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Use the orthogonal change of coordinatesIt preserves distances and angles and gives $x^{T}Mx=x^{\prime T}D{x}^{\prime }$. Thus the quadric surface is expressed along its principal axes asThe following three cases are the subparts of (ii) in the original PDF.

The spectral theorem for real symmetric matrices supplies real eigenvalues and an orthonormal eigenbasis. Compute eigenvectors, normalize them, and within each repeated-eigenvalue eigenspace choose an orthonormal basis. Distinct eigenspaces are already orthogonal because symmetry gives $(\lambda -\mu )v^{T}w=0$ for eigenvectors with eigenvalues $\lambda ,\mu$.

Use the resulting basis vectors as columns of $O$. ThenThe order and signs of the columns can be chosen freely; rotations within a repeated-eigenvalue eigenspace are also allowed.

Taking the adjoint of $U=M+iN$ gives $U^{†}=M-iN$. In particular, the Hermitian parts of a unitary matrix are uniquely determined:Expanding both unitary identities givesSubtracting shows $MN=NM$, and adding gives $M^2+N^2=I$. If $Mv=\mu v$, commutation implies $M(Nv)=\mu Nv$. The assumed simple spectrum makes the eigenspace one-dimensional, so $Nv=\nu v$. Thus $M,N$ have a common orthonormal eigenbasis.

On such a vector, $Uv=(\mu +i\nu )v$. If the corresponding unitary eigenvalue is $u=e^{i\theta }$, thenThe formulas for $M,N$ also show the same eigenvalue correspondence directly from $Uv=uv$ and $U^{†}v=u^{\ast }v$. Degeneracies do not prevent existence of a common basis, although the simple-spectrum assumption makes the elementary proof immediate.

The absolutely convergent matrix exponential series givesfor Hermitian $H$. The exponents $iH$ and $-iH$ commute, so their exponentials multiply to the exponential of their sum:Hence $U$ is unitary. Alternatively, an orthonormal eigenbasis of $H$ turns the exponential eigenvalues into $e^{i{\lambda }_j}$, all on the unit circle.

A Hermitian matrix satisfies $H^{†}=H$, where $†$ denotes the conjugate transpose. For a nonzero eigenvector $v$ with $Hv=\lambda v$, the scalar $v^{†}Hv$ equals its own complex conjugate, andThus every Hermitian eigenvalue is real.

A unitary matrix satisfies $U^{†}U=U{U}^{†}=I$. If $Uv=\mu v$, preservation of the squared norm givesSince $v\ne 0$, $\boxed{|\mu |=1}$. These statements concern complex eigenvectors, even when the matrix entries happen to be real.

Differentiating $\int e^{-a{x}^2}dx=\sqrt{\pi /a}$ with respect to $a>0$ gives $\int x^2{e}^{-a{x}^2}dx=\sqrt{\pi }/(2a^{3/2})$. ThereforeFrom part (iii), $|\tilde{f}(k){|}^2=(\pi /4)k^2{e}^{-k^2/2}$. Its frequency-side energy isin agreement with Parseval's identity.

For sufficiently rapidly decaying functions, substitute the inverse Fourier transform of $g$ and interchange the integrals:The bracketed integral is the complex conjugate of $\tilde{f}(k)$. This is Parseval's identity for the chosen normalization. One may first prove it for Schwartz functions, where the interchanges are justified, and then extend it to $L^2$ by the Plancherel theorem.

Let $A(k)$ be the Fourier transform of $e^{-x^2}$. Since $(e^{-x^2}{)}^{\prime }=-2x{e}^{-x^2}$, integration by parts gives $\mathcal{F}[x{e}^{-x^2}]=-(ik/2)A(k)$. Differentiation under the integral then yields $A^{\prime }(k)=-(k/2)A(k)$. The Gaussian integral supplies $A(0)=\sqrt{\pi }$, soApplying part (ii) with $n=1$ givesIt is imaginary and odd, as expected for the transform of a real odd function.

Differentiate under the Fourier transform integral:Hence the Fourier transform of a polynomial multiple isAbsolute integrability of the required moments justifies differentiation under the integral. The identity also extends to tempered distributions.

With the stated negative sign in the forward exponential, the inverse Fourier transform isFor example, this holds pointwise for Schwartz functions; more general square-integrable functions use the corresponding mean-square interpretation.

Separate variables as $u=X(x)T(t)$. The zero values at both plates give the Fourier sine basisExpand the initial profile using orthogonality. With $s=x/L$, its coefficients areConsequentlyOnly odd modes occur because the initial profile is symmetric about the midpoint. The coefficients are absolutely summable, so dominated convergence gives $\boxed{u(x,t)\to 0}$ uniformly as $t\to \infty$. The slowest decay rate is $\nu {\pi }^2/L^2$, associated with $j=1$.

For the established time-periodic response, take a complex separated solution $v=X(x)e^{i\omega t}$ and set $u=\mathrm{Re}v$. The diffusion equation gives $X^{\prime \prime }=(i\omega /\nu )X$. For $\omega >0$, define the penetration depth $\delta =\sqrt{2\nu /\omega }$; the spatial roots are $\pm (1+i)/\delta$. Decay at infinity selectsIts real part is the solution of Stokes' second problem:The velocity amplitude decays exponentially, while the phase lags by $x/\delta$. Direct substitution verifies the diffusion equation, and the two boundary conditions follow immediately.

These data select a unique periodic response, but do not specify an arbitrary initial transient. With no initial velocity supplied, one may add any suitable solution of the homogeneous-boundary diffusion problem. The displayed solution is the usual established oscillatory flow after transients have decayed.

The boundary conditions on the positive half-line, absent from the transcription, select the half-harmonic oscillator. The trial state $x{e}^{-\alpha x}$ with $\alpha >0$ satisfies both endpoints and has finite norm and energy. Integration by parts, with vanishing endpoint term, givesUsing ${\int }_0^{\infty }{x}^n{e}^{-2\alpha x}dx=n!/(2\alpha {)}^{n+1}$, the three required integrals areThus the exponential trial state for a half-harmonic oscillator gives $\Lambda (\alpha )={\alpha }^2+3/{\alpha }^2$. Its derivative vanishes at ${\alpha }^4=3$, and the arithmetic-geometric mean inequality confirms the global minimum:For comparison, the exact lowest half-line state is proportional to $x{e}^{-x^2/2}$ and has eigenvalue three. The exponential trial family gives a valid upper bound rather than an exact eigenvalue.

Write $N[\psi ]={\int }_a^b(p{\psi }^{\prime 2}+q{\psi }^2)dx$ and $D[\psi ]={\int }_a^{b}w{\psi }^{2}dx>0$. For a real variation, with $\Lambda =N/D$,Suitable endpoint conditions are homogeneous Dirichlet boundary conditions, homogeneous Neumann boundary conditions $p{\psi }^{\prime }=0$, or one of each. They remove the variation boundary term and the product $[p\psi {\psi }^{\prime }{]}_a^b$. Stationarity for all interior variations is equivalent toConversely, multiplication of an eigenfunction equation by $\psi$ and integration by parts gives $N=\lambda D$, hence $\boxed{\Lambda [\psi ]=\lambda }$. General Robin data require their boundary-energy contributions unless these endpoint products happen to vanish.

Use an eigenbasis orthonormal with respect to the weight: ${\int }_a^{b}w{\psi }_i^{\ast }{\psi }_{j}dx={\delta }_{ij}$. For complex coefficients use the Hermitian version of the quotient, with absolute squares. Self-adjointness givesWhen $c_0\ne 0$, dividing numerator and denominator by $|c_0{|}^2$ yieldsIf $c_0=0$, these ratios are undefined, but the unscaled formula and upper-bound conclusion remain valid. The assumptions are a problem with a self-adjoint operator bounded below and a complete eigenbasis, with the trial function in its energy-form domain. The Rayleigh-Ritz method minimizes this quotient over an admissible trial family, obtaining an upper bound on ${\lambda }_0$. Higher eigenfunctions are stationary directions and need not be unrestricted extrema.

Use the full field in the PDF, including the exponential term missing from the transcription. Its divergence isThe exponential in the third component has zero $z$ derivative. In cylindrical coordinates, the solid has $0\le r\le 2$ and $0\le z\le 4-r^2$. Therefore the divergence theorem gives the outward fluxThe integration includes the entire closed boundary, both cap and base.

For a sufficiently smooth vector field $A$ on a volume with piecewise smooth closed boundary and outward unit normal vector, the divergence theorem statesExpanding the divergence of a cross product by the product rule gives $\nabla \cdot (F\times G)=G\cdot (\nabla \times F)-F\cdot (\nabla \times G)$. Apply the theorem to $A=F\times G$ to obtain

Work in a Cartesian coordinate system, write ${\partial }_j=\partial /\partial x_j$, and use the Einstein summation convention. For a twice continuously differentiable vector field, the contraction of two Levi-Civita symbols givesHere the Kronecker deltas select components, and equality of mixed partial derivatives gives the curl of the curl identityFor the second identity, the same contraction followed by the product rule yieldsThus the curl of a cross product isThe directional derivatives in the last two terms act on the vector field immediately to their right.