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Write the Lorentz transformation as $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$. Acting on a scalar field, the chain rule gives the four-gradientFor the specified Lorentz boost, the inverse is obtained by changing $\beta$ to $-\beta$, soThese are precisely the covector component rules for the lowered coordinates. Raising the index with the Minkowski metric ${\eta }^{\mu \nu }=\mathrm{diag}(-1,1,1,1)$ reverses the time component, yieldingIndeed ${\partial }^{\prime 0}=\gamma ({\partial }^0-\beta {\partial }^1)$ and ${\partial }^{\prime 1}=\gamma ({\partial }^1-\beta {\partial }^0)$, the contravariant four-vector rules. Thus the index on a partial derivative with respect to an upper coordinate is lower.

With constant rest mass, $\mathbf{p}=m\gamma \mathbf{v}$ and the chain rule gives $\dot{\gamma }={\gamma }^3(\mathbf{v}\cdot \mathbf{a})/c^2$. Hence the laboratory relativistic force isUse the vector triple product identity $\mathbf{v}\times (\mathbf{v}\times \mathbf{a})=\mathbf{v}(\mathbf{v}\cdot \mathbf{a})-v^2\mathbf{a}$, and ${\gamma }^3(1-v^2/c^2)=\gamma$, to obtainThe relativistic energy is $E=m\gamma c^2$, whose time derivative is $m{\gamma }^3\mathbf{v}\cdot \mathbf{a}$. Taking the dot product of the first expression for $\mathbf{F}$ with $\mathbf{v}$ gives the same quantity, because $\gamma +{\gamma }^3{v}^2/c^2={\gamma }^3$. Thus $dE/dt=\mathbf{F}\cdot \mathbf{v}$. Since $d/d\tau =\gamma d/dt$, the four-force isIts tensor contraction with the four-velocity vanishes: $F^{\alpha }{U}_{\alpha }=-{\gamma }^2\mathbf{F}\cdot \mathbf{v}+{\gamma }^2\mathbf{F}\cdot \mathbf{v}=0$. This orthogonality also follows by differentiating $p^{\alpha }{p}_{\alpha }=-m^2{c}^2$. Resolving the laboratory acceleration relative to $\mathbf{v}$ gives ${\mathbf{F}}_{\parallel }=m{\gamma }^3{\mathbf{a}}_{\parallel }$ and ${\mathbf{F}}_{\perp }=m\gamma {\mathbf{a}}_{\perp }$.

Label fluid particles by $\mathbf{a}$, with flow map $\mathbf{x}=\chi (\mathbf{a},t)$ satisfying ${\chi }_t=\mathbf{u}(\chi ,t)$. If $J=\det D_{\mathbf{a}}\chi$, differentiation of the determinant gives $J_t=J\nabla \cdot \mathbf{u}$. Pull the material-volume integral back to a fixed initial volume:The chain rule and the Jacobian identity give the integrand $J(F_t+\mathbf{u}\cdot \nabla F+F\nabla \cdot \mathbf{u})$. Transforming back yields the Reynolds transport theoremEquivalently the additional surface flux is ${\int }_{\partial V(t)}F\mathbf{u}\cdot \mathbf{n}dS$. The material boundary moves with the same normal velocity as the fluid, which is the key distinction from a fixed control volume.

Use $u={\psi }_Y$, $V=-{\psi }_x$, and put $\eta =Y/x^{\gamma }$, $\psi =x^{\beta }f(\eta )$. ThenMatching $u\sim x^m$ requires $\beta -\gamma =m$. Equating the powers in streamwise inertia $x^{2m-1}$ and transverse diffusion $x^{m-2\gamma }$ givesThe two convective terms combine to $x^{2m-1}[m(f^{\prime }{)}^2-\beta f{f}^{\prime \prime }]$, while $U{U}^{\prime }=m{x}^{2m-1}$. Hence the Falkner-Skan equation in this scaling isThe additive stream-function constant is fixed by $f(0)=0$; when $\beta \ne 0$ this also encodes no penetration. For $\beta =0$ it is a harmless choice of additive constant choice. Existence of an attached physical branch for a given $m$ is a separate issue from deriving the similarity reduction.

On an attached branch with $w=f^{\prime }>0$ away from the wall, use $f$ itself as the independent variable. The chain rule gives $f^{\prime \prime }=w{w}_f$ and $f^{(3)}=w(w_f^2+w{w}_{ff})$. ThereforeIf the wall shear $s=f^{\prime \prime }(0)>0$, then $f\sim s{\eta }^2/2$, so $w(f)\sim \sqrt{2sf}$. Thus $w_f$ is singular at zero while $w{w}_f\to s$ is finite. Imposing a finite arbitrary $w_f(0)$ would remove the usual attached branch. This is the velocity as dependent variable in the Falkner-Skan equation reduction.

The Von Mises boundary-layer transformation uses $\Psi$ as transverse coordinate. With $u=U(x,\Psi )$ and ${\Psi }_x=-V$, ${\Psi }_Y=u=U$, the chain rule givesThe streamwise advection terms reduce to $u{u}_x+V{u}_Y=U{U}_x$, so, for $Y>0$ where $U>0$, division by $U$ givesContinuity was already enforced by the stream function. Substitute $U=g(\eta )$ with $\eta =\Psi /\sqrt{x}$. Then $U_x=-\eta g^{\prime }/(2x)$ and ${\partial }_{\Psi }(U{U}_{\Psi })=(g{g}^{\prime }{)}^{\prime }/x$, yieldingThe wall coordinate is zero because $\Psi$ was set to zero there, and the far condition matches unit external speed. This is the transformed Blasius similarity problem. One must not additionally prescribe a finite $g^{\prime }(0)$: the coordinate change degenerates at the no-slip wall, and the regular physical shear corresponds to $g\propto \sqrt{\eta }$ near zero. This similarity reduction is obtained for the attached monotone boundary layer assumed in the question.

Existence of a positive normalized profile can also be seen directly. Put $q=g{g}^{\prime }$ and choose $q(0)=a>0$. In $g$ coordinates the regular initial-value system is $dq/dg=-\eta /2$, $d\eta /dg=g/q$, with $\eta (0)=0$. It starts as $g\sim \sqrt{2a\eta }$. For finite positive $\eta$, $q^{\prime }/q=-\eta /(2g)$ keeps $q$ positive and $q\le a$, while $(g^2{)}^{\prime }=2q$ gives $g\le \sqrt{2a\eta }$. Consequently $q^{\prime }/q\le -\sqrt{\eta }/(2\sqrt{2a})$, so $q$ is integrable at infinity. It follows that $g$ increases to a finite positive limit $M$. The rescaled profile $\tilde{g}(\eta )=M^{-1}g(\sqrt{M}\eta )$ satisfies the same equation and has limit one. This supplies the requested similarity solution with both boundary conditions, rather than only its formal differential equation.

Use autonomous third-order reduction using the dependent variable. On a monotone branch where $f$ can serve as the independent variable, the chain rule givesTherefore the reduced second-order equation isThe wall and matching conditions becomeFor the usual positive outer flow this is the limit $f\to +\infty$. The reduced equation is singular at $G=0$: if $f^{\prime \prime }(0)=b\ne 0$, then $f\sim b{\eta }^2/2$ and $G\sim b\eta$, so $G$ has square-root behavior as a function of $f$ near the wall. A finite value for $G_f(0)$ should not be imposed as an extra boundary condition.

In canonical coordinates, the Poisson bracket isIn particular, $\{q_j,g\}={\partial }_{p_j}g$, $\{p_j,g\}=-{\partial }_{q_j}g$, $\{f,q_j\}=-{\partial }_{p_j}f$ and $\{f,p_j\}={\partial }_{q_j}f$. Hence $\{q_i,p_j\}={\delta }_{ij}$ and the coordinate–coordinate and momentum–momentum brackets vanish.

The chain rule along a Hamiltonian trajectory givesThe Poisson bracket of time-dependent conserved quantities follows by defining $D_t={\partial }_t+\{\cdot ,H\}$. The ordinary product rule and Jacobi identity implyTo see the Hamiltonian part explicitly, Jacobi rearranges to $\{\{f,g\},H\}=\{\{f,H\},g\}+\{f,\{g,H\}\}$. Therefore $D_{t}f=D_{t}g=0$ implies $D_t\{f,g\}=0$, even when the conserved functions depend explicitly on time.

The Euler-Lagrange field equation is $F^{\prime }(\varphi )-{\partial }_t^2\varphi +{\partial }_x^2\varphi =0$. ThereforePut $c=\cosh \beta$, $s=\sinh \beta$, $X=cx+st$ and $T=ct+sx$. The chain rule gives ${\partial }_t=s{\partial }_X+c{\partial }_T$ and ${\partial }_x=c{\partial }_X+s{\partial }_T$. Cross terms cancel and $c^2-s^2=1$, soThus the boosted scalar field is another solution for any constant rapidity. The nonlinear term in the nonlinear Klein-Gordon equation is unchanged because it is a local function of the scalar itself.

An invertible change $(q,p)\mapsto (Q,P)$ is canonical when it preserves the Poisson bracket. Its fundamental brackets must beThe chain rule then makes the bracket of any two observables identical in the old and new coordinates. Consequently Hamilton's equations retain their canonical form. A time-dependent canonical transformation also requires the corresponding extra term in the transformed Hamiltonian; bracket preservation by itself concerns each fixed time.

The Lagrangian density is Lorentz invariant, so a scalar-valued static solution can be composed with a Lorentz transformation. Let $X=\gamma (x-vt-x_0)$, with $\gamma =(1-v^2{)}^{-1/2}$ and $|v|<1$. If $\varphi (x,t)=f(X)$, the chain rule givesThe boosted equation of motion is therefore exactly the static one already verified in solution, andis a solution. The center moves as $x=x_0+vt$: $v$ is the physical velocity in units where the limiting speed is one. The kink width contracts from $k^{-1}$ to $(\gamma k{)}^{-1}$, while its topological charge remains unchanged.

Use the Gibbs-Duhem equation for a homogeneous, extensive, single-component thermodynamic phase:At fixed temperature, this becomes $dp=nd\mu$, with number density $n=N/V$. Separately, at fixed particle number,The chain rule along the same isothermal process in the equation of state now givesConsequentlyThis is isothermal compressibility from chemical-potential response. Since chemical potential has units of energy per particle, the right-hand side has units of volume per energy, or $\mathrm{P}{\mathrm{a}}^{-1}$, as required. The derivation assumes a smooth bulk equation of state; changing the fractions of coexisting thermodynamic phases at fixed pressure needs separate treatment.

A fixed point of $F$ is also a fixed point of $F\circ F$. By the chain rule,If the original fixed-point multiplier has modulus greater than one, its square also exceeds one, so the same fixed point is unstable for the twice-iterated map. New stable fixed points of the twice-iterated map, if present, must be different points; they can represent a genuine two-cycle of the original map.

Choose a positive reference biological growth factor concentration $G_{ref}$, for example its concentration ahead of the travelling wave. SetwhereThe chain rule then transforms the equations intoEach displayed coefficient is one with the chosen scales. Hence the dimensionless system isThe free choice of $G_{ref}$ expresses the freedom to normalize the as-yet unspecified concentration scale.

Take profiles $b=b(y)$ and $g=g(y)$ with $y=x-ct$, and use a prime for differentiation with respect to $y$. The chain rule gives $b_t=-c{b}^{\prime }$, $b_{xx}=b^{\prime \prime }$ and $g_t=-c{g}^{\prime }$. ThereforeThese are coupled ordinary differential equations for a right-moving travelling wave with speed $c>0$. A first-order formulation, using $v=b^{\prime }$, is $b^{\prime }=v$, $v^{\prime }=-cv-gb$, $g^{\prime }=gb/c$. In particular a positive profile has $g^{\prime }>0$, consistent with depleted substrate behind the travelling front and fresh biological growth factor ahead.

A fixed point $x_{\ast }$ satisfies $f(x_{\ast },\mu )=x_{\ast }$. A perturbation obeys $u_{n+1}=m{u}_n+O(u_n^2)$, where the periodic-orbit multiplier is $m=f_x(x_{\ast },\mu )$. It is linearly attracting for $|m|<1$ and repelling for $|m|>1$; at $m=\pm 1$ nonlinear terms decide Lyapunov stability. A crossing through $+1$ can produce a saddle-node, transcritical or pitchfork bifurcation, depending on the mathematical model parameter dependence and symmetry. A crossing through $-1$ generically gives a period-doubling bifurcation. A real one-dimensional function cannot have the complex conjugate pair needed for a Neimark–Sacker bifurcation.

For a genuine period-two orbit, study the return map $f\circ f$ at either point. By the chain rule its periodic-orbit multiplier isIts modulus determines linear stability by the same criterion, now per two iterates.

For the real quadratic iteration map, the fixed-point equation givesThe derivative is $-2x$. The lower branch $x_{-}$ is repelling whenever $\mu >-1/4$. The upper branch satisfies $|-2x_{+}|<1$ exactly forAt $\mu =-1/4$ the two fixed points merge in a saddle-node bifurcation. With $u=x+1/2$ at that mathematical model parameter, $u^{\prime }=u-u^2$, so the point attracts from the right and is repelling from the left. At $\mu =3/4$, $x_{+}=1/2$ has periodic-orbit multiplier $-1$ and undergoes a period-doubling bifurcation. At the threshold itself $u^{\prime }=-u-u^2$ and the two-step return is $u^{\prime \prime }=u-2u^3-u^4$, proving weak asymptotic stability there despite the inconclusive linear test.

Factor the second-iterate equation:Removing the fixed-point factor yields the periodic orbit pointsThey are distinct and exchanged by $f$ exactly when $\mu >3/4$; at equality they coincide with the fixed point. Their product is $1-\mu$, hence their periodic-orbit multiplier is $4(1-\mu )$. ThusIt is repelling for $\mu >5/4$ and reaches its next period-doubling bifurcation threshold at $\mu =5/4$. At that exact endpoint it is weakly attracting: if $f^2$ is expanded about either periodic orbit point as $v^{\prime }=-v+A{v}^2+B{v}^3+O(v^4)$, then $A=2\mu -6x_i^2$, $B=-4x_i$, and $A^2+B=20\pm 10\sqrt{2}>0$. Its second return is $v^{\prime \prime }=v-2(A^2+B)v^3+O(v^4)$.

At $\mu =2$ put $h(y)=4{\sin }^2(\pi y)-2=-2\cos (2\pi y)$. The double-angle identity givesThis verifies the printed relation to the Bernoulli shift. Precisely, it is a semiconjugacy: $h(y)=h(1-y)$, so it is not invertible on the circle. To obtain a genuine interval topological conjugacy, use the homeomorphism $H(u)=-2\cos (\pi u)$ from $[0,1]$ onto $[-2,2]$, and the full tent mapThe same identity proves $f(H(u),2)=H(T(u))$.

The full tent map has dense periodic points, is topologically mixing and has sensitive dependence on initial conditions. To see these properties directly, its $n$th iterate has $2^n$ monotone branches on dyadic intervals of length $2^{-n}$, each mapping onto $[0,1]$. Every nonempty open interval contains such a branch domain, and a later iterate covers all of $[0,1]$, proving topological mixing. On any branch, the function graph of the iterate has a fixed point by the intermediate value theorem, so periodic points occur in every open interval. Given any initial point and any neighborhood, some iterate of that neighborhood covers the interval; one can choose a nearby point whose image differs from the initial point's image by at least a fixed positive distance. This proves sensitive dependence on initial conditions. Uniform continuity of the inverse homeomorphism transfers these properties to $f$ on $[-2,2]$.

Consequently the real quadratic iteration map is chaotic on its invariant interval in the Devaney sense: topological transitivity, dense periodic points and sensitive dependence on initial conditions, with all these orbits bounded. There are also aperiodic orbits: periodic points form a countable union of finite sets of roots of an equation of the equations $f^n(x)=x$, whereas the interval is uncountable. The semiconjugacy alone would not justify treating $h$ as invertible; the tent-map argument supplies the missing rigorous transfer.

Let $L=\log r$. The given expansion implies $u_{\ast }=2L+O(L/r)$ and thereforeUsing $H^{\prime }(u)=r{e}^{-u}(1-u)$ gives preciselyThe paired low value is $v_{\ast }=H(u_{\ast })=(2L/r)[1+O(L/r)]$. It is small, so $e^{-v_{\ast }}(1-v_{\ast })=1+O(L/r)$ andBy the chain rule the second-iterate periodic-orbit multiplier is their product:Since $L^2/r\to 0$ while $L\to \infty$, this tends to $-\infty$ and eventually has magnitude above one. Thus the large second-iterate fixed point, and hence its associated two-cycle, is unstable for sufficiently large $r$. The high state's small derivative alone does not imply stability: the low state's derivative supplies the compensating factor of order $r$.

For a nearby orbit, linearized separation satisfies ${\delta }_{i+1}=f^{\prime }(x_i){\delta }_i$. Repeated use of the chain rule gives ${\delta }_n={\delta }_0{\prod }_{i=0}^{n-1}{f}^{\prime }(x_i)$ to first order. Taking the logarithm of its magnitude and dividing by $n$ gives the exponential growth rateThe infinitesimal-perturbation limit is taken before interpreting long-time growth; finite separations eventually leave the linear regime. The limit is assumed to exist, with a limsup used when needed.

On a $p$-cycle, blocks of $p$ successive derivatives repeat. If $M=(f^p{)}^{\prime }(x_0)={\prod }_{i=0}^{p-1}{f}^{\prime }(x_i)$, the remainder of fewer than $p$ terms has vanishing contribution to the long-time average andLinear attraction means $|M|<1$, giving $\lambda <0$. A superstable cycle has $M=0$, meaning at least one derivative in the cycle vanishes. Its infinitesimal return multiplier is zero and the corresponding logarithmic exponent is $-\infty$. This concerns the derivative cocycle; actual finite perturbations can converge by higher-order rather than literally vanishing in one step.

The logarithm requires $y>0$, and $x>0$ is given. Set $z=\log (y/x)$, so $y=x{e}^z$ and the chain rule yields $x{y}^{\prime }/y=1+x{z}^{\prime }$. Dividing the ordinary differential equation by $y$ therefore givesSince $(z/x{)}^{\prime }=(x{z}^{\prime }-z)/x^2=0$, we obtain $z=Cx$, including the case $z=0$. Thus the general positive solution is $y=x{e}^{Cx}$. The initial condition $y(1)=e$ gives $C=1$:For the separate decay requirement, $x{e}^{Cx}\to 0$ precisely when $C<0$: if $C=0$, the result is $x$, and if $C>0$, it grows. Consequently the full family with the stated limit isThis latter family does not also satisfy $y(1)=e$; the solution selected by that initial condition grows. The identically zero function is excluded by the original logarithm.

For $r>0$, the chain rule gives $r_x=x/r$ and $r_y=y/r$. Applying it twice to the radial function $u=f(r)$ givesUsing $x^2+y^2=r^2$, Laplace's equation becomesTwo integrations now giveThese are the general solutions of the specified radial function form on an annulus or a punctured plane. If the solution must extend continuously to the origin, the logarithm term forces $A=0$, leaving only a constant function.

Differentiate the product $4(x-1{)}^{-2}(x+1{)}^{-1}$ using the product rule and chain rule:Combining terms over a common denominator gives