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Multiplying the defining equation on the right by $I+A$ gives $(I+U)A=I-U$. Consequently the only possible answer isThe factors commute because they are rational functions of $U$. Its conjugate transpose, using the unitarity $U^{†}=U^{-1}$, isThus $A$ is Skew-Hermitian. Moreover $I+A=2(I+U{)}^{-1}$ and $I-A=2U(I+U{)}^{-1}$, so substitution recovers $U$ and establishes existence as well as uniqueness. The unitary matrix $U=-I$ is a counterexample to invertibility of $I+U$; more generally the obstruction is an eigenvalue $-1$.

Using part (i) and $A^{†}=-A$ in the Cayley transform with argument $-A$ givesThe factors commute because $I-A$ and $I+A$ are polynomials in the same matrix, and an invertible commuting matrix also commutes with the other factor's matrix inverse. Both inverses exist by part (ii). Thus $U$ is unitary.

Since $A$ is Skew-Hermitian, $H=iA$ is a Hermitian matrix. The spectral theorem gives a unitary matrix $Q$ with $Q^{†}HQ=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_N)$ and real eigenvalues ${\lambda }_j$. Hence $Q^{†}AQ=\mathrm{diag}(-i{\lambda }_j)$ andIn particular the determinant is nonzero, so $I+A$ is invertible. Applying the same argument to $-A$ proves that $I-A$ is also invertible. Equality holds precisely when all ${\lambda }_j=0$, equivalently $A=0$.

Taking the conjugate transpose of $M{M}^{-1}=I$ and $M^{-1}M=I$ reverses the product order and givesThus $(M^{-1}{)}^{†}$ is the two-sided matrix inverse of $M^{†}$, proving

The direct squared L2 norm is the length of the interval, namely $2$. The Fourier transform isFor the transformed norm, let $I={\int }_0^{\infty }{\sin }^{2}k/k^{2}dk$. Integration by parts, with vanishing endpoint term $-{\sin }^{2}k/k$, givesThe supplied oscillatory integral is understood as an improper integral. Thereforewhich agrees with the direct norm and verifies the Parseval identity with the specified normalization.

Set $g=f$. The cross-correlation at zero is $h(0)={\int }_{\mathbb{R}}|f(x){|}^{2}dx$, while part (i) gives $\hat{h}=|\hat{f}{|}^2$. Evaluating the inverse Fourier transform at zero yields the Parseval identityThis derivation is immediate for Schwartz functions, for which all integrations and Fourier inversion are justified. Approximation by such functions and completeness then extend the identity to L2 space, defining its Fourier transform by the corresponding norm limit. This is the Plancherel theorem extension; no absolute integrability of an arbitrary square-integrable function is presumed.

First take $f,g\in L^1(\mathbb{R})$, so Fubini's theorem applies to the cross-correlation double integral. With $s=x+y$, the Fourier transform isThe complex conjugation and the sign in $x+y$ are both essential to this convention.

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 $B=0$.

Use the boundary-adapted homogeneous solutionsFor a source point $\xi \ne 0$, the Green function isIt vanishes at both endpoints and is continuous at $x=\xi$. Its first derivative has the jumpas required for a monic second-order differential operator with a Dirac delta source. Away from that source it is a homogeneous solution and extends smoothly through $x=0$.

The interior singular point requires a qualification. This is a Green kernel with a vanishing Wronskian at an interior singular point, with a $1/\xi$ singularity in its source variable. It is not an ordinary integral inverse for unrestricted forcing. For smooth forcing $f(\xi )=\xi q(\xi )$, however, the product $Gf$ extends continuously at zero and the integral is well defined. This includes part (iii). We use solutions with two continuous derivatives across zero, interpreting the equation there after multiplication by $x$. For such a solution, expansion at zero forces $f(0)=0$; arbitrary nonzero forcing at zero cannot satisfy this regularity. Merely imposing one continuous derivative would leave a piecewise homogeneous ambiguity, so the original singular coefficients alone do not specify an unrestricted boundary-value inverse.

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.

Let $d_n$ be the even Fourier coefficient of the initial displacement, with the normalization ${\int }_{-L}^L{\cos }^2(k_{n}x)dx=L$. Writing $s_n=\sin (n\pi /2)=(-1{)}^{(n-1)/2}$ givesZero initial velocity requires the real modal time factorOne consistent choice in the question's complex representation is thereforeFor real ${\omega }_n$, $B_n=\overline{A_n}$; the real-part notation otherwise makes the complex coefficients nonunique. For imaginary ${\omega }_n$, the same formula is interpreted using hyperbolic functions. Its limit at ${\omega }_n=0$ is $d_n{e}^{-\alpha t/2}(1+\alpha t/2)$.

For $l/L\to 0$ at fixed mode number,which can be substituted into $A_n,B_n$ above. The profile specified at $t=0$ is actually nonzero at $x=\pm L$. Thus for finite $l/L$ it cannot be the initial data of a solution continuous at the initial endpoint corners with fixed endpoints. The Fourier series reproduces it in the open interval, with zero endpoint values; it is an L2 space initial-data interpretation. The assumption $l\ll L$ makes this compatibility error exponentially small, rather than exactly zero.

The angular frequency of an underdamped mode isFor ordinary damping, $\alpha \ge 0$, mode $n$ oscillates precisely when $\alpha <cn\pi /L$. In particular, all modes, including the fundamental, oscillate when $\alpha <c\pi /L$. There are some oscillatory modes for every finite $\alpha$, since the allowed $n$ are unbounded; no finite damping coefficient overdamps the entire ideal string spectrum.

If $\alpha <0$, the same mode has negative damping: oscillatory amplitudes grow as $e^{|\alpha |t/2}$. For modes with real characteristic roots, both roots are positive, since $|\alpha |/2>\sqrt{{\alpha }^2/4-c^2{k}_n^2}$. Hence those modes also grow. Negative damping makes the equilibrium unstable.

An even function satisfying the spatial equation has $X=C\cos (kx)$. The Dirichlet boundary conditions require $\cos (kL)=0$, so the allowed wavenumbers areThere is no nonzero zero-wavenumber mode. With $T=e^{-\alpha t/2}S$, the time equation becomes $S^{\prime \prime }+{\omega }_n^{2}S=0$, where ${\omega }_n^2=c^2{k}_n^2-{\alpha }^2/4$. Thus superposition of the even normal modes givesThis uses completeness of the even Fourier series modes. Imaginary ${\omega }_n$ describe overdamping. At critical damping, ${\omega }_n=0$, the two displayed exponentials coincide: the complete time factor must instead be $e^{-\alpha t/2}(C_n+D_{n}t)$. This repeated-root exception is necessary for a genuinely general solution.

For separation of variables, take $u(x,t)=X(x)T(t)$ in the damped wave equation. Dividing by $c^{2}XT$, with $c>0$, gives $X^{\prime \prime }/X=(T^{\prime \prime }+\alpha T^{\prime })/(c^{2}T)=-k^2$. The separated ordinary differential equations are thereforeThe negative separation constant anticipates the positive spatial eigenvalues selected by the fixed endpoints.

The Rayleigh-Ritz variational principle bounds the lowest eigenvalue by the Rayleigh quotient of any nonzero admissible trial function. For $f=1-(r/a{)}^p$, take $p>0$, which ensures boundedness, the outer Dirichlet boundary condition, and finite radial Dirichlet energy. Direct integration givesThereforeThe derivative is $1-2/p^2$, so the unique minimizing positive exponent is $p=\sqrt{2}$. This yields the power-law variational bound for a circular membraneThe optimizing function has zero derivative at the origin and is admissible for the energy principle despite its unbounded second derivative there. For comparison, the exact radial fundamental has $\sqrt{{\lambda }_1}a\simeq 2.404826$, giving ${\nu }_1\simeq 0.382740c/a$. The bound is about $0.39\%$ above this value.

Use the weighted inner productMultiply the equation for $R_n$ by ${\overline{R}}_m$, subtract the complex conjugate equation for $R_m$ times $R_n$, and integrate. Since the eigenvalues are real,The boundary term vanishes as before. Thus eigenfunctions with distinct eigenvalues are orthogonal. The full disk inner product differs only by the angular factor $2\pi$.

The conditions of regularity at $r=0$ and $R(a)=0$ select the radial vibrations of a circular membrane as an eigenvalue problem. Multiply $-(r{R}^{\prime }{)}^{\prime }=\lambda rR$ by $\overline{R}$ and integrate. Integration by parts givesThe boundary term vanishes by regularity and $R(a)=0$. For a nonzero eigenfunction the denominator is positive; the numerator could vanish only for a constant function, and the outer condition would then make that function zero. ConsequentlyThis also proves reality without presuming real-valued eigenfunctions. Explicitly, regular solutions are Bessel functions of the first kind $R=J_0(\sqrt{\lambda }r)$, and the positive zeros of $J_0(\sqrt{\lambda }a)$ give the discrete spectrum.

The time dependence gives $u_{tt}=-{\omega }^{2}u$, and the angular derivative vanishes for these radial normal modes. The Laplacian in polar coordinates therefore yieldsMultiplication by $r$ puts the Sturm-Liouville problem in the formThe relevant self-adjoint operator is $\mathcal{A}R=-r^{-1}(r{R}^{\prime }{)}^{\prime }$ in the weighted inner product with weight $r$. A physical displacement is regular at the origin, excluding the logarithmic singular solution. In particular $R$ is bounded and $r{R}^{\prime }\to 0$ there; regular eigenfunctions also have $R^{\prime }(0)=0$. This origin condition is needed in addition to the fixed outer boundary.

Direct differentiation givesOn the unit disk $D$, polar coordinates give ${\int }_D{y}^{2}dA=\pi /4$, ${\int }_{D}xdA=0$ and ${\int }_{D}dA=\pi$. Consequently the volume integral isFor an independent surface integral calculation, the relevant cross product isThe bottom face contributes ${\int }_{D}xydA=0$ and the top contributes ${\int }_D(-h{y}^2-xy)dA=-\pi h/4$. On the cylindrical side, put $x=\cos \theta$, $y=\sin \theta$, $\hat{\mathbf{n}}=(x,y,0)$ and $dS=d\theta dz$. The flux density is $x^3+2xy{z}^2+2y^{2}z-x{y}^2$. All terms except $2y^{2}z$ integrate to zero because they are odd functions of $x$ or $y$, givingThus minus the total outward flux is $-(\pi h^2-\pi h/4)$, in agreement with the volume integral.

The divergence theorem states that a continuously differentiable vector field $\mathbf{F}$ on a volume $V$ with piecewise smooth closed boundary $S$ satisfieswhere $\hat{\mathbf{n}}$ is the outward unit normal. Take $\mathbf{F}=\nabla a\times \mathbf{B}$. Since the curl of a gradient is zero for a twice continuously differentiable scalar field, part (i) givesIntegrating this identity and applying the divergence theorem proves