Close the real segment $[-R,R]$ by an anticlockwise semicircle in the upper half-plane. For $F(z)=1/[(z^2+1)(z^2+4)]$, the contour integral over the arc is $O(R^{-3})$: its length is $\pi R$, whereas $|z^2+1|\ge R^2-1$ and $|z^2+4|\ge R^2-4$. The enclosed simple poles are $i$ and $2i$, with residuesLetting $R\to \infty$ in the residue theorem yields

Place the interface at $y=0$, with source $(x_A,h_1)$ and observer $(x_B,-h_2)$ in the two uniform media. Let $(s,0)$ be the crossing point. The optical length isDifferentiating with respect to $s$ and setting the result to zero givesRecognizing the ratios as the sines of the angles measured from the interface normal proves Snell's law:The stationarity condition is the continuity of the tangential optical momentum across the planar interface.

Choose the mirror as the line $y=0$, the source as $(x_A,h_A)$ and the observer as $(x_B,h_B)$, with positive heights. Within the plane of incidence, let the reflection point be $(s,0)$. In a uniform medium the optical length is a constant index timesStationarity with respect to the point on the mirror givesThese two ratios are the sines of the angles to the normal, with the stationary point between the two projected endpoints. ThereforeThe angles lie in $[0,\pi /2)$, where sine is injective. Varying a second tangential coordinate in three dimensions also enforces that the source, normal and reflected ray lie in one plane. This derives the law of reflection from Fermat principle.

The complex exponential never vanishes. The zeros of $\sinh z$ occur at $z=n\pi i$, $n\in \mathbb{Z}$, and they are simple because $\cosh (n\pi i)=(-1{)}^n\ne 0$. The numerator is $e^{2n\pi i}=1$ there, so no cancellation occurs. ThereforeFor example their residues are $1/\cosh (n\pi i)=(-1{)}^n$.

Infinity requires care: the poles themselves tend to infinity. Equivalently $f(1/w)$ has poles at $w=1/(n\pi i)$ accumulating at zero. Thus $\boxed{\infty \text{ is a non-isolated singularity}}$, and it is not a zero. In the usual classification of isolated singularities, it cannot be called a removable singularity, pole or isolated essential singularity. The non-isolated singularity distinction is necessary even though the function is meromorphic at every finite point.

The denominator factors as $z^2+iz+6=(z-2i)(z+3i)$, with no cancellation against the numerator. Thus there areTo inspect infinity, put $w=1/z$:Hence infinity is a removable singularity at infinity, with limiting value zero. After this natural extension to the Riemann sphere it is also a simple zero. There are no other zeros or singularities.

The original PDF shows $x^2$ in the nonzero source branch. Inserting that source into $G_0$ gives, for $x<1/2$,For $x\ge 1/2$, split the integral at $z=x$:Because of the homogeneous freedom established in part (i), the full answer to the stated endpoint problem isAt the interface, both branches have value $-5/96$ and derivative $-5/48$, so no extra delta or delta derivative occurs there. Direct substitution gives zero source on the left and $x^2$ on the right; the endpoint values are zero for every $C$. If the extra condition $y/x\to 0$ is imposed, it selects $C=-5/48$, but that is additional information absent from the printed question.

Choose $y_a=x$ and $y_b=x^3-x$, satisfying the left and right conditions respectively. Their Wronskian isPart (a) therefore supplies the valid kernelIt is continuous at the source, its first derivative has jump one, and it vanishes at both endpoints. However, it is not unique: $x^3-x$ is a nonzero homogeneous solution satisfying both stated endpoint conditions. Thus $G_0+(x^3-x)h(z)$ has the same defining properties for any suitable $h$.

This is a nonunique Green kernel at a singular endpoint with two vanishing modes. Choosing $y_a=x^3$ instead also gives a valid kernel, corresponding to the stronger selection $y(x)/x\to 0$ at the left endpoint. The original problem states no such selection, so the family must be acknowledged.

With the stated parameters, the mass and restoring matrices areThe generalized eigenvalue problem givesTherefore the two normal angular frequencies areThe second row of the modal equation is $-{\omega }^{2}x+(1-{\omega }^2)\varphi =0$. It gives the mode shapesEach normal mode has arbitrary small amplitude $a$ and phase $\delta$. The bob's horizontal displacement is $X=x+\varphi$ to first order. Thus in the lower mode, bob and cart move in phase, with bob amplitude $1+\sqrt{2}$ times the cart amplitude. In the higher mode they move oppositely, because $X=(1-\sqrt{2})x$; the bob amplitude is $\sqrt{2}-1$ times the cart amplitude. Its vertical displacement is quadratic in the angle and is absent at this order.

Choose positive $\varphi$ so the bob is displaced to the right relative to the cart. Its Cartesian coordinates areDifferentiation gives ${\dot{X}}^2+{\dot{Y}}^2={\dot{x}}^2+2L\cos \varphi \dot{x}\dot{\varphi }+L^2{\dot{\varphi }}^2$. Thus the exact kinetic energy and potential energy areRetain quadratic terms, using $\cos \varphi =1+O({\varphi }^2)$ in the kinetic cross term and $1-\cos \varphi ={\varphi }^2/2+O({\varphi }^4)$ in the potential. The Lagrangian isIts Euler-Lagrange equations giveThe kinetic cross term is essential: acceleration of the support couples the pendulum and cart. These equations define the normal modes of a spring-mounted pendulum cart.

The divergence of a cross product identity and the vacuum Maxwell equations implyThe divergence theorem therefore gives the required balance withAn additive time-independent constant in $\lambda$ is immaterial. This is the vacuum Poynting theorem in the normalization of the given equations: equivalently ${\partial }_t\lambda +\nabla \cdot (\mathbf{E}\times \mathbf{B})=0$. The order of the cross product accounts for the plus sign in the stated integral identity.

The same Maxwell equations givewhere the last equality uses $\nabla \cdot \mathbf{E}=0$. Combining with part (ii),The divergences of the symmetric gradients satisfyThus the local identity is ${\partial }_t(u_i^E+u_i^B)={\partial }_j(S_{ij}^E-c^2{S}_{ij}^B)$. Integrate over the fixed volume and apply the divergence theorem componentwise:Here $d{A}_j=n_{j}dA$ uses the outward normal vector.

Differentiate ${\mathbf{u}}^E=\frac{1}{2}\nabla \times \mathbf{E}$ with respect to time. The vacuum Maxwell equations give ${\partial }_t\mathbf{E}=c^2\nabla \times \mathbf{B}$, soThe curl of the curl identity is $\nabla \times (\nabla \times \mathbf{B})=\nabla (\nabla \cdot \mathbf{B})-{\nabla }^2\mathbf{B}$. Since $\nabla \cdot \mathbf{B}=0$,

Apply the symmetric gradient and curl decomposition separately to the electric field and the magnetic field:The corresponding axial parts areThe second equalities use the vacuum Maxwell equations. Their divergence constraints also show that both symmetric gradient tensors are traceless.

At $\Omega =\kappa$, the inverse-transform denominator is $-(\omega -i\kappa {)}^2$. For $t>0$, its upper-half-plane double-pole residue is the derivative of the exponential:It is also the continuous limit of both answers above as $\beta \to 0$ or $\eta \to 0$. This critically damped response tends to zero with a linear factor multiplying the exponential.

In all three cases, for $t<0$ close the contour in the lower half-plane, where $e^{i\omega t}$ decays. There are no poles inside, soAt $t=0$ its continuous value is zero and $G^{\prime }(0^{+})=1$, producing the unit derivative jump required by $(D_t^2+2\kappa D_t+{\Omega }^2)G=\delta (t)$. Thus all three formulas are causal Green's functions and all decay to zero at late times.

For $\kappa >\Omega$, put $\eta =\sqrt{{\kappa }^2-{\Omega }^2}$, so $0<\eta <\kappa$ and the poles are $i(\kappa \pm \eta )$. Closing in the upper half-plane for $t>0$ again includes both simple poles. The residue theorem givesThis overdamped response has no oscillation. Its slow tail is $e^{-(\kappa -\eta )t}/(2\eta )$, and it tends to zero because $\eta <\kappa$.

For $t>0$, close the inverse-transform contour in the upper half-plane. There $|e^{i\omega t}|\le 1$; the rational factor is $O(|\omega {|}^{-2})$, so the arc integral vanishes. For $\Omega >\kappa$, write $\beta =\sqrt{{\Omega }^2-{\kappa }^2}>0$. The two simple poles are $i\kappa \pm \beta$, and the residue theorem yieldsThis causal Green function of a damped oscillator has damped oscillations with envelope $e^{-\kappa t}$ and tends to zero as $t\to \infty$.

Write the integral's multiplier in the Chebyshev polynomial basis:At ${\gamma }^2=5$, part (i) givesBy orthogonality, the $y_1$ term and all cross terms vanish. Using $I_0=\pi$ and $I_2=\pi /2$, the integral isThe remainder can be integrated against this integrable weight in the regular expansion, so it retains the stated order.

Let $L=-(1-x^2)D^2+xD$, so $L{y}_n=n^2{y}_n$ and the forced equation is $({\gamma }^2-L)Z=e^{\epsilon x}$. The small-parameter Taylor expansion of the forcing isEach Chebyshev polynomial can be inverted separately under ${\gamma }^2-L$. None of the required denominators vanishes because $\gamma$ is a real noninteger. Thus the regular solution of a forced Chebyshev equation hasHere we select the same regular endpoint domain as in part (a), with finite function values and finite endpoint derivatives. This qualification is needed: the printed forced equation alone does not state endpoint conditions. Without this selection one can add $c_1\cos (\gamma \arccos x)+c_2\sin (\gamma \arccos x)$, which remain finite but generally have singular endpoint derivatives. With $x=\cos \theta$, regularity gives $Z_{\theta }(0)=Z_{\theta }(\pi )=0$; the corresponding homogeneous equation has only the zero solution for noninteger $\gamma$. This makes the displayed expansion unique.

Retain the question's common crossing condition $y^{\prime }(0)=1$, and replace the slope-ratio condition of part (iii) by the specified initial slope. Then part (i) givesso $\alpha =7$. Substitution in part (ii) gives the rayIt starts at $(-7,0)$ with slope $1/2$, passes through $(0,9/2)$ with slope one, and reaches the receiver with slope $1/\sqrt{8}$. The final height isThe left-hand curvature is $y^{\prime \prime }=\frac{1}{3}(1-x{)}^{-4/3}>0$; the right-hand curvature is $y^{\prime \prime }=-\frac{1}{2}(1+x{)}^{-3/2}<0$. Thus the graph is increasing throughout, with a change from upward to downward curvature at the interface.

If the word “instead” were also taken to remove the common crossing condition, the initial slope alone would leave $\alpha$ undetermined: it would give $p=\mu (-\alpha )/\sqrt{5}$ for any selected source position. The numerical answer above uses the crossing condition stated before all four subparts.

The initial and final slopes from part (i) are $(1+\alpha {)}^{-1/3}$ and $(1+\alpha {)}^{-1/2}$. The prescribed ratio givesConsequently