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For $N=1$, writeThe given evolution of the discrete scattering data givesSolving the resulting two scalar linear equations for the components of ${\psi }_1$ and substituting in the reconstruction formula givesChoosing the additive multiple of $2\pi$ so that $u\to 0$ as $X\to +\infty$ and integrating in $X$ yields the One-soliton solution of the sine-Gordon equation in light-cone coordinatesIt depends only on $X+T$ exactly when the two positive coefficients in the exponent agree:The unique positive solution is $l=1/2$, and then

The transformations satisfy $g^s{g}^r=g^{s+r}$, $g^0$ is the identity map, and $(g^s{)}^{-1}=g^{-s}$, so they form a one-parameter group. Ifthen the chain rule givesThus $g^s$ is a Lie point symmetry. Applied to the one-soliton family, it replaces $l$ by $l^{\prime }=e^{-s}l$. Taking $e^s=2l$ gives $l^{\prime }=1/2$, so every member is transformed to the function $F(X+T)$ found above.

For the stated $N=2$ solution, set $x=X+T$ and $t=T-X$. When $l^2+m^2=1/4$, its two arguments reduce toAt fixed $x$, only $\sin (2mt)$ varies, so the Sine-Gordon breather has fundamental period

Finally put $r^2=l^2+m^2$. Under the same symmetry, the parameters becomeand hence $(l^{\prime }{)}^2+(m^{\prime }{)}^2=e^{-2s}{r}^2$. Choosing $e^s=2r$ makes this sum $1/4$, proving that every solution in the family is symmetry-equivalent to the normalized breather.

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Fréchet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that

Differentiate the energy and use the chain rule:After integration by parts, the middle term is $-\int (\Delta u)u_t$. Thereforeby the defocusing semilinear wave equation. Thus the total energy is conserved.

Write $X=\mu +\sigma Z$ with $Z$ standard normal and set $g(z)=f(\mu +\sigma z)$. The Gaussian Poincaré inequality and the chain rule giveHence $N(\mu ,{\sigma }^2)$ satisfies a $\sigma$-Poincaré inequality in the question's convention, where the constant is squared on the right-hand side.

Let $g:\mathbb{R}\to \mathbb{R}$ be continuously differentiable. Applying the Poincaré inequality for $p$ to $g\circ \varphi$ and using the chain rule givesThus the Pushforward of a Poincaré inequality by a Lipschitz function gives a $cL$-Poincaré inequality for $q$.

For these increasing warps, the square-root velocity function is $Q(h)(t)=\sqrt{h^{\prime }(t)}$. The chain rule givesThereforewhere the last equality again uses the change of variables formula. Taking square roots proves invariance under common right composition by $\gamma$.

Substitute $m_4=e^{\Psi }$ into the given equation. The chain rule givesThe incident reduced field is the constant one, so $m_4(-\Delta x)=1$ and $\Psi (-\Delta x)=0$. During a thin-screen crossing, $\Psi =O(\Delta x)$. Its mixed-derivative contribution integrates to $O(\Delta x)$. More precisely, its contribution to $\Psi$ is $O((\Delta x{)}^2)$, while the product of its first derivatives contributes only at still higher order. Thereforeand hence

The chain rule andgiveApplying the same rule to $L_1$ and $L_2$ in the Euler-Lagrange field equation yields the exact modulated equation

Multiply this equation by ${\Phi }_{\theta }$. Since $k$ and $\omega$ are independent of the fast phase,The product rule then rearranges the field equation into the exact modulated-wave first integralAverage this identity over one $2\pi$ period in $\theta$. Periodicity kills the first term, while differentiation of the averaged Lagrangian givesConsequently

The same two equations follow directly from the modulated variational principle. Varying $\Phi$ giveswhich is the exact field equation above. For a variation of $\Theta$, use $\delta k={\partial }_X\delta \Theta$ and $\delta \omega =-{\partial }_T\delta \Theta$. Integration by parts in $X$ and $T$ gives the averaged equation. Thus variation with respect to the periodic profile reproduces the local wave equation, whereas variation with respect to its slow phase gives the Whitham modulation equation for wave action.

Choose the orientation of a surface represented by a smooth Gauss map $N:S\to {\mathbb{S}}^2$. The shape operator at $p$ isIf its principal curvatures are $k_1,k_2$, then by definition

Let $E(x)=Ax+b$ be a proper Euclidean motion of Euclidean three-space, so $A\in SO(3)$. Orient $\tilde{S}=E(S)$ by transporting the normal:The chain rule gives the orthogonal conjugacyThe determinant and trace are invariant under similarity, so the Euclidean invariance of the shape operator yields

If the surface orientation is reversed, then $N$ and $S_p$ are multiplied by $-1$. Therefore the orientation reversal of surface curvature says

Hamilton's equations and the multivariable chain rule giveThus the Hamiltonian conservation law gives

For the phase-space dilation generator $D=q\cdot p$,Therefore the condition $\{f,D\}=0$ is the first-order equation

If $f(\lambda q,{\lambda }^{-1}p)=f(q,p)$ for every positive real $\lambda$, differentiate with respect to $\lambda$ at $\lambda =1$. The chain rule gives exactlyso every smooth scale-invariant $f$ satisfies the constraint. The same conclusion for negative $\lambda$ follows wherever the stated invariance is defined; differentiation only needs the positive component containing $\lambda =1$.

To say that $f$ is twice differentiable at $a$ means that $f^{\prime }$ is defined in a neighbourhood of $a$ and is differentiable at $a$; its derivative is $f^{\prime \prime }(a)$.

By the chain rule, near $a$,Since $f$ is differentiable and $g^{\prime }$ is differentiable at $f(a)$, another use of the chain rule shows that $g^{\prime }\circ f$ is differentiable at $a$. The product rule then proves that the displayed product is differentiable. Hence $g\circ f$ is twice differentiable, with the second derivative chain rule

Differentiability lets us writeand, with $k\to 0$,Substitute $k=f(a+h)-f(a)=f^{\prime }(a)h+r(h)$. Since $k=O(h)$, the remainder $s(k)$ is $o(h)$. Thusand the chain rule is

The space of test functions isA sequence ${\phi }_j$ converges to $\phi$ in $\mathcal{D}$ when all supports lie eventually in one compact set $K$ andfor every multi-index $\alpha$. A distribution is a linear functional $u:\mathcal{D}\to \mathbb{C}$ such that, for every compact $K$, there are $C_K>0$ and $m_K\in {\mathbb{N}}_0$ withwhenever $\mathrm{supp}\phi \subseteq K$. Convergence in ${\mathcal{D}}^{\prime }$ is pointwise convergence on test functions.

Continuity plainly implies sequential continuity. Conversely, suppose the linear form is sequentially continuous but the displayed estimate fails for some compact $K$. For every $j$, choose ${\phi }_j$ supported in $K$ such thatThen ${\phi }_j\to 0$ in $\mathcal{D}$, while $u({\phi }_j)\nrightarrow 0$, a contradiction. Hence the seminorm estimate holds on every compact set and $u\in {\mathcal{D}}^{\prime }$.

For a translation vector $h$ and a multi-index $\alpha$, defineThese definitions extend ordinary translation and differentiation to distributions.

If ${\tau }_{t{e}_i}u=u$ for every $t$, differentiating its pairing at $t=0$ gives ${\partial }_{i}u=0$. Conversely, if ${\partial }_{i}u=0$, then for every test functionThe pairing is constant in $t$, hence ${\tau }_{t{e}_i}u=u$. This proves both directions of translation invariance and vanishing distributional derivative.

Finally, the distributional differentiation obeys the linear chain rule under the linear coordinates $s=x-y$ and $t=x+y$. Thusand likewiseas distributions. Adding the two identities givesthe one-dimensional wave equation; this is the low-regularity form of the travelling waves in the D'Alembert formula.

Use the reflection extension from a half-spaceIt is plainly linear and restricts to $u$ on $U$. For a smooth $u$, the chain rule givesA change of variables therefore givesfor $p<\infty$, with the evident equality of essential suprema for $p=\infty$. Approximate a general $u$ by the smooth functions from part c. The estimate makes their reflections Cauchy in $W^{1,p}({\mathbb{R}}^2)$, and their limit defines a bounded Sobolev extension operator with $\Vert E\Vert \le 2^{1/p}$.

For $s\ge 0$ and $\xi \in \mathbb{R}$, define the characteristic curve $X(t;s,\xi )$ byThe bounded derivative $F_x$ makes $F(t,\cdot )$ globally Lipschitz, uniformly in $t$. On each finite time interval, $|F(t,x)|\le |F(t,0)|+L|x|$, so Gronwall inequality prevents finite-time escape. The Picard-Lindelöf theorem therefore gives a unique trajectory for every finite $t\ge 0$. Differentiation in $\xi$ givesso the characteristic flow map is a $C^1$ increasing diffeomorphism.

Along a characteristic, the chain rule changes the equation intoTracing $(t,x)$ backward by the flow therefore givesThe regularity of the flow makes this a $C^1$ classical solution. Conversely, every classical solution obeys the same ordinary differential equation along every characteristic, so the formula also proves uniqueness.

At $x\in \Sigma$, keep the unit normal $N(x)$ fixed and define the $j$th normal derivative byFor $j=1$, the chain rule and $N=\nabla \varphi /|\nabla \varphi |$ give

The characteristic flow map solves the ordinary differential equationand henceAlong this characteristic curve, the chain rule givesThe value is therefore constant, and tracing $(t,x)$ back to time zero gives the classical solutionDirect differentiation verifies both the linear transport equation and its initial value.

On $(0,1)$, the product rule and chain rule give $f^{\prime }(x)=2x\sin (1/x)-\cos (1/x)$. On $(-1,0)$, the function is constant and has derivative zero. At zero,Its absolute value is at most $|h|$, so the squeeze theorem gives $f^{\prime }(0)=0$. ThereforeFor $x_n=1/(2\pi n)$, we have $x_n\to 0$ but $f^{\prime }(x_n)=-1$ for every $n$. This differs from $f^{\prime }(0)=0$, so $f^{\prime }$ is not continuous at zero. Indeed, $y_n=1/((2n+1)\pi )$ gives $f^{\prime }(y_n)=1$, so the right-hand limit of a function does not exist. Thus $f$ is differentiable everywhere on $(-1,1)$, while $f^{\prime }$ is discontinuous at zero.

Using the first partial derivatives from part (b),Also $uvg=yf$. Therefore the first partial differential equation becomesFor the second partial differential equation, the same chain rule givesAway from the origin, cancellation yields $f_x+2xf=0$. Treating $y$ as fixed, an integrating factor gives ${\partial }_x(e^{x^2}f)=0$. Thuswhere $F$ is an arbitrary sufficiently differentiable function. This is the local general solution, and the global form when $f$ is defined on the whole plane. If $f$ and its first partial derivatives are continuous at the origin, the reduced partial differential equation also holds there by continuity.