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Divide the recurrence by $2k$ and substitute a smooth exact solution. Centered Taylor expansions in time and space giveThus the leapfrog advection scheme is consistent with normalized residual $O(k^2+{\delta }^2)$. Its unnormalized defect is $O(k^3+k{\delta }^2)$. For a fixed $0<\nu <1/2$, the uniform two-level bound proved above and accumulation over $O(1/k)$ steps yield, for a sufficiently smooth solution,Both starting levels must be supplied consistently; errors of order $k^2+{\delta }^2$ give second-order convergence. For general square-integrable functions as initial data, point sampling is not defined on equivalence classes. Use cell averages or another bounded grid projection, together with a bounded consistent startup. Approximate the initial data by smooth functions, apply the smooth-data convergence estimate, and use the uniform numerical and exact translation bounds to pass to the L2 space limit. This is the content of Lax equivalence theorem in the present setting.

At $\nu >1/2$ the growing Fourier mode prevents general convergence. At the endpoint, initial data supported in increasingly narrow neighborhoods of the double root can have a second-level error of norm $O(k)$ but a final error of order one after $O(1/k)$ steps. These are vanishing starting errors, so the endpoint fails general two-level convergence as well. Consequentlyis also the general convergence range, with a consistent stable initialization. Claims for a particular specially chosen endpoint startup would be a different, restricted assertion.

Fourier transform on the infinite spatial lattice giveswith polynomial roots $G_{\pm }=iA\pm \sqrt{1-A^2}$. When $|A|$ is uniformly less than one, the polynomial roots have modulus one and a uniform positive separation. Diagonalizing the companion matrix then bounds all its powers independently of $\theta$ and $n$. Parseval's identity proves stability in the discrete $L^2$ norm for arbitrary square-summable starting perturbations at both levels.

At $\mu =1/2$ the coefficients giveConsequently $\mu =1/2$ is stable.

At $\mu =3/2$,The growing polynomial root at this phase has modulusTo turn this into a Cauchy problem linear instability proof, rather than rely on a plane wave of infinite norm, choose a small interval around $\pi /3$ on which the growing polynomial root has modulus at least $1+\delta$. Choose an initial Fourier transform amplitude that is a nonzero square-integrable function supported there, and set the level-one amplitude equal to that polynomial root times the level-zero amplitude. The recurrence then multiplies by the same growing polynomial root at every step on the support. Parseval identity gives $\Vert u^n{\Vert }_2\ge (1+\delta {)}^n\Vert u^0{\Vert }_2$, whereas both initial norms are bounded independently of $h$. Real data can be obtained by adding the conjugate interval around $-\pi /3$. Since $n=T/(\mu h)$ tends to infinity at fixed $T$, $\mu =3/2$ is unstable.

A statistical path through $P$ is a family $t\mapsto P_t\in \mathcal{P}$, defined on an open interval containing zero, with $P_0=P$. Write $p_t=d{P}_t/d\mu$ and $p=p_0$. Its differentiability in quadratic mean at zero means that there is a square-integrable function $g\in L^2(P)$ such thatThe function $g$ is the score function of the statistical path. Thus it is the derivative of the square-root probability density function, multiplied by two and divided by $\sqrt{p}$ where $p>0$. The score function is determined only $P$-almost everywhere. The displayed definition also controls any probability mass entering a region where $p=0$; a pointwise derivative of the log probability density function alone would not do that. The required derivative is in the square-root-density $L^2$ norm.

The Parseval identity for a complex Fourier series isIt holds for every square-integrable function, with complex conjugation expressed by the absolute squares. It is the Pythagorean theorem extended to the complete orthonormal system of complex exponentials.

The complex Fourier series of a $2L$-periodic function isThe coefficient formula follows from orthogonality of the exponential functions. For an arbitrary square-integrable function, this representation means convergence in the $L^2$ norm. With appropriate additional regularity it also gives pointwise convergence, to the mean of the one-sided limits at a jump. No pointwise representation is asserted for an unrestricted function merely from periodicity.

Use the time convention $e^{-i\omega t}$. The free-space Helmholtz equation, with $\psi =e^{ikx}E$, gives the exact reduced equationFor a forward plane wave at angle $\alpha$, $E_x=O(k{\alpha }^{2}E)$ and $E_{zz}=O(k^2{\alpha }^{2}E)$, whereas $E_{xx}=O(k^2{\alpha }^{4}E)$. Thus the paraxial approximation neglects $E_{xx}$ relative to $2ik{E}_x$. The reduced field satisfies the parabolic wave equationWith the Fourier transform convention used below and its inverse, this becomes an ordinary differential equation for each transverse wavenumber:Consequently the initial-value solution isEquivalently, evaluating the oscillatory Gaussian integral gives the one-dimensional transverse Fresnel propagation formulaThe square-root branch has $\sqrt{1/i}=e^{-i\pi /4}$. The formula is an oscillatory integral for general data, or the Fresnel propagator acting on square-integrable functions. It approaches the initial field as $x\downarrow 0$. A plane wave has transverse wavenumber $\nu =k\sin \alpha$; the reduced axial phase $-{\nu }^{2}x/(2k)$ agrees with $k(\cos \alpha -1)x$ through order ${\alpha }^2$. This also checks the sign. The paraxial approximation applies to $|\nu |\ll k$.

Since the initial velocity vanishes, D'Alembert's formula reduces toEach translated copy has half the original amplitude. To identify each requested cross-section, track the translated support intervals and add the values where they overlap.

At $t=\pi$, $f(x-\pi )$ is supported on $[-\pi ,0]\cup [2\pi ,3\pi ]$ and $f(x+\pi )$ on $[-3\pi ,-2\pi ]\cup [0,\pi ]$. On these intervals $\sin (x\pm \pi )=-\sin x$, soThe pieces meet continuously at zero, where both contributions vanish.

At $t=3\pi /2$, the support of $f(x-3\pi /2)$ is $[-\pi /2,\pi /2]\cup [5\pi /2,7\pi /2]$; that of $f(x+3\pi /2)$ is $[-7\pi /2,-5\pi /2]\cup [-\pi /2,\pi /2]$. In the central overlap,so the two contributions cancel. The remaining wave profile isFor the time cross-section at the origin, the initial profile is an odd function, soFor the time cross-section at $x=\pi /2$, the left-travelling contribution is nonzero for $\pi /2\le t\le 3\pi /2$ and equals $\cos t/2$. The right-travelling contribution is nonzero for $3\pi /2\le t\le 5\pi /2$ and also equals $\cos t/2$. At the common endpoint both vanish. HenceThe following graphs show all four cross-sections. In particular, the last trace reaches $-1/2$ at $t=\pi$ and $1/2$ at $t=2\pi$.

Because $u(0,t)=0$, restriction to the half-line solves the following initial-boundary value problem with a homogeneous Dirichlet boundary condition:This is wave reflection by odd extension. If a condition at infinity is desired, take the solution with finite wave energy; here the profile has bounded support at every finite time.

The initial profile is continuous and piecewise continuously differentiable, but its first derivative jumps at the support endpoints. Thus the formula gives a weak solution globally and a piecewise classical solution away from the characteristic lines for the one-dimensional wave equation issuing from those endpoints. The initial velocity is zero in the square-integrable function sense and pointwise wherever the initial profile is differentiable; at the kink points a pointwise right time derivative need not equal zero. This regularity qualification is needed for the given data.

For the period-two Fourier series of a square-integrable function $w$, Parseval's identity statesFor $w=f-g$, write $s=\sinh 1$. Its coefficients are $a_0=2s$, $a_n=2(-1{)}^{n}s/(1+n^2{\pi }^2)$ and $b_n=n\pi a_n$, because $g$ is subtracted. ThereforeThe sum was obtained in part (a). As a separate check, $f-g=e^{-x}$ in the interval, whose square has integral ${\int }_{-1}^1{e}^{-2x}dx=\sinh 2$. Endpoint jumps have no effect on this integral.

With the standard unweighted inner product on real square-integrable functions, orthogonality on $[-L,L]$ meansMultiplying this inner product by a common nonzero normalising constant does not change orthogonality.

Take the inner product of the Fourier series with $\sin (n\pi x/L)$. Fourier orthogonality eliminates the constant, all cosine terms, and all sine terms except the $n$th. The remaining squared norm issoFor a square-integrable function, the calculation is justified by taking inner products of finite Fourier partial sums and passing to their mean-square limit.

Use the Fourier transform convention $\hat{f}(k)={\int }_{\mathbb{R}}f(t)e^{-ikt}dt$, with inverse factor $1/(2\pi )$. For functions in L1 space, their convolution is $h(t)={\int }_{\mathbb{R}}f(s)g(t-s)ds$. Fubini's theorem applies because the absolute double integral is $\Vert f{\Vert }_1\Vert g{\Vert }_1$. Substituting $r=t-s$ therefore proves the convolution theorem:For the specified rectangular pulse, the support condition is $|t|<1$. Its Fourier transform isThe convolution $f\ast f$ is the length of the overlap of $(-1,1)$ and $(t-1,t+1)$. Thus it is the triangular pulseThe transform's value at zero is four, given by its continuous extension. In this convention, the Parseval identity for square-integrable functions isFor the rectangular pulse the left side is two, giving the sinc-squared integralFor the triangular pulse, $\int |f\ast f{|}^{2}dt=2{\int }_0^2(2-t{)}^{2}dt=16/3$. Apply the same Parseval identity with transform $4{\sin }^{2}k/k^2$ to obtain the sinc fourth-power integralBoth integrands have removable singularities at zero and integrable tails.

Projecting the Sturm-Liouville eigenfunction expansion onto $y_m$ using the weighted inner product givesFor the partial sum $f_N={\sum }_{n=0}^N{a}_n{y}_n$, orthonormality therefore givesFor $f$ of finite weighted norm, the given completeness means that $f_N\to f$ in that norm. The triangle inequality implies $|\Vert f_N{\Vert }_w-\Vert f{\Vert }_w|\le \Vert f_N-f{\Vert }_w\to 0$. Taking the limit of the finite identity proves Parseval's identity:This argument justifies the limiting step without assuming that a double infinite series can be integrated term by term. As usual, the expansion is understood for square-integrable functions with respect to the weighted measure; the boundary and smoothness assumptions of the regular Sturm-Liouville problem provide the usual stronger interpretations when needed.

Insert the inverse Fourier transforms of $f$ and its complex conjugate, initially under enough decay to justify interchanging integrals:The middle step uses the stated Dirac delta distribution identity, and the last uses its sampling property. This is the Parseval identity for the chosen normalization. Approximation extends the result to square-integrable functions, including the function used below.

For the trial function, $y_T^{\prime }=-ax{y}_T$. The required Gaussian integrals areThe second formula follows by differentiating the first with respect to $a$. Thereforeand division by $G[y_T]$ gives the Gaussian variational bound for an attractive Gaussian well:This is a good trial family because it consists of smooth, positive, square-integrable functions decaying at both infinities. Each is an even function, matching the symmetry of the lowest bound state in this even potential, and the parameter varies its width. It also permits exact Gaussian integrals. Its Gaussian tail need not equal the exact exponential tail of a negative-energy bound state for the Rayleigh-Ritz variational principle to give a valid bound.

Differentiating the Gaussian variational bound for an attractive Gaussian well givesThus an interior stationary point obeysThe left side of the first equation increases strictly from zero to infinity, so there is exactly one such point. More explicitly,so $E$ is a strictly convex function on $a>0$ and $a_{\ast }$ is its global minimum. At that minimum the stationarity equation givesThere is a continuous limiting zero at $a=0$, although $y_T=1$ at exactly $a=0$ is not a square-integrable function and hence is not an admissible trial. For a positive zero, division by $\sqrt{a}$ and squaring gives $a(1+a)=4V_0^2$, sois the unique positive zero. The other quadratic root is negative and inadmissible. Near zero,so the graph leaves the origin with slope $-\infty$, falls to its unique negative minimum, rises through $a_z$, and grows as $a/2-V_0+O(a^{-1})$ at infinity.

By the Rayleigh-Ritz variational principle,Thus the attractive well supports a negative-energy bound state for every $V_0>0$.

Let $\chi =(\sqrt{5}/3,2/3{)}^{\mathsf{T}}$, so ${\chi }^{†}\chi =1$. The Pauli identity $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=|\mathbf{p}{|}^{2}I$ givesusing $E^2=|\mathbf{p}{|}^2+m^2$. Thus unit spinor probability density is obtained withAn arbitrary constant phase is immaterial. For the alternative Lorentz normalization $u^{†}u=2E$, choose $N=\sqrt{E+m}$. Neither convention makes an infinite plane wave square-integrable; a box-normalized state includes an additional $V^{-1/2}$, or one forms a normalizable wave packet.

Multiply the complex Fourier series by $e^{-i\omega rt}$ and integrate over one period. Fourier orthogonality removes every term except the one with index $r$, givingFor general square-integrable functions, this is an orthogonal projection onto the corresponding Fourier mode. It remains valid when the expansion converges only in L2 space: integrate finite Fourier partial sums first, then pass to the limit using the Cauchy-Schwarz inequality. No unjustified pointwise exchange of an infinite series and an integral is needed.

A vanishing moment is a cancellation condition: positive and negative parts of an analyzing function balance when weighted by a monomial. Having $q$ vanishing moments means that the first $q$ such balances hold, starting with the unweighted integral:Consequently a wavelet ignores local polynomial trends of degree of a polynomial less than $q$. A smooth feature is close to its Taylor polynomial, so only the small Taylor remainder contributes to its detail coefficient. A jump discontinuity cannot be approximated this way across its edge; its jump response of a localized analyzing function can remain large relative to details from a smooth region.^[ref]

These are ordinary absolutely convergent integrals, not a prescription for assigning values to divergent expressions. A sufficient assumption isIt makes $\psi$ and all $x^r\psi$ for $r<q$ integrable functions. Being square-integrable alone does not ensure these weighted integrals exist. For a square-integrable function with compact support, the Cauchy-Schwarz inequality guarantees every such weighted integral is finite. Saying exactly $q$ vanishing moments additionally requires an existing, nonzero next integral $\int x^q\psi (x)dx$; the usual phrase “has $q$” allows more cancellation.