Fourier sine series

A Fourier sine series expands a function on an interval in the eigenfunctions $\sin (n\pi x/L)$ of the Dirichlet Laplacian. It is the odd Fourier-series extension across the interval endpoints.

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.

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.

On $(0,1)$, integration by parts gives the displayed Fourier sine series. For normalized Dirichlet eigenfunctions $y_i=\sqrt{2}\sin [(i+1)\pi x]$, indexed from $i=0$, it has the form $A[y_0+{\sum }_{i\ge 1}{c}_i{y}_i]$, with $A=4\sqrt{2}/{\pi }^3$ and $c_i=[1-(-1{)}^{i+1}]/[2(i+1{)}^3]$. Its Rayleigh quotient for $-d^2/d{x}^2$ is $10$, since ${\int }_0^1(u^{\prime }{)}^{2}dx=1/3$ and ${\int }_0^1{u}^{2}dx=1/30$. Reflection symmetry forces $c_1=0$, because that first excited eigenfunction is antisymmetric about $1/2$. Thus the first excited state's contribution to the error is exactly zero; the first nonzero correction is $c_2=1/27$.

The period-$L$ extension of $f(x)=Kx/L$ on $-L/2<x<L/2$ has Fourier coefficients $a_0=a_n=0$ and $b_n=K(-1{)}^{n+1}/(\pi n)$. ConsequentlyThe Fourier series converges to the ramp away from its jumps and to zero at every jump, the average of the one-sided limits. For $K\ne 0$, Termwise differentiation of a Fourier series fails here as an ordinary convergent series: the differentiated summands have nondecaying amplitude $2K/L$. The classical derivative equals $K/L$ between jumps, whereas the periodic distributional derivative includes negative point masses of weight $K$ at the jumps.

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.

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

For $0<t\le \pi$, split the sum at $\lfloor 1/t\rfloor$. The first part is bounded using $|\sin kt|\le kt$; summation by parts and the geometric bound on interval sine sums control the rest. The harmonic coefficient sum grows like $\log m$, but cancellations keep the whole trigonometric polynomial uniformly bounded. Modulating this polynomial lets a Fourier partial sum expose the large uncancelled half.

The functions $\sqrt{2/\pi }\sin (nx)$, $n\ge 1$, form a complete orthonormal basis of $L^2(0,\pi )$. An odd extension to $(-\pi ,\pi )$ reduces completeness to that of the Fourier series. Equality of all sine coefficients therefore implies equality almost everywhere; if both functions are continuous, it implies equality everywhere.