. HenceAt $(x,t)=(0,2)$ this becomesIt vanishes exactly whenThis is the cancellation of two heat-kernel impulses.

The causal Green function for the heat operator iswhere $H$ is the Heaviside step function. It vanishes for $t<\tau$, satisfies the homogeneous heat equation away from the source, and approaches $\delta (x-\xi )$ as $t\downarrow \tau$. Superposing the responses to all infinitesimal sources gives Duhamel principle:The lower limit and causality give the required homogeneous initial data.

Take the Fourier transform in $x$. Since $\widetilde{{\theta }_{xx}}=-k^2\tilde{\theta }$, the transformed heat equation isThereforeThe given transform pair and the convolution theorem yield the heat-kernel solution

Let $u\mapsto u+\epsilon \eta$ and $v\mapsto v+\epsilon \xi$, where the variations vanish on the boundary. The first variation isApplying integration by parts to the four derivative terms and discarding the boundary contributions givesThe variations $\eta$ and $\xi$ are independent and arbitrary in the interior. The fundamental lemma of the calculus of variations therefore gives the two Euler-Lagrange equations for two fields

The coefficient matrix has eigenvector $u_{+}=(1,1{)}^T$ with eigenvalue $3$ and eigenvector $u_{-}=(1,-1{)}^T$ with eigenvalue $-1$. WriteSinceand $y(0)=(1,-2{)}^T=-\frac{1}{2}{u}_{+}+\frac{3}{2}{u}_{-}$, the system diagonalises toUsing an integrating factor givesThe unique fastest term is thereforeConsequently $e^{-nt}y(t)$ diverges for $n<3$, tends to zero for $n>3$, and for $n=3$ tends to the finite nonzero vectorThus the only requested integer isThis is an instance of dominant eigenmode in a forced linear system.

Taking the Laplace transform of the differential equation and usinggivesHenceWhenever $s$ is not an eigenvalue of $A$, the matrix $sI-A$ is invertible, and thusThis is the Laplace-transform solution of a constant-coefficient vector ODE.

Set $g=0$. The given homogeneous solution is $y(t)=e^{tA}{y}_0$, so the preceding formula saysfor every initial vector $y_0$. Equality on every vector gives the matrix identityon the common domain of convergence and in particular away from the eigenvalues of $A$. This is the Laplace transform of a matrix exponential.

Because the matrix-valued Laplace transform is taken componentwise, for every $i,j$ we haveThe sum is finite and $A$ is constant, so it may be moved outside the integral. Therefore

Use the hyperboloid model with Minkowski inner productPut the vertex with angle $\theta$ at $v=(1,0,0)$, and place the endpoints of its adjacent sides of lengths $a$ and $b$ atThe geodesic through $P$ perpendicular to $vP$ has spacelike unit normalwhile the geodesic through $Q$ perpendicular to $vQ$ has spacelike unit normal

These two geodesics form the remaining two sides of the quadrilateral. Their angle equals the angle between their normals in the tangent plane at their intersection. The third right angle therefore givesDividing by $\cosh a\cosh b$ proves the Lambert quadrilateral identity

Rotate the disc so that $P=r\in (0,1)$ lies on the positive real axis. From part (b), the perpendicular hyperbolic line $\ell$ is a Euclidean circle orthogonal to the unit circle, with centre $c>1$ on the real axis and radius $R=c-r$. Orthogonality of the two circles givesCombining the two equations yields

The radial distance formula from part (b) saysThe hyperbolic double-angle identities now giveThereforeThis is the Euclidean circle representing a perpendicular hyperbolic line.

The Poincare disc model isIts geodesics through the origin are the Euclidean diameters. To see directly that the radial segment from $0$ to $r{e}^{i{\theta }_0}$ minimizes length, write an arbitrary joining curve as $z(t)=r(t)e^{i\theta (t)}$. Its hyperbolic length satisfiesThe radial segment has constant $\theta$ and monotone $r$, so equality holds. Henceand every radial diameter is length minimizing. Rotational symmetry and uniqueness for the geodesic equation show that these are all geodesics through $0$.

Given any geodesic and a point $p$ on it, a disc-preserving maptakes $p$ to the origin. By part (a), $T$ is a hyperbolic isometry and commutes with $J(z)=1/\overline{z}$. The transformed geodesic is a diameter, hence a generalized circle invariant under $J$. Its inverse image is therefore also a Generalized circle under a Möbius transformation invariant under $J$. Thus every hyperbolic geodesic is the part in $\mathbb{D}$ of a Euclidean line or circle preserved by reflection in the unit circle; equivalently, it is a diameter or a circle orthogonal to the unit circle. This is the Geodesics of the Poincare disc description.