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Let $T_q\phi (x)={\sum }_{k=0}^q{\phi }^{(k)}(0)x^k/k!$ denote the Taylor polynomial at zero. On $0<|x|<1$, the integrand defining the Hadamard finite-part reciprocal-power distribution isThe first term is odd, so it cancels under symmetric truncation. The Taylor remainder in the second term has absolute value at most $|x{|}^m\Vert {\phi }^{(m)}{\Vert }_{\infty }/m!$. The integral near zero therefore has a finite limit. For $|x|\ge 1$, the original terms are integrable: $x^{-m}\phi$ has compact support, and each subtracted power has exponent $k-m\le -2$.

This proves existence and linearity. It also gives the global estimateThusIn particular the order of a distribution is bounded by one integer on all compact sets, not merely separately on each of them.

For the recurrence, write the truncated Hadamard finite-part integral aswhere parity evaluates all subtraction integrals explicitly:Set $C_1=0$ and ${\Lambda }_1=\mathrm{pv}(1/x)$. On the two truncated intervals, integration by parts givesThere are no boundary contributions at infinity. The coefficients of the subtraction terms satisfyThis is exactly the negative of the boundary term's Taylor polynomial through degree $m$; the degree-$m$ term itself vanishes by parity. HenceTaking the limit proves the full distributional identityThe same calculation includes $m=1$. Starting with the distributional derivative of the logarithmic modulus from part (i), induction yieldsThis identity holds on all of $\mathbb{R}$ as a distributional identity. It is stronger than matching the ordinary derivatives away from zero, which would leave possible differentiated Dirac delta distributions at zero undetermined.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$. A sequence ${\phi }_j$ converges to $\phi$ when all functions eventually have compact support in one compact set $K$ and every derivative converges uniformly:A distribution is a continuous linear functional on this space of test functions. Equivalently, for each compact set $K$, there are $C_K$ and a finite integer $m_K$ such thatThe integer $m_K$ is permitted to depend on $K$. The usual weak convergence of distributions isThus the convergence convention on the distribution space is its weak dual topology.

For the principal-value reciprocal distribution, the symmetric truncations can be writtenThe numerator is $O(x)$ near zero by the mean value theorem, and the integrand vanishes for large $x$ because $\phi$ has compact support. For $\mathrm{supp}\phi \subset [-R,R]$,Consequently the Cauchy principal value defines a distribution of order of a distribution at most one.

The function $\log |x|$ has local integrability, since ${\int }_0^1|\log x|dx<\infty$, and therefore defines a distribution. For its distributional derivative, remove $(-\epsilon ,\epsilon )$ and use integration by parts on both remaining intervals:The boundary term is $O(\epsilon |\log \epsilon |)$ and tends to zero. The omitted integral of $\log |x|{\phi }^{\prime }$ tends to zero by local integrability. Hence the distributional derivative of the logarithmic modulus satisfiesSymmetric truncation is essential to this normalization of the principal-value reciprocal distribution.

First suppose $c_i\ne 0$ and put $F=\hat{w}/(U-c)$. The wall condition gives $F(\pm L)=0$. Substitution into the Rayleigh equation for inviscid shear flow givesMultiply by $\overline{F}$ and use integration by parts. The resulting weighted identity for Howard's semicircle theorem isIts imaginary part gives $\int UQ=c_{r}I$, and its real part then gives $\int U^{2}Q=(c_r^2+c_i^2)I$. Set $a=U_{\min }$ and $b=U_{\max }$. Since $(U-a)(b-U)\ge 0$,Completing the square proves Howard's semicircle theorem:In particular every growing normal mode lies in the upper semicircle for $k>0$. If $c$ is real and outside $[a,b]$, $F$ is again regular, but the same identity has a strictly positive integrand unless the mode vanishes. Such a neutral mode is impossible. Consequently neutral regular modes also satisfy the bound; no division by $c_i$ is needed for this last conclusion. The nonreal proof avoids critical-level singularities altogether.

An Orr-Sommerfeld mode has $\hat{v}\ne 0$ satisfying the Orr-Sommerfeld equation, with accompanying vorticity satisfying the forced Squire equation. A Squire mode has $\hat{v}=0$ and $\hat{\eta }\ne 0$, so its Squire equation is homogeneous.

For a Squire mode multiply that homogeneous equation by $\overline{\eta }$ and integrate over the channel. Under the no-slip condition $\eta =0$ at the walls, integration by parts givesSince $U,\alpha$ are real, taking the real part provesThe numerator is strictly positive for a nontrivial finite-channel mode, and $Re>0$. Thus Squire modes are exponentially damped in the convention $e^{-i\omega t}$. This does not rule out transient growth from non-normal modes in the coupled system.

For a sufficiently regular solution with zero Dirichlet boundary conditions, differentiate its squared L2 norm. Since $u_t=-i(\Delta u-Vu)$,Integration by parts and the zero boundary trace give $\int \overline{u}\Delta u=-\int |\nabla u{|}^2$. The potential contribution is real because $V$ is real. The quantity in parentheses is consequently real, and its product with $-i$ has zero real part. HenceThis is the energy method for a skew-adjoint evolution. For general L2 space data, the same result follows from the strongly continuous unitary group generated by a self-adjoint realization of $\Delta -V$. A sufficient assumption is a bounded real potential with the Dirichlet Laplacian; arbitrary real potentials require appropriate domain and self-adjointness assumptions. The differentiated proof presumes the regularity needed for the displayed integrals, rather than asserting it for every pointwise real function $V$.

Work in the real zero-boundary Sobolev space $X=H_0^1(0,1)$, whose members have square-integrable weak derivatives and zero endpoint trace. Define the bounded bilinear form and linear functionalThe weak formulation is to find $u\in X$ with $a(u,v)=\ell (v)$ for every $v\in X$. Integration by parts recovers the differential equation for a smooth solution; conversely, this identity is the definition of its weak solution.

Use the full $H^1$ norm. The Cauchy-Schwarz inequality gives $|a(u,v)|\le \Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$ and $|\ell (v)|\le \Vert v{\Vert }_{H^1}/\sqrt{3}$. Also $a(v,v)=\Vert v{\Vert }_{H^1}^2$, so $a$ is a coercive bilinear form with constant one. The Lax-Milgram theorem states that a bounded, coercive bilinear form on a real Hilbert space determines a unique weak solution for every bounded linear functional. It therefore applies here.

The equivalent minimization problem isIf $u$ is the weak solution, symmetry and $a(u,v-u)=\ell (v-u)$ giveThis is positive unless $v=u$, proving a unique global minimizer. Conversely, differentiating $\mathcal{E}(u+tv)$ at $t=0$ gives the weak identity. Thus the unique energy minimizer is exactly the unique weak solution. As a check, the classical representative isIt satisfies both endpoint conditions and $-u^{\prime \prime }+u=x$.

The starting point of a conforming finite element method is a weak formulation in a Hilbert space $X$. For example, a homogeneous Dirichlet boundary condition on a bounded Lipschitz domain gives $X=H_0^1(\Omega )$. One seekswhere $a$ is a bounded bilinear form, $|a(w,v)|\le L\Vert w{\Vert }_X\Vert v{\Vert }_X$, and $\ell \in X^{\ast }$ is a bounded linear functional. A conforming finite element space $X_h\subset X$ is finite dimensional and is typically built from functions that are polynomial on each element of a finite element mesh. With a basis ${\phi }_j$, the Galerkin method imposes the same identity only for $v_h\in X_h$, givingThe stiffness matrix is sparse when basis supports overlap only locally. Local element integrals assemble its entries; the mass matrix represents an $L^2$ term. This separates the choice of trial functions from the variational principle determining their coefficients.

The fundamental well-posedness theorem is the Lax-Milgram theorem: if $a(v,v)\ge \gamma \Vert v{\Vert }_X^2$ for some $\gamma >0$, there is a unique solution for each $\ell \in X^{\ast }$, with $\Vert u{\Vert }_X\le \Vert \ell {\Vert }_{X^{\ast }}/\gamma$. The same theorem on $X_h$ gives a unique discrete solution. The constants must be independent of the mesh for uniform stability of a numerical method. Boundary conditions are built into the trial space when they are essential constraints; flux conditions arise through boundary terms in integration by parts and are natural constraints in the weak formulation.

When $a$ is symmetric and a coercive bilinear form, the Ritz method minimizesover $X_h$. Its first variation is exactly the Galerkin method, so Ritz minimization and Galerkin testing coincide for symmetric coercive problems. The energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ turns the approximation into an orthogonal projection. Indeed, subtracting the exact and discrete weak equations gives Galerkin orthogonality, $a(u-u_h,v_h)=0$. For every $v_h\in X_h$ the Pythagorean identity becomessoThis exact best-approximation property is the central advantage of the Ritz method.

Symmetry is unnecessary for the Galerkin method. For any bounded, coercive form, the Céa lemma givesTo prove it, put $e=u-u_h$. Galerkin orthogonality yields $a(e,e)=a(e,u-v_h)$, so coercivity and boundedness give $\gamma \Vert e{\Vert }_X^2\le L\Vert e{\Vert }_X\Vert u-v_h{\Vert }_X$. Divide when $e\ne 0$ and take the infimum. Thus approximation capability plus uniform coercivity gives convergence. In contrast, minimizing $\frac{1}{2}a(v,v)-\ell (v)$ for a nonsymmetric form differentiates its symmetric part and generally does not solve the original weak equation.

A concrete symmetric example is the one-dimensional reaction-diffusion problem from Question 5. Continuous piecewise-linear hat functions give the diffusion stiffness matrix $S$ with diagonal $2/h$ and neighbouring entries $-1/h$, and the mass matrix $M$ with diagonal $2h/3$ and neighbouring entries $h/6$. Therefore $K=S+M$ and $F_i=i{h}^2$. Its positive-definite matrix is the coordinate form of the continuous energy. The exact smooth solution $u(x)=x-\sinh (x)/\sinh 1$ makes this an explicit test of the method, not just an abstract existence result.

For a nonsymmetric example, take positive $\epsilon$, constant real $\beta$, andIts form is $a(u,v)=\epsilon \int u^{\prime }{v}^{\prime }+\beta \int u^{\prime }v+\int uv$. Since $\int v^{\prime }v=0$, it has coercivity constant $\min (\epsilon ,1)$ in the full $H^1$ norm, although it is not symmetric when $\beta \ne 0$. The Galerkin method remains well posed. On the uniform hat basis, $K=\epsilon S+\beta C+M$, where $C_{i,i+1}=1/2$, $C_{i,i-1}=-1/2$ and $C_{ii}=0$. The skew-symmetric matrix $C$ cancels from $c^{T}Kc$, proving nonsingularity. A Ritz energy would omit precisely this advection contribution. Small $\epsilon$ may nevertheless make the approximation constant large and produce poorly resolved layers; algebraic solvability is not an accuracy guarantee.

Quantitative convergence uses a finite element interpolation estimate. On a shape-regular mesh, for piecewise-linear conforming elements and $u\in H^2(\Omega )$ in the usual low-dimensional setting,The Céa lemma therefore gives an $O(h)$ energy error. The Aubin–Nitsche duality argument improves the L2 norm error when the dual elliptic problem has $H^2$ regularity: solve $a(v,z)=(e,v{)}_{L^2}$ and useConsequentlywith constants depending on the solution and regularity bounds. The second estimate is conditional on dual regularity; corners or insufficient data regularity can reduce the rate.

The same Galerkin method handles evolution equations. For the homogeneous heat equation, the semidiscrete finite element heat equation is $M\dot{c}+Sc=0$. Because $M$ is a positive-definite matrix and $S$ is a positive semidefinite matrix,The decreasing quadratic form is exactly the L2 norm squared of the finite-element function. This separates spatial stability of a numerical method from the subsequent time integrator, just as in the Schrodinger equation example. Conformity, coercivity and approximation estimates together explain convergence.

Let $h_i=\delta F/\delta p_i$ be the polar molecular field, with the positive functional derivative convention used in the paper. During a pure advective displacement $u_j$, the polar order parameter changes byThe free energy change is $\delta F=\int h_i\delta p_i{d}^{d}r$. Assume periodic boundaries, or boundary conditions that eliminate the surface work, and incompressibility ${\partial }_j{u}_j=0$. Integration by parts in the advective term givesRelabelling the dummy indices in the rotational and flow alignment of a polar order parameter terms givesTherefore $\delta F=\int {\Sigma }_{ij}^p{\partial }_i{u}_j{d}^{d}r$, with the reversible stress of a polar liquid crystalFor a local free-energy density $f(\mathbf{p},\nabla \mathbf{p})$, an explicit choice isIts divergence is exactly $-p_k{\partial }_j{h}_k$, by the Euler-Lagrange equation for $h_k$. A pressure term can be reassigned in an incompressible flow; the chosen representative realizes the printed divergence without that ambiguity. The force density ${\partial }_i{\Sigma }_{ij}^p$ has mechanical power $-\int {\Sigma }_{ij}^p{\partial }_i{v}_j$, the negative of the advective free energy rate, which checks the stress sign.

Insertinto the equation. Since $w^{\prime }={\int }_{\gamma }t{e}^{zt}fdt$ and $w^{\prime \prime }={\int }_{\gamma }{t}^2{e}^{zt}fdt$, use $z{e}^{zt}={\partial }_t{e}^{zt}$ and integration by parts:The integral vanishes whenso, up to a constant and on a consistently chosen branch,The remaining contour condition isThis is the Laplace contour-integral method for a linear differential equation.

Write $D=d/dx$. For rapidly decaying smooth functions, integration by parts gives $D^{\ast }=-D$, while multiplication by the real functions $p$ and $q$ is self-adjoint. Thereforeand henceLikewise $D^2$ and multiplication by $q$ are self-adjoint, so their purely imaginary multiples in $A=3i{D}^2-4iq$ are anti-self-adjoint:

If $L\psi =\lambda \psi$ with $\psi \ne 0$, self-adjointness givesConsequently every eigenvalue of $L$ is real.

SetThe ground-state transform for a weighted Dirichlet energy, obtained by expanding the square and using integration by parts, isThe boundary term vanishes because $h$, and hence $g$, has compact support.

Since $V(x)\to \ell >0$, choose $R_0$ so that $V\ge \ell /2$ when $|x|\ge R_0$, and setFor every $R\ge R_0$, discard the nonnegative $|g^{\prime }{|}^2$ term and the nonnegative contribution from $R_0\le |x|\le R$ to obtainBecause $|g{|}^2=|h{|}^2{e}^{-\Phi }$, this is exactly the required estimate.

The intended integral representation of the modified Bessel function isDifferentiation under the integral sign givesSinceintegration by parts and the vanishing endpoint term yield $x{I}_0^{\prime \prime }+I_0^{\prime }-x{I}_0=0$.

For large positive $x$, Laplace's method localizes the integral near the endpoint $\theta =0$. Put $\theta =u/\sqrt{x}$ and useExtending the exponentially small tail to infinity and evaluating the resulting Gaussian integrals givesand therefore

Now write $y=x^{-1/2}w$. Direct differentiation givesso substitution into the ordinary differential equation givesFor the Liouville–Green approximation, put $Q(x)=1-1/(4x^2)$. The two approximate solutions areso at leading order $y_{\pm }(x)\sim x^{-1/2}{e}^{\pm x}$. The growing solution agrees with the preceding asymptotic expansion, including its leading normalization $I_0(x)\sim e^x/\sqrt{2\pi x}$.

For square-integrable wavefunctions with the stated decay, multiplication by the real coordinate $x$ is Hermitian becauseFor $p_x=-i\hslash {\partial }_x$, integration by parts givesand the boundary term vanishes. The same arguments apply to $y$ and $p_y$.

If $F$ and $G$ are Hermitian on a common suitable domain, then $(FG{)}^{†}=GF$, soBecause $x$ commutes with $p_y$ and $y$ with $p_x$,is Hermitian. Using $p_{x}x=x{p}_x-i\hslash$ and its $y$ analogue,so $D$ is Hermitian as well.

Finally set $R=x{\partial }_y-y{\partial }_x$ and $S=x{\partial }_x+y{\partial }_y$. Direct evaluation on a smooth test function gives $[R,S]=0$: rotations commute with radial dilations. The scalar term in $D$ also commutes with everything, and therefore

Write the admissible variation as $y_{\epsilon }=y+\epsilon \eta$, where $\eta (x)\to 0$ at both ends. Differentiating the functional under the integral sign gives the first variationAfter integration by parts, with the boundary term vanishing,Thus the Euler-Lagrange equation isA second differentiation gives the second variation

Let $D=d/dx$. Repeated integration by parts gives the formal adjoint rulesThereforeEquality of the coefficients of $D$ in $L$ and $L^{\ast }$ requiresso $q=p^{\prime }$. This also makes the zeroth-order coefficients equal because $p^{\prime \prime }-q^{\prime }=0$. Hence, under boundary conditions that remove the boundary terms,This is the self-adjoint fourth-order scalar differential operator $D^4+p{D}^2+p^{\prime }D+r$.

Introduce the Cauchy approximate identityIt is an approximate identity converging to the Dirac delta function $\delta$, and $g_{\epsilon }={\delta }_{\epsilon }^{\prime }$. For a smooth rapidly decaying test function $\varphi$, integration by parts givesBy the definition of the derivative of the Dirac delta, $⟨{\delta }^{\prime },\varphi ⟩=-{\varphi }^{\prime }(0)$. Therefore the limit as a distribution is

The Dirichlet boundary condition is built into the first component's space, while the Neumann boundary condition is natural. Thus a weak solution is a pairsuch that for every $(v,z)\in H_0^1(U)\times H^1(U)$,If $u,w$ are $C^2$ up to the boundary, taking compactly supported test functions and applying the fundamental lemma of the calculus of variations gives both differential equations pointwise in $U$. Membership of $H_0^1(U)$ gives $u=0$ on $\partial U$. Applying integration by parts to the second identity and using its differential equation leavesfor every smooth boundary trace $z$. Hence $\partial w/\partial \nu =0$ on $\partial U$, so the equations and both boundary conditions hold classically.

Put $a(y)=(1-y^2{)}^2$. Multiply $u_{xx}={\partial }_y(a{u}_y)$ by $2u_x$ and integrate over $-1<y<1$. An integration by parts givesbecause $a(1)=a(-1)=0$. Therefore the energy estimate is in fact the conservation lawIf $u(0,y)=u_x(0,y)=0$, then differentiating the first identity in $y$ also gives $u_y(0,y)=0$, so the conserved nonnegative energy is zero. Hence $u_x=0$ and $(1-y^2)u_y=0$ throughout the open strip. There $|y|<1$, so both derivatives vanish; connectedness and the initial value now give