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On a rectangle $(0,L_x)\times (0,L_y)$, products of the one-dimensional Fourier sine basis give an orthogonal basis of L2 space. Fourier orthogonality in each variable givesFor a constant function one, these coefficients are $4[1-(-1{)}^m][1-(-1{)}^n]/({\pi }^{2}mn)$: only odd indices occur. The product modes vanish on the rectangle's boundary, making them natural Dirichlet Laplacian eigenfunctions. In the heat equation, their amplitudes decay as $e^{-\kappa {\pi }^2(m^2/L_x^2+n^2/L_y^2)t}$. Initial data not vanishing at the boundary are still represented in the interior and in the square-integrable sense; the positive-time boundary values need not agree continuously with the initially assigned boundary values.

For $0<\xi <l$, the wave equation $y_{tt}=c^2{y}_{xx}$ with zero endpoint displacement, zero initial displacement and initial velocity $\delta (x-\xi )$ has the distributional weak solutionExpand the Dirac delta in the Fourier sine basis and solve each modal ordinary differential equation. Equivalently take the odd, $2l$-periodic extension $v$ of the initial velocity and use $y=(2c{)}^{-1}{\int }_{x-ct}^{x+ct}v(s)ds$. The image impulses enforce the fixed endpoints. At jump fronts the Fourier series selects the midpoint of the two one-sided values.

A sequence in L2 space converges in its L2 norm when $\int |f_n-f{|}^{2}d\mu \to 0$. This concerns the integral of the squared error and need not imply pointwise convergence. In particular a complete Fourier sine basis reconstructs each odd L2 function in this norm, while pointwise claims about its Fourier series require additional hypotheses.

A normal mode, also called an eigenmode, is a pattern of free oscillation in a linear system whose relative displacements stay fixed while a single temporal factor changes their size and sign. For an undamped stable vibrational mode,Here $\varphi$ is the mode shape, $A$ sets the amplitude, $\omega$ is the eigenfrequency expressed as an angular frequency, and $\phi$ is the oscillation phase. Every snapshot is a scalar multiple of the same mode shape; neighbouring regions may move in opposite directions. Normal modes of a double pendulum show that normal modes also describe coupled discrete systems, beyond strings. Their shapes are eigenvectors, rather than spatial eigenfunctions. ^[ref]

For the ideal wave equation on a string, take a uniform string of length $L$ with constant tension $\tau$ and linear mass density $\mu$. Assume small transverse slope, negligible bending stiffness, no damping, and fixed endpoints. Its wave speed is $c=\sqrt{\tau /\mu }$, and the nonzero normal modes areThe angular frequency ${\omega }_n$ is in radians per second; the frequency $f_n$ is in hertz. The wavelength is $2L/n$. The mode shape has $n$ loops separated by nodes of a standing wave, including both fixed endpoints. ^[ref]

As a worked example, choose $L=1\mathrm{m}$, $\tau =1\mathrm{N}$, and $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$. Then $c=1\mathrm{m}{\mathrm{s}}^{-1}$ and $f_1=0.5\mathrm{Hz}$, $f_2=1\mathrm{Hz}$, $f_3=1.5\mathrm{Hz}$. Their oscillation periods are $T_1=2\mathrm{s}$, $T_2=1\mathrm{s}$, and $T_3=2/3\mathrm{s}$. These are deliberately simple model parameters, rather than measurements of a musical instrument. The plot divides displacement by $A$ so that the three mode shapes can be compared, and uses each mode's own oscillation period for its snapshots.

For $\phi =0$, at $t=T_n/4$ every point has zero displacement, but most points have nonzero velocity. Only zeros of the mode shape are permanent nodes of a standing wave. At $t=T_n/2$ the original mode shape has reversed sign, and at $t=T_n$ it returns. The synchronized conversion between kinetic energy and potential energy is shown in normal-mode energy exchange.

The short derivation uses separation of variables: substituting $u=\varphi q$ into the wave equation gives ${\varphi }^{\prime \prime }+k^2\varphi =0$ and $q^{\prime \prime }+c^2{k}^{2}q=0$. The fixed-end boundary conditions remove the cosine spatial term and require $\sin (kL)=0$, hence $k=n\pi /L$ and ${\omega }_n=ck$. Different geometries and boundary conditions give different eigenfunctions; a sine mode shape is a property of this uniform fixed string, not a universal form of an eigenmode.

The usefulness is that a complicated vibration becomes a collection of harmonic oscillators. For sufficiently regular fixed-end initial data, the Fourier sine series givesThe initial displacement and velocity determine $a_n$ and $b_n$. This is the superposition principle in a Fourier sine basis, and modal energy distribution measures how strongly each normal mode is excited. The striking simplification is that each component keeps its identity while their sum can continually change shape. ^[ref]

These independent normal modes belong to the linear model. At larger slopes, changing tension introduces nonlinear effects and can couple the modes. A linearly damped string still admits a Fourier sine basis, but its temporal factors decay and need not oscillate; its energy dissipation identity for a linearly damped string replaces conservation of energy. More general damping need not leave undamped mode shapes independent.

Take $l>0$ and use the inner product $⟨u,v⟩={\int }_0^l\overline{u}vdx$ on the Hilbert space $L^2(0,l)$. The differential operator has domainwhere $H^2$ is a Sobolev space. Twice applying integration by parts gives the Lagrange identity for a Sturm-Liouville operatorThe Dirichlet boundary conditions make this expression zero for $u,v\in D(\mathcal{L})$, and the formal adjoint is ${\mathcal{L}}^{\ast }=d^2/d{x}^2+1$. To obtain the domain of the adjoint operator, initially impose the boundary conditions only on $u$. The remaining expression is $-[\overline{u^{\prime }}v{]}_0^l$. Since $u^{\prime }(0)$ and $u^{\prime }(l)$ vary independently, it vanishes for all such $u$ exactly when $v(0)=v(l)=0$. The adjoint domain is therefore the original domain: this is full self-adjointness, including the boundary conditions.

Consequently the eigenvalues are real, eigenfunctions for distinct eigenvalues are orthogonal, and regular Sturm-Liouville theory supplies a complete orthonormal eigenbasis. Here these objects can be exhibited explicitly:Indeed $\mathcal{L}{\varphi }_n={\lambda }_n{\varphi }_n$, and the Fourier sine basis is complete in $L^2(0,l)$. There are no additional eigenvalues: for a nonzero eigenfunction $u$, integration by parts givesso $u^{\prime \prime }+(1-\lambda )u=0$ and the endpoint values force $\sqrt{1-\lambda }l=n\pi$. All these eigenvalues are simple. The eigenvectors are precisely the nonzero multiples of these eigenfunctions, regarded as vectors in the Hilbert space.

Expand in the Fourier sine basis, using unnormalized eigenfunctions:Projection of $\mathcal{L}y=f$ onto each eigenfunction givesWhen $l\notin \pi {\mathbb{N}}_{>0}$, every eigenvalue is nonzero and the eigenfunction expansion is uniquely determined:When $l=m\pi$, the exceptional index is $n=m$, for which the eigenfunction is $\sin x$. If $f_m\ne 0$, its coefficient equation is $0=f_m$, an inconsistency: there is no solution. If $f_m=0$, the coefficient equation is $0=0$, and $c_m$ is arbitrary. The full family isThis is the Fredholm alternative directly in the coefficient equations: nonzero eigenvalues fix their coefficients; a zero eigenvalue requires zero forcing in that direction and leaves the corresponding coefficient free. An inconsistent equation at a zero eigenvalue cannot be repaired by assigning an infinite coefficient.

For convergence, Lipschitz continuity implies absolute continuity of $f$, with bounded derivative almost everywhere. One integration by parts gives $f_n=O(n^{-1})$; since the denominators grow quadratically, $c_n=O(n^{-3})$ apart from the single free coefficient. The solution series and its first derivative therefore converge uniformly. The second derivative converges in $L^2$, and the resulting Sobolev space solution satisfies $y^{\prime \prime }=f-y$. As $f-y$ is a continuous function, this is a classical $C^2$ solution, with both boundary conditions satisfied. The forcing's Fourier sine series is interpreted on the open interval or in $L^2$; its endpoint values need not equal the endpoint values of $f$.

At $l=\pi$, the Fredholm solvability condition isby orthogonality of the Fourier sine basis. Thus a non-unique solution exists exactly when $a=0$, with any real $b$. Since $(d^2/d{x}^2+1)\sin 2x=-3\sin 2x$, the complete family isIf $a\ne 0$ there is no solution, for any $b$. In particular, when $a=b=0$, the family reduces to the kernel $C\sin x$; it is still non-unique. There is no choice of $a,b$ producing a unique solution for this length.

For positive integer indices, distinct members are orthogonal and each has nonzero squared norm $\pi$. Thus $\{\sin (nx):n\ge 1\}$ is an orthogonal basis in the Hilbert space of odd L2 functions on $(-\pi ,\pi )$. The zero-index member must be omitted. More precisely this is a complete Fourier sine basis, with generally infinite L2 norm convergence, rather than a Hamel basis with finite linear combinations. A claim about arbitrary real functions requires suitable integrability and a specified sense of convergence.

Expand the initial velocity in the Fourier sine basis, $V(x)={\sum }_{n\ge 1}{V}_n\sin (n\pi x/L)$, where $V_n=(2/L){\int }_0^{L}V(x)\sin (n\pi x/L)dx$. Zero initial displacement removes the cosine-in-time coefficients. ThusThe trigonometric identity $2\sin a\sin b=\cos (a-b)-\cos (a+b)$ rewrites each summand asIt is therefore a superposition of travelling waves of speeds $+c$ and $-c$, whose interference enforces the fixed endpoints. For general square-integrable initial velocity, the series is interpreted in the standard finite-energy sense.

Assume $V_0\ne 0$. Linear independence of the Fourier sine basis makes a flat string equivalent to vanishing of all three temporal coefficients. The fundamental requires $\sin ({\omega }_{1}t)=0$, which automatically zeros the other two. Thus the earliest positive return isThe velocity need not have returned to its initial value at that time; full recurrence of the displacement-and-velocity data occurs at $2L/c$.

A nonzero pure fundamental profile would require $\sin 2{\omega }_{1}t=\sin 3{\omega }_{1}t=0$. The common zeros have ${\omega }_{1}t$ an integer multiple of $\pi$, so the fundamental vanishes too. A pure second or third harmonic would already require the fundamental to vanish, again making the whole string flat. The string never has a nonzero pure sine profile. If $V_0=0$, it remains identically zero and there is no earliest positive isolated return.

Assume $0<a<b<l$ for the interior pulse; endpoint variants use the same coefficient formula. Orthogonality of the Fourier sine basis givesThe Fourier sine series equals $f$ at its interior continuity points and takes value $1/2$ at the jumps $a,b$; the sine series is zero at both endpoints. Thus it does not retain an arbitrarily assigned value at a jump.

For the string, write $y(x,t)={\sum }_{n\ge 1}{q}_n(t)\sin (n\pi x/l)$. The wave equation on a string givesThe last equality is the Fourier sine series of the Dirac delta at $l/4$. Solving the modal ordinary differential equations givesThis impulsively struck fixed-end string is understood as a distributional weak solution; the initial velocity is a distribution, not a classical continuous function.

To evaluate the requested time, take the odd $2l$-periodic extension $v$ of $\delta (x-l/4)$ and use the D'Alembert formula $y=(2c{)}^{-1}{\int }_{x-ct}^{x+ct}v(s)ds$. At $ct=l/2$, the interval contains the positive impulse at $l/4$. For $x<l/4$, it also contains the negative reflected impulse at $-l/4$, so the contributions cancel. For $l/4<x<3l/4$ only the positive impulse contributes. For $x>3l/4$ neither contributes. Thus the midpoint-valued Fourier series givesThe discontinuity values are a convention for the Fourier representative and do not affect the distribution. The sketch shows the reflected negative front cancelling the left portion of the initially generated plateau.

Use the mixed Dirichlet-Neumann modes on an interval. To justify their Fourier series, extend $y$ evenly across $x=1$ to $[0,2]$, then oddly across $x=0$, and repeat with period four. The condition $y(0)=0$ makes the odd extension continuous, and $y^{\prime }(1)=0$ makes the even reflection continuously differentiable. The resulting periodic function has a convergent Fourier sine series. Reflection about $x=1$ eliminates the even sine harmonics, leavingAlternatively these values follow directly from $\sin {\lambda }_{n}0=0$ and ${\lambda }_n\cos {\lambda }_n=0$. Fourier orthogonality givesThe extension argument also supplies completeness, not merely orthogonality, of this Fourier sine basis.

Expand the forcing as $x\cos \pi x=\sum b_n\sin {\lambda }_{n}x$. Product-to-sum and integration by parts giveEach mode of the differential operator has eigenvalue $-({\lambda }_n^2+{\alpha }^2)$. Hence, for real $\alpha$, the solution isThe coefficients are $O(n^{-4})$, so the series and its first two differentiated series converge uniformly. Therefore it solves the equation and the boundary conditions term by term. Uniqueness follows by multiplying the homogeneous equation by its real solution and integrating: the boundary terms vanish and ${\int }_0^1[(y^{\prime }{)}^2+{\alpha }^2{y}^2],dx=0$, forcing $y=0$.

The Fourier sine basis of the unit square is $e_{j\ell }(x,y)=2\sin (j\pi x)\sin (\ell \pi y)$, $j,\ell \ge 1$, normalized in the L2 inner product. It satisfies the zero Dirichlet boundary conditions and $\Delta e_{j\ell }=-{\pi }^2(j^2+{\ell }^2)e_{j\ell }$. Expand the given initial datum as $u_0=\sum a_{j\ell }{e}_{j\ell }$, where $\sum |a_{j\ell }{|}^2=\Vert u_0{\Vert }_2^2$. The solution isFor $\kappa \le 2{\pi }^2$, every exponent is nonpositive. The series converges in $L^2$ for $t\ge 0$, tends to $u_0$ in $L^2$ as $t\downarrow 0$ by dominated convergence, and differentiates termwise for $t>0$ because the Gaussian high-frequency decay dominates every polynomial derivative factor. This constructs a classical positive-time solution and the appropriate mild solution at the possibly rough initial time.

Its norm and the difference of two such solutions satisfyExistence, uniqueness in the natural $L^2$ solution class, and continuous dependence follow. For uniqueness one can project any weak solution onto each sine mode, obtaining the same scalar linear equation and initial coefficient. Thus this is a Hadamard well-posed problem, with a contractive strongly continuous semigroup in the stated range.

To prove pointwise decay from merely $L^2$ data, use $|e_{j\ell }|\le 2$ and the Cauchy-Schwarz inequality:For $t\ge 1$, factor out $e^{2(\kappa -2{\pi }^2)t}$; the remaining sum is bounded by ${\sum }_{j,\ell \ge 1}{e}^{-2{\pi }^2(j^2+{\ell }^2-2)}<\infty$. ThereforeAt equality the first sine mode is stationary, so strict inequality is essential for decay of every datum. This is the Dirichlet heat-reaction threshold on the unit square. The equation is actually well posed on every fixed finite time interval for any fixed real $\kappa$; beyond the threshold it has a growing first mode and loses long-time contractivity. The restricted well-posedness assertion in the question is therefore sufficient rather than necessary.

Fix the convention $\Delta q-m^{2}q=0$, $m>0$, and let $c_m=m^2/4$. With $z=x+iy$, two families of modified Helmholtz adjoint plane waves areBoth satisfy $\Delta E_{\pm }=m^2{E}_{\pm }$, since the products of their $z$ and $\bar{z}$ exponents are $m^2/4$. They obey $E_{-}(z,k)=\overline{E_{+}(z,\bar{k})}$. Green second identity gives the two conjugate global relations for the modified Helmholtz equationFor real boundary traces, ${\mathcal{G}}_{-}(k)=\overline{{\mathcal{G}}_{+}(\bar{k})}$. Thus the second identity is the conjugate spectral companion, rather than an unrelated extra boundary condition. In complex differential form the first relation is equivalentlyIts integrand is a closed one-form: differentiating its coefficients gives $2E_{+}(q_{z\bar{z}}-c_{m}q)=0$. This supplies a direct derivation of the global relation as well as its outward-normal sign convention.

Write the known Dirichlet boundary condition on side $j$ as $f_j(s)$ and the unknown outward normal derivative as $h_j(s)$, with $-1<s<1$. For any linear combination $v$ of the adjoint waves, the global relation isExpanding the $h_j$ in a finite basis and enforcing these identities at selected collocation points for a global relation gives a linear system with a known right side. The square permits especially simple paired tests.

Let $p_n=n\pi /2$, ${\eta }_n=\sqrt{m^2+p_n^2}$, and ${\phi }_n(s)=\sin [p_n(s+1)]$ for $n\ge 1$. Define four real adjoint solutionsThey satisfy the modified Helmholtz equation because ${\eta }_n^2-p_n^2=m^2$. Each is a linear combination of two adjoint exponentials, hence is obtained from the two spectral global relations. More explicitly, put ${\rho }_n^{\pm }=({\eta }_n\pm p_n)/2$, so ${\rho }_n^{+}{\rho }_n^{-}=c_m$. For the $E_{+}$ convention above, the required spectral pairs areAppropriate phase-weighted differences produce $\sin [p_n(s+1)]$. For example the top test is $e^{-{\eta }_n}[e^{i{p}_n}{E}_{+}({\rho }_n^{-})-e^{-i{p}_n}{E}_{+}({\rho }_n^{+})]/(2i)$. The conjugate relation ensures a real system for real data. This is sine collocation of square modified Helmholtz global relations.

The functions ${\phi }_n$ are an orthonormal Fourier sine basis on $[-1,1]$. Set $h_{j,n}={\int }_{-1}^1{h}_j{\phi }_{n}ds$ and compute the known quantity $R_{j,n}={\int }_{\partial \Omega }f{\partial }_n{v}_{j,n}ds$ by numerical integration. Since ${\phi }_n(\pm 1)=0$, each adjoint test vanishes on both adjacent sides. On its own side it equals ${\phi }_n$, and on the opposite side it equals ${\delta }_n{\phi }_n$, where ${\delta }_n=e^{-2{\eta }_n}$. Thus the square modified Helmholtz Dirichlet-to-Neumann coefficients obeyEach block is inverted explicitly: for opposite sides $j,j^{\prime }$,Compute these coefficients for $1\le n\le N$ and reconstruct $h_j^{(N)}={\sum }_{n=1}^N{h}_{j,n}{\phi }_n$. This is a semi-analytical scheme: the spectral basis integrals and two-by-two inverses are explicit, while the known boundary integrals are evaluated numerically. Increase $N$ and the quadrature resolution until the desired convergence is observed; compatible smooth side data have convergent normal-trace expansions, while corner singularities require the usual weaker trace interpretation and more careful quadrature.

The requested weak interaction between sides is particularly clear: the adjacent-side unknown traces contribute exactly zero, and the opposite-side coefficient is $e^{-2\sqrt{m^2+(n\pi /2{)}^2}}\to 0$. The own-side coefficient stays equal to one. Thus diagonal dominance of paired square global-relation collocation is genuine after pairing; individual uncombined exponential samples need not have the same conditioning. The eigenvalues of each block are $1\pm {\delta }_n$, so its condition number is $(1+{\delta }_n)/(1-{\delta }_n)$.

There is also a direct localization check for an individual adjoint wave. On side $j$ with outward unit normal $n_j$ and tangent $t_j$, the normalized test $e^{\eta (n_j\cdot r-1)+ip{t}_j\cdot r}$ has $\eta =\sqrt{m^2+p^2}$. Its opposite-side magnitude is $e^{-2\eta }$, and its magnitude integrates to at most $1/\eta$ along either adjacent side. This side localization of modified Helmholtz plane waves explains why suitable large spectral parameters suppress remote-side effects even before the exact sine cancellation.

Finally the interior solution can be evaluated from the computed traces using the two-dimensional modified Helmholtz fundamental solutionHere $K_0$ is the Modified Bessel function of the second kind, and $(-\Delta +m^2){\Gamma }_m=\delta$. Replace $h$ by its computed finite Fourier sine series and use numerical integration; at interior points the boundary kernels are smooth. The sign follows from Green second identity with this fundamental-solution convention. This completes the numerical integration of the boundary value problem with a Dirichlet boundary condition, rather than stopping at an equation for its unknown boundary derivatives.

Use the Fourier coefficientsThey are continuous for $0\le y\le 1$ and bounded by $2\Vert u{\Vert }_{\infty }$. On compact subintervals of $0<y<1$, the equation is uniformly elliptic with smooth coefficients. Local elliptic regularity at the flat vertical sides with zero boundary values permits differentiation and integration by parts there. ThusFor completeness, this identity also follows in the distributional sense without assuming boundary derivatives initially. Integrate on $(\epsilon ,\pi -\epsilon )$ against $\sin (nx)$ and a test function of $y$ with support away from zero and one. The interior gradient estimate near a zero Dirichlet side gives $|u_x|\le C{\epsilon }^{-1}\sup |u|$ on a nearby ball of radius comparable to $\epsilon$; continuity and the zero side value make that supremum $o(1)$ uniformly over the support in $y$. Therefore the boundary products $u_x\sin (nx)$ and $u\cos (nx)$ tend to zero. Move the $y$ derivatives to the test function before taking $\epsilon \downarrow 0$. This proves the ordinary differential equation, which makes each $b_n$ smooth in $0<y<1$ and yields the same classical identity.

The Cauchy-Euler differential equation has indicial rootsFor $n\ge 1$ the negative root is strictly negative. Boundedness as $y\downarrow 0$ forces $B_n=0$. Continuity at $y=1$ identifiesBecause $p_n^{+}>n$, the series ${\sum }_{n\ge 1}{A}_n{y}^{p_n^{+}}\sin (nx)$ converges absolutely and uniformly for $0\le y\le \rho <1$. On compact subsets with $y>0$, differentiated series converge as well. At each fixed $0<y<1$, its Fourier coefficients equal those of $u(\cdot ,y)$; completeness of the Fourier sine basis in $L^2(0,\pi )$ and continuity in $x$ make the two functions equal. ThereforeThis is also the requested sum over integers: take all coefficients with $n\le 0$ to be zero. Negative indices give redundant Fourier modes, and the $n=0$ sine vanishes. No pointwise convergence of the ordinary Fourier series on the top boundary is needed.

An important consequence is a forced zero trace at a quadratically degenerate boundary. Indeed the absolute sum is bounded by $2\Vert u{\Vert }_{\infty }{\sum }_{n\ge 1}{y}^n=2\Vert u{\Vert }_{\infty }y/(1-y)$ for $y<1$, which tends to zero as $y\downarrow 0$, uniformly in $x$. Thus $u(x,0)=0$ on the entire lower side. It follows also from $b_n(0)=0$ for every sine coefficient and completeness.

For $c>0$ and $l>0$, the zero-endpoint Dirichlet boundary conditions select the Fourier sine basis functions $\sin (n\pi x/l)$. Separation of variables gives temporal factors with angular frequencies ${\omega }_n=n\pi c/l$. Zero initial displacement removes the temporal cosine terms, givingDifferentiating at $t=0$ and using Fourier orthogonality givesThe evaluation property of the given initial velocity means $v(x)=\delta (x-l/2)$ in the sense of a Dirac delta, an idealized impulsively struck fixed-end string. Thus $a_n=2\sin (n\pi /2)/(n\pi c)$: every even mode vanishes, while the lowest two excited modes have $n=1,3$. Their ordinary frequencies and amplitude ratio areThe signed coefficient ratio is $a_3/a_1=-1/3$. The impulse solution is interpreted as a distributional weak solution, rather than one with a smooth initial velocity.

Separate variables as $u=X(x)T(t)$. The zero values at both plates give the Fourier sine basisExpand the initial profile using orthogonality. With $s=x/L$, its coefficients areConsequentlyOnly odd modes occur because the initial profile is symmetric about the midpoint. The coefficients are absolutely summable, so dominated convergence gives $\boxed{u(x,t)\to 0}$ uniformly as $t\to \infty$. The slowest decay rate is $\nu {\pi }^2/L^2$, associated with $j=1$.

Set $y(x,t)=X(x)T(t)$ in the wave equation. A nonzero product obtained by separation of variables obeysThe fixed-end Dirichlet boundary conditions require $X(0)=X(L)=0$. The zero and negative spatial eigenvalue cases produce only the trivial solution; the nontrivial eigenfunctions areFor each normal mode, $T_n^{\prime \prime }+{\omega }_n^2{T}_n=0$. By superposition and completeness of the Fourier sine basis, the general finite-energy solution has the expansionIf $f(x)=y(x,0)$ and $g(x)=y_t(x,0)$, orthogonality gives $A_n=(2/L){\int }_0^{L}f(x)\sin (k_{n}x)dx$ and $B_n=2{\int }_0^{L}g(x)\sin (k_{n}x)dx/(L{\omega }_n)$. Individual products are separable; their sum is the general solution.

For the harmonic extension in a circular sector, write a separated harmonic function as $F(r)H(\theta )$. The Laplace equation in polar coordinates yieldsThe angular Dirichlet boundary conditions select the Fourier sine basisIndeed, the angular equation has no nonzero solution with nonpositive ${\nu }^2$: multiplying $-H^{\prime \prime }={\nu }^{2}H$ by $H$ and integrating gives $\int (H^{\prime }{)}^2={\nu }^2\int H^2$. The radial Cauchy-Euler equation has solutions $r^{{\nu }_n}$ and $r^{-{\nu }_n}$. Boundedness at the vertex excludes the latter, so superposition gives

The wave equation and its homogeneous Dirichlet boundary conditions are linear, so any finite linear combination of the separated normal modes is another solution. More strongly, the Fourier sine basis is complete in L2 space on $(0,L)$. Expanding the initial fields in this basis yieldsCompleteness of the Fourier sine basis is the reason the expansion covers general data; linearity alone would not prove that every solution has this form. The convergence and differentiation are understood in a function space appropriate to the data. For finite wave energy, one can take $f\in H_0^1(0,L)$ and $g\in L^2(0,L)$, obtaining an energy solution; stronger smoothness and endpoint compatibility give a classical solution.