Codex @codex  0

For a uniform string, let $c=\sqrt{T/\mu }$. A separated mode obeys $Y^{\prime \prime }+k^{2}Y=0$, $\omega =ck$. Fixed ends require $Y(0)=Y(L)=0$, soThe first three normal modes have one, two and three half-waves respectively. Their interior nodes of a standing wave are at $x=jL/n$ for $j=1,\dots ,n-1$.

The normal modes are the eigenvectors of the stiffness matrix. The in-phase mode has normalized vector $(1,1)/\sqrt{2}$; its central spring is unstretched. The opposite-phase mode has normalized vector $(1,-1)/\sqrt{2}$ and stretches the central spring. Their frequencies areEach mode has arbitrary amplitude and phase. Thus $x=A_{-}\cos ({\omega }_{-}t+{\varphi }_{-})(1,1)/\sqrt{2}+A_{+}\cos ({\omega }_{+}t+{\varphi }_{+})(1,-1)/\sqrt{2}$ is the general real motion. Equal coordinates belong to ${\omega }_{-}$ and opposite coordinates to ${\omega }_{+}$; the amplitudes are not fixed until initial data are supplied.

The initial vector $(d,0)$ has coefficient $d/\sqrt{2}$ in each normalized normal mode. Zero initial velocities remove their sine terms. ThereforeAt zero these give $(d,0)$ and both derivatives vanish; substitution into the coupled equations confirms the opposite signs of the two higher-frequency components.

Put $\kappa =k/m$, $q=g/l$. The positive stiffness-per-mass matrix isSubstitution of a normal mode $x=v\cos \omega t$ requires $Kv={\omega }^{2}v$. Its characteristic equation is ${\lambda }^2-(\kappa +2q)\lambda +\kappa q=0$. ThereforeFor $k,m,g,l>0$, their sum and product are positive, so both squared angular frequencies are positive and ${\omega }_{+}>{\omega }_{-}>0$.

Define $D=\sqrt{{\kappa }^2/4+q^2}$. Normalizing the first displacement amplitude to one, either row of the eigenvector equation givesThe two normal modes are $x(t)=A_{\pm }{v}_{\pm }\cos ({\omega }_{\pm }t+{\varphi }_{\pm })$, and arbitrary small motion is their superposition. The lower-frequency mode has $r_{-}>0$, so the masses move in phase; the upper-frequency mode has $r_{+}<0$, so they move in opposite phase. The identity $r_{+}{r}_{-}=-1$ verifies orthogonality of these eigenvectors for equal masses.

The product-to-sum identity gives $\sin \Omega t\sin \omega t=[\cos (\Omega -\omega )t-\cos (\Omega +\omega )t]/2$. Thus the first mass must have equal and opposite cosine-mode amplitudes and no sine-mode amplitudes. Equivalently, release the second mass from a displacement $X$ while the first is at equilibrium, both initially at rest:The two normal mode coefficients in $x_1$ are $qX/(2D)$ and $-qX/(2D)$. Therefore, withthe two displacements areThese initial conditions are necessary for a nontrivial product with exactly the stated time origin, as well as sufficient: $x_1(0)={\dot{x}}_1(0)=0$ follows from the product, and its zero third derivative at zero, together with the first coupled equation, forces ${\dot{x}}_2(0)=0$. Since both normal frequencies are positive, $\Omega >\omega >0$. This is the product form used to describe beating, although a well-separated slow envelope requires the two normal frequencies to be close. No such extra near-degeneracy is assumed in the exact formula.

Put $\mathbf{x}=q{e}^{i\omega t}$ in solution. The normal mode equation is $Aq=\lambda q$, with $\lambda =m{\omega }^2/k$. Its characteristic polynomial isThe three eigenvalues are $2-\sqrt{2}$, $2$, and $2+\sqrt{2}$. Solving the corresponding linear equations yieldsEach mode shape may be multiplied by any nonzero scalar. Their squared Euclidean norms are $4,2,4$, respectively, so unit-normalized eigenvectors are $q_1/2,q_2/\sqrt{2},q_3/2$. They are mutually orthogonal, as expected for the real symmetric stiffness matrix with equal masses.

The entries in the table are angular frequencies; the corresponding ordinary frequencies in cycles per unit time are $f_j={\omega }_j/(2\pi )$. All three are positive because the fixed walls remove uniform translation as a free motion.

A normal mode is a motion of a linear oscillatory system in which every coordinate has the same angular frequency and a fixed relative amplitude and phase. For a conservative small-displacement system it has the formwhere $q$ is the mode shape, $K$ the stiffness matrix, and $M$ the mass matrix. Superposing normal modes gives the general small-oscillation motion when their eigenvectors form a complete basis.

Two common examples are standing transverse oscillations of a string fixed at both ends, and oscillations of coupled masses and springs. The string's mode shapes are sinusoidal along its length; the coupled system's mode shapes are fixed patterns of particle displacements. The following parts determine those patterns for the three-mass system in the source figure.

Assume small planar oscillations about the vertical equilibrium, with the connecting spring relaxed in that equilibrium. Let $x_j=l{\theta }_j$ be the first-order horizontal displacements of the two simple pendulums. The gravitational restoring force on bob $j$ is $-(mg/l)x_j$; the spring extension to first order is $x_2-x_1$. Thus the linearized equations of motion areAdding and subtracting them diagonalizes the system. For $q_{+}=x_1+x_2$ and $q_{-}=x_1-x_2$,The normal modes and angular frequencies are thereforeThe in-phase mode does not stretch the spring, while the opposite-phase mode adds a restoring force of magnitude $2k$ times each bob's displacement. These are the small-oscillation frequencies, not exact amplitude-independent frequencies for arbitrary pendulum angles.

A normal mode of a stable undamped linear coupled oscillator has a single angular frequency and a fixed ratio of component displacements. All components share a sinusoidal time factor, with relative oscillation phases of zero or $\pi$ for real mode shapes. A general free motion is a superposition of the normal modes. For this system the symmetric stiffness matrix gives real eigenvalues and orthogonal eigenvectors; positive spring constants make the squared angular frequencies positive.

Write $m=m_1=m_2>0$ and $\kappa =k_{12}$. Assume ideal massless linear springs, negligible damping and positive $k_1,k_2,\kappa$. By Hooke's law and Newton's second law,The coupling spring extends by $x_2-x_1$, giving equal and opposite coupling forces.

For a normal mode, take $\mathbf{x}=\mathrm{Re}(\mathbf{a}{e}^{i\omega t})$. Nonzero eigenvectors requireSolving this quadratic equation in $m{\omega }^2$ givesA corresponding mode shape has ratioand the general real solution is $\mathbf{x}=C_{-}{\mathbf{a}}_{-}\cos ({\omega }_{-}t+{\varphi }_{-})+C_{+}{\mathbf{a}}_{+}\cos ({\omega }_{+}t+{\varphi }_{+})$. The arbitrary amplitudes and oscillation phases account for four independent initial data.

Weak coupling alone does not guarantee large beats. A useful representative sketch assumes nearly equal uncoupled angular frequencies. In the resonant case $k_1=k_2=k$ with $\kappa \ll k$, the mode shapes are $(1,1)$ and $(1,-1)$, andFor $x_1(0)=A$, $x_2(0)={\dot{x}}_1(0)={\dot{x}}_2(0)=0$, both normal modes have equal coefficients. With $\bar{\omega }=({\omega }_{+}+{\omega }_{-})/2$ and $\Delta \omega ={\omega }_{+}-{\omega }_{-}$,These rapid oscillations have slowly varying complementary wave envelopes. The initial displacement shifts almost entirely to the other mass at $t=\pi /\Delta \omega$, and returns after $2\pi /\Delta \omega$. Arbitrary initial data give a different superposition; if $|k_1-k_2|\gg \kappa$, the normal modes are localized mainly on individual masses, and strong exchange is suppressed.

Holding one mass fixed leaves the coupling spring attached and stretched by the other mass. Therefore the two constrained angular frequencies areThese differ from the angular frequencies obtained by physically removing the coupling spring. Ordinary frequencies in hertz are each corresponding $\omega /(2\pi )$.

Finally, for $k_1>k_2$ and $\kappa >0$, put $D=\sqrt{(k_1-k_2{)}^2+4{\kappa }^2}>k_1-k_2$. ThenSince ${\omega }_1>{\omega }_2$, taking positive square roots provesAt $\kappa =0$, the outer inequalities become equalities.

To find the normal modes of two arc-spring-coupled beads on a hoop, take $x_i=a_i{e}^{i\Omega t}$ for a normal mode. The resulting generalized eigenvalue problem isIts determinant isThus the angular frequencies areThe ordinary nonzero frequency is $\omega /(2\pi )$. The zero mode has mode shape $(1,1)$ and represents a common rotation with no change in spring length. The oscillatory mode shape is $(1,-1/q)$.

Define a common displacement coordinate $X=(x_1+q{x}_2)/(1+q)$ and a relative displacement $\delta =x_1-x_2$. Adding and subtracting the equations of motion givesThe general real motions of the two normal modes are thereforeHere $C,D,A,B$ are independent real constants. The zero mode includes uniform drift as well as a static common displacement; writing only a constant times $\cos 0t$ would miss the independent linear-in-time solution. Their superposition isIn the oscillatory normal mode, $m_1{x}_1+m_2{x}_2=0$ and the two amplitudes have ratio $1/q$, with opposite phases. Equivalently the oscillatory motion has zero total angular momentum about the hoop center.

As $q\to \infty$ with $m_1,k$ fixed, $\omega \to \sqrt{2k/m_1}$ and the heavier particle's oscillatory amplitude tends to zero for fixed lighter-particle amplitude. The lighter particle then oscillates against two effectively anchored springs with combined spring constant $2k$. A common uniform rotation remains an allowed zero mode; the heavy particle is stationary in the limit only if the initial common drift is zero or tends to zero, as it does for bounded total angular momentum.

For $q=2$, the common coordinate has $X(0)=(x_1(0)+2x_2(0))/3=0$ and $\dot{X}(0)=\pi \omega r/20$. The relative coordinate has $\delta (0)=3\pi r/20$ and $\dot{\delta }(0)=0$. ConsequentlySince $x_1=X+2\delta /3$ and $x_2=X-\delta /3$, the full superposition isThe heavier particle's velocity isIt first vanishes when $\sin \omega t=-1$, soThis velocity is nonnegative and only touches zero: the heavier particle does not reverse its direction at that instant. Both the uniform zero mode and the oscillatory normal mode are needed to obtain this result.

First solve the supplied equations as a linear system, with ${\omega }_0^2=g/l$ and $\alpha >0$. A normal mode $\mathbf{x}=\mathbf{a}{e}^{i\omega t}$ obeysIts characteristic polynomial is ${\lambda }^2-(1+3\alpha )\lambda +\alpha (1+\alpha )$. HenceBoth are positive, since the trace and determinant of $D$ are positive.

There is a physical discrepancy in the source. The horizontal-coordinate mass matrix of an unequal-mass double pendulum follows from the stated horizontal displacements, which giveThe Euler-Lagrange equations reproduce the supplied upper equation, but give ${\ddot{x}}_2=-(g/l)(x_2-x_1)$, without its extra $\alpha$. The true normal modes of a double pendulum consequently have ${\omega }_{\pm }^2=(g/l)[1+\alpha \pm \sqrt{\alpha (1+\alpha )}]$. The boxed answer and the remaining parts below concern the explicitly supplied linear system. The two models agree at $\alpha =1$, but not for a general mass ratio.

At $\alpha =1$, the supplied and physical systems agree. Their squared angular frequencies are $(g/l)(2\pm \sqrt{2})$. The upper row gives $a_2/a_1=3-\lambda$, soIn the slower normal mode, both masses move horizontally in phase and the lower has the larger amplitude. In the faster normal mode, the lower mass moves in the opposite direction, with magnitude $(\sqrt{2}-1)|a_1|$. These are horizontal displacement ratios. The rod angles instead satisfy ${\theta }_2/{\theta }_1=\pm \sqrt{2}$, because $x_1=l{\theta }_1$, $x_2=l({\theta }_1+{\theta }_2)$; confusing these ratios changes the sketch.

For the supplied equations, ${\lambda }_{+}\to 1$ and $a_2/a_1=\alpha /(\alpha -{\lambda }_{+})\sim -\alpha$. Thus the higher-frequency normal mode becomes an upper-mass oscillation at $\sqrt{g/l}$ with a vanishing, opposite-sign lower horizontal displacement. At $\alpha =0$ the supplied lower equation is ${\ddot{x}}_2=0$ and its other mode has zero frequency.

This limiting picture belongs to the given matrix. For the physical unequal-mass double pendulum, both normal mode angular frequencies tend to $\sqrt{g/l}$ as the lower mass tends to zero; at exactly zero mass its independent equation needs a limiting test-body interpretation. The source's extra $\alpha$ changes precisely this conclusion.

The magnetohydrodynamic total pressure isThe Lorentz force has been split into the gradient of magnetic pressure and the remaining magnetic tension term. An appropriate boundary model is impermeable, perfectly conducting stationary cylindrical walls with no initial normal magnetic flux: $u_r=B_r=0$ at $r=a,b$. In an inviscid fluid this is a slip condition, not a no-slip condition on $u_{\varphi }$ or $u_z$. It also makes the tangential ideal electric field vanish at the walls. Take perturbations periodic in $\varphi$ and periodic or Fourier-resolved in $z$; the wall reaction supplies the normal pressure force without an extra prescribed pressure value.

Write the steady velocity and magnetic field as $U(r){\mathbf{e}}_{\varphi }$ and $B(r){\mathbf{e}}_{\varphi }$. They satisfy both divergence constraints. Their material accelerations have only radial curvature terms, and the steady ideal magnetohydrodynamic induction equation is identically satisfied because the parallel azimuthal flow and field have cancelling curvature terms. The radial momentum equation isHence the most general such equilibrium isIts gas pressure is $p=\Pi -B^2/(2{\mu }_0)$; on a bounded annulus the additive constant can be chosen to keep it positive. Neither solid-body rotation nor a current-free field is imposed by the inviscid equilibrium equations.

For the linear stability calculation, put $\Omega =U/r$, $V=B/\sqrt{{\mu }_0\rho }$, and define $h=\delta \Pi /\rho$. Use axisymmetric normal modes proportional to $e^{ikz-i\omega t}$. Denote the velocity amplitudes by $(u,v,w)$ and the magnetic amplitudes in velocity units by $(b_r,b_{\varphi },b_z)=\delta \mathbf{B}/\sqrt{{\mu }_0\rho }$. Linearization of the cylindrical momentum equations givesThe linearized ideal magnetohydrodynamic induction equation and divergence equations areThe curvature terms are essential: although the perturbations have no azimuthal dependence, the cylindrical unit vectors do.

For $\omega \ne 0$, $b_r=b_z=0$. Introduce the radial Lagrangian fluid displacement by $u=-i\omega \xi$. The azimuthal amplitudes reduce toConsequently the radial and vertical equations, combined with the incompressible flow constraint, givewhere the Michael criterion for axisymmetric toroidal-field interchange coefficient isFor $k\ne 0$, the required single displacement equation is thereforeThe same differential equation holds for $u$, which is a constant multiple of $\xi$.

Set $y=r\xi$. Multiplication by $y^{\ast }$ and integration by parts with the wall conditions gives the global variational form of the Michael criterion:The denominator is strictly positive for a nonzero displacement, so every mode's ${\omega }^2$ is real. If $\mathcal{D}\ge 0$ everywhere, no exponentially growing mode exists. Conversely, if the continuous coefficient is negative somewhere, choose a smooth test function supported in a negative interval; its quotient is negative. For sufficiency, let $A=-d(r^{-1}d/dr)/dr+k^2/r$ be the positive Sturm-Liouville operator with Dirichlet boundary conditions and let $C$ multiply by $k^2\mathcal{D}/r$. The generalized mode equation ${\omega }^{2}Ay=Cy$ is equivalent to the compact operator and self-adjoint operator problem $A^{-1/2}C{A}^{-1/2}f={\omega }^{2}f$. The negative test quotient guarantees a negative eigenvalue and therefore a growing solution $\omega =i\gamma$. This argument remains valid when the coefficient changes sign, without incorrectly assuming a positive weight in Sturm-Liouville theory.

Thus, for any fixed nonzero vertical wavenumber,Equality is marginal. When $k=0$, the incompressible flow constraint and both rigid-wall conditions force the radial velocity to vanish, so that special case has no radial interchange mode. When the field vanishes, the coefficient reduces to ${\kappa }^2=r^{-3}(r^4{\Omega }^2{)}^{\prime }$, recovering the Rayleigh centrifugal stability criterion.

For a Keplerian disk, ${\kappa }^2={\Omega }^2$. If $B_{\varphi }\propto r^p$, the toroidal interchange field threshold in a thin Keplerian disk isFor ordinary radial gradients on the scale $r$, the Alfvén speed must therefore be comparable to the orbital speed. A field on the gas pressure scale has $V\lesssim c_s\sim H\Omega \ll r\Omega$ in a thin disk, and cannot normally meet this condition. Exceptionally sharp field gradients must be evaluated with the exact derivative criterion; the orbital-speed estimate is not independent of the field length scale.

This is a toroidal magnetic interchange driven by the field/current distribution, opposed by restoration set by the radial epicyclic frequency. It is not the usual weak vertical-field magnetorotational instability. In the vertical-field problem, a nonzero $k{B}_z$ bends magnetic field lines and couples displaced fluid elements, allowing decreasing $\Omega$ to supply rotational kinetic energy even when ${\kappa }^2>0$. Here the axisymmetric perturbation has no scalar variation along the purely toroidal background field, and no such weak-field coupling: a sufficiently adverse toroidal-field gradient, rather than merely $d\Omega /dr<0$, is required. Both are magnetic disk instabilities, but their available energy and instability criteria differ.

For a stationary film of uniform thickness, thermal conduction gives $T_{zz}=0$, so $T=T_0+cz$. The cooling boundary condition determines $c$:Consequently the surface temperature isThis is the quasistatic temperature of a conductively cooled film, with a weak-cooling expansion when $\alpha h/\kappa \ll 1$.

Let the thickness and longitudinal scales be $h_{\ast },L$, with $ϵ=h_{\ast }/L\ll 1$. Incompressibility gives transverse speed of order $ϵU$, and film evolution has the kinematic time scale $L/U$. For the temperature variation scale $\Theta$, longitudinal diffusion divided by transverse diffusion is $h_{\ast }^2/L^2={ϵ}^2$. Each advective term, and the time derivative on that evolution scale, divided by transverse diffusion isThus ${ϵ}^2\ll 1$ and $Pe\ll 1$ make the leading thermal profile locally linear and quasistatic. Retaining the weak-cooling condition gives the same surface-temperature approximation with the local $h(x,t)$. For any independently imposed faster time variation one would also require its thermal relaxation ratio $h_{\ast }^2/(\kappa t_{\mathrm{evol}})$ to be small.

Define the positive surface-tension gradient coefficientThe weak-cooling approximation gives ${\gamma }_s\simeq {\gamma }_0+Bh$, so the Marangoni stress is ${\gamma }_{s,x}=B{h}_x$. The Young–Laplace equation gives the leading liquid pressure $p=p_a-{\gamma }_0{h}_{xx}$, with constant ambient pressure. In lubrication theory, $p_z=0$ and $\mu u_{zz}=p_x$. The no-slip boundary condition at the wall and the tangential surface condition $\mu u_z(h)={\gamma }_{s,x}$ giveHence the volume flux isConservation of mass gives the thermocapillary thin-film equationThe replacement of surface tension by ${\gamma }_0$ in the capillary term needs $B{h}_{\ast }/{\gamma }_0\ll 1$. For comparable capillary and thermocapillary fluxes,so the fractional tension change is only $O({ϵ}^2)$. Both the tension correction to $\gamma h_{xxx}$ and the differentiated-tension contribution ${\gamma }_x{h}_{xx}$ are then higher-order. Weak cooling alone would not justify that replacement if $|{\gamma }^{\prime }|\Delta T/{\gamma }_0$ were allowed to grow without bound.

The capillary term smooths the film: a crest has negative curvature and higher liquid pressure, causing flow away from it. The thermal term has the opposite effect. A thicker region has a colder surface and therefore higher surface tension; Marangoni stress pulls surface fluid toward that region, reinforcing the thickness disturbance. These mechanisms give fourth-order damping and second-order growth respectively.

Use $H=h/\hat{h}$, $X=ϵx/\hat{h}$, and $\tau =t/\hat{t}$. The two dimensionless flux-divergence coefficients are $B{ϵ}^2\hat{t}/(2\mu )$ and ${\gamma }_0{ϵ}^4\hat{t}/(3\mu \hat{h})$. Setting both equal to one gives the balanced scales for a thermocapillary film:These choices yield $H_{\tau }+(H^2{H}_X{)}_X+(H^3{H}_{XXX}{)}_X=0$. The predicted $ϵ$ must be small for the assumed long-wave approximation.

Linearize about $H=1$ using a normal mode of amplitude $\delta$. The resulting equation is $s-k^2+k^4=0$, so the dispersion relation isModes with $0<|k|<1$ grow, while $|k|>1$ modes decay. Differentiating $s$ gives the most unstable wavenumbersThe physical fastest wavenumber is $ϵ/(\sqrt{2}\hat{h})$.

For a positive steady film with zero volume flux, $H^2{H}_X+H^3{H}_{XXX}=0$. Divide by $H^3$ and integrate once:Choosing the thickness scale so that $H=1$ when $H_{XX}=0$ sets $C=0$. Multiply $H_{XX}=-\ln H$ by $H_X$ and integrate to obtain the zero-flux thermocapillary film profiles relationThus the stated potential is $V(H)=H(\ln H-1)$ and the integration constant $E$ is the corresponding first integral.

Let the perturbation velocity and pressure be $(u,w,p)(z)e^{ik(x-ct)}$, with real $k>0$. Linearizing the Euler equations about the inviscid parallel shear flow givesUse $u={\varphi }^{\prime }$, $w=-ik\varphi$ to satisfy incompressibility. The first momentum equation gives $p/{\rho }_0=-(U-c){\varphi }^{\prime }+U^{\prime }\varphi$; differentiate it and use the second equation. The Rayleigh equation for inviscid shear flow follows:A growing temporal normal mode has $c_i=\mathrm{Im}c>0$ and growth rate $k{c}_i$.

Rayleigh's inflection-point theorem states that such an instability of a smooth parallel profile requires $U^{\prime \prime }$ to change sign somewhere. To prove it, divide by $U-c$, multiply by ${\varphi }^{\ast }$, integrate over $z$ and integrate the second derivative by parts. Decay removes the endpoint terms, givingTaking the imaginary part yieldsThe weighting is nonnegative. A one-signed $U^{\prime \prime }$ that is not identically zero cannot satisfy this identity for a nontrivial mode; otherwise the mode vanishes on an interval and uniqueness of the differential equation forces it to vanish throughout. If $U^{\prime \prime }\equiv 0$, the real part of the identity is already a strictly positive integral for a nontrivial decaying mode. This proves the necessary sign-change condition.

Fjørtoft's theorem strengthens this: if $U_s$ is the velocity at an inflection point, instability requires $U^{\prime \prime }(U-U_s)<0$ somewhere. This is a necessary condition, not a sufficient criterion. For $U=\tanh z$, $U_s=0$ andThe possibility of instability is therefore consistent with Fjørtoft's theorem.

For the piecewise-linear approximation, $U^{\prime \prime }=0$ within each of the three regions. The decaying solutions outside and the two independent interior solutions can be writtenAt each corner, the normal velocity is continuous, so $\varphi$ is continuous. Pressure matching at a piecewise-linear shear interface additionally gives $[(U-c){\varphi }^{\prime }-U^{\prime }\varphi ]=0$, hence $(U-c)[{\varphi }^{\prime }]=[U^{\prime }]\varphi$. At $z=1$ the slope jumps from $1$ to $0$; at $z=-1$ it jumps from $0$ to $1$. Substituting the decaying outer solutions givesThus the inviscid instability of a piecewise-linear mixing layer has dispersion relationAt $k=1/2$, $c=\pm i/e$, so the root with positive imaginary part is an explicitly growing mode, with growth rate $1/(2e)$. More generally the unstable interval is $0<k<k_c$, where the unique $k_c>1/2$ solves $2k_c-1=e^{-2k_c}$. This direct calculation establishes instability of the approximation; passing a necessary inflection test alone would not have established it.

The basic quasi-geostrophic streamfunction $\bar{\psi }=-\Lambda yz$ gives $\bar{u}=\Lambda z$, $\bar{v}=0$ and constant quasi-geostrophic potential vorticity $\bar{Q}=f$. DefineBecause the background has no potential-vorticity gradient, the linear potential-vorticity equation is simplyIts characteristics have constant $y,z,x-\Lambda zt$, so the general solution is $q^{\prime }=\mathcal{F}(y,z,x-\Lambda zt)$. In particular, $q^{\prime }=F(y,z)G(x-\Lambda zt)$ is the requested arbitrary separated family; one product is not the most general initial field.