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After integration by parts, define the Euclidean quadratic operatorThe action can then be written asEach component of $\mathbf{n}$ gives the same bosonic Gaussian functional integral, soDiscarding a $\lambda$-independent normalization, the remaining functional determinant gives

Take the real $L^2$ inner product of the equation with $u$. The homogeneous Dirichlet boundary conditions and integration by parts giveThus $\Vert u(t){\Vert }_{L^2}\le \Vert u_0{\Vert }_{L^2}$, and applying the same estimate to the difference of two solutions gives continuous dependence on the initial data. The constant-advection term is skew-symmetric under these boundary conditions and contributes no energy. Hence

Assume $p\in L^{\infty }(-1,1)$ with $p(x)\ge p_0>0$, $q\in L^{\infty }(-1,1)$ with $q\ge 0$, and $f\in L^2(-1,1)$. The natural energy space is the Sobolev space $V=H_0^1(-1,1)$. Multiplyingby $v\in V$ and using integration by parts gives the weak formulationwhereThe form is bounded, and the Poincaré inequality together with $p\ge p_0$ makes it coercive:The functional $\ell$ is bounded by the Cauchy-Schwarz inequality. The Lax-Milgram theorem therefore gives a unique weak solution.

Define the energyIf $y$ is the weak solution, then for every $w\in V$,Thus $y$ is the unique minimizer of $J$. Conversely, differentiating $J(y+tw)$ at $t=0$ recovers $a(y,w)=\ell (w)$, so the minimization and weak problems are equivalent.

The Ritz method chooses a finite-dimensional conforming space $V_h\subset V$ and minimizes $J$ over $V_h$. Its minimizer $y_h$ satisfiesFor a basis $\{{\varphi }_j\}$ this becomes $Kc=F$, withCoercivity makes the stiffness matrix $K$ symmetric positive definite, so the discrete minimizer exists and is unique.

Subtracting the continuous and discrete equations gives Galerkin orthogonality,In the energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$, for any $v_h\in V_h$,After cancellation and minimization over $v_h$, this proves the sharp symmetric form of the Céa lemma:

For a mesh $-1=x_0<x_1<\cdots <x_N=1$, take $V_h$ to be the continuous functions that are affine on each interval and vanish at the endpoints. The interior piecewise-linear hat functions form a basis, and each overlaps only its two neighbours, so $K$ is symmetric tridiagonal. In the illustrative case $p=1$ and $q=0$ on a uniform mesh, its diagonal and neighbouring entries areIf $y\in H^2(-1,1)$, its nodal interpolant $I_{h}y\in V_h$ satisfies $\Vert y-I_{h}y{\Vert }_{H^1}=O(h)$. The best-approximation estimate therefore gives $\Vert y-y_h{\Vert }_{H^1}=O(h)$, and a standard duality argument improves the $L^2$ error to $O(h^2)$ under the corresponding elliptic regularity. This completes the route from the boundary-value problem through its variational principle to a sparse, stable finite-element approximation.

Equality of the mixed derivatives of the auxiliary vector $\Psi$ givesThus the zero-curvature condition for this convention isSubstituting the displayed matrices and collecting powers of the spectral parameter $\lambda$ makes every diagonal entry and every $\lambda$-dependent term cancel. The remaining matrix isConsequently compatibility is equivalent toso the requested constant isThis is the AKNS Lax pair for the nonlinear Schrodinger equation.

The reductionis preserved by the two equations, which become complex conjugates and reduce to the Focusing nonlinear Schrodinger equationThe analogous reductiongives the Defocusing nonlinear Schrödinger equation

For a complex field $r$ and rapidly decreasing boundary conditions, defineTheir variational derivatives, after integration by parts, areBoth equations therefore have the Hamiltonian field equationor, including the conjugate equation,This is the Hamiltonian form of the cubic nonlinear Schrodinger equations.

Now putwhere $E$ and $f$ are real and $f$ is smooth and rapidly decreasing. For the focusing equation,Multiplication by $f^{\prime }$ and use of the decay at infinity gives the first integralA nonzero solution requires $E<0$. Writing $E=-{\kappa }^2$ with $\kappa >0$, separation of variables, or direct substitution, givesHenceis the Bright standing soliton of the focusing nonlinear Schrodinger equation.

For the defocusing equation the profile instead satisfieswith first integralIf a nonzero rapidly decreasing profile existed, $|f|$ would attain a positive maximum $A$. At that point $f^{\prime }=0$, so the identity forces $E=A^2>0$. But along either tail, where $0<|f|<A$, the right-hand side $f^2(f^2-E)$ is negative, which is impossible. Thus there is no nonzero solution of the prescribed form, as recorded by No rapidly decaying standing wave for the defocusing cubic nonlinear Schrodinger equation.

For $u(x)=\sin |x|$, the ordinary derivative away from zero isThis bounded function is locally integrable. For a test function $\phi$, integration by parts on $(-\pi ,0)$ and $(0,\pi )$ produces boundary terms at zero which cancel because $u$ is continuous there and $u(0)=0$. HenceTherefore $u$ has the weak derivativeThe jump in the ordinary derivative creates no Dirac delta function; such a term would arise from a jump in the function itself.

For $L=d^2/d{x}^2+1$ with homogeneous Dirichlet data at $0$ and $2\pi$, integration by parts shows that $L^{\ast }=L$. Its homogeneous kernel isbecause $A\cos x+B\sin x$ vanishes at both endpoints exactly when $A=0$.

The forcing obeys the orthogonality conditionThe Fredholm alternative for an elliptic Dirichlet problem therefore says that solutions exist, though they are not unique. Indeed,satisfies $u^{\prime \prime }+u=\cos x$ and both boundary conditions for every constant $C$.

SetThe Holder inequality givesUsing $x^{-p}dx=-(p-1{)}^{-1}d(x^{1-p})$ and integration by parts, while discarding the nonpositive boundary term at $x=1$, yieldsAnother application of Hölder's inequality givesAfter cancellation, with the zero case immediate,This is the Hardy averaging inequality.

Multiply the differential equation by $v\in H$ and integrate. The Hardy inequality on an interval makes the terms containing $u/x$ and $v/x$ integrable. For smooth $v$, integration by parts givesbecause $v(0)=0$ and the Neumann boundary condition is $u^{\prime }(1)=0$. ThereforeThe density of smooth functions in a Sobolev space and continuity of all four terms extend this weak formulation to every $v\in H$.

For compactly supported smooth variations $(\varphi ,\psi )$, differentiating at $t=0$ givesThis is the weak formulation. Taking $\psi =0$ or $\varphi =0$ separates its two equations. When $X,Y$ are smooth, integration by parts transfers each derivative from the test function; the fundamental lemma of the calculus of variations then recovers exactly the two pointwise equations in part (a).

Multiply $\Delta u+qu\le 0$ by ${\zeta }^2/u_{\epsilon }$ and integrate. Since $\zeta$ is compactly supported, integration by parts and Young's inequality giveAs $\epsilon \downarrow 0$, $u/u_{\epsilon }$ converges to the indicator of $\{u>0\}$. The dominated convergence theorem therefore gives

A cubic spline with knots $x_1<\cdots <x_n$ is twice continuously differentiable and restricts to a polynomial of degree at most three between consecutive knots. It is a natural cubic spline when it is linear outside $[x_1,x_n]$, equivalently when its second derivative vanishes at the two outer knots and on the exterior intervals.

Let $s$ be the natural cubic spline interpolating prescribed values and let $\tilde{g}=s+h$ be any other interpolant in $S_2[a,b]$. Then $h(x_i)=0$. Piecewise integration by parts, using that $s^{\prime \prime \prime \prime }=0$ between knots, $s^{\prime \prime }$ is continuous, $s^{\prime \prime }=0$ on the exterior intervals, and every jump of $s^{\prime \prime \prime }$ is multiplied by $h(x_i)=0$, givesConsequentlyEquality forces $h^{\prime \prime }=0$ almost everywhere. Then $h$ is affine and its zeros at at least two distinct knots force $h=0$, proving uniqueness.

For the penalized problem, fix a vector $u=(u_1,\dots ,u_n{)}^T$ of fitted values. The preceding variational result says that its natural spline interpolant has the least roughness among all functions taking those values, and by assumption that roughness is $u^{T}Ku$. The infinite-dimensional problem therefore reduces toBecause $K$ is positive semidefinite, $I+\lambda K$ is positive definite. The unique fitted-value vector isand ${\hat{g}}_{\lambda }$ is its unique natural cubic spline interpolant.

For Leave-one-out cross-validation, let ${\hat{g}}_{-i,\lambda }$ minimizeand defineIf $u_j^{(-i)}={\hat{g}}_{-i,\lambda }(x_j)$, its normal equations sayThus $u^{(-i)}=A_{\lambda }{\tilde{Y}}^{(i)}$, whereTaking the $i$th coordinate and writing $\hat{u}=A_{\lambda }Y$ givesTherefore one spline fit and the diagonal of its smoothing matrix give

Vary the gravitational volume factor and use part (a)(ii):Twice applying integration by parts moves both derivatives in $\delta R$ from the compactly supported $h_{ab}$ onto $f^{\prime }(R)$. Combining the result with the stated matter variation and requiring every coefficient of $h_{ab}$ to vanish gives the f(R) gravity equationTherefore

For $0<\lambda <1$, the preceding change of variables and one integration by parts giveSplit the last improper integral at one and add and subtract one on $(\lambda ,1)$. Since ${\int }_{\lambda }^{1}dx/x=-\log \lambda$,

The definition also gives the complex conjugate relationConsequently, for $-1<\lambda <0$,

LetThe integral is absolutely convergent. For a Schwartz function $\phi$, Fubini's theorem and integration by parts in $\lambda$ giveEquivalently, $v^{\prime }=-\hat{u}$ as a distributional derivative, and hence

For a test function $f\in \mathcal{D}(\mathbb{R})$, define the proposed integral by reversing the order of integration:The Fourier transform of a test function is a Schwartz function, while $\sqrt{{\theta }^2+1}\asymp 1+|\theta |$. The final integral is consequently absolutely convergent. Repeated integration by parts in $x$ bounds it by finitely many seminorms of the test function, so it defines a continuous linear functional on $\mathcal{D}(\mathbb{R})$.

Equivalently, put $\lambda =\sqrt{{\theta }^2+1}$ on the two half-lines. ThenThe density has only an integrable inverse-square-root singularity at one and is bounded at infinity, so it is a regular tempered distribution. Its inverse Fourier transform is exactly the proposed oscillatory integral. Thus

The centrifugal criterion concerns axisymmetric disturbances, so set $m=0$. The azimuthal equation givesEliminating $w^{\prime }$ with incompressibility and then $p^{\prime }$ with the axial equation yieldswhere the Rayleigh discriminant isImpermeability at the two solid walls gives $u^{\prime }(r_i)=u^{\prime }(r_o)=0$.

Multiply the equation by $r\overline{u^{\prime }}$, integrate between the walls, and use integration by parts. The boundary terms vanish and one obtainsThe denominator is positive. Therefore $\Phi \ge 0$ throughout the annulus excludes positive real ${\sigma }^2$ and gives centrifugal stability. Since $r^4{\Omega }^2=(r^2\Omega {)}^2$ is the square of the specific angular momentum, Rayleigh's circulation criterion isAn outward decrease of squared specific angular momentum permits an axisymmetric centrifugal instability.

Multiplyby $\hat{w}$ and integrate from the rigid bottom to an upper endpoint $z$ at which $\hat{w}{\hat{w}}_z=0$. The bottom term also vanishes because $\hat{w}(0)=0$. Integration by parts givesTherefore the horizontal wavenumber has the Rayleigh quotientFor a trapped wave the upper endpoint may be taken to infinity because the eigenfunction decays.

The clamped energy space isTwice applying integration by parts, with the boundary terms killed by the clamped conditions, gives the symmetric bilinear formConsequentlyfor every nonzero $u\in \mathcal{H}$: equality forces $u^{\prime \prime }=0$ almost everywhere, and the clamped boundary values then force $u=0$. Hence $L$ is symmetric and positive definite.

Strictly under the stated assumption $p\in L^2$ rather than $L^{\infty }$, the first integral need not be finite for every $u\in H_0^2$. The literal energy domain is therefore $\{u\in H_0^2:\sqrt{p}{u}^{\prime \prime }\in L^2\}$; under the usual coefficient assumption $0<p_0\le p\le p_1<\infty$, it is exactly $H_0^2$ and the form is coercive there.

Steady mass conservation gives $\dot{m}=-2\pi R\Sigma u_R$. With specific angular momentum $l=R^2\Omega$, multiply the angular-momentum equation by $2\pi$ and define the signed viscous torque in an accretion diskThenIntegration from $R_1$ to $R_2$ gives

The viscous power generated in an annulus is $2\pi R{F}_{\mathrm{diss}}dR=\mathcal{G}d\Omega$. Taking the inner torque to vanish and writing $l_{\mathrm{in}}=l(R_{\mathrm{in}})$ gives $\mathcal{G}=-\dot{m}(l-l_{\mathrm{in}})$, henceIntegration by parts yieldsFor steady circular force balance, $de=\Omega dl$; equivalently, the stated equality of gravitational- and rotational-potential differences makes the integral $e_{\mathrm{out}}-e_{\mathrm{in}}$. Therefore

When $R_{\mathrm{out}}\gg R_{\mathrm{in}}$, the outer energy and boundary term vanish. For an approximately Keplerian inner orbit, $e_{\mathrm{in}}=-R_{\mathrm{in}}^2{\Omega }_{\mathrm{in}}^2/2$, soComparison with part (c) givesA Keplerian inner flow generates the standard thin-disk power. A pressure-supported sub-Keplerian slim disk generates less through shear, and its emergent luminosity can be smaller still because photon trapping in an accretion flow carries part of that generated energy through the inner edge.

On the boundary the integral is ${\int }_{-\pi }^{\pi }\theta dy$. Integration by parts has zero endpoint term and leaves