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Subtract a known function with the required boundary trace so that the remaining unknown has homogeneous boundary conditions. For an interval with Dirichlet boundary conditions $g(t),h(t)$, $\ell =(1-x/L)g+(x/L)h$ is a simple lifting. In $w_t=w_{xx}-cw$, $z=w-\ell$ has forcing $-{\ell }_t-c\ell$ and zero endpoint values. This improves Fourier sine series convergence.

For $y_{tt}-y_{xx}=F(t)\delta (x-x_0)$ with zero endpoint displacement, a Fourier sine series gives $q_n^{\prime \prime }+k_n^2{q}_n=2F(t)\sin (k_n{x}_0)/L$. With the $e^{-i\omega t}$ Fourier transform convention, the causal inverse of this oscillator uses the displayed denominator. Its inverse transform is $H(t)\sin (k_{n}t)/k_n$, giving $q_n(t)=2\sin (k_n{x}_0)/(L{k}_n){\int }_{-\infty }^{t}F(s)\sin [k_n(t-s)]ds$ when the past integral exists. The plain denominator $k_n^2-{\omega }^2$ requires a pole prescription: its Cauchy principal value gives the time-symmetric kernel $\sin (k_n|t|)/(2k_n)$ instead. Homogeneous sine and cosine modes are additional freedom unless temporal data or causality are specified.

On $0<x<L_x$, $0<y<L_y$ with positive side lengths and zero Dirichlet boundary conditions, separation of variables for $-\Delta u=\lambda u$ giveswith the displayed eigenvalues. The one-dimensional endpoint conditions exclude constant and hyperbolic factors and force the positive integer sine frequencies. The product Fourier sine series basis is complete in the square-integrable functions on the rectangle. Consequently an eigenfunction with prescribed $\lambda$ is a linear combination of exactly those modes with ${\lambda }_{mn}=\lambda$; if none match, only the zero solution exists. Distinct pairs with the same sum give an eigenvalue with geometric multiplicity greater than one. On a square of side $\pi$, for instance, $\lambda =25$ has precisely the pairs $(3,4)$ and $(4,3)$, so its eigenspace has dimension two.

For an integrable function on $[-L,L]$, with $L>0$ and integer $n\ge 1$, the displayed Fourier coefficient multiplies the mode $\sin (n\pi x/L)$ in its real Fourier series. Fourier orthogonality fixes the normalization because the squared mode has integral $L$. If $f$ is odd, the product with the sine is even, so $b_n=(2/L){\int }_0^{L}f(x)\sin (n\pi x/L)dx$. These are also the coefficients of the Fourier sine series of the restriction to $(0,L)$.

Extend $F_L(x)=x(L-|x|)$ from $[-L,L]$ to a periodic function of period $2L$, with $L>0$. It is odd and continuous, so its Fourier series contains only sine terms. Its Fourier coefficients areTwo applications of integration by parts prove the coefficient formula. The first has no endpoint contribution because $x(L-x)$ vanishes at both endpoints; the second leaves a multiple of $1-\cos (n\pi )$. Thus only odd frequencies survive, giving the displayed Fourier sine series. Its absolutely summable coefficients give uniform convergence, and Fourier orthogonality identifies its sum with $F_L$.

Taking $L=1$, ${\int }_{-1}^1{F}_1(x{)}^{2}dx=2{\int }_0^1{x}^2(1-x{)}^{2}dx=1/15$. Parseval's identity therefore gives

On $0\le x\le L$, the Fourier sine series of ${\sin }^2(\pi x/L)$ has zero even-index coefficients and odd-index coefficients $b_n=8/[\pi n(4-n^2)]$. Substitute $u=\pi x/L$ in the coefficient integral and use ${\sin }^{2}u\sin (nu)=\frac{1}{2}\sin (nu)-\frac{1}{4}\sin ((n+2)u)-\frac{1}{4}\sin ((n-2)u)$. For odd $n$, integrating gives $b_n=(2/\pi )[1/n-\frac{1}{2}(1/(n+2)+1/(n-2))]$; reflection about the midpoint makes the even coefficients zero. The $O(n^{-3})$ decay gives absolute and uniform convergence, and Fourier orthogonality identifies the sum. In the heat equation with zero endpoint temperatures, multiply each coefficient by $\exp [-\kappa (n\pi /L{)}^{2}t]$. With insulated endpoints, the same initial profile is instead just the constant and second cosine modes of a Fourier cosine series.

On $0\le x\le L$, with $L>0$, the tent $f(x)=\min (x,L-x)$ has Fourier sine series coefficients $b_n=4L\sin (n\pi /2)/({\pi }^2{n}^2)$. Split the coefficient integral at $L/2$ and integrate its two linear pieces by parts; their cosine endpoint terms cancel and their sine terms add. Reflection about the midpoint makes every even-index coefficient zero, and the remaining odd-index coefficients alternate in sign. The $n^{-2}$ bound gives absolute convergence and uniform convergence; the series equals the continuous tent, including its zero endpoints.

For a unidirectional incompressible flow between stationary plates a distance $H>0$ apart, removing the driving pressure gradient leaves $u_t=\nu u_{yy}$ when the kinematic viscosity $\nu >0$ is constant. The no-slip boundary condition gives $u(0,t)=u(H,t)=0$. For an initial profile $g$ in L2 space, separation of variables and the Fourier sine series give the displayed solution, withEach normal mode decays at a rate proportional to $n^2$. The lowest mode present dominates at late times; when $b_1\ne 0$, its time constant is $H^2/(\nu {\pi }^2)$, consistent with the viscous diffusion time. The solution approaches the initial data in L2 space; compatible smooth profiles also give the familiar pointwise initial condition. This is a heat equation for momentum diffusion, rather than temperature.

On $0<x<a$, $0<y<b$, the bounded harmonic function with value $F$ on the open upper edge and zero on the other three open edges has the sine-hyperbolic series in the title. Separation of variables gives the modes, and the Fourier sine series of a constant selects only odd $n$. The series converges exponentially on compact subsets of the interior, justifying its derivatives in Laplace's equation.

For $F\ne 0$, the upper corners have incompatible one-sided boundary limits, so the solution cannot extend continuously to them. Assigning isolated corner values does not alter the bounded interior Dirichlet problem. For numerical evaluation, the stable mode ratio isThis avoids overflow from separately evaluating large hyperbolic sines.

On $0<x<L$ with zero-temperature endpoints, let the initial temperature rise linearly from zero to height $H$ at $L/2$ and then fall linearly to zero. Integration by parts on the two halves gives Fourier sine series coefficients $8H\sin (n\pi /2)/({\pi }^2{n}^2)$. Under the heat equation $T_t=D{T}_{xx}$ with $D>0$, each coefficient decays by $e^{-D(n\pi /L{)}^{2}t}$, giving the displayed solution. Only odd spatial modes occur; absolute coefficient summability proves uniform convergence for $t\ge 0$ via the Weierstrass M-test. This spatial profile also occurs in a midpoint-plucked fixed-end string, where the time factors oscillate instead of decaying.

A string of length $L$ with fixed endpoints, initially displaced into a triangular tent of height $k$ at its midpoint and released from rest, has a Fourier sine series with the coefficients in the title. Integrating the two linear halves gives $a_n$, and symmetry cancels the even modes. Each coefficient evolves as $a_n\cos ({\omega }_{n}t)$ under the wave equation. The displacement vanishes everywhere at $t=L/(2c)$, reverses sign at $t=L/c$, and returns after $2L/c$. A straight configuration at an intermediate instant does not mean the string is at rest: its kinetic energy is then nonzero. The initial slope corner gives a finite-energy weak solution whose corner lines propagate and reflect at the ends.

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.

Project the expansion onto one normalized basis function. Orthogonality removes every other coefficient:Consequently,This is a Fourier sine series in normalized form. For a square-integrable $f$, completeness gives convergence in the norm of the function space; pointwise convergence requires the usual additional regularity. If $f$ is a normalized wavefunction, the Born rule gives probabilities $|a_n{|}^2$ and ${\sum }_n|a_n{|}^2=1$.

Infinite walls impose the Dirichlet boundary conditions $\psi (-a)=\psi (a)=0$. Inside the infinite square well, the time-independent Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2m)=E\psi$. Its nonzero solutions satisfying both endpoint conditions have wave numbers $k_n=n\pi /(2a)$, $n\ge 1$. Sincethe normalized energy eigenstates are the sine modes with prefactor $1/\sqrt{a}$. They are orthonormal by sine orthogonality and complete by the Fourier sine series on an interval of length $2a$. Differentiating twice gives

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.

The Fourier sine series has coefficientsby integration by parts. ThusIts odd extension is $x$ on $(-L,L)$, continued with period $2L$. The usual Fourier convergence result for a piecewise smooth periodic function gives the value at a continuous point and the mean of one-sided limits at a jump. At $x=3L/2$, subtracting one period gives $-L/2$, a continuous point of this extension. Therefore the sine series converges to $-L/2$ there.

For the Fourier cosine series, the even extension is $|x|$ on $[-L,L]$, continued with period $2L$. Its triangular wave is continuous even at the joining points, so the cosine series converges to that extension everywhere. On the requested interval its graph isThe vertices are $(-2L,0),(-L,L),(0,0),(L,L),(2L,0)$. In particular, its value at $3L/2$ is $L/2$.

Assume the impulse point is in the interior, $0<a<l$. Expand the displacement in the fixed-end Fourier sine seriesOrthogonality gives the initial velocity coefficient by applying the Dirac delta to the sine function:Substitution in the wave equation yields $T_n^{\prime \prime }+2k{T}_n^{\prime }+{\alpha }_n^2{T}_n=0$. For $k<{\alpha }_n$, its roots are $-k\pm i{\omega }_n$, where ${\omega }_n=\sqrt{{\alpha }_n^2-k^2}$. The two initial conditions determineHence the normalized velocity-impulse Green function for a damped string isFor a mode with critical damping, use the limit $e^{-kt}t$ in place of the time factor. For $k>{\alpha }_n$, put ${\eta }_n=\sqrt{k^2-{\alpha }_n^2}$ and use $e^{-kt}\sinh ({\eta }_{n}t)/{\eta }_n$. These choices solve the same modal ordinary differential equation and initial conditions, so the formula covers arbitrary $k>0$. The impulse initial velocity is a distribution, so initial differentiation of the series is interpreted distributionally.

The printed formula requires correction: the factor depending on $n$ must lie inside the sum, and sine-mode normalization requires $2/l$, rather than $2$. The printed slash after the final sine is also stray. Direct differentiation at $t=0$ verifies the corrected normalization: the initial sine coefficients are exactly those of $\delta (x-a)$.

Irrotational flow means $\nabla \times \mathbf{u}=0$; incompressible flow means $\nabla \cdot \mathbf{u}=0$. The given streamfunction automatically satisfies the second condition, since ${\partial }_x{\psi }_y-{\partial }_y{\psi }_x=0$. For an oriented path with unit tangent $(dx/ds,dy/ds)$, choose its right-hand unit normal $\mathbf{n}=(dy/ds,-dx/ds)$. ThenThe opposite normal reverses the sign; the source's unspecified normal orientation must be chosen consistently. The volume flux between two streamlines is thus their streamfunction difference. A closed contour has zero net flux, expressing mass conservation for constant-density flow without a source inside.

For the channel, a rigid boundary has zero normal velocity and hence constant streamfunction. Choose zero on the lower wall. Uniform inflow at infinity must have horizontal velocity $-m$, so there $\psi =-my$. The upper wall then has constant $-m$. The unbroken left wall connects to it and also has constant $-m$. The jump by $m$ between the walls at the corner represents the withdrawn flux. Finally the vorticity is $v_x-u_y=-\Delta \psi$, so irrotationality gives $\Delta \psi =0$.

Write $\psi =-my+w$. Then $w$ vanishes at $y=0,1$ and at infinity, while $w(0,y)=-m(1-y)$ for $0<y<1$. Separation of variables gives the harmonic sine mode in a half-strip $e^{-n\pi x}\sin n\pi y$. The Fourier sine series coefficients of the left boundary areTherefore the corner sink in a semi-infinite channel isFor $x>0$ the series and its differentiated series converge on compact subsets, making the sum harmonic. At $x=0$, ${\sum }_{n\ge 1}\sin (n\pi y)/n=\pi (1-y)/2$ for $0<y<1$, so the left-wall value is exactly $-m$. The other wall values and far-field limit follow directly.

The plus sign printed in the final series is inconsistent with the printed boundary conditions. It would give $m-2my$ on the left wall, not $-m$. The negative sign above is required. As a local flux check, near the corner the wall values give $\psi \sim -2m\theta /\pi$; the radial velocity is $-2m/(\pi r)$ and its outward flux across a quarter-circle is $-m$, precisely withdrawal.

The fixed-end wave equation has spatial modes $\sin (n\pi x)$ and temporal factors $\cos (n\pi t),\sin (n\pi t)$. The zero initial velocity removes every temporal sine term. Expand the initial displacement in a Fourier sine series, whose coefficients areTwo integrations by parts give the last equality; the endpoint terms from $x(1-x)$ vanish and the remaining derivative endpoints distinguish even from odd $n$. ConsequentlyThis solves the initial-boundary problem in the finite-energy sense, with convergent first-derivative series. At $t=0$ the energy in the question is ${\int }_0^1(1-2x{)}^{2}dx=1/3$. At any time, sine/cosine orthogonality gives the contribution of mode $n$ as $(n\pi {)}^2{b}_n^2/2$: its kinetic and elastic pieces have factors ${\sin }^2(n\pi t)$ and ${\cos }^2(n\pi t)$ summing to one. ThereforeThe square summability of the derivative coefficients justifies the orthogonality calculation and its limit.

For the Fourier sine series on $(0,1)$, orthogonality givesThereforeThe series is that of the odd, two-periodic extension. At $x=0$ its value is zero, agreeing with $f(0)$; at $x=1$ every summand is zero and the series converges to zero, the average of the two one-sided extension limits. Thus it represents the function in the interior and in the square-integrable sense, not the prescribed value one at the endpoint $x=1$. This endpoint issue matters when using sine expansions with homogeneous boundary conditions.