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Laplace transform. Use $Y(p)={\int }_0^{\infty }{e}^{-pt}y(t)dt$. The initial value is zero and the Dirac delta contributes $e^{-p}$, so the transformed equation is $pY+Y=e^{-p}$. Therefore

Bromwich inversion. The Bromwich inversion formula iswith the vertical Bromwich contour traversed upwards and its symmetric improper limit understood. For $t>1$, close to the left, where the exponential decays; the enclosed simple pole at $p=-1$ has residue $e^{-(t-1)}$. For $t<1$, close to the right, which contains no pole. Thuswhere $\Theta$ is the Heaviside step function. This agrees with the Laplace transform time-shift rule. At $t=1$, inversion returns the average $1/2$ of the two one-sided values. The value assigned at that single point does not affect the differential equation in the distributional sense.

Fourier transform. State the convention explicitly:The Fourier transform of the distributional derivative is $ik\tilde{y}(k)$, including the jump at $t=1$. The Dirac delta has transform $e^{-ik}$, giving

Inverse Fourier integral. Substitute into the inverse Fourier transform:The only pole is at $k=i$. For $t>1$, close in the upper half-plane. Jordan's lemma removes the large arc and the residue theorem gives $i{\mathrm{Res}}_{k=i}=i[e^{-(t-1)}/i]=e^{-(t-1)}$. For $t<1$, close in the lower half-plane, where there are no poles, giving zero. This reproducesIntegrating the differential equation across the impulse checks the unit jump, $y(1^{+})-y(1^{-})=1$. Away from the impulse the solution obeys $y^{\prime }=-y$, and it satisfies the stipulated zero past.

Poles and residues. An isolated singularity $z_0$ is a pole of order $m$ if its Laurent series has finitely many negative powers, the lowest being a nonzero multiple of $(z-z_0{)}^{-m}$. The residue is the coefficient of $(z-z_0{)}^{-1}$.

If $h(z_0)=0$ and $h^{\prime }(z_0)\ne 0$, factor $h(z)=(z-z_0)H(z)$, where $H$ is holomorphic near $z_0$ and $H(z_0)=h^{\prime }(z_0)$. For $g(z_0)\ne 0$,The last quotient is holomorphic and nonzero at $z_0$, proving a simple pole with

The given function. For noninteger $a$, specify a branch of a multivalued function before using $z^a$. Choose $z^a=\exp (a\log z)$ with $-\pi /2<\arg z<3\pi /2$, a cut along the negative imaginary axis. This leaves all three roots of $1+z^3$ off the cut. They are $z_{+}=e^{i\pi /3}$, $z_{-}=e^{-i\pi /3}$ and $z_0=-1$, all simple. The numerator is nonzero at these points. Since the denominator has derivative $3z^2$, the residues areFor noninteger $a$, zero is a branch point, rather than an isolated pole. This remains true when $-1<a<0$ and the numerator diverges there. For nonnegative integer $a$ it is a regular point. Other logarithm branches change the phases of the residues consistently.

Integral and convergence. Near zero the integrand behaves as $x^a$, requiring $a>-1$. At infinity it behaves as $x^{a-3}$, requiring $a<2$. Therefore the exact convergence range isUse a counterclockwise sector contour with rays at angles $0$ and $2\pi /3$, inner radius $\epsilon$ and outer radius $R$. The same complex logarithm branch is valid throughout this sector. Only $z_{+}=e^{i\pi /3}$ lies inside. The small circular arc contributes $O({\epsilon }^{a+1})$ and the large one contributes $O(R^{a-2})$, so both vanish precisely in the stated range.

The outward positive-real ray contributes $I$. On the returning ray set $z=x{e}^{2\pi i/3}$: the denominator is still $1+x^3$, while $z^{a}dz=e^{2\pi i(a+1)/3}{x}^{a}dx$, with reversed limits. The residue theorem givesPut $\alpha =\pi (a+1)/3$. The right-hand phase is $-e^{i\alpha }$ and $1-e^{2i\alpha }=-2i{e}^{i\alpha }\sin \alpha$. HenceThe sine is positive in this range, consistent with a positive real integral. This is a special case of the Mellin integral of a reciprocal binomial.

Harmonic parts. Write the holomorphic function as $f=u+iv$. The Cauchy-Riemann equations say $u_x=v_y$ and $u_y=-v_x$. Equality of mixed partial derivatives therefore givesThus both the real and imaginary parts are harmonic functions and satisfy two-dimensional Laplace's equation.

Single-valued logarithm. A convenient choice isusing the principal logarithm for each term. Its branch cuts in the $z$ plane are $(-\infty ,-1]$ and $[1,\infty )$. The branch points are $-1$ and $1$: going once around the first adds $2\pi i$, while going once around the second subtracts $2\pi i$. Removing the two exterior rays makes this a consistent branch of a multivalued function and in particular leaves the whole open unit disk available.

Potential inside the cylinder. Put $z=r{e}^{i\theta }$ with $r<1$. Multiplication by $1-\overline{z}$ givesThe denominator is positive and the real part of the numerator is positive. Thus its argument lies strictly between $-\pi /2$ and $\pi /2$, andAs $r\uparrow 1$, this tends to $\pi /2$ on the upper open semicircle and $-\pi /2$ on the lower open semicircle. Scale the imaginary part to obtainIt is a harmonic function, has the required boundary values, and gives $\Phi (0)=0$. The two joining points of the boundary arcs have discontinuous boundary data, so no common continuous boundary value is demanded there. Along the real diameter, including radial approaches to either joining point, the displayed solution is zero. This is the disk potential with two semicircular boundary values.

For a second check, the Taylor series of the logarithms gives $f(z)=2{\sum }_{j\ge 0}{z}^{2j+1}/(2j+1)$. Taking the imaginary part yields the Fourier serieswhose boundary limit is exactly the required alternating pair of constant semicircle values.

Green identity. With $n$ the outward unit normal, the product rule gives $\nabla \cdot (u\nabla v)=u{\nabla }^{2}v+\nabla u\cdot \nabla v$. Integrating over $V$ and applying the divergence theorem yields Green's first identity:

Uniqueness for nonzero mass. Suppose two real solutions have the same prescribed boundary data, and let $w$ be their difference. The spatial equation called the Klein-Gordon equation in the question is the modified Helmholtz equation, so ${\nabla }^{2}w=m^{2}w$. With $u=v=w$, Green's first identity becomesFor identical Dirichlet boundary conditions, $w=0$ on $S$; for identical Neumann boundary conditions, ${\partial }_{n}w=0$ there. Either makes the right-hand side zero. When $m$ is real and nonzero, both integrands on the left are nonnegative, forcing $w=0$. ThusThis is the required uniqueness assertion; existence is understood for admissible smooth data on a suitable bounded region. Complex-valued solutions obey the same result by treating their real and imaginary parts separately.

Zero mass. For $m=0$, the argument gives $\nabla w=0$, hence $w=C$ on the connected region. Dirichlet boundary conditions still force $C=0$, but Neumann boundary conditions leave an arbitrary additive constant:The prescribed normal data in the zero-mass case must also satisfy ${\int }_S{\partial }_n\Phi dS=0$, by the divergence theorem. For nonzero $m$, that flux can instead equal $m^2{\int }_V\Phi dV$.

Energy minimization. Let $\eta =w-v$ for the two functions in the last part. They share their boundary values, so $\eta =0$ on $S$. Since $v$ is a harmonic function, Green's first identity givesExpand the square in the Dirichlet energy of $w=v+\eta$:The cross term vanishes, soEquality requires $w-v$ to be constant, and the shared boundary values then make it zero. This is the Dirichlet principle: among functions with fixed boundary values, the harmonic function minimizes Dirichlet energy.

Fourth-order spatial formula. For a sufficiently smooth function, pair the Taylor expansions at opposite shifts:Multiply the first sum by sixteen, subtract the second, and subtract $30u(x)$. The constant and fourth-derivative terms cancel, leavingHence the fourth-order centered second derivative satisfiesRearranging proves the requested $u^{\prime \prime }=D_{4,h}u+O(h^4)$, including its leading error coefficient.

Explicit diffusion scheme. Take $x_j=jh$, $h=1/N$, $t_n=n\tau$, and let $U_j^n$ approximate $\psi (x_j,t_n)$. Combining the forward Euler method in time with $D_{4,h}$ in space givesTo check the error, insert the exact solution. The heat equation gives ${\psi }_t={\psi }_{xx}$, while the time Taylor expansion and the spatial formula give the one-step defectThus its unnormalized local truncation error isas requested. Dividing this defect by $\tau$ gives the normalized residual $O(\tau )+O(h^4)$; these are distinct error conventions.

Boundary implementation. The centered stencil requires two ghost values at each end. An even-reflection closure for Neumann diffusion imposesThe reflected first centered difference at either endpoint is zero. Apply the same evolution formula at the endpoints after substituting these ghost values; for example,The ghost values also close the stencil at $j=1$ and $j=N-1$. They are determined auxiliary values, not new physical data. For a smooth solution compatible with the Neumann boundary conditions, differentiating those conditions in time and using the heat equation gives ${\psi }_{xxx}={\psi }_{xxxxx}=\cdots =0$ at the endpoints. Thus even reflection preserves the smoothness needed for fourth-order spatial accuracy. If the initial data fail the required higher compatibility conditions, a uniform smooth error expansion through $t=0$ cannot be assumed.

For completeness, the reflected spatial operator has cosine eigenvectors $\cos (\ell \pi j/N)$ and eigenvaluesThe most negative one is $-16/(3h^2)$. Requiring $|1+\tau {\lambda }_{\ell }|\le 1$ gives the forward Euler method stability restriction $\boxed{\tau /h^2\le 3/8}$ in the natural trapezoid-weighted discrete norm. A fourth-order spatial stencil does not remove the explicit diffusion time-step restriction.

Implicit alternative. The Crank-Nicolson diffusion scheme using the same spatial operator and boundary closure isIt has second-order time accuracy and fourth-order spatial accuracy for smooth compatible solutions; its one-step defect is $O({\tau }^3)+O(\tau h^4)$. Each spatial eigenvalue $\lambda \le 0$ has amplification factor $(1+\tau \lambda /2)/(1-\tau \lambda /2)$ of magnitude at most one, so it removes the explicit restriction for this boundary closure.

The price is solving a banded linear system at every step; the same matrix factorization can be reused for fixed $h$ and $\tau$. Large time steps still reduce accuracy. Moreover, the amplification factor tends to $-1$ as $\tau |\lambda |\to \infty$, so short spatial wavelengths can alternate sign with little numerical damping. Thus unconditional linear stability does not guarantee a monotone or well-resolved transient. Boundary accuracy and initial compatibility remain necessary with either time discretization.

Representations. A complex group representation is a group homomorphism $\rho :G\to \mathrm{GL}(W)$ to the invertible linear maps on a vector space. It is an irreducible representation when $W\ne 0$ and the only invariant subspaces are $0$ and $W$.

Group table. The supplied matrices satisfy $A=-I$, $C=-B$, and $A^2=B^2=C^2=I$. Their multiplication table isThe table proves closure and gives identity $I$ and an inverse for every element; associativity is inherited from matrix multiplication. The table is symmetric, so the group is abelian. It is the Klein four-group $V\cong C_2\times C_2$.

Dimensions and conjugacy. The sum of squares of irreducible representation dimensions is $|V|=4$. A two-dimensional irreducible would alone contribute four, leaving no room for the ever-present one-dimensional trivial representation. Hence the given two-dimensional representation is reducible. Since the group is abelian, every conjugacy class is a singleton. There are therefore four inequivalent complex irreducible representations, each of dimension one:Taking the trace of the given matrices gives their character of a representation:

Irreducible characters. In a one-dimensional representation, the values on $A$ and $B$ must each be $\pm 1$, since both elements square to $I$. Their values determine the value on $C=AB$. The four choices give the complete character table:Each row is a character because it respects the generating relations, and the rows have unit norm and pairwise zero character inner product. The multiplicities of these rows in $(2,-2,0,0)$ are respectively $0,0,1,1$, by character orthogonality. Thus

Diagonal form. The common invariant lines are spanned by $(1,1{)}^T$ and $(1,-1{)}^T$. Use their normalized vectors as columns ofThen simultaneous diagonalization givesReading the first and second diagonal entries across the four matrices recovers ${\chi }_3$ and ${\chi }_4$, respectively, verifying the decomposition explicitly.

For the nonconstant boundary data, use separation of variables in polar coordinates and put $\varphi =R(r)\cos 2\theta$. Laplace's equation reduces toThis Cauchy-Euler equation has independent solutions $r^2$ and $r^{-2}$. Unlike a disk containing the origin, the annulus permits both. The zero inner boundary value requires $R=A(r^2-a^4{r}^{-2})$, and the outer value requires $A=V{b}^2/(b^4-a^4)$. HenceSubstituting $r=a$ gives zero, while $r=b$ gives $V\cos 2\theta$. This is the Dirichlet problem on an annulus for one Fourier mode; no other angular mode is needed.

For the changed inner boundary value, use superposition to add a radial harmonic function that is $-V$ at $a$ and zero at $b$. The radial zero-mode equation is $(r{R}^{\prime }{)}^{\prime }=0$, giving $R=C+D\log r$. Imposing the two constant boundary values givesTherefore the second answer isThe logarithmic term is independent of $\theta$, so it changes only the angular average of the electric potential. Both answers are unique for these Dirichlet boundary conditions, by uniqueness for Laplace's equation.

Variational formulation. Take real $p,q,w$ with $p,w>0$ in the interval, and impose homogeneous Dirichlet boundary conditions, or other suitable conditions that cancel the boundary terms. The functionalhas Euler-Lagrange equationwhich is equivalent to $-(p{y}^{\prime }{)}^{\prime }+qy-\lambda wy=0$. The same construction is constrained stationarity of the energy $\int [p(y^{\prime }{)}^2+q{y}^2]$ subject to $\int w{y}^2=1$, with $\lambda$ a Lagrange multiplier. For fixed zero endpoint values, the variation vanishes at both ends; homogeneous zero-flux conditions also cancel the boundary term $[2p{y}^{\prime }\delta y{]}_a^b$.

Multiplying the differential equation by $y$ and using integration by parts gives the Rayleigh quotientFor a nonzero admissible trial function, $R[y]\ge {\lambda }_0$ by the Rayleigh-Ritz variational principle. The Rayleigh-Ritz method chooses trial functions satisfying the endpoint conditions, often with adjustable coefficients, and minimizes this quotient over that trial family. Expanding $y=\sum c_n{y}_n$ in a normalized eigenbasis shows why the lower bound holds: $R[y]=\sum {\lambda }_n{c}_n^2/\sum c_n^2$, a weighted average of eigenvalues no smaller than the least one.

Radial quadratic trial. The radial problem becomeswhere the printed endpoint notation $y$ refers to $\psi$. Thus $p=w=r^2$ and $q=0$. At the origin the vanishing $r^2$ and regularity of $\psi$ make the boundary term vanish; the weight must be retained in both numerator and denominator.

For a quadratic $\psi =A+Br+C{r}^2$, the derivative condition sets $B=0$ and the outer condition gives $C=-A$. Overall amplitude cancels in the Rayleigh quotient, so take $\tilde{\psi }=1-r^2$. ThenwhileConsequently

Why the estimate is higher. For a regular radial function put $u=r\psi$. Then $u(0)=u(1)=0$, and direct differentiation transforms the equation to $u^{\prime \prime }+\lambda u=0$. Its first mode is $u=\sin (\pi r)$, corresponding to $\psi =\sin (\pi r)/r$ with its continuous value at the origin. Thus ${\lambda }_0={\pi }^2$. The quadratic trial is not proportional to that Spherical Bessel function, so it has nonzero components in higher modes. Its Rayleigh quotient is therefore strictly larger than ${\pi }^2$, in agreement with $21/2>{\pi }^2$.

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.

Boundary form. For real functions and $\mathcal{L}y=-(p{y}^{\prime }{)}^{\prime }+qy$, integration by parts givesThe integral terms containing $p{y}_1^{\prime }{y}_2^{\prime }$ cancel, as do the $q$ terms. This is the Lagrange identity for a Sturm-Liouville operator. The boundary conditions must make this boundary form zero for every admissible pair. Several suitable choices are homogeneous Dirichlet boundary conditions at both endpoints; homogeneous Neumann boundary conditions at both endpoints; one of each; or separated homogeneous Robin boundary conditionswhere each fixed real pair $(\alpha ,\beta )$ is nonzero. At each endpoint the two boundary data vectors are then proportional, making their antisymmetric product zero. Another choice is periodic data $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$, which cancel the two endpoint contributions. These domains give the required self-adjoint differential operator in the regular setting.

Eigenfunctions. Set $x=\cos \theta$, $0\le \theta \le \pi$, and $Y(\theta )=y(\cos \theta )$. The chain rule givesThe Chebyshev differential equation therefore becomes $Y^{\prime \prime }+n^{2}Y=0$, with $Y(0)=Y(\pi )=0$. For a positive integer $n$, the first condition excludes the cosine term and the second is satisfied by $\sin (n\theta )$. ThusThere is no nonzero $n=0$ eigenfunction: the transformed solution is linear in $\theta$ and vanishes at both endpoints only if identically zero. Negative $n$ merely repeats a positive mode with the opposite sign, so it is excluded from the indexing of an independent eigenbasis.

Weight and normalization. Multiplying the differential equation by $-1/\sqrt{1-x^2}$ puts it into Sturm-Liouville form:Hence $p=\sqrt{1-x^2}$, $q=0$ and $w=(1-x^2{)}^{-1/2}$. Under $x=\cos \theta$, the measure $w(x)dx$ becomes $-d\theta$, soThereforeThese endpoints are singular because $p$ vanishes and $w$ diverges there, but the weight is integrable. Moreover $p{y}_n^{\prime }=-N_{n}n\cos (n\theta )$ remains finite, so the vanishing values of $y_n$ make the boundary form zero. These are the Dirichlet eigenfunctions of the Chebyshev differential operator, not the first-kind Chebyshev polynomials $\cos (n\arccos x)$, which have nonzero endpoint values.

Classification and existence. For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, an ordinary point $x_0$ is one where $p,q$ are analytic. There are two independent analytic power series solutions, with arbitrary initial values $y(x_0),y^{\prime }(x_0)$. A regular singular point is a singular point where $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ are analytic. The Frobenius method then constructs solutions beginning with $(x-x_0{)}^{\sigma }$, where $\sigma$ solves the indicial equation. At least one Frobenius solution exists; two such solutions exist without logarithms when the two exponents differ by a noninteger. Repeated exponents or an integer difference can require a logarithmic second solution.

For this equation,Both are analytic at zero, while $p,q$ themselves have poles there. Thus zero is a regular singular point.

Indicial equation and recurrence. Multiply by $4x$ and insert $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$:The lowest power $x^{\sigma -1}$ has coefficient $[4\sigma (\sigma -1)+2\sigma ]a_0$, givingThe coefficient of $x^{n+\sigma }$ isFor either exponent the common factor is nonzero for $n\ge 0$, so the recurrence relation reduces toNormalizing each leading coefficient to one, the two Frobenius solutions areMore explicitly, $y_2=\sqrt{x}{\sum }_{n\ge 0}{2}^{n}n!x^n/(2n+1)!$. The distinct leading powers establish linear independence, so $y=C_1{y}_1+C_2{y}_2$ on the positive real side of the singular point. The displayed exponential function is the requested closed-form solution; substitution gives $y_1^{\prime }=y_1/2,y_1^{\prime \prime }=y_1/4$, which makes the differential expression zero.

Convergence. For either exponent the ratio of successive terms in the coefficient power series isThus both coefficient series have $\boxed{R=\infty }$ by the ratio test. This does not remove the $\sqrt{x}$ branch point of $y_2$: its analytic coefficient factor is entire, while the full Frobenius solution requires a square-root branch away from zero in the complex plane. The differential equation has no other finite singular points to limit continuation of these coefficient series.

Laplace identities. The one-sided Laplace transform isfor $\mathrm{Re}p$ sufficiently large that the integral converges. For a piecewise smooth function of exponential order, integration by parts givesDifferentiating under the integral givesIteration also yields $\mathcal{L}[f^{\prime \prime }]=p^{2}F-pf(0)-f^{\prime }(0)$. For real $k$ and $p>0$, taking the imaginary part of $(p-ik{)}^{-1}$ gives

Transformed differential equation. Let $G=\mathcal{L}[g]$, using the initial values $g(0)=1,g^{\prime }(0)=0$. ThenSumming the three contributions givesFor $\alpha \ne 0$, the condition $G(p)\to 0$ as $p\to \infty$ fixes the integration constant:It is simpler to invert the derivative: $\mathcal{L}[tg]=-G^{\prime }=1/(p^2+{\alpha }^2)=\mathcal{L}[\sin (\alpha t)/\alpha ]$. ThusThis is the sinc function. Its Taylor series begins $1-{\alpha }^2{t}^2/6+O(t^4)$, verifying both initial values. For a direct differential check set $h=tg=\sin (\alpha t)/\alpha$. Then $h^{\prime \prime }+{\alpha }^{2}h=t{g}^{\prime \prime }+2g^{\prime }+{\alpha }^{2}tg=0$. For $\alpha =0$, take the continuous limit $\boxed{g(t)=1}$, which also follows from $G=1/p$.