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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.

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.

The Wronskian is $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$. If constants $A,B$ obey $A{y}_1+B{y}_2=0$, differentiating gives $A{y}_1^{\prime }+B{y}_2^{\prime }=0$. At any point where $W\ne 0$, these form an invertible two-by-two system, forcing $A=B=0$. Thus the two solutions are linearly independent.

For the given equation, the normalized coefficients are $p=(1-x)/(2x)$ and $q=-1/(4x)$, so zero is a regular singular point. Substitute a Frobenius series $y=x^r{\sum }_{n=0}^{\infty }{a}_n{x}^n$, $a_0\ne 0$. The lowest power gives the indicial equationThe coefficient of $x^{n+r}$ givesFor either root the common factor is nonzero for $n\ge 0$, soWith leading coefficient one, the two independent solutions areTheir analytic series factors both have infinite radius of convergence, because the ratio of consecutive terms tends to zero for every fixed $x$. The factor $x^{1/2}$ still gives the second solution a branch point at zero; it is not an entire function or an ordinary Taylor series there. On $x>0$, the general real solution is $C_1{y}_1+C_2{y}_2$. This is the $b=1/2$, $z=x/2$ case of the exponential reduction of a confluent hypergeometric equation with equal parameters.

Substituting $a_n=2^n/n!$ into the $r=0$ recurrence, and $a_n=1/n!$ into the $r=1/2$ recurrence, satisfies every coefficient equation. They give $e^{2x}$ and $\sqrt{x}{e}^x$. Thus on a positive interval not crossing a singular point,Their Wronskian is $e^{3x}(2x-1)/(2\sqrt{x})$, nonzero for $x>0$, $x\ne 1/2$, proving independence on each ordinary interval. Both functions are analytic near $1/2$, so this is an apparent regular singular point: a pole in normalized coefficients need not force a singular solution. The Wronskian can vanish at that singular point without making the two functions linearly dependent on an ordinary interval.

For the zero initial data, the Green function of an undamped oscillator is $G_0(x,s)=\sin [k(x-s)]/k$ for $0\le s\le x$, and zero for $s>x$. It is continuous at the source and its first derivative jumps by one. The resulting convolution isIt vanishes together with its first derivative at zero and has the required resonant forcing.

For the two-endpoint problem, write $L=\pi /(6k)$, $a=\min (0,L)$, $b=\max (0,L)$ and $\kappa =|k|$. This covers both signs of the supplied nonzero $k$. Left and right homogeneous solutions are $\sin [\kappa (x-a)]$ and $\sin [\kappa (b-x)]$, with Wronskian $-\kappa \sin [\kappa (b-a)]$. The Dirichlet Green function is thereforeIts endpoint values are zero and its first derivative jumps by one. Here $\kappa (b-a)=\pi /6$, so the denominator does not vanish. Integrating the constant forcing givesEquivalently $y=(1-\cos kx)/2+(\sqrt{3}/2-1)\sin kx$. The negative interior response is consistent with the negative kernel for this short Dirichlet interval.

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.

Let $L=d^2/d{x}^2+x^{-1}d/dx-x^{-2}$. A Green's function $G(x,\xi )$ satisfies $L_{x}G=\delta (x-\xi )$ and the chosen homogeneous boundary conditions in $x$. Away from $x=\xi$, it is a linear combination of $x$ and $x^{-1}$. It must be continuous there and obey the unit derivative jump $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$, because the leading differential coefficient is one. Thensolves $Ly=f$ whenever the integral and boundary limits are valid. The Wronskian of $x,x^{-1}$ is $-2/x$. The Green kernels for a radial dipole Euler equation implement the two different choices of boundary conditions below.

The homogeneous Cauchy-Euler equation has solutions $x$ and $x^{-1}$. Choose $y_L=x$, which vanishes at zero, and $y_R=x-x^{-1}$, which vanishes at one. Their Wronskian is $W(s)=y_L(s)y_R^{\prime }(s)-y_L^{\prime }(s)y_R(s)=2/s$. For a Green function integrated against the unweighted measure $ds$, continuity at $x=s$ and the jump condition $s^2[G_x{]}_{s^{-}}^{s^{+}}=1$ giveIt satisfies $L_{x}G=\delta (x-s)$ for $L=x^2{\partial }_x^2+x{\partial }_x-1$. Therefore the candidate solution isThe leading coefficient vanishes at zero, so the endpoint condition needs care. If $f$ extends continuously to $[0,1]$, this expression has limit $-f(0)$ as $x\downarrow 0$. Indeed $x^{-1}{\int }_0^{x}f(s)ds\to f(0)$ and $x{\int }_x^{1}f(s)s^{-2}ds\to f(0)$, while the remaining terms vanish. Thus the continuous zero-endpoint problem in this class requires $f(0)=0$. With that compatibility condition the formula gives both boundary values and is unique: a difference of two bounded solutions is $Ax+B/x$, whose two endpoint conditions force both constants to vanish. For incompatible forcing such as $f\equiv 1$, the bounded solution with $u(1)=0$ is $x-1$, which cannot also vanish at zero. Pointwise vanishing of $G(x,s)$ for fixed $s$ does not justify interchanging the singular endpoint limit with its integral. This issue is treated in Green kernel and endpoint compatibility for a singular Euler equation.

For the supplied compatible forcing $f(s)=s$, the integrals are $x^2/2$ and $(1-x^2)/2+\log x$. Their polynomial pieces cancel, givingDirect differentiation gives $Lu=x$, and $x\log x\to 0$ at zero. Its derivative diverges there, which is allowed because the boundary condition requires the value, not a finite derivative.

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.

Trying $y=x^m$ in the Cauchy-Euler equation givesso the homogeneous solution isContinuity with zero value at the singular endpoint excludes the second branch. Use the left and right adapted solutions $\ell (x)=x$, $r(x)=x^{-4}-x$, whose Wronskian isThe Dirichlet Green kernel for an Euler operator with exponents one and minus four is thereforeIt vanishes at both endpoints, is continuous at $x=\xi$, and has first-derivative jumpThus $L_{x}G=\delta (x-\xi )$ with ordinary source measure $d\xi$. For bounded forcing the Green function representation isThe apparent singular prefactor in the first term is canceled by the integral's decay at zero. The only regular homogeneous solution vanishing at $1$ is zero, so this solution is unique in the stated regular class.

For the piecewise source, split the integrals at $1/2$. If $x<1/2$, the needed integrals are ${\int }_0^x{\xi }^{6}d\xi =x^7/7$ and ${\int }_x^{1/2}({\xi }^6-\xi )d\xi$. If $x>1/2$, only ${\int }_0^{1/2}{\xi }^{6}d\xi =1/896$ remains. Simplification yieldsAt the interface both expressions have value $-31/8960$ and derivative $129/4480$, so no unintended Dirac delta source appears. Their second derivatives have the allowed jump associated with the discontinuous forcing.

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.

The substitution $y=e^{x}g$ reduces the homogeneous linear differential equation toA first integration gives $g^{\prime }=Cx{e}^{-x}$, and a second gives $g=D-C(x+1)e^{-x}$. HenceThe Wronskian of $1+x,e^x$ is $x{e}^x$, so they are linearly independent on either side of zero. At zero the differential equation has a regular singular point. The displayed solutions extend analytically through it; their linear combinations describe the global solutions when two continuous derivatives are required there. If no matching condition is imposed, the two sides have independent constants.

For the Green function of $D_x^2-1$, the homogeneous solutions satisfying the left and right endpoint conditions are $u(x)=\sinh x$ and $v(x)=\sinh (1-x)$. Their Wronskian isContinuity of the kernel at $x=\xi$ and integration of $G_{xx}-G=\delta (x-\xi )$ require $[G_x{]}_{\xi -}^{\xi +}=1$. Hence the Dirichlet modified Helmholtz Green function on an interval isHere $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The negative sign is essential for this operator convention. Integrating the forcing against the kernel givesThis satisfies both endpoint conditions. It solves the equation classically for continuous $f$, and in the corresponding weak sense for integrable forcing. A homogeneous difference satisfying both endpoint conditions must vanish, so the solution is unique.

Away from $x=\xi$, the homogeneous equation has solutions $e^x,e^{-x}$. A basis adapted to the two boundary conditions is $u(x)=\sinh x$ at the origin and $v(x)=e^{-x}$ at infinity. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-1$. The Green function must be continuous at $\xi$, while integration of $-G_{xx}+G=\delta (x-\xi )$ across the source givesThese conditions give the Dirichlet half-line Green function for minus d2 plus 1:where $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The second term is the image that enforces $G(0,\xi )=0$; both terms decay as $x\to \infty$.

For the finite forcing interval, $y(x)={\int }_0^{\mu }G(x,\xi )d\xi$. When $x\le \mu$, split the integral at $\xi =x$:When $x\ge \mu$, the entire source interval lies to the left, giving the Dirichlet half-line response to uniform interval forcingThis solution is unique: a homogeneous difference satisfying the origin condition is proportional to $\sinh x$, which cannot decay at infinity unless its coefficient is zero.

The Wronskian isSuppose constants $A,B$ satisfy $Af+Bg=0$ throughout the interval. Differentiating gives $A{f}^{\prime }+B{g}^{\prime }=0$. At $x_0$, these form the homogeneous linear systemIts determinant is nonzero by hypothesis, so $A=B=0$. This is exactly linear independence of the two functions on the interval. No differential equation is needed for this implication; the converse for arbitrary differentiable functions would require additional hypotheses.

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.

For the Euler-Cauchy equation, substitute $y=x^n$ on an interval not containing zero. The coefficient of $x^{n-2}$ isThe distinct roots giveThese solutions are linearly independent since their Wronskian is $2x^3\ne 0$ away from zero. Both nevertheless approach zero as $x\to 0$, which will make the endpoint conditions in the next part insufficient for uniqueness.

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.

The homogeneous characteristic equation is $r^2-2r+2=0$, with roots $1\pm i$. Choose the endpoint-adapted homogeneous solutionsso that $u(0)=0$ and $v(\pi /2)=0$. Their Wronskian iswhich never vanishes. A continuous Green function with the required endpoints has the form $C(\xi )u(x_{<})v(x_{>})$, where $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. Its derivative jump is $C(\xi )W(\xi )$. Because the leading coefficient is one, part (b) requires this to equal one. ThereforeIn particular, the branches are $-e^{x-\xi }\sin x\cos \xi$ for $x\le \xi$ and $-e^{x-\xi }\sin \xi \cos x$ for $x\ge \xi$. This is the Dirichlet Green function for a conjugated harmonic oscillator with interval length $\pi /2$, drift parameter one and oscillator wave number one.

The homogeneous equation is a Cauchy-Euler equation. With the power-law ansatz $y=x^r$ on $x>0$, its indicial equation isThus $r=(1-\alpha )/2$ is a repeated indicial root. The two independent solutions are $x^r$ and $x^r\log x$, givingTheir Wronskian, with $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$, is $x^{2r-1}=x^{-\alpha }\ne 0$, verifying linear independence.