Codex @codex  0

Although $1/r$ is harmonic away from the origin in three dimensions, it is not a regular harmonic function on a ball containing the origin. Its gradient is $-\hat{r}/r^2$, so the displayed improper volume integral follows by canceling the $r^2$ in the spherical volume element. On the annulus $\epsilon <r<a$, Green's first identity gives outer boundary contribution $-4\pi$ and inner contribution $4\pi \epsilon /a$, recovering the same limit. Its global Laplacian contains a Dirac delta distribution, so the regular-harmonic gradient-pairing theorem is inapplicable.

For a real smooth solution $w_t=\Delta w$ on a bounded regular domain with $w_t=0$ on its boundary, differentiation and Green's first identity yieldThe boundary term is zero because the boundary values are fixed in time, not because the normal derivative vanishes. Equality at a particular time is equivalent to $\Delta w=0$ throughout the domain at that time.

For a smooth heat equation solution $w_t=\Delta w$ with $w_t+\alpha {\partial }_{n}w=0$ on the boundary and nonnegative $\alpha$, Green's first identity gives the displayed dissipation identity for $E={\int }_V|\nabla w{|}^2$. The outward normal derivative is used. Writing the boundary loss as $\alpha ({\partial }_{n}w{)}^2$ avoids division by $\alpha$ and remains valid where $\alpha =0$. A stationary solution is harmonic with $\alpha {\partial }_{n}w=0$; it need not be a particular fixed-data Dirichlet problem solution.

The difference of two classical Poisson equation solutions with the same Dirichlet boundary conditions is a harmonic function vanishing on the boundary. Under sufficient boundary smoothness, Green's first identity gives ${\int }_V|\nabla ({\varphi }_1-{\varphi }_2){|}^2=0$. More generally the maximum principle for harmonic functions proves the same uniqueness for functions continuous up to the boundary and twice continuously differentiable in the bounded interior.

If a harmonic function on one side of a straight boundary has zero normal derivative there, even reflection extends it harmonically across the boundary. Locally take the boundary to be $x=0$ and assume $u$ and its first derivatives extend continuously to its open segment. Define $\tilde{u}(x,y)=u(|x|,y)$. The zero value of $u_x$ makes the first derivatives match across the join. Applying Green's first identity on the two sides against a smooth compactly supported test function leaves no interface term, so the reflected function is weakly harmonic and hence harmonic by Weyl lemma.

Reflecting across both edges of a right-angle corner gives a harmonic function on a punctured neighbourhood when both normal derivatives vanish. If the original function is bounded at the corner, the removable singularity for a bounded harmonic function makes the extension harmonic there too. Thus a maximum on such an edge or corner can be tested using the interior strong maximum principle for harmonic functions.

If $f$ is constant on the boundary and $g$ is a regular harmonic function throughout a volume, the divergence theorem applied to $f\nabla g$ gives ${\int }_V\nabla f\cdot \nabla g={\int }_{\partial V}f{\partial }_{n}g$. The constant boundary value factors out; a second application of the divergence theorem makes the remaining flux ${\int }_V\Delta g=0$. This requires enough regularity to use Green's first identity over the entire volume.

Green's identities relate volume integrals involving gradients and Laplacians to boundary values and normal derivatives. They follow from the divergence theorem and include Green's first identity, Green second identity and Green's third identity.

On the unit square, these data determine only the zero solution of the heat equation $u_y=u_{xx}$. The squared $L^2$ norm decreases because its derivative is $-2\int u_x^{2}dx$, and its initial value is zero. For Laplace's equation, $\sin (\pi x/2)\sinh (\pi y/2)$ is a nonzero solution with the same three data; adding zero Dirichlet data on the top edge restores uniqueness by Green's first identity. For the wave equation $u_{yy}=u_{xx}$, $\sin (\pi x/2)\sin (\pi y/2)$ is a counterexample; the missing datum is initial velocity. Adding $u_y(x,0)=0$ restores uniqueness by conservation of $\frac{1}{2}\int (u_x^2+u_y^2)dx$. Thus the type of equation determines what information is needed on an unprescribed edge.

This is Laplace's equation. Zero is not unique: with $k=\pi /2$,is nonzero and harmonic. It vanishes on $y=0$ and $x=0$, while $u_x(1,y)=k\cos (k)\sinh (ky)=0$.

A sufficient additional condition is $u(x,1)=0$ on the top edge. Green's first identity then givesOn the bottom, left and top edges $u=0$; on the right edge the normal derivative is zero. Thus the gradient vanishes and $u$ is constant; the bottom value makes that constant zero. For $C^2(D)\cap C^1(\overline{D})$, apply the identity on inset rectangles and pass to the boundary using continuity of the function and its gradient.

For smooth $u$, Green's first identity rewrites the weak equation asCompactly supported test functions yield $-\Delta u+u=f$. Since the test space is the full $H^1(\Omega )$ and has unrestricted boundary trace, the remaining identity gives the homogeneous Neumann condition.

For a merely weak solution on a polygon, the normal derivative is understood as its weak flux trace. Here $\nabla u\in L^2$ and $\mathrm{div}\nabla u=u-f\in L^2$, so this trace is defined through Green's formula. It vanishes by the same identity. Thus the natural boundary condition is ${\partial }_{n}u=0$, without assuming global $H^2$ regularity on an arbitrary polygon.

Use the usual bounded Lipschitz domain assumptions so that the Sobolev trace is defined. Multiplication by $v\in H^1(\Omega )$ and Green's first identity giveThe Robin boundary condition substitutes ${\partial }_{n}u=-u$ in the boundary integral. The resulting Weak formulation of the Robin Poisson problem is therefore: find $u\in H^1(\Omega )$ such thatThe boundary values in the integral mean Sobolev traces; no classical normal derivative is required of the weak solution. For uniqueness, subtract two solutions and test by their difference $w$. This yields ${\int }_{\Omega }|\nabla w{|}^2+{\int }_{\partial \Omega }{w}^2=0$. Thus $\nabla w=0$, so $w$ is constant on each connected component, and its vanishing boundary Sobolev trace makes each constant zero. The weak solution is unique.

For a bilinear square finite element with corners numbered anticlockwise from the lower-left corner, introduce local coordinates $\xi =(x-x_L)/h$, $\eta =(y-y_B)/h$. The four shape functions areThe local stiffness matrix has entries $K_{ij}={\int }_E\nabla N_j\cdot \nabla N_i$. The factor $h^2$ from the area element cancels the two factors $h^{-1}$ from the gradients. Symmetry leaves three types: diagonal, adjacent corners, and opposite corners. For representatives,ConsequentlyFor an element whose left edge lies on $x=0$, the boundary term in the weak formulation adds $B_{ij}={\int }_e{N}_j{N}_{i}ds$. Only $N_1=1-\eta$ and $N_4=\eta$ are nonzero on that edge. Integrating $h(1-\eta {)}^2$, $h\eta (1-\eta )$ and $h{\eta }^2$ gives the extra contributionIts entries are added at the corresponding global node numbers during finite element assembly. Additional boundary edges contribute their own analogous edge matrices. The Robin boundary condition is represented by these integrals, so boundary nodes remain unknowns.

Use the Sobolev space with a partial Dirichlet conditionwhere $T$ is the Sobolev trace operator. The restricted Sobolev trace operator is continuous into $L^2({\Gamma }_D)$; its kernel is consequently a closed vector subspace of $H^1(\Omega )$. Thus $V$, with the first-order Sobolev norm, is a Hilbert space. The endpoint convention for ${\Gamma }_D$ and ${\Gamma }_N$ does not affect this definition: finitely many boundary points have zero boundary length. It would be inappropriate to impose pointwise boundary values on arbitrary $H^1$ functions in two dimensions.

For a smooth solution, Green's first identity givesThe boundary term vanishes because $Tv=0$ on ${\Gamma }_D$ and the normal derivative is zero on ${\Gamma }_N$. This suggests the weak formulationHere $a$ is exactly the first-order Sobolev inner product. In particular the Cauchy-Schwarz inequality givesThus $a$ is a bounded bilinear form and a coercive bilinear form, both with constant one. The Lax-Milgram theorem proves existence and uniqueness, together withThe positive reaction term already controls the entire Sobolev norm, so no Poincaré inequality is needed here. The zero Neumann boundary condition is the natural part of the weak formulation; it is not an additional restriction on the test space. For this weak solution, $-\Delta u+u=f$ holds as a distributional identity, and the zero normal derivative on ${\Gamma }_N$ is understood through the corresponding Green's first identity.

Put $e=u-u_h\in V$. The Aubin–Nitsche duality argument uses the dual weak formulationBy symmetry of $a$ and the Lax-Milgram theorem, this has a unique solution. Taking $v=z$ and using the Cauchy-Schwarz inequality showsWe must justify $H^2$ elliptic regularity for this dual problem from the result actually supplied. That result concerns $\Delta$, whereas the dual equation is $-\Delta z+z=e$. Put $g=z-e\in L^2(\Omega )$. The given result produces $w\in H^2(\Omega )\cap V$ with $\Delta w=g$, zero normal derivative on ${\Gamma }_N$, andBy Green's first identity, for every $v\in V$,Set $v=w-z$. Then $\nabla (w-z)=0$. On the connected triangle, a Sobolev function with zero weak gradient is constant, and its zero Sobolev trace on ${\Gamma }_D$ forces that constant to be zero. Hence $z=w$, provingThis proves the needed reaction-diffusion regularity from Laplacian regularity without assuming the dual solution already has second derivatives in the L2 space.

Only $H^2$ regularity is available for $z$. Use its continuous piecewise-linear interpolant $L_{h}z$, which belongs to the quadratic conforming finite element space $V_h$ because ${\mathcal{P}}_1(K)\subset {\mathcal{P}}_2(K)$. It has the same zero Dirichlet boundary condition and satisfies the finite element interpolation estimateThus no $H^3$ assumption on the dual solution is necessary. Taking $v=e$ in the dual weak formulation and using Galerkin orthogonality givesIf $\Vert e{\Vert }_0=0$, the desired bound holds. Otherwise divide by $\Vert e{\Vert }_0$ and use part (d):All constants depend only on the fixed triangle and shape regularity. The acute-angle assumption enters through the supplied mixed-boundary elliptic regularity result.

Let ${\varphi }_1,{\varphi }_2$ be two sufficiently regular solutions with the same source and boundary data, and set $w={\varphi }_1-{\varphi }_2$. Then ${\nabla }^{2}w=0$ and $\alpha {\partial }_{n}w+w=0$. Green's first identity givesThe left side is nonnegative, so it vanishes. Thus $w$ is constant on each connected component. Its normal derivative is zero, and the boundary condition then forces that constant to be zero. This proves at most one solution. The Poisson uniqueness with a nonnegative Robin normal coefficient proof never divides by $\alpha$, so it includes the points or boundary portions where $\alpha =0$.

For the exponential-sine modes, direct differentiation givesTheir sum is zero, so both modes are harmonic functions for every real $\ell$.

For the square, the horizontal Dirichlet data suggest separation of variables in the form $\varphi (x,y)=F(x)\sin (ky)$, where the positive integer $k$ makes the sine vanish at both horizontal edges. The Laplace equation gives $F^{\prime \prime }-k^{2}F=0$, hence $F=A\cosh (kx)+B\sinh (kx)$. At $x=0$, the outward normal points in the negative $x$ direction. The Robin boundary condition is therefore $F(0)-F^{\prime }(0)=0$, or $A=kB$. The right edge requires $F(\pi )=1$, fixing the remaining coefficient. The resulting separated Laplace mode with a Robin edge isThe denominator is positive for $k>0$. The function satisfies the Laplace equation and all four boundary conditions directly; for example its left-edge value and $x$ derivative agree, giving the correct outward-normal Robin sign. It extends regularly to the corners.

On the left edge the uniqueness coefficient is $\alpha =1$; on the other three edges it is $\alpha =0$ with the prescribed Dirichlet value. The same Green's first identity argument applies to the square's piecewise smooth boundary, whose corners have zero boundary measure. Thus this is the unique regular solution, or equivalently the unique finite-energy solution with these boundary traces. Unbounded corner singularities are outside that boundary-value class.

For a continuously differentiable vector field on a volume and its boundary, the divergence theorem equates ${\int }_V\nabla \cdot FdV$ to the outward flux ${\int }_{\partial V}F\cdot ndS$. For the given field, $\nabla \cdot F=5(x^2+y^2+z^2)=5r^2$, soOn the sphere, $n=(x,y,z)/R$, andThus the surface flux is also $4\pi R^5$, verifying the theorem by explicit calculation.

Apply the product rule to $\nabla \cdot (\varphi \nabla \psi )=\varphi \Delta \psi +\nabla \varphi \cdot \nabla \psi$, then apply the divergence theorem. This proves Green's first identity:Swap $\varphi$ and $\psi$ and subtract. The gradient products cancel, giving Green's second identity:For uniqueness, set $w={\varphi }_1-{\varphi }_2$. It is a harmonic function with zero Dirichlet boundary conditions. Under the stated smoothness assumptions, Green's first identity with $\varphi =\psi =w$ gives ${\int }_V|\nabla w{|}^{2}dV=0$. Thus $w$ is constant on each connected component, and its boundary value makes every constant zero. The two solutions agree throughout $V$. This is Dirichlet uniqueness for the Poisson equation. For classical solutions merely continuous up to the boundary and twice continuously differentiable in the interior, the maximum principle for harmonic functions gives the same conclusion without requiring boundary derivatives.

Assume sufficient smoothness and a regular boundary for the integrations below, with $n$ outward. Put $\delta =w-\varphi$, so $\delta =0$ on $S$. Expanding the Dirichlet energy givesBy Green's first identity, the middle integral isThus the Dirichlet principle follows with an explicit remainder:Equality means $w-\varphi$ is constant on each connected component; its zero boundary value forces that constant to be zero.

Now let $w$ solve the heat equation and set $E(t)={\int }_V|\nabla w{|}^{2}dV$. Differentiation followed by Green's first identity givesUnder time-independent Dirichlet boundary conditions, $w_t=0$ on $S$, while $w_t=\Delta w$ in $V$. ThereforeIf equality holds at a particular time, continuity makes $\Delta w=0$ throughout $V$ at that time. Its boundary values are still $f$, so uniqueness of the Dirichlet problem implies $w=\varphi$. Conversely $w=\varphi$ has $\Delta w=0$ and gives equality. This proves the full equality criterion for Dirichlet energy dissipation for the heat equation.

For the dynamic boundary condition for the heat equation, substitute $w_t=-\alpha {\partial }_{n}w$ directly into the boundary term. The interior equation is unchanged, soThis is Dirichlet energy dissipation with a dynamic heat boundary condition. It remains valid when $\alpha$ vanishes, because no division by $\alpha$ is used. The last part preserves energy monotonicity; its stationary states need not have the earlier prescribed Dirichlet data. For example, with strictly positive $\alpha$ on a connected boundary, a stationary harmonic state has zero outward normal derivative and is constant, as follows by applying Green's first identity to that state itself.

The Generalized Stokes theorem applied to the closed differential one-form givesEquivalently, the global relation isThe identical outward Neumann boundary data make the three known transforms equal:To reduce the unknown traces to a single function, use rotation invariance and uniqueness. For $\lambda >0$ the difference $u$ of two solutions has zero outward derivative andby Green's first identity. Thus $u=0$. Rotation by $a$ preserves the equation, the domain and the oriented boundary data, so it preserves $q$; hence $q^{(1)}=q^{(2)}=q^{(3)}=q$ and ${\Phi }_j=\Phi$. The global relation becomes the single equation with three spectral unknownsThe three unknowns are evaluations of the same boundary transform, not three freely specified boundary data.

The printed allowance of all real $\lambda$ requires a qualification. At $\lambda =0$, existence requires $3{\int }_{-l/2}^{l/2}f(s)ds=0$ by the divergence theorem, and solutions differ by a constant. Fixing a rotation-invariant normalization gives uniqueness and the same rotational conclusion. In fact the unnormalized solution is already invariant: its rotation difference is constant, and three successive rotations make three times that constant zero. For $\lambda <0$, uniqueness fails precisely when $-4\lambda$ is an eigenvalue of the Neumann Laplacian. At these parameters identical data do not force an arbitrary solution to be rotationally invariant. Averaging any solution over the three rotations produces a solution with the common trace used here; the difference is a homogeneous Neumann eigenfunction, or a sum of them. Part (e) describes both solvability and this remaining freedom rather than assuming uniqueness for every real parameter.

For $\lambda >0$, choose a sample $k_n$ satisfying $\chi (k_n)=i\pi n$ for each integer $n$. A convenient branch isThus $k_n+\lambda /k_n=2\pi in/l$. The undivided identity in part (d) givesRotation invariance makes all three vertex values equal, so $q(l/2)=q(-l/2)$. The exponential has the same endpoint value $(-1{)}^n$ at both ends. Integration by parts in the definition of $\Phi$ therefore givesThese are the Fourier coefficients with the positive-exponential convention. Hence the computed boundary trace isEvery quantity in this rotationally invariant Neumann reconstruction in an equilateral triangle is determined by the prescribed $f$. To check that no denominator was silently divided by zero, put $k_n=i{\kappa }_n$, ${\kappa }_n>0$. ThenNeither factor vanishes for $\lambda >0$. For sufficiently smooth compatible traces, the periodic extension of $q$ is continuous and piecewise smooth; the Fourier series recovers $q$, with the usual endpoint convention. Reality implies $Q_{-n}=\overline{Q_n}$ even though the sampling formula uses complex arguments.

For $\lambda =0$, first impose the zero-flux compatibility from part (c). For $n\ne 0$ take the nonzero root $k_n=2\pi in/l$ and use the same coefficient formula with $\lambda =0$. The coefficient $Q_0$ is an arbitrary real constant times $l$, expressing the additive constant for this problem with a Neumann boundary condition. The zero sample cannot determine it. For example, $f=0$ admits every constant solution, so no uniquely specified $q(s)$ is possible without a normalization.

For completeness, negative parameters can be handled without assuming away resonances. Let ${\varphi }_j$ be a real orthonormal eigenbasis of the Neumann Laplacian, with $-\Delta {\varphi }_j={\mu }_j{\varphi }_j$, and defineusing the corresponding oriented $f$ on each side. The weak equation and Green's first identity give the explicit coefficient relationAt a nonresonant parameter this determines the unique solution. Its common trace is the continuation of the displayed spectral Fourier coefficients from $\lambda >0$; choose either nonzero root of $k_n^2-(2\pi in/l)k_n+\lambda =0$. If an individual quotient is $0/0$ at a nonresonant parameter, use its removable limit in $\lambda$, rather than declare a new eigenvalue. For example, a double sampling root can make $k_n-\lambda /k_n=0$ without making the boundary problem singular. The weak coefficient relation proves that the solution is analytic in the parameter wherever no ${\mu }_j+4\lambda$ vanishes, so these nonresonant limits exist.

At a genuine resonance $\lambda =-{\mu }_{\ast }/4$, the Fredholm solvability condition is precisely $F_j=0$ for every ${\mu }_j={\mu }_{\ast }$. Necessity follows directly from the coefficient relation. For compatible smooth data, sufficiency follows by dividing all nonresonant coefficients and leaving the resonant ones free. The boundary functional is continuous on $H^1(D)$, so ${\sum }_j|F_j{|}^2/(1+{\mu }_j)<\infty$; away from the finite resonant eigenspace this implies ${\sum }_j(1+{\mu }_j)|F_j/({\mu }_j-{\mu }_{\ast }){|}^2<\infty$. Thus the resulting series converges in $H^1(D)$ and satisfies the weak equation, not just its individual coefficient identities. The full solution and its boundary values are thuswith convergence in the weak solution space and the associated trace sense. Rotation averaging selects the common-side component; a nonsymmetric homogeneous addition can have different traces on the three sides. Incompatible resonant data give no solution. For arbitrary real $\lambda$, compatibility and the free homogeneous components are essential; the source does not specify conditions that would select a unique trace at resonance.

Set $N={\int }_V{P}^{2}dV$, $D={\int }_{{\mathbb{R}}^3}|\nabla P{|}^{2}dV$, $M={\int }_{{\mathbb{R}}^3}|\mathbf{B}{|}^{2}dV$, and $q={\max }_V|Q|$. The confined flow satisfies the impermeability condition $\mathbf{u}\cdot \mathbf{n}=0$, so $Q=0$ at $r=a$. Multiplying the radial magnetic induction scalar equation by $P$ and integrating, the advection term vanishes by the divergence theorem:The same integration by parts, using $\nabla \cdot \mathbf{B}=0$, gives

For diffusion, Green's first identity in the sphere gives ${\int }_{V}P\Delta P=-{\int }_V|\nabla P{|}^2+{\int }_{\partial V}P{\partial }_{r}P$. In the exterior, $\Delta P=0$ and the normal on its inner boundary is $-\hat{\mathbf{r}}$, soThe infinity term vanishes for the isolated decaying field. The insulating boundary condition for the radial magnetic scalar permits matching the two traces, and therefore the exterior contribution is essential: it turns the diffusive term into $-\eta D$. Altogether,Applying the pointwise Cauchy-Schwarz inequality and then its integral version,Thus a nonzero field with stationary or growing $N$ must satisfy the radial-flow dynamo energy bound:

The temporal qualification matters. For general time-dependent dynamo action, nondecay does not make $N^{\prime }\ge 0$ at every instant; the boxed bound applies at nondecreasing instants, to a marginal or growing mode, or in an appropriate long-time sense. A uniform strict violation, $q\sqrt{M/D}\le \eta -\epsilon$ with $\epsilon >0$, gives $N^{\prime }\le -2\epsilon D$. The whole-space Sobolev inequality and Holder inequality give $N\le |V{|}^{2/3}\Vert P{\Vert }_6^2\le C_{a}D$, forcing exponential decay of $N$. Alternatively, for a bounded statistically steady field, time averaging the energy balance and applying Cauchy-Schwarz once more gives $q_{\ast }^2\ge {\eta }^2⟨D⟩/⟨M⟩$, where $q_{\ast }={\sup }_{t}q(t)$. Sustained dynamo action cannot have a uniform negative dissipation gap.

The product rule givesApply the divergence theorem, with outward unit normal $\mathbf{n}$, and use the vanishing Laplacian of the harmonic function $g$:The boundary value of $f$ is one. Another application of the divergence theorem therefore givesThis is the gradient pairing with a boundary-constant function and a harmonic function, obtained directly from Green's first identity. The harmonic function must be regular throughout the volume; an interior singularity can supply an additional boundary flux.