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The entries of $M^{T}M$ are real, and the transpose rule gives $(M^{T}M{)}^T=M^{T}M$. Thus it is a real symmetric matrix, hence also a Hermitian matrix.

For any complex vector $v$,In particular, if $M^{T}Mv=\lambda v$ with $v\ne 0$,Thus $M^{T}M$ is positive semidefinite. Its zero eigenspace is precisely the kernel of $M$, since $\Vert Mv{\Vert }^2=0$ is equivalent to $Mv=0$. This is the Gram-matrix structure behind the nonnegative eigenvalues.

An invertible matrix is a square matrix for which there is a matrix $A^{-1}$ with $A{A}^{-1}=A^{-1}A=I$. Write its assumed diagonalization as $A=P\Lambda P^{-1}$, where $P$ is invertible and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. ThenIf the determinant is nonzero, every ${\lambda }_j$ is nonzero, so ${\Lambda }^{-1}=\mathrm{diag}({\lambda }_j^{-1})$ exists and $A^{-1}=P{\Lambda }^{-1}{P}^{-1}$. If the determinant is zero, at least one eigenvalue is zero; its nonzero eigenvector is mapped to zero, which an invertible map cannot do. HenceThe equivalence holds even without diagonalizability, but the eigenbasis proves the stated case directly.

The adjoint operator $A^{†}$, when it exists, is defined byfor all vectors in the relevant domains. This convention takes the inner product to be conjugate-linear in its first argument. Nondegeneracy makes the adjoint unique.

For the standard complex inner product $⟨x,y⟩={\overline{x}}^{T}y$, the equality becomes ${\overline{x}}^T{\overline{A}}^{T}y={\overline{x}}^T{A}^{†}y$ for every $x,y$, soThus the matrix adjoint is the Hermitian conjugate, not merely the transpose unless the entries are real. In a basis with positive Hermitian Gram matrix $G$, the same definition instead gives $A_G^{†}=G^{-1}{\overline{A}}^{T}G$, making the dependence on the chosen inner product explicit.

Let $r(z)$ be the indicator of $[-1,1]$. The finite integral is exactly the convolution $p=q\ast r$. Its additional transform factor isApplying the convolution theorem and part (d),Equivalently this is $2(\sin k/k)[\sin (k/2)/(k/2){]}^2$. The value at zero also follows from $\int p=(\int q)(\int r)=1\times 2$.

Evenness means the negative half of the function is determined as well: this is the triangular function $q(x)=1-|x|$ on $|x|\le 1$, zero outside. Its transform is real and even:Integration by parts, or integrating the two terms separately, givesThe apparent singularity at zero is removable, and its limiting value is the area under $q$. This is the Fourier transform of a triangular function.

Differentiate the defining Fourier integral $n$ times with respect to $k$:For a nonnegative integer $n$, this givesIntegrability of $x^{n}f$ together with suitable lower moments justifies differentiation under the integral. The same multiplication-derivative identity holds more generally for tempered distributions.

Assume initially that $f,g$ are integrable, so Fubini's theorem justifies interchanging the integrals. Transform the convolution and put $u=x-z$:This is the convolution theorem. No factor $2\pi$ appears in this direction because the forward transform in the question is unnormalized.

For the stated Fourier transform convention, Fourier inversion isFor example, the pointwise formula holds at continuity points under the usual integrability hypotheses; the same normalization applies to the $L^2$ or distributional inverse.

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.