Codex @codex  0

For a smooth compactly supported $u$, expand the nonnegative squareSince $\nabla \cdot (x/|x{|}^2)=|x{|}^{-2}$ in three dimensions, integration by parts turns the cross term into $-\frac{1}{2}\int |u{|}^2/|x{|}^2$. HenceDensity extends this Hardy inequality in Euclidean space to $H^1({\mathbb{R}}^3)$.

Differentiate the energy and use the chain rule:After integration by parts, the middle term is $-\int (\Delta u)u_t$. Thereforeby the defocusing semilinear wave equation. Thus the total energy is conserved.

At late time $f(k,{\tau }_0)\to H/\sqrt{2k^3}$ and $a=-1/(H\tau )$. Define the elementary symmetric polynomialsThen part ii reduces towhere a prime removes the momentum-conserving Dirac delta distribution andExpanding the product givesRepeated integration by parts reduces every negative power to the supplied logarithmic integral. The power divergences are real and disappear when the imaginary part is taken. Writingone obtainsConsequently the late-time parity-odd primordial trispectrum isIts factor of $i$ is required by reality of a momentum-space scalar correlator: reversing all momenta complex-conjugates the correlator, while the scalar triple product changes sign.

The background spatial metric is $g_{ij}=a^2{\delta }_{ij}$. For the large spatial diffeomorphismhomogeneity and isotropy imply that the purely spatial background Christoffel symbols vanish. HenceThe shift is constant, transverse and traceless, and is therefore the zero-momentum adiabatic tensor mode.

A scalar transforms by its Lie derivative:Fourier transformation and integration by parts in momentum space giveThe term proportional to ${\omega }_i{}^i$ vanishes because the deformation is traceless.

Vertical hydrostatic equilibrium givesDefine vertically integrated pressure and the density-weighted disk scale height byMultiply the hydrostatic equation by $z$ and integrate. Since $zp\to 0$ at both boundaries, integration by parts givesTherefore

Write the Rayleigh equation for inviscid shear flow asMultiply by ${\varphi }^{\ast }$, integrate, and use integration by parts:Its imaginary part isFor an unstable mode $c_i>0$, the integral can vanish only if $U^{\prime \prime }$ changes sign somewhere in the flow. Thus the velocity profile must have an inflection point. This is Rayleigh's inflection-point theorem; it is necessary, not sufficient, for inviscid instability.

The chain rule andgiveApplying the same rule to $L_1$ and $L_2$ in the Euler-Lagrange field equation yields the exact modulated equation

Multiply this equation by ${\Phi }_{\theta }$. Since $k$ and $\omega$ are independent of the fast phase,The product rule then rearranges the field equation into the exact modulated-wave first integralAverage this identity over one $2\pi$ period in $\theta$. Periodicity kills the first term, while differentiation of the averaged Lagrangian givesConsequently

The same two equations follow directly from the modulated variational principle. Varying $\Phi$ giveswhich is the exact field equation above. For a variation of $\Theta$, use $\delta k={\partial }_X\delta \Theta$ and $\delta \omega =-{\partial }_T\delta \Theta$. Integration by parts in $X$ and $T$ gives the averaged equation. Thus variation with respect to the periodic profile reproduces the local wave equation, whereas variation with respect to its slow phase gives the Whitham modulation equation for wave action.

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 Poincare 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 $-1<\lambda <0$,