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A three-dimensional Bravais lattice is the set of all integer combinations of three linearly independent primitive vectors. Its reciprocal lattice isand the first Brillouin zone is the Wigner-Seitz cell of the origin in reciprocal space.

For the stated FCC lattice, the primitive-cell volume isThe reciprocal-basis formula givesA reciprocal primitive cell, and hence the Brillouin zone, has volume

In the first Born approximation, with momentum transfer $q=k-k^{\prime }$, the scattering amplitude isFor the spherical shell, align the polar axis with $q$ and use its radial Dirac delta function:Therefore

Translation of each atomic potential multiplies its transform by a phase, so the crystal lattice structure factor givesFor an infinite lattice, the lattice Poisson summation formula turns the sum intoThus the amplitude vanishes away from reciprocal-lattice momentum transfers.

In elastic scattering, $|k|=|k^{\prime }|$ and henceThe two shortest relevant reciprocal-vector lengths of this FCC lattice areBoth are kinematically available when $k>2\pi /a$. Assigning their scattering angles ${\theta }_2$ and ${\theta }_1$, respectively, gives

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

Expand $g$ in its Fourier sine seriesBy separation of variables, the decaying solution of Laplace equation in each half-strip has $n$th mode proportional to $e^{-k_n|x|}\sin (k_{n}y)$. Continuity at $x=0$ and integration of the equation across the Dirac delta function require the normal-derivative jumpThe jump of $C_n{e}^{-k_n|x|}$ is $-2k_n{C}_n$, so

For $g(y)=\delta (y-c)$, its Fourier coefficients areThusTranslation in $x$ gives the Dirichlet Green function for an infinite strip with source $(b,c)$:

Fordirect differentiation givesandThus ${\theta }_t=D{\theta }_{xx}$.

Its total mass is found from the Gaussian integral:The unit-mass solution converging to the Dirac delta function as $t\downarrow 0$ is therefore the heat kernel

For the second initial condition, the heat-kernel solution is the convolutionWith $u=(x-y)/(2\sqrt{Dt})$ and the definition of the error function, this becomesThis is the heat equation with interval-indicator initial data for $a=1$.

At $t=0$ the graph is the rectangle of height one on $[-1,1]$. For every $t>0$ it is smooth, positive and even, with its maximum at $x=0$. As time increases the graph broadens and its maximum falls, while its total area remains $2$; pointwise it tends to zero as $t\to \infty$.

The exact mode functions satisfy the Wronskian normalizationUsing the Dirac delta function generated by the ladder-operator commutator therefore gives the exact canonical commutation relationThis remains true on superhorizon scales, so the exact operators do not literally commute. A claim of classical behavior must instead compare this fixed commutator with the growing statistical fluctuations, or appeal to quantum decoherence.

The solution $x+\lambda$ satisfies the left boundary condition, while $e^{-x}$ satisfies decay at infinity. The proposed expression for $x<\xi$ is therefore the correct left homogeneous solution. For $x>\xi$, write $G=C{e}^{-x}$. Continuity at $x=\xi$ givesHence the Green function isIndeed, its derivative has the required unit jump,which produces the Dirac delta distribution in $L[G]$.

The Fourier transform of a derivative givesSince $p^{\prime }=g$,An arbitrary additive constant in $p$ affects only the zero-frequency Dirac delta function when transforms are interpreted as distributions.

The values of $\log i$ are $i(\pi /2+2\pi n)$. HenceThese are infinitely many distinct positive real numbers.

Using Euler's formula and the Inverse Fourier transforms of phase factors,andThese are distributional identities, with $\delta$ the Dirac delta function.

LetConvolution with the initial Dirac delta function translates the heat kernel, while part (b) propagates the impulsive source from time one. HenceAt $(x,t)=(0,2)$ this becomesIt vanishes exactly whenThis is the cancellation of two heat-kernel impulses.

For a point charge, the charge density and current density areInsert these into the retarded electromagnetic potential. The argument of the Dirac delta function depends on $x^{\prime }$ throughAt its unique zero $x^{\prime }=y(t_{\mathrm{ret}})$, the Jacobian determinant contributes the factorSince the source is subluminal, this quantity is positive. Writing $\mathbf{R}=x-y(t_{\mathrm{ret}})$, $R=|\mathbf{R}|$, and evaluating $\mathbf{v}$ at $t_{\mathrm{ret}}$ gives the Lienard-Wiechert potentialsThe identity ${\mu }_0{ϵ}_0{c}^2=1$ equivalently gives $\mathbf{A}=\mathbf{v}\varphi /c^2$.

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.

Letwhere $\mathcal{H}$ is the Heaviside step function. Thenso the Eliassen equation for residual circulation isTake no normal residual flow at $y=0,L$ and $z=0$, decay as $z\to \infty$, and choose the streamfunction constant on the connected rigid boundary to be zero. Thus

The required Fourier series in sine modes isDefineFor each mode, the vertical equation isThe Dirac delta function requiresThe solution satisfying the boundary conditions is thereforewhere

The momentum equation givesThe jump of ${\overline{\chi }}_a^{\ast }$ supplies an equal positive delta function in $f_0{\overline{\chi }}_{a,z}^{\ast }$, so the singular terms cancel. The regular acceleration iswhereFinally, the transformed density equation givesThese exponentially decaying modes are the balanced mean response to wave-activity deposition at $z=H$.

The two-dimensional Fokker-Planck equation iswith Fokker-Planck probability currentThe point initial condition iswhere $\delta$ is the Dirac delta function. The reflecting boundary condition for a diffusion iswith $\mathbf{n}$ the outward unit normal. This zero-flux condition conserves the integral of the probability density function over the square.

False. The constant distribution $u=1$ is tempered, but its Fourier transform is a multiple of the Dirac delta function, which is not an $L^2$ function after multiplication by any Sobolev weight. Thus $1\notin H^t({\mathbb{R}}^n)$ for every $t$.

Change variables $x={\lambda }_{n}y$ and use the profiles from part 5:Since $v_n\to Q{e}^{i\gamma }$ in $L^2$, the difference made by replacing $|v_n{|}^2$ with $|Q{|}^2$ tends to zero against the bounded function $\varphi ({\lambda }_n\cdot )$. For each fixed $y$, continuity gives $\varphi ({\lambda }_{n}y)\to \varphi (0)$, and dominated convergence then givesThus the mass measures $|u(t_n,x){|}^{2}dx$ converge weakly to the Dirac delta function $\Vert Q{\Vert }_2^2{\delta }_0$.

Adapted homogeneous solutions. First assume real $\kappa \ne 0$. The functions $u(x)=\cosh (\kappa x)$ and $v(x)=\cosh (\kappa (1-x))$ solve $z^{\prime \prime }-{\kappa }^{2}z=0$ and obey $u^{\prime }(0)=v^{\prime }(1)=0$. Their Wronskian isIt is independent of $x$ and nonzero.

Green function. For $0<\xi <1$, continuity at $x=\xi$ and the jump condition $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$ follow by matching the Dirac delta function source. In particular a jump in $G$ itself would produce a derivative of the Dirac delta function, which is absent. Write $G=A(\xi )u(x)$ for $x<\xi$ and $G=B(\xi )v(x)$ for $x>\xi$. Solving the continuity and derivative-jump equations gives $A=v(\xi )/W$, $B=u(\xi )/W$. Hence the Neumann Green function for the modified Helmholtz equation on an interval isThe two pieces coincide at the source. Their derivative difference is $W/W=1$ and their endpoint derivatives vanish. The same expression has continuous one-sided limits as the source approaches an endpoint; the interior-source construction is what is needed in the integral solution.

Solution for a general source. By superposition, $y(x)={\int }_0^{1}G(x,\xi )f(\xi )d\xi$. Splitting the integral at $\xi =x$ givesNo nonzero homogeneous solution satisfies both endpoint conditions when $\kappa \ne 0$: the condition at zero eliminates $\sinh (\kappa x)$, and the condition at one then eliminates $\cosh (\kappa x)$. Thus the Green function gives the unique solution.

The printed zero-parameter case. The statement permits real $\kappa$, but $\kappa =0$ is exceptional. Integrating a putative equation $G^{\prime \prime }=\delta (x-\xi )$ would give $G^{\prime }(1)-G^{\prime }(0)=1$, contrary to the two zero derivatives. Thus this Green function does not exist at zero. For $y^{\prime \prime }=f$ with homogeneous Neumann boundary conditions, solvability instead requires ${\int }_0^{1}f=0$, and a compatible solution has an arbitrary additive constant. This is the Fredholm solvability condition associated with the constant zero mode.

Explicit forcing. For $f(x)=x$ and $\kappa =1$, the particular solution is $-x$. Write $y=C\cosh x+D\sinh x-x$. Then $y^{\prime }(0)=0$ gives $D=1$ and $y^{\prime }(1)=0$ gives $C=(1-\cosh 1)/\sinh 1=-\tanh (1/2)$. ThereforeDirect differentiation gives $y^{\prime }=\cosh (x-1/2)/\cosh (1/2)-1$, which vanishes at both endpoints, and $y^{\prime \prime }-y=x$, verifying the solution independently of the Green function integration.

Transform convention and derivatives. Use the Fourier transform pairFor sufficiently decaying $f,f^{\prime }$ with integrable second derivative, two applications of integration by parts giveThe identity also holds for tempered distributions, which is the setting needed for the Dirac delta function boundary data.

Upper half-plane solution. Take the Fourier transform in $x$ only, writing $\Phi (k,y)=\tilde{\varphi }(k,y)$. Laplace's equation becomes ${\Phi }_{yy}-k^2\Phi =0$. For $k\ne 0$, discard the exponentially growing solution and use $\Phi (k,0)=1$, the Fourier transform of the Dirac delta function. This gives $\Phi (k,y)=e^{-|k|y}$. The value at $k=0$ can be set to its continuous limit one. Pointwise decay of $\varphi$ as $y\to \infty$ does not require this individual transformed value to decay: the total integral in $x$ remains one while the profile spreads.

The inverse Fourier transform now yields, for $y>0$,Evaluate the last integral as the real part of ${\int }_0^{\infty }{e}^{-(y-ix)k}dk=(y-ix{)}^{-1}$:This is the Poisson kernel for the upper half-plane. Its boundary condition is a distributional convergence, rather than an ordinary value at $(0,0)$: for a smooth compactly supported test function $g$,The dominated convergence theorem applies because $g$ is bounded and $(1+t^2{)}^{-1}$ is integrable. Also $0\le \varphi (x,y)\le 1/(\pi y)$, proving the required decay as $y\to \infty$.

Direct verification. Differentiating the rational expression givesTheir sum is zero throughout $y>0$, so the solution is a harmonic function there.

Choose the homogeneous solutions $u(x)=x\ln x$, which vanishes at $1$, and $v(x)=x(\ln x-1)$, which vanishes at $e$. Their Wronskian in the order $u{v}^{\prime }-u^{\prime }v$ is $x$. For a source point $1<\xi <e$, seek the Green's function asIt is continuous at $x=\xi$; otherwise its second distributional derivative would contain a derivative of the Dirac delta function, which the source does not contain. Integrating the differential equation across the source, using continuity and the leading coefficient one, gives the jump conditionContinuity and this derivative jump requireSolving with the Wronskian $\xi$ gives $A=v(\xi )/\xi =\ln \xi -1$ and $B=u(\xi )/\xi =\ln \xi$. ThereforeThe two values at $x=\xi$ are identical. The derivative jump is explicitlyBoth endpoint values vanish and the differential expression is zero off the diagonal, completing the verification of the Green's function.

Take $x\ge 0$ and a source position $\xi >0$. Before the source, the zero initial data require $G=0$. After it, set $G=e^{-2(x-\xi )}h(x-\xi )$; the homogeneous equation becomes $h^{\prime \prime }+a^{2}h=0$. Continuity of $G$ and a unit jump in $G_x$ at $x=\xi$ give $h(0)=0$, $h^{\prime }(0)=1$. Thus the causal Green function of a damped oscillator isHere $H$ is the Heaviside step function. For $a=0$, take the continuous limit $\sin (as)/a\to s$, giving $G=H(x-\xi )(x-\xi )e^{-2(x-\xi )}$. The formula is also unchanged when a nonzero $a$ changes sign. Its continuity removes a delta derivative, and its unit first-derivative jump supplies the required Dirac delta function.

For the forced equation use $a=2$. The zero-initial-data Green function representation isThe sine-product identity reduces this to elementary exponential-trigonometric integrals. Evaluating givesFor an independent check, the steady particular part is $\sin 2x/20-\cos 2x/10$, while the decaying homogeneous part has coefficients $1/10$ and $1/20$ that impose $y(0)=y^{\prime }(0)=0$.

For a bounded region $R$ with piecewise smooth boundary and a continuously differentiable vector field, the divergence theorem gives ${\int }_R\nabla \cdot \mathbf{F}dV={\int }_{\partial R}\mathbf{F}\cdot \mathbf{n}dS$, with outward unit normal $\mathbf{n}$. Apply it to $\mathbf{F}=u\nabla v-v\nabla u$. The cross terms cancel, giving the Green's second identityThe $m^{2}uv$ terms cancel identically. Sending $R\to \infty$ proves the whole-space identity provided the boundary integral tends to zero. The boundary term at infinity in Green's second identity requires adequate derivative decay as well as field decay; mere pointwise convergence $u,v\to 0$ does not suffice. The screened kernels used below satisfy the required decay.

Take $m>0$. Apply the identity with $u=\varphi$ and $v=G(\cdot ,\mathbf{r})$, treating the source through its Dirac delta function, or excising a small source sphere and taking its radius to zero. It gives $\varphi (\mathbf{r})-\int G({\mathbf{r}}^{\prime },\mathbf{r})\rho ({\mathbf{r}}^{\prime })d{V}^{\prime }=0$. Symmetry of the radial Green function then yields the requested representation.

For separation $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|>0$, the modified Helmholtz equation reduces to $(sG{)}^{\prime \prime }-m^2(sG)=0$. Decay selects $G=C{e}^{-ms}/s$. Integrating the equation over a small sphere gives $4\pi s^2{G}^{\prime }\to 1$, so $C=-1/(4\pi )$. ThusFor $\rho =e^{-\beta r}/r$ with $\beta >0$, put $h=r\varphi$. Then $h^{\prime \prime }-m^{2}h=e^{-\beta r}$. When $\beta \ne m$, decay removes $e^{mr}$, while $h(0)=0$ removes an additional point source at the origin. HenceSince $G$ has a minus sign, the positive Yukawa kernel convolution isThe coincident-rate limit $\beta =m$ is $e^{-mr}/(2m)$ for the convolution, and the origin limit for general rates is $1/(m+\beta )$. Both are finite; the apparent denominators in the distinct-rate formula are removable limits.