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Expand both sides through second order as $ϵ\to 0$, holding $X,Y$ fixed. The product isPut $Z=ϵ(X+Y)+{ϵ}^2[X,Y]/2$. Its exponential isThusThis is the quadratic part of the Baker--Campbell--Hausdorff formula. The remainder estimate is in any fixed finite-dimensional matrix norm, and is an asymptotic statement rather than an exact formula for arbitrary finite $ϵ$.

Let $A{v}_j=a_j{v}_j$. Commutation gives $A(B{v}_j)=B(A{v}_j)=a_{j}B{v}_j$, so $B$ preserves each eigenspace of $A$. Since the eigenvalues of $A$ are nondegenerate, each eigenspace is one-dimensional and $B{v}_j=b_j{v}_j$. The spectral theorem for a Hermitian matrix lets us choose these vectors as an orthonormal eigenbasis. With $U=(v_1,\dots ,v_n)$,This is simple spectrum gives simultaneous diagonalization. The common eigenvectors can be chosen up to their phases and ordering. In fact degeneracy does not prevent a common eigenbasis for commuting Hermitian matrices, but the one-dimensional argument already suffices here.

The convergent matrix exponential series respects similarity, so $e^A=U{e}^{{\Lambda }_A}{U}^{†}$ and likewise for $B$. Scalar exponentials multiply on each diagonal entry, givingThe essential hypothesis is commutation; Hermiticity supplies this convenient diagonalization proof.

The Fourier transform of a Gaussian givesThe Fourier transform of a derivative obeys $\mathcal{F}[q^{\prime \prime }]=-{\omega }^2\tilde{q}$, so multiplication of the spectrum by $1+ϵ{\omega }^2$ corresponds to $q-ϵq^{\prime \prime }$. Since $q^{\prime \prime }=(2/\sqrt{\pi })(4t^2-2)e^{-t^2}$,This is real and even, as required by the given spectrum. Its time integral is two for every $ϵ$, consistently with $\tilde{s}(0)=2$.

By the Fourier modulation and translation identity, delaying the pulse by $\tau$ multiplies its transform by $e^{-i\omega \tau }$. HenceThe removable value at zero is $2(1+\alpha )$. For real $\alpha$, the squared spectral magnitude is multiplied by $1+{\alpha }^2+2\alpha \cos (\omega \tau )$, showing the interference between the direct and reflected signals.

The filter is a real even triangular function. Its Fourier transform iswith $\tilde{f}(0)=1$. Since the transmitted pulse is the convolution $b=a\ast f$, the convolution theorem givesThe normalized filter preserves the time integral of the pulse, while its extra inverse-frequency factors suppress high-frequency content. It is a smoothing filter rather than an exact compact-bandwidth cutoff.

Use the Fourier transform convention $\tilde{a}(\omega )={\int }_{-\infty }^{\infty }a(t)e^{-i\omega t}dt$, with inverse factor $1/(2\pi )$. Direct integration over the pulse givesThe apparent singularity at zero is removable. Changing the values of the pulse at its two endpoints does not affect the transform.

For the Green function of $D_x^2-1$, the homogeneous solutions satisfying the left and right endpoint conditions are $u(x)=\sinh x$ and $v(x)=\sinh (1-x)$. Their Wronskian isContinuity of the kernel at $x=\xi$ and integration of $G_{xx}-G=\delta (x-\xi )$ require $[G_x{]}_{\xi -}^{\xi +}=1$. Hence the Dirichlet modified Helmholtz Green function on an interval isHere $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The negative sign is essential for this operator convention. Integrating the forcing against the kernel givesThis satisfies both endpoint conditions. It solves the equation classically for continuous $f$, and in the corresponding weak sense for integrable forcing. A homogeneous difference satisfying both endpoint conditions must vanish, so the solution is unique.

After multiplying the initial field by $e^{\beta x/2}$, the $y$ dependence is exactly the $n=2$ cosine mode. Its orthogonality therefore gives $A_{mn}=0$ when $n\ne 2$. For $n=2$, the remaining Fourier coefficient in the Fourier sine series isThusThe coefficients are independent of $\beta$, because the initial exponential exactly cancels the drift-removing weight. Substitution into the preceding expansion gives the full subsequent solution, with decay rates $m^2{\pi }^2+9{\pi }^2/4+{\beta }^2/4$.

Remove the first spatial derivative by the Dirichlet gauge transform for constant drift:The nonzero exponential factor preserves the zero side and top values; it also preserves $v_y=0$ on the bottom. A separated product $v=X(x)Y(y)T(t)$ has the spatial eigenvalue problemsThe nontrivial solutions are $X_m=\sin (m\pi x)$ for $m\ge 1$ and $Y_n=\cos ((n-1/2)\pi y)$ for $n\ge 1$. There is no zero spatial mode in either direction. The temporal equation gives $T=e^{-{\lambda }_{mn}t}$, whereTherefore the mixed-boundary drift-diffusion eigenmodes on a rectangle giveAt $t=0$ the temporal factors are one, giving the required expansion. The sine and mixed-end cosine bases are complete and orthogonal for $v$. In particular,The factor four comes from the product of the two squared basis norms, each $1/2$.

Let $R[y]=F[y]/G[y]$, where $G[y]>0$ for a nonzero admissible real function. Writing $\lambda =R[y]$, its first variation isChoose boundary conditions making the boundary term vanish, for example homogeneous Dirichlet boundary conditions, homogeneous Neumann boundary conditions, or a corresponding mixed pair. Arbitrary interior variations then give the Sturm-Liouville problemMultiplying this equation by $y$ and integrating by parts, with $[py{y}^{\prime }{]}_{\alpha }^{\beta }=0$, yieldsThus eigenfunctions are stationary directions of the quotient. The lowest eigenvalue is its global minimum; higher eigenfunctions need not be unrestricted maxima or minima. General Robin boundary conditions require including their boundary energy in the quotient unless the displayed endpoint products also vanish.

The Rayleigh-Ritz method chooses admissible trial functions depending on finitely many coefficients and minimizes this quotient over them. Expansion in a complete orthonormal basis with weight $w$ shows $R[y]\ge {\lambda }_0$, so the trial minimum is an upper bound on the true lowest eigenvalue. In a finite basis, the minimization becomes a generalized eigenvalue problem for the stiffness and weighted mass matrices. Enlarging the trial space can only lower or preserve this upper bound.

Only the quartic term contributes to the Laplacian. The mixed cubic has zero second derivative in each coordinate, while the exponential–trigonometric pairs are harmonic:Using $z=r\cos \alpha$ and the Jacobian determinant from part (a), the divergence theorem givesThe term containing $\sin \alpha {\cos }^2\alpha$ integrates to zero. Equivalently, the second vertical moment of a solid ring torus with major radius one is ${\int }_V{z}^{2}dV={\pi }^2{R}^4/2$.

Take the closed surface with its outward orientation. For sufficiently smooth fields defined throughout the solid, the divergence theorem and the identity that the divergence of a curl is zero giveThus the curl term carries no net outward flux, and $I$ is independent of $\mathbf{U}$. The hole in the solid torus does not invalidate the divergence theorem; singularities of $\mathbf{U}$ inside the integration region would invalidate the smoothness argument and must be excluded.

The coordinate tangent vectors are the partial derivatives of the position vector:Their pairwise dot products vanish. Their lengths, the scale factors of orthogonal coordinates, are $1,r,1+r\sin \alpha$. Since $R<1$, the last is positive throughout the body. Thus these circular torus coordinates are orthogonal wherever $r>0$; at $r=0$ the cross-section angle degenerates, as in ordinary polar coordinates.

The ordered Jacobian determinant isIts sign is positive for $r>0$, so the volume element is $dV=r(1+r\sin \alpha )drd\alpha d\beta$. The coordinate degeneracy on the central circle has zero volume and does not obstruct volume integration.