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The canonical Poisson-bracket relations are

The Poisson bracket of two phase-space functions is

For canonical position and momentum coordinates $(q_i,p_i)$, Hamilton's equations are

The first model alone would suggest that more cycling is beneficial, but cycling was not randomized in that model: its coefficient can be distorted by confounding or reverse causality. The randomized incentive both substantially increased cycling and substantially increased asthma episodes.

If the incentive affects asthma only through cycling, it is an instrumental variable, and the corresponding Wald estimator of the effect per additional weekly mile isadditional episodes. Under that exclusion assumption the evidence therefore advises against cycling to work more often for this population. Without the exclusion assumption, the experiment establishes the harmful effect of the incentive itself but cannot by itself distinguish an effect of cycling from a direct effect of the monetary incentive.

For participants with the same recorded history, the first linear regression estimates that one additional average mile cycled per week is associated with $0.04917$ fewer asthma episodes. The corresponding $p$-value is $0.00761$, giving evidence against a zero conditional association.

The second model estimates that, among participants with the same history, receiving the incentive is associated with $0.91387$ more episodes. The third estimates that receiving the incentive is associated with $1.73282$ more miles cycled per week. Both incentive coefficients have very small p-values. Because the incentive was assigned in a randomized controlled trial, these last two comparisons estimate a causal effect of the incentive.

The function is the predecessor function. It has the primitive recursion definitionThe initial function is the zero function and the recursion step is a projection function, both of which are primitive recursive functions. Closure under primitive recursion therefore proves that $f$ is primitive recursive.

Encode each $k$-tuple of nonnegative integers by one nonnegative integer using a computable pairing function. Define a unary partial procedure $P$ on input $m$ as follows: use dovetailing on the computations of $f_{n,k}$ over all encoded $k$-tuples, and halt as soon as one of them halts with output $m$. ThenBy the Church–Turing thesis, this effective procedure is implemented by a register machine with some fixed code $j$, soThus $E$ is the domain of a partial computable function and is recursively enumerable.

A recursive set $E\subseteq {\mathbb{N}}_0$ has an indicator function computed by a register machine that halts on every input. A recursively enumerable set is the exact halting set of some register machine; equivalently, it is the domain of a partial computable function.

If $E$ is recursive, its decider immediately gives machines that halt exactly on $E$ and on ${\mathbb{N}}_0\setminus E$, so both sets are recursively enumerable. Conversely, suppose machines $M_E$ and $M_{E^c}$ recognize the two sets. On input $x$, use dovetailing to run one step of $M_E(x)$ and one step of $M_{E^c}(x)$ alternately. Exactly one must eventually halt. Return one if $M_E$ halts first and zero if $M_{E^c}$ halts first. This is a total decider for $E$, so $E$ is recursive.

Each coordinate of the transformed vectors evolves independently because $D$ and $E$ are diagonal matrices. Reorder the unknowns by collecting the $k$th components across all grid columns:Then system (3) splits into the $m$ uncoupled systemswhereThus the original $m^2\times m^2$ problem reduces to one discrete sine transform followed by $m$ independent tridiagonal solves and an inverse sine transform.

The simultaneous diagonalization from part (a) applies to both $B$ and $C$. With ${\theta }_k=k\pi h$,andMultiplying the $j$th block equation by $Q^{-1}$ and writing $v_j=Q^{-1}{u}_j$ and $c_j=Q^{-1}{b}_j$ giveswith the absent boundary terms omitted. This is exactly system (3).

Let $u_j=(u_{1,j},\dots ,u_{m,j}{)}^T$ collect the unknowns in grid column $j$, and define $b_j=h^2(f_{1,j},\dots ,f_{m,j}{)}^T$. The terms within one column form the blockwhile coupling to either adjacent column contributesThe zero boundary values remove all blocks outside the displayed range, so the nine-point finite-difference stencil becomes the block tridiagonal matrix system

Let ${\theta }_k=k\pi h=k\pi /(m+1)$ and define $q_j^{(k)}=\sin (j{\theta }_k)$. The artificial boundary values are $q_0^{(k)}=q_{m+1}^{(k)}=0$. For the symmetric tridiagonal Toeplitz matrix $H=[\beta ,\alpha ,\beta ]$, the sine addition formula givesThus all such matrices have the same discrete sine transform eigenvectors, with eigenvaluesThe sine vectors are mutually orthogonal vectors and nonzero, so the matrix $Q=(\sin (ij\pi h){)}_{i,j=1}^m$ is invertible. If $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$, then $HQ=QD$ and therefore

Use the real modeThe kinetic energy density and isotropic linear-elastic energy density areAverage over one period and integrate across the layer. Since ${\int }_0^h{\sin }^2(q_{n}y)dy={\int }_0^h{\cos }^2(q_{n}y)dy=h/2$, the mean energy per unit horizontal area isThe $x$-directed elastic-energy flux is $S_x=-{\sigma }_{zx}{w}_t=-\mu w_x{w}_t$. Its corresponding average isUsing ${\omega }^2=c_S^2(k^2+q_n^2)$ givesand therefore

The cutoff occurs at $k=0$:For $k>0$, the phase velocity and group velocity areTheir product is $c_{p,n}{c}_{g,n}=c_S^2$. On a plot against positive $k$, $c_{p,n}$ decreases from $+\infty$ to $c_S$, while $c_{g,n}$ increases from zero to $c_S$; larger mode number shifts both transitions to larger wavenumber.

Take a shear-horizontal wave polarized in the $z$ direction,Then $\nabla \cdot \mathbf{u}=0$, and the elastic-wave equation reduces toFor $w=Y(y)e^{i(kx-\omega t)}$, the rigid boundary gives $Y(0)=0$, while vanishing shear traction at the free surface gives $Y^{\prime }(h)=0$. ThusThe dispersion relation of each guided shear-horizontal mode is

Suppose $U_{+}=U$ and $U_{-}=-U$. Integrating the two velocity profiles gives the volume fluxes per unit spanHereConsequentlyso all dependence on the interface position cancels from the ratio:

Write $d_{+}=\sigma +h$ and $d_{-}=h-\sigma$ for the lower and upper layer thicknesses. A steady unidirectional velocity $\mathbf{u}=u(y){\mathbf{e}}_x$ with no pressure gradient satisfies the Navier-Stokes equations through $u^{\prime \prime }=0$ in each layer. Continuity of velocity and tangential stress at the interface gives one constant shear stressThe two-layer Couette flow is therefore