For the Euler-Cauchy equation, substitute $y=x^n$ on an interval not containing zero. The coefficient of $x^{n-2}$ isThe distinct roots giveThese solutions are linearly independent since their Wronskian is $2x^3\ne 0$ away from zero. Both nevertheless approach zero as $x\to 0$, which will make the endpoint conditions in the next part insufficient for uniqueness.

The Green-function representation isThe delta identity gives $L{y}_p=f$ and the endpoint conditions follow from those of $G$. If a nonzero homogeneous solution also satisfies both endpoint conditions, it may be added; uniqueness is not automatic in that case.

For each source point $z$, construct the Green function from homogeneous solutions on either side:The endpoint conditions are then automatic. The equation $L_{x}G=\delta (x-z)$ requires continuity of $G$ and a unit jump in its first derivative: a discontinuity in $G$ would introduce an unwanted delta derivative, while integrating across the source fixes $G_x(z^{+},z)-G_x(z^{-},z)=1$. ThusSolving this two-by-two system, with the Wronskian $W=y_a{y}_b^{\prime }-y_b{y}_a^{\prime }\ne 0$, givesThis derivation also remains valid if one of the solution values at the matching point vanishes: only the Wronskian must be nonzero. For a regular nonsingular two-point problem it supplies the inverse kernel; at a singular endpoint, uniqueness requires separate checking, as part (d) illustrates.

The slowest surviving mode has $n=1,m=0$. HenceIt is independent of $y$: insulation allows the constant transverse mode, while every nonconstant transverse mode decays faster. The next possible relative decay rates are $8\lambda {\pi }^2/a^2$ from $n=3,m=0$ and $16\lambda {\pi }^2/b^2$ from $n=1,m=4$.

First decompose the initial $y$ dependence using ${\sin }^2(2\pi y/b)=[1-\cos (4\pi y/b)]/2$. Only the cosine indices $m=0$ and $m=4$ occur. Expand the remaining factor in a Fourier sine series:For $k_n=n\pi /a$, two integrations by parts giveThus even $n$ vanish, while $b_n=8a^2/(n^3{\pi }^3)$ for odd $n$. The nonzero double-series coefficients are $A_{n0}=b_n/2$ and $A_{n4}=-b_n/2$. Their separate heat-mode decay factors giveAt $t=0$ the bracket is $2{\sin }^2(2\pi y/b)$, which recovers the initial profile with the displayed sine coefficients. Each term satisfies the mixed Dirichlet boundary conditions and Neumann boundary conditions.

Write a nonzero separated solution as $T=X(x)Y(y)S(t)$. Division of the heat equation by this product givesThe two spatial factors solve independent constant-coefficient eigenvalue problems. Zero endpoint values in $x$ requireThe insulating conditions requireThe zero cosine mode is the nonzero constant function and must be retained; there is no nonzero zero-index sine mode. The temporal equation is $S^{\prime }=-\lambda {\pi }^2(n^2/a^2+m^2/b^2)S$. Superposing the complete sine-cosine basis yieldsThe constants are fixed by expansion of the initial profile. Orthogonality gives sine norm $a/2$; the cosine norm is $b$ at $m=0$ and $b/2$ otherwise. This is the heat equation on a rectangle with two insulated sides.

For any real coupling, the same normalized Gaussian family has energyIf $ϵ<0$, then as $\alpha \downarrow 0$ the negative ${\alpha }^{-2}$ term dominates the positive ${\alpha }^{-1}$ and $\alpha$ terms. Each trial still has finite energy, is normalized and decays at infinity, but its energy tends to $-\infty$. ThereforeThis variational sequence makes the instability of the negative quartic potential explicit, without mistaking any local well near the origin for a global ground-state bound.

Enlarge the trial space while retaining the optimized Gaussian. For example, usenormalize it and minimize the Rayleigh quotient over its coefficients, optionally optimizing $\alpha$ too. This even trial space respects the even ground-state parity of the confining symmetric potential and adds independent shape freedom beyond a Gaussian width.

The original Gaussian remains included by setting the later coefficients to zero, so the optimized upper bound cannot increase. Finite-basis optimization is the Rayleigh-Ritz method, with matrix entries given by the Hamiltonian and overlap integrals. Increasing the basis size systematically improves the available approximation.

Normalize the Gaussian trial first. The Gaussian integral givesIts second and fourth moments are $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$. Since ${\psi }^{\prime }=-\alpha x\psi$, the kinetic term is $\int ({\psi }^{\prime }{)}^2={\alpha }^2⟨x^2⟩=\alpha /2$. At the specified coupling, the energy is thereforeThe stationary condition isThe quadratic factor is positive for real $\alpha$, so the unique positive stationary width is $\alpha =2$. Also $E^{\prime \prime }={\alpha }^{-3}+9{\alpha }^{-4}>0$ and $E$ diverges at both ends of $\alpha >0$, proving a global minimum within this trial family. Its value isThe associated Sturm-Liouville problem isHere $\rho =1$, $\sigma =x^2+2x^4$ and $w=1$. The Rayleigh-Ritz variational principle makes the exact lowest eigenvalue the minimum over all normalized admissible states. Restricting to Gaussians can only raise this minimum, hence $\boxed{E_0\le 13/8}$. The inequality is strict because $H{\psi }_{\ast }/{\psi }_{\ast }=2-3x^2+2x^4$ is not constant: the trial is not an exact eigenfunction. This is the Gaussian trial for a quadratic-plus-quartic oscillator.

Take real admissible functions with finite $F,G$, positive $G$, and a self-adjoint domain for the Sturm-Liouville operator. Compactly supported variations $\eta$ avoid any boundary contribution at infinity. Formal self-adjointness givesFor $\Lambda =F/G$, the quotient rule therefore givesVanishing for arbitrary variations requiresConversely an eigenfunction in the self-adjoint domain makes this variation zero. Multiplying its equation by $y$ and integrating also shows directly that its stationary quotient is $\boxed{\Lambda [y_n]={\lambda }_n}$.

Thus the stationary values are the admissible eigenvalues. Under the usual positive-weight, semibounded hypotheses the lowest is the global minimum, while higher eigenfunctions generally give saddle points. The assumption $y\to 0$ alone is not a complete domain specification for arbitrary $\rho ,\sigma$: the required integrals and boundary form must exist. The rapidly decaying oscillator trials in part (b) meet these conditions.

The ellipse bounding the surface is $u=1$. With the upward normal, its positively oriented boundary is traversed by increasing $v$ from zero to $2\pi$:The given components are Cartesian components, so their ordinary dot product with this tangent isTherefore Stokes theorem givesA direct check uses $F_x=2y/a$ and $F_y=-2x/a$, hence $(\nabla \times \mathbf{F}{)}_z=-4/a$, and multiplies by the ellipse area $\pi a^2\sinh 1\cosh 1$. The third component cannot affect either calculation because the surface and its boundary lie at fixed $z$. Equivalently one may apply Stokes to the smooth planar field $(2y/a,-2x/a,0)$, avoiding any need to differentiate the elliptic-coordinate expression for the third component at its foci.

For $u>0$, eliminating $v$ givesThus constant $u$ surfaces are confocal elliptic cylinders parallel to the $z$ axis. The limiting $u=0$ surface is the vertical strip $y=0$, $|x|\le a$.

For generic $v$, eliminating $u$ givesThese are hyperbolic cylinders with the same focal lines. A specified angular value selects the half-branch with $\mathrm{sgn}x=\mathrm{sgn}(\cos v)$ and $\mathrm{sgn}y=\mathrm{sgn}(\sin v)$, because $u\ge 0$. At $v=0,\pi$ the surfaces are the vertical half-planes $y=0$, $x\ge a$ or $x\le -a$; at $v=\pi /2,3\pi /2$ they are $x=0$, $y\ge 0$ or $y\le 0$.

Finally, constant $z$ gives horizontal planes. The three surface families express the orthogonality of elliptic cylindrical coordinates.

Differentiate the position vector, keeping the other coordinates fixed in each derivative:The first two have zero dot product, since their two planar contributions cancel, and both are orthogonal to ${\mathbf{h}}_z$. Hence these are orthogonal coordinates away from coordinate degeneracies. Their scale factors, which are the lengths of these coordinate basis vectors, areFor example $|{\mathbf{h}}_u{|}^2=a^2[{\sinh }^{2}u{\cos }^{2}v+{\cosh }^{2}u{\sin }^{2}v]=a^2({\sinh }^{2}u+{\sin }^{2}v)$. The unit basis vectors are ${\mathbf{h}}_i/h_i$ when $h_i\ne 0$; the given coordinate basis vectors themselves are not normalized. The Jacobian determinant is $a^2({\sinh }^{2}u+{\sin }^{2}v)>0$ except on the focal lines. These are elliptic cylindrical coordinates.

For an oriented piecewise smooth surface $S$ with boundary curve $C=\partial S$, and a continuously differentiable vector field $\mathbf{F}$ near $S$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal, $dS$ is positive surface area, and $d\mathbf{r}$ is the directed tangent line element. The positive traversal of $C$ is fixed by the right-hand rule relative to $\mathbf{n}$. Thus the flux of the curl through the surface equals the circulation along its oriented boundary.

Use the Levi-Civita symbol and the contraction of two Levi-Civita symbols in Cartesian coordinates. With repeated indices summed,The last line is the product rule. Reading its four terms in vector notation proves the curl of a cross product identity:

The two diagonal generators are independent involutions, soIt is a Klein four-group. Write the swapping generator as $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Then $J^2=I$, so $H_2=\{I,J\}$ has order two. Moreover,Thus $J$ normalizes $H_1$. The setis a group: its products reduce usingand inverses reduce in the same way. It contains both generating subgroups, so minimality gives $G\subseteq K$; conversely every displayed product is in $G$, giving $G=K$.

There are four diagonal matrices $D$ and four off-diagonal matrices $DJ$, all distinct. ConsequentlyAlso $H_1\cap H_2=\{I\}$, since $J$ is not diagonal, and every element of $G$ has the form $D{J}^{\eta }\in H_1{H}_2$. This proves condition (I), hence the exact factorization of a group. Here $H_1$ is a normal subgroup and $G=H_1⋊H_2$ is a semidirect product. Its matrices are the two-dimensional signed permutation matrices, also realizing the dihedral group of the square.

Assume (I). Existence of a decomposition is part of the hypothesis. If $h_1{h}_2=k_1{k}_2$ are two decompositions, multiply on the left by $k_1^{-1}$ and on the right by $h_2^{-1}$ to obtainThe left side belongs to $H_1$ and the right side to $H_2$, since they are subgroups. Their common value belongs to $H_1\cap H_2=\{e\}$, so $h_1=k_1$ and $h_2=k_2$. This proves uniqueness, hence (II).

Conversely, assume (II). Existence immediately gives $G=H_1{H}_2$. If $a\in H_1\cap H_2$, thenare two allowed factorizations, because the identity is in each subgroup. Uniqueness makes their first components equal, so $a=e$. Thus $H_1\cap H_2=\{e\}$ and (I) follows.

Therefore (I) and (II) are equivalent. This is an exact factorization of a group. For finite groups it gives $|G|=|H_1||H_2|$, but it does not by itself identify $G$ with a direct product of groups; multiplication between the two factors can be noncommutative.

The hypothesis implies $g^2=e$ for every $g\in G$, including the identity, so $g^{-1}=g$. For arbitrary $a,b\in G$, their product also equals its own inverse. The inverse-of-a-product rule therefore givesThus $G$ is an abelian group. Finiteness is not needed for this implication; it holds for every group in which each element squares to the identity.