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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.

By superposition, $y(x)={\int }_1^{e}G(x,\xi )f(\xi )d\xi$. Split the integral at $\xi =x$: the portion $\xi >x$ uses the first branch of the Green's function, and the portion $\xi <x$ uses the second. This yieldsFor $f(\xi )=1/\xi$, put $t=\ln x$. The two integrals areSubstitution and cancellation giveIndeed $y(1)=y(e)=0$. For a direct differential check,and thereforeThe homogeneous uniqueness shown in part (a) makes this the unique solution.

Branches and cuts. For the square root of z squared minus one, a branch point is a point around which analytic continuation along an arbitrarily small closed loop can change the value of a chosen branch of a multivalued function. Near $z=1$, the square root behaves as $\sqrt{2}(z-1{)}^{1/2}$; near $z=-1$ it is a nonzero analytic factor times $(z+1{)}^{1/2}$. One circuit around either point reverses its sign. Thus the finite branch points are $\boxed{z=\pm 1}$. Infinity is not a branch point: in $w=1/z$ the expression is $w^{-1}\sqrt{1-w^2}$, which has two single-valued meromorphic choices near zero.

A single branch cut can join $-1$ to $1$; its shortest choice is the real segment $[-1,1]$. Any surviving loop then surrounds either both branch points or neither, so it produces no sign change. Alternatively, use two branch cuts, $(-\infty ,-1]$ and $[1,\infty )$, each joining a branch point to infinity. On the resulting slit region one can choose a consistent branch of a multivalued function.

For the finite cut, fix $f(z)\sim z$ at infinity, equivalently $f(x)>0$ for $x>1$. To give its actual values on both sides, putFor $y\ne 0$, the selected square root isHere $\mathrm{sgn}(0)=0$; when $x=0$, $U=0$. Squaring verifies the formula because $UV=|xy|$. Its real part has the sign of $x$, its imaginary part has the sign of $y$, and it matches the chosen value at infinity. In particular, the two boundary values on the branch cut areOutside the cut, $f(x)=\mathrm{sgn}(x)\sqrt{x^2-1}$. Reversing the global branch of a multivalued function reverses every displayed value.

Differential equation. Write $f(\zeta )=\sqrt{{\zeta }^2-1}$ with this finite cut, and orient $C$ counterclockwise with one circuit around it. Since $C$ is fixed and avoids the branch points, differentiation under the integral sign givesOn the contour $f^{\prime }=\zeta /f$. ThereforeThe last contour integral vanishes because the expression in brackets returns to its original value after circling both branch points. This proves $\boxed{z{u}^{\prime \prime }+u^{\prime }-zu=0}$, including at $z=0$; division by $z$ is unnecessary.

Value at zero. Near either endpoint ${\zeta }_0=\pm 1$, the integrand at zero is $O(|\zeta -{\zeta }_0{|}^{-1/2})$. A small arc of radius $\epsilon$ has length $O(\epsilon )$ and hence contributes $O(\sqrt{\epsilon })\to 0$. The upper lip runs from $1$ to $-1$, and the lower lip from $-1$ to $1$. ThusUsing $x=\sin t$ evaluates the final integral as $\pi$, and soThe same deformation gives $u(z)=2i{\int }_{-1}^1{e}^{zx}(1-x^2{)}^{-1/2}dx$. The sign of $u(0)$ reverses if either the global square-root choice or the contour orientation is reversed; the differential equation is unaffected.

Choose $\gamma >1/2$. The inverse integrand is ${\lambda }^{-s}/[2(s-1/2)]$. Because $-\log \lambda >0$, the exponential factor decays towards the left. Closing the Bromwich contour to the left and using the residue theorem picks up just the simple pole at $s=1/2$, with residue ${\lambda }^{-1/2}/2$. HenceIts defining integral checks both the answer and its convergence condition:

Use the Fourier transform convention $\hat{f}(k)={\int }_{\mathbb{R}}f(t)e^{-ikt}dt$, with inverse factor $1/(2\pi )$. For functions in L1 space, their convolution is $h(t)={\int }_{\mathbb{R}}f(s)g(t-s)ds$. Fubini's theorem applies because the absolute double integral is $\Vert f{\Vert }_1\Vert g{\Vert }_1$. Substituting $r=t-s$ therefore proves the convolution theorem:For the specified rectangular pulse, the support condition is $|t|<1$. Its Fourier transform isThe convolution $f\ast f$ is the length of the overlap of $(-1,1)$ and $(t-1,t+1)$. Thus it is the triangular pulseThe transform's value at zero is four, given by its continuous extension. In this convention, the Parseval identity for square-integrable functions isFor the rectangular pulse the left side is two, giving the sinc-squared integralFor the triangular pulse, $\int |f\ast f{|}^{2}dt=2{\int }_0^2(2-t{)}^{2}dt=16/3$. Apply the same Parseval identity with transform $4{\sin }^{2}k/k^2$ to obtain the sinc fourth-power integralBoth integrands have removable singularities at zero and integrable tails.

Substitution of $y=x^p$ into the homogeneous Euler-Cauchy equation gives $(p(p-1)+p-1)x^{p-2}=(p^2-1)x^{p-2}$, so $\boxed{p=1,-1}$. Use the solution $u(x)=x$ that vanishes at zero and $v(x)=1/x$ that decays at infinity. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-2/x$. The half-line Green function for the first cylindrical harmonic is thereforeIt is continuous at $x=s$, its first derivative has jump $1/2-(-1/2)=1$, and it solves the homogeneous equation elsewhere. Thus $L_{x}G=\delta (x-s)$ for integration with ordinary measure $ds$. The boundary conditions hold for each fixed $s>0$.

Integrating against the forcing givesSince the two integrals are $2-(x^2+2x+2)e^{-x}$ and $e^{-x}$, respectively,A Taylor expansion gives $y=-x/2+O(x^2)$ near zero, while $y\sim -1/x$ at infinity. This verifies the two endpoint limits; direct differentiation verifies $Ly=e^{-x}$ on $x>0$. The differential expression itself is singular at zero, so the endpoint condition is imposed by continuity. No nonzero combination of $x$ and $1/x$ satisfies both homogeneous endpoint conditions, proving uniqueness.

Expand the divergence of $\Phi \nabla \Psi -\Psi \nabla \Phi$. The cross terms cancel, leaving $\Phi \Delta \Psi -\Psi \Delta \Phi$. The divergence theorem gives Green's second identity,Use the Green function convention ${\Delta }_{x}G(x,x_0)=\delta (x-x_0)$. Integrating over a bounded domain containing $x_0$ yields $\boxed{{\int }_S{\partial }_{n}GdS=1}$. Thus zero Neumann boundary data would be incompatible with this unit Dirac delta source. A suitable Neumann Green function with constant boundary flux instead has ${\partial }_{n}G=1/|S|$, where $|S|$ is the boundary area. Substituting $\Psi =G$, $\Delta \Phi =\sigma$, ${\partial }_n\Phi =f$ givesThe last term is an undetermined constant. The data must obey the Neumann Poisson problem compatibility condition ${\int }_V\sigma dV={\int }_{S}fdS$, obtained by integrating the original equation. Under this condition an additive constant in $G$ cancels from the first two terms.

For the upper half-space, reflect the source to $x_0^{\ast }=(x_{0,1},x_{0,2},-x_{0,3})$. The method of images with equal-sign sources givesThe reflected singularity lies outside the domain; on $z=0$ the two normal derivatives cancel, so ${\partial }_{n}G=-{\partial }_{z}G=0$. Its total unit flux passes through a large hemisphere at infinity, where $G\sim -1/(2\pi R)$. This is why the bounded-domain condition on the boundary flux cannot be applied to the infinite plane alone.

With zero interior source and the given point boundary source, the boundary representation is $\Phi (x_0)=-G((0,0,0),x_0)$. The terms at infinity vanish for this resulting $1/R$ field. ThereforeTo check the singular boundary condition, $-{\partial }_z\Phi =z/[2\pi (x^2+y^2+z^2{)}^{3/2}]$ is the Poisson kernel for the upper half-space. Its integral over a horizontal plane is one and its mass concentrates at the origin as $z\downarrow 0$, giving ${\partial }_n\Phi =\delta (x)\delta (y)$ in the distributional sense. This is the Neumann point source on a half-space boundary.

Choose the square root of one minus z squared with $f(0)=1$: explicitly $f(z)=\sqrt{1-z}\sqrt{1+z}$ with principal square roots. It is single valued and holomorphic away from the two specified exterior real branch cuts. If $q(x)=\sqrt{x^2-1}>0$ for $|x|>1$, its boundary values areFor example, above the right cut $1-z$ approaches the negative real axis from below, selecting $-i\sqrt{x-1}$; above the left cut $1+z$ approaches it from above, selecting $+i\sqrt{-x-1}$. Continuing once around either branch point changes the sign, which explains the need for cuts.

Apply the residue theorem to $F(z)=1/[zf(z)]$ on a large disk slit along both cuts, with small arcs excluding $\pm 1$. The only enclosed pole is at zero, and its residue is $1/f(0)=1$. On the positively oriented boundary of the slit domain, the upper right lip runs from one to $R$, and the lower right lip returns from $R$ to one. The jump in $F$ gives their combined integral as $2i{I}_R$, where $I_R={\int }_1^{R}dx/[x\sqrt{x^2-1}]$. The left lips similarly give $2i{I}_R$ after the substitution $x=-t$.

The small arcs contribute $O(\sqrt{\epsilon })$ because the branch singularities have order $-1/2$. The large arc contributes $O(R^{-1})$ because $F(z)=O(|z{|}^{-2})$. Taking limits therefore yields $4iI=2\pi i$, and henceThe real substitution $x=\sec \theta$ independently reduces the integral to ${\int }_0^{\pi /2}d\theta$, confirming its value and sign.

Choose the left homogeneous solution $v(x)=\sinh x$ and the right one $w(x)=\sinh (1-x)$. Their Wronskian is $v{w}^{\prime }-v^{\prime }w=-\sinh 1$. The Dirichlet modified Helmholtz Green function on an interval, with operator sign $D_x^2-1$, is thereforeIt vanishes at both endpoints, is continuous at $x=\xi$, and its first derivative jumps by $+1$, giving $(D_x^2-1)G=\delta (x-\xi )$. The Green-function representation then givesThe two integrals are $(e^{2x}-1)/4-x/2$ and $e(1-x)/2-(e-e^{2x-1})/4$, respectively. Substituting them and simplifying givesDirect differentiation verifies $u^{\prime \prime }-u=e^x$ and both zero boundary values. The negative kernel also explains why the solution is negative inside the interval despite the positive right-hand side.

Take a variation $y_{\epsilon }=y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. Differentiating the functional at zero and using integration by parts givesThe boundary term vanishes. Since the remaining integral must vanish for every smooth endpoint-zero variation, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation $d(f_{y^{\prime }})/dx=f_y$.

For the mechanical problem put $A(r)=1-2m/r>0$. The Lagrangian is independent of $\theta$, a cyclic coordinate, so its conjugate momentum $\partial L/\partial \dot{\theta }=2r^2\dot{\theta }=2J$ is conserved. Therefore $J$ is constant, including the factor of two between $J$ and the canonical momentum for this normalization of $L$.

For the radial coordinate, $\partial L/\partial \dot{r}=2\dot{r}/A$ and $\partial L/\partial r=-{\dot{r}}^2{A}^{\prime }/A^2+2r{\dot{\theta }}^2-\mu A^{\prime }/A^2$. Its Euler-Lagrange equation isFor a circular orbit $r=R>2m$, the radial velocity and acceleration vanish. Using $A^{\prime }(R)=2m/R^2$ and $\dot{\theta }=J/R^2$ leaves $2J^2/R^3=2m\mu /[R^{2}A(R{)}^2]$, henceThis is the circular orbit condition for a radial kinetic Lagrangian applied to the coefficients here.

Expand the radial modified Helmholtz equation as $Ly=-y^{\prime \prime }-2y^{\prime }/x+y$. Setting $y=z/x$ gives $Ly=(-z^{\prime \prime }+z)/x$, soBoundedness at zero requires $A+B=0$, whereas boundedness at infinity requires $A=0$. The respective adapted homogeneous solutions are thereforeThe first has limit one at zero. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-x^{-2}$. Since the highest-derivative coefficient of $L$ is $-1$, the Green function must have first-derivative jump $-1$. The radial modified Helmholtz Green function on the half-line isIt is continuous at the source, and its derivative jump is $-1$, giving $L_{x}G=\delta (x-\xi )$. For fixed source it tends to $\xi e^{-\xi }$ as $x\downarrow 0$ and decays at infinity. This kernel is for the original operator and ordinary measure $d\xi$; multiplying the equation by $x^2$ would change the source normalization.

For the compact source, split the integral at $\xi =x$ when $x<R$. The needed antiderivatives areThuswhile for $x>R$ the result is $e^{-x}{x}^{-1}{\int }_0^R\xi \sinh \xi d\xi$. Simplifying givesThe continuous value at zero is $1-(R+1)e^{-R}$, and $y^{\prime }(0)=0$. Both $y$ and $y^{\prime }$ match at $R$, so no extra Dirac delta appears there. The only homogeneous solution bounded at both ends is zero, establishing uniqueness.

Put $d=\sqrt{a^2-b^2-c^2}>0$ and $h=b-ic$. The denominator of the complex rational function becomesThe quadratic formula gives its two zeros, hence the two poles:Their product is $(b+ic)/(b-ic)$, so $|z_{\mathrm{in}}{z}_{\mathrm{out}}|=1$. To locate them one also needs to exclude the possibility of both lying on the unit circle. Here their moduli are explicit:Both are simple poles: $P^{\prime }(z)=hz+a$ takes the nonzero values $d$ and $-d$ at them. In particular their residues are $1/d$ and $-1/d$.

For the real integral, use the change of variables $z=e^{i\theta }$. As $\theta$ runs from $-\pi$ to $\pi$, this traverses the unit circle once counterclockwise, andThus $I=(1/i){\oint }_{|z|=1}f(z)dz$. Only $z_{\mathrm{in}}$ is enclosed. The residue theorem, or Cauchy integral formula after subtracting its simple-pole part, givesThe real denominator is positive because $a>\sqrt{b^2+c^2}$. As an independent check, write $b\cos \theta +c\sin \theta =\sqrt{b^2+c^2}\cos (\theta -\varphi )$; a period shift reduces the answer to the existing reciprocal affine sine integral.

First prove the convolution theorem with the stated negative-exponential Fourier transform. Assuming absolute integrability sufficient for Fubini's theorem, change variables $u=t-s$:

For the differential equation, integration by parts givesThe boundary terms vanish by the assumed decay of $y,y^{\prime }$. As with the convolution step, the transforms can first be justified for integrable functions and derivatives; the resulting kernel gives the desired decaying response. ThusThe convolution theorem and Fourier inversion theorem imply $y=f\ast G$.

For the inverse transform of $\tilde{G}$, useThe poles are at $ib,ia$. For $t>0$, close the inverse-transform contour above; the residues of $e^{ikt}\tilde{G}(k)$ there are $-i{e}^{-bt}/(a-b)$ and $i{e}^{-at}/(a-b)$. The inverse-transform factor $1/(2\pi )$ makes the answer $i$ times their sum. For $t<0$, close below, enclosing no poles. The large arcs vanish: the complete rational factor is $O(|k{|}^{-2})$ and the exponential is bounded on the appropriate semicircle. Therefore the causal Green function of an overdamped oscillator isIt is continuous, with $G(0)=0$ and $G^{\prime }(0^{+})=1$, so $(D_t+a)(D_t+b)G=\delta (t)$. The response can also be written $y(t)={\int }_{-\infty }^{t}G(t-s)f(s)ds$.

For the bilateral exponential forcing, split the real integral at zero:ConsequentlyFor negative time the lower half-plane contains only $k=-i$. The residue of $e^{ikt}\tilde{y}$ there is $i{e}^t/[(a+1)(b+1)]$. Clockwise closure makes the inverse transform $-i$ times that residue. For positive time the three upper residues, at $i,ia,ib$ respectively, areMultiplication of their sum by $i$ gives the bilateral exponential forcing of an overdamped oscillator:Both expressions have $y(0)=y^{\prime }(0)=1/[(a+1)(b+1)]$. This verifies the matching that prevents an extra Dirac delta function at zero. Direct substitution gives the forcing on either side, and all terms decay at their corresponding infinities. A causal Green function does not imply zero response before $t=0$ when the forcing has already been acting at negative times. The assumptions $a,b\ne 1$ avoid repeated poles; their limiting cases would contain polynomial times exponential terms.

The homogeneous equation is a Cauchy-Euler equation. Trying $y=x^m$ gives the repeated indicial rootHence, for $x>0$,Both independent solutions diverge at the origin, so no nonzero homogeneous combination satisfies the zero-origin conditions by continuous extension.

For a source point $\xi >0$, finiteness on $0<x<\xi$ forces the Green function to vanish on that side. On $x>\xi$, continuity and the unit first-derivative jump requireUsing the homogeneous solutions gives the causal Green function for a repeated-root Euler operatorIndeed, the Wronskian of $x^{-1}$ and $x^{-1}\log x$ is $x^{-3}$, and direct differentiation gives $G_x({\xi }^{+},\xi )=1$. Since $G$ is continuous, its second derivative produces precisely a unit Dirac delta, with no derivative of a delta. For each positive source it is identically zero near the origin, so both required origin values vanish. The differential operator itself is interpreted on $x>0$; no value of its singular coefficients at zero is needed.

Integrating against the forcing yieldsThe last integral equals $1/16$, by integration by parts or differentiating ${\int }_0^1{s}^{p}ds=1/(p+1)$. ConsequentlySubstitution gives $y^{\prime \prime }+3y^{\prime }/x+y/x^2=(6+9+1)x/16=x$, while $y(0)=y^{\prime }(0)=0$. The divergent homogeneous solutions prove uniqueness in the regular class. No boundary condition at infinity was imposed or required.

Complete the square in the Fourier transform exponent:The supplied shifted Gaussian integral therefore givesThis is the Fourier transform of a Gaussian, with angular-wavenumber convention $e^{-ikx}$.

Take the spatial Fourier transform of the heat equation. Two integrations by parts, with decaying fields and first derivatives at infinity, give $\mathcal{F}[{\psi }_{xx}]=-k^2\tilde{\psi }$. Thus each mode satisfies the ordinary differential equationInserting the transformed initial condition and integrating the time equation givesFinally set $A=t+1/(4\lambda )>0$. The inverse Fourier transform, evaluated by the same Gaussian identity, givesEquivalently the Gaussian initial data for the heat equation evolve asThe profile broadens and its peak decreases, while its total integral remains $\sqrt{\pi /\lambda }$, as required by diffusion without loss on the whole line. Direct differentiation verifies the equation and the initial limit.

The source factor cancels the singular denominator of the Green function. Splitting the integral at $\xi =x$ givesUsing $\int (1+\xi )e^{-\xi }d\xi =-(\xi +2)e^{-\xi }$, these two integrals are $e-(x+2)e^{-x}$ and $5/2-x-e(x+2)e^{-x}/2$, respectively. Simplification givesThe exponential and linear terms are homogeneous solutions, while direct differentiation gives $L(-x^2)=x$ for $x\ne 0$. Also $y(-1)=y(1)=0$, and the expression is analytic through zero. Uniqueness in this regular class follows because any difference is a global linear combination $A{e}^x+B(1+x)$: the left boundary forces $A=0$, and the right then forces $B=0$.

The boundary conditions on the positive half-line, absent from the transcription, select the half-harmonic oscillator. The trial state $x{e}^{-\alpha x}$ with $\alpha >0$ satisfies both endpoints and has finite norm and energy. Integration by parts, with vanishing endpoint term, givesUsing ${\int }_0^{\infty }{x}^n{e}^{-2\alpha x}dx=n!/(2\alpha {)}^{n+1}$, the three required integrals areThus the exponential trial state for a half-harmonic oscillator gives $\Lambda (\alpha )={\alpha }^2+3/{\alpha }^2$. Its derivative vanishes at ${\alpha }^4=3$, and the arithmetic-geometric mean inequality confirms the global minimum:For comparison, the exact lowest half-line state is proportional to $x{e}^{-x^2/2}$ and has eigenvalue three. The exponential trial family gives a valid upper bound rather than an exact eigenvalue.