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For $x\ne \xi$, the homogeneous solutions are $e^x,e^{-x}$. The left boundary condition selects $\sinh x$, while boundedness at infinity selects $e^{-x}$ on the right. Thus write $G=A(\xi )\sinh x$ for $x<\xi$ and $G=B(\xi )e^{-x}$ for $x>\xi$.

A unit Dirac delta distribution in the second derivative requires continuity of $G$ and a derivative jump of one:The second equation and the first give $-A{e}^{\xi }=1$, so the Dirichlet half-line Green function for d2 minus 1 isThe second expression is also a method of images construction. Applying the Green's function to the forcing givesFor bounded forcing this is bounded, vanishes at zero and solves the equation; a bounded homogeneous solution vanishing at zero is zero, proving uniqueness. More general forcing is allowed when the integrals converge and the resulting function has the required growth. Arbitrary unbounded forcing need not admit a bounded solution; this distinction is relevant in part (b).

The reduced forcing is $f(\xi )=\xi 1_{[1,\infty )}(\xi )$. Its exponential-weighted integrals converge, even though the resulting reduced function is not bounded. For $0\le x<1$, the Green's function formula givesFor $x\ge 1$ it givesUse $\int \xi \sinh \xi d\xi =\xi \cosh \xi -\sinh \xi$ and ${\int }_x^{\infty }\xi e^{-\xi }d\xi =(x+1)e^{-x}$ to simplify this to $y(x)=-x+e^{-x-1}$. Dividing by $x$ yieldsThus $a(x)=-(2/e)\sinh x/x$ with its continuous value at zero. At $x=1$, both pieces give $u=-1+e^{-2}$ and $u^{\prime }=-2e^{-2}$, so neither a Dirac delta distribution nor its derivative is introduced at the forcing jump. The solution is bounded at both ends, with $u\to -1$ at infinity. Its reduced function $y=xu\sim -x$ confirms the growth qualification in part (i).

The derivative of an analytic function must be independent of the direction of approach. Taking a real increment gives $f^{\prime }(z)=u_x+i{v}_x$; taking an imaginary increment givesEquality of real and imaginary parts proves the Cauchy-Riemann equations:A real harmonic function is a twice continuously differentiable function whose Laplacian vanishes. For the given logarithmic function and $x^2+y^2>0$,It is therefore harmonic everywhere on the punctured complex plane.

On any domain admitting a branch of the complex logarithm, chooseIts derivatives are $v_x=-2y/(x^2+y^2)$ and $v_y=2x/(x^2+y^2)$, as required. For example, the principal complex logarithm gives such a harmonic conjugate on the plane cut along the negative real axis.

There is no single-valued harmonic conjugate on the whole punctured plane. Indeed its putative differential would satisfywhereas a globally defined function has zero integral of its differential around a closed curve. This is the log modulus has no global harmonic conjugate on the punctured plane obstruction. Thus the conjugate requested in the question must be interpreted locally or on a cut plane.

Seek a positive weight such that the differential operator becomes $-w^{-1}(p{y}^{\prime }{)}^{\prime }$. Matching the second derivative gives $p=w(1-x^2)$, while matching the first derivative requires $p^{\prime }=-xw$. ConsequentlyChoosing the multiplicative constant to be one givesThis is Sturm-Liouville form with zero potential term. Multiply the equations for $y_n,y_m$ by $y_m,y_n$, subtract, and integrate. The Lagrange identity for a Sturm-Liouville operator givesThe boundary expression vanishes because $p\to 0$ and the functions and their derivatives are finite. Since $n,m\ge 0$ are distinct, their eigenvalues $n^2,m^2$ are distinct, provingThe endpoint singularities of the weight are integrable, so these weighted inner products exist.

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.

For $\Delta G=+{\delta }^{(3)}$, the fundamental solution of the Laplace equation in three dimensions isIts sign follows from the flux: on a small sphere of radius $\epsilon$, its outward radial derivative is $1/(4\pi {\epsilon }^2)$, with total flux one.

Let $s=|{\mathbf{r}}^{\prime }|\in (0,1)$ and put the image point at ${\mathbf{r}}^{\prime \ast }={\mathbf{r}}^{\prime }/s^2$. The method of images gives the Dirichlet Green function inside a sphere:The image lies outside the ball, so its potential is harmonic inside. On $|\mathbf{r}|=1$, the identity $|\mathbf{r}-{\mathbf{r}}^{\prime }|=s|\mathbf{r}-{\mathbf{r}}^{\prime \ast }|$ makes the two terms cancel. The original source retains unit strength. For a source at the centre, take the limit:

The only denominator zero is $z_0=-1$, which lies strictly inside the given circle. The numerator is an entire function. Apply the derivative form of the Cauchy integral formula with $n=6$:The sixth derivative is $-\sin z$, so its value at $-1$ is $\sin 1$. Consequently

A sufficient assumption is that $f$ is absolutely continuous, $f,f^{\prime }\in L^1(\mathbb{R})$, and $f(t)\to 0$ at both infinities. Integration by parts in the Fourier transform givesThe decay assumption removes the boundary term, and integrability makes both transforms ordinary convergent integrals. The same derivative identity extends to tempered distributions.

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.

Differentiate the position vector, keeping the other coordinates fixed in each derivative:The first two have zero dot product, since their two planar contributions cancel, and both are orthogonal to ${\mathbf{h}}_z$. Hence these are orthogonal coordinates away from coordinate degeneracies. Their scale factors, which are the lengths of these coordinate basis vectors, areFor example $|{\mathbf{h}}_u{|}^2=a^2[{\sinh }^{2}u{\cos }^{2}v+{\cosh }^{2}u{\sin }^{2}v]=a^2({\sinh }^{2}u+{\sin }^{2}v)$. The unit basis vectors are ${\mathbf{h}}_i/h_i$ when $h_i\ne 0$; the given coordinate basis vectors themselves are not normalized. The Jacobian determinant is $a^2({\sinh }^{2}u+{\sin }^{2}v)>0$ except on the focal lines. These are elliptic cylindrical coordinates.

For each source point $z$, construct the Green function from homogeneous solutions on either side:The endpoint conditions are then automatic. The equation $L_{x}G=\delta (x-z)$ requires continuity of $G$ and a unit jump in its first derivative: a discontinuity in $G$ would introduce an unwanted delta derivative, while integrating across the source fixes $G_x(z^{+},z)-G_x(z^{-},z)=1$. ThusSolving this two-by-two system, with the Wronskian $W=y_a{y}_b^{\prime }-y_b{y}_a^{\prime }\ne 0$, givesThis derivation also remains valid if one of the solution values at the matching point vanishes: only the Wronskian must be nonzero. For a regular nonsingular two-point problem it supplies the inverse kernel; at a singular endpoint, uniqueness requires separate checking, as part (d) illustrates.

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.