We transform the global solution of the equivalent system (2.10) into the global conservative solution of the original system (2.1) in this section. It suffices to establish the correspondence between the Lagrangian equivalent system and the original system.

We first introduce a set

*G* as the set of relabeling functions defined by

$G=\{f\text{is invertible}|f-Id\text{and}{f}^{-1}-Id\text{both belong to}{W}^{1,\mathrm{\infty}}(R)\},$

where Id denotes the identity function. For any

$\alpha >1$, we define the subsets

${G}_{\alpha}$ of

*G* as

${G}_{\alpha}=\{f\in G|{\parallel f-Id\parallel}_{{W}^{1,\mathrm{\infty}}(R)}+{\parallel {f}^{-1}-Id\parallel}_{{W}^{1,\mathrm{\infty}}(R)}\le \alpha \},$

with a useful property: If

$f\in {G}_{\alpha}$ (

$\alpha \ge 0$), then

$1/(1+\alpha )\le {f}_{\xi}\le 1+\alpha $ almost everywhere. Conversely, if

*f* is absolutely continuous,

$f-Id\in {L}^{\mathrm{\infty}}(R)$ and there exists

$c\ge 1$ such that

$1/c\le {f}_{\xi}\le c$ almost everywhere, then

$f\in {G}_{\alpha}$ for some

*α* depending only on

*c* and

${\parallel f-Id\parallel}_{{L}^{\mathrm{\infty}}(R)}$. We now define the subsets

*F* and

${F}_{\alpha}$ of Γ such that

$\begin{array}{c}F=\{X=(y,U,V,H)\in \mathrm{\Gamma}|y+H\in G\},\hfill \\ {F}_{\alpha}=\{X=(y,U,V,H)\in \mathrm{\Gamma}|y+H\in {G}_{\alpha}\}.\hfill \end{array}$

With the above useful property of ${G}_{\alpha}$, it is not hard to prove that the space *F* is preserved by the governing equation (2.10).

Notice that the map $\mathrm{\Phi}:G\times F\to F$ given by $\mathrm{\Phi}(f,X)=X\circ f$ defines a group action of *G* on *F*, we then consider the quotient space $F/G$ of *F* w.r.t. the group action. The equivalence relation on *F* is defined as: for any $X,{X}^{\prime}\in F$, if there exists $f\in G$ such that ${X}^{\prime}=X\circ f$, we claim that *X* and ${X}^{\prime}$ are equivalent. We denote the projection $\mathrm{\Pi}:F\to F/G$ by $\mathrm{\Pi}(X)=[X]$. For any $X=(y,U,V,N,H)\in F$, we introduce the map $K:F\to {F}_{0}$ given by $K(X)=X\circ {(y+H)}^{-1}$. It is not hard to prove that $K(X)=X$ when $X\in {F}_{0}$, and $K(X\circ f)=K(X)$ for any $X\in F$ and $f\in G$. Hence, we can define the map $\tilde{K}:F/G\to {F}_{0}$ as $\tilde{K}([X])=K(X)$ for any representative $[X]\in F/G$ of $X\in F$. For any $X\in {F}_{0}$, we have $\tilde{K}\circ \mathrm{\Pi}(X)=K(X)=X$. Hence, $\tilde{K}\circ {\mathrm{\Pi}}_{|{F}_{0}}={Id}_{|{F}_{0}}$. Note that any topology defined on ${F}_{0}$ is naturally transported into $F/G$ by this isomorphism, that is, if we equip ${F}_{0}$ with the metric induced by the *E*-norm, *i.e.*, ${d}_{{F}_{0}}(X,{X}^{\prime})={\parallel X-{X}^{\prime}\parallel}_{E}$ for all $X,{X}^{\prime}\in {F}_{0}$, which is complete, then for any $[X],[{X}^{\prime}]\in F/G$, the topology on $F/G$ is defined by a complete metric given by ${d}_{F/G}([X],[{X}^{\prime}])={\parallel K(X)-K({X}^{\prime})\parallel}_{E}$.

For any initial data $\overline{X}\in F$, we denote the continuous semigroup with the solution $X(t)$ of system (2.10) by $S:F\times {R}^{+}\to F$. As we indicated earlier, Eq. (2.1) is invariant w.r.t. relabeling. That is, $t>0$, ${S}_{t}(X\circ f)={S}_{t}(X)\circ f$ for any $X\in F$ and $f\in G$. Thus, the map ${\tilde{S}}_{t}:F/G\to F/G$ defined by ${\tilde{S}}_{t}([X])=[{S}_{t}X]$ is valid, which generates a continuous semigroup.

To derive the correspondence between the Lagrangian equivalent system and the original system, we have to consider the space

*D*, which characterizes the solutions in the original system:

$D=\{(z,\mu )|z\in {H}^{1}(R)\times [{L}^{2}(R)\times {L}^{\mathrm{\infty}}(R)]\text{and}{\mu}_{\mathrm{ac}}=({u}^{2}+{u}_{x}^{2}+{v}^{2})\phantom{\rule{0.2em}{0ex}}dx\},$

where $z=(u,v)$ and *μ* is a positive finite Radon measure with ${\mu}_{\mathrm{ac}}$ as its absolute continuous part.

We now establish a bijection between $F/G$ and *D* to transport the continuous semigroup obtained in the Lagrangian equivalent system (functions in $F/G$) into the original system (functions in *D*).

We first introduce the mapping

*M*, which corresponds to the transformation from the Lagrangian equivalent system into the original system. In the other direction, we obtain the energy density

*μ* in the original system, by pushing forward by

*y* the energy density

${H}_{\xi}\phantom{\rule{0.2em}{0ex}}d\xi $ in the Lagrangian equivalent system, where the push-forward

${f}_{\mathrm{\#}}\nu $ of a measure

*ν* by a measurable function

*f* is defined by

${f}_{\mathrm{\#}}\nu (B)=\nu ({f}^{-1}(B))$

for all Borel set

*B*. Let

$(z,\mu )$ be defined as

$z(x)=Z(\xi )\phantom{\rule{1em}{0ex}}\text{for any}\xi \text{such that}x=y(\xi ),$

(4.1a)

$\mu ={y}_{\mathrm{\#}}({H}_{\xi}\phantom{\rule{0.2em}{0ex}}d\xi ),$

(4.1b)

where $z(x)=(u,v)(x)$, $Z(\xi )=(U,V)(\xi )$. We have that $(z,\mu )\in D$, which does not depend on the representative $X=(y,U,V,N,H)\in F$ of $[X]$ we choose. We denote by $M:F/G\to D$ the mapping to any $[X]\in F/G$ and $(z,\mu )\in D$ given by (4.1a) and (4.1b), which transforms the Lagrangian equivalent system into the original system.

We are led to the mapping $L:D\to F/G$, which conversely transforms the original system into the Lagrangian equivalent system defined as follows.

**Definition 4.1** For any

$(z,\mu )\in D$, let

$y(\xi )=sup\{y|\mu (-\mathrm{\infty},y)+y<\xi \},$

(4.2a)

$U(\xi )=u\circ y(\xi ),\phantom{\rule{2em}{0ex}}V(\xi )=v\circ y(\xi ),\phantom{\rule{2em}{0ex}}N(\xi )={u}_{x}\circ y(\xi ),$

(4.2b)

$H(\xi )=\xi -y(\xi ),$

(4.2c)

where $z=(u,v)$. We define $L(z,\mu )\in F/G$ as the equivalence class of $(y,U,V,N,H)$.

**Remark 4.1** Note that $X=(y,U,V,N,H)\in E$, which satisfies (3.5a)-(3.5c) from the definition of *y*, *U*, *V*, *N*, *H* in (4.2a)-(4.2c). Moreover, by the definition (4.2c), we have that $y+H=Id$. Thus, $X=(y,U,V,N,H)\in {F}_{0}$.

We claim that the transformation from the original system into the Lagrangian equivalent system is a bijection.

**Theorem 4.1** *The maps* *M* *and* *L* *are invertible*,

*that is*,

$L\circ M={Id}_{F/G},\phantom{\rule{2em}{0ex}}M\circ L={Id}_{D}.$

*Proof* Let

$[X]\in F/G$ be given. We consider

$X=(y,U,V,N,H)=\tilde{K}([X])$ for a representative of

$[X]$ and

$(z,\mu )$ given by (4.1a) and (4.1b) for this particular

*X*. From the definition of

$\tilde{K}$, we have

$X\in {F}_{0}$. Let

$\overline{X}=(\overline{y},\overline{U},\overline{V},\overline{N},\overline{H})$ be the representative of

$L(z,\mu )$ in

${F}_{0}$ given by (4.2a)-(4.2c). To derive

$L\circ M={Id}_{F/G}$, it suffices to show that

$(\overline{y},\overline{U},\overline{V},\overline{N},\overline{H})=(y,U,V,N,H)$. Let

$g(x)=sup\{\xi \in R|y(\xi )<x\}.$

(4.3)

Using the fact that

*y* is increasing and continuous, it follows that

and

${y}^{-1}((-\mathrm{\infty},x))=(-\mathrm{\infty},g(x))$. From (4.1b) and since

$H(-\mathrm{\infty})=0$, for any

$x\in R$, we get

$\mu ((-\mathrm{\infty},x))={\int}_{{y}^{-1}((-\mathrm{\infty},x))}{H}_{\xi}\phantom{\rule{0.2em}{0ex}}d\xi ={\int}_{-\mathrm{\infty}}^{g(x)}{H}_{\xi}\phantom{\rule{0.2em}{0ex}}d\xi =H(g(x)).$

Since

$X\in {F}_{0}$ and

$y+H=Id$, we have

$\mu ((-\mathrm{\infty},x))+x=g(x).$

(4.5)

From the definition of

$\overline{y}$, it follows that

$\overline{y}(\xi )=sup\{x\in R|g(x)<\xi \}.$

(4.6)

For any given $\xi \in R$, using the fact that *y* is increasing and (4.4), it follows that $\overline{y}(\xi )\le y(\xi )$. If $\overline{y}(\xi )<y(\xi )$, there then exists *x* such that $\overline{y}(\xi )<x<y(\xi )$ and (4.6) implies that $g(x)\ge \xi $. Conversely, since *y* is increasing, we have $x=y(g(x))<y(\xi )$, which implies that $g(x)<\xi $. This is a contradiction. Hence, we have that $\overline{y}=y$. Since $y+H=Id$, it follows directly from the definitions that $\overline{H}=H$, $\overline{U}=U$, $\overline{V}=V$ and $\overline{N}=N$. Hence, $L\circ M={Id}_{F/G}$.

Let

$(z,\mu )\in D$ be given and

$(y,U,V,N,H)$ be the representative of

$L(z,\mu )$ in

${F}_{0}$ given by (4.2a)-(4.2c). Then, let

$(\overline{z},\overline{\mu})=M\circ L(z,\mu )$. Let

*g* be defined as before by (4.3). The same computation that leads to (4.5) now gives

$\overline{\mu}((-\mathrm{\infty},x))+x=g(x).$

(4.7)

Given

$\xi \in R$, we consider an increasing sequence

${x}_{i}$ converging to

$y(\xi )$, which is guaranteed by (4.2a), and such that

$\mu ((-\mathrm{\infty},{x}_{i}))+{x}_{i}<\xi $. Let

*i* tend to infinity. Since

$F(x)=\mu ((-\mathrm{\infty},x))$ is lower semi-continuous, we have

$\mu ((-\mathrm{\infty},y(\xi )))+y(\xi )\le \xi $. Take

$\xi =g(x)$ and then we get

$\mu ((-\mathrm{\infty},x))+x\le g(x).$

(4.8)

By the definition of

*g*, there exists an increasing sequence

${\xi}_{i}$ converging to

$g(x)$ such that

$y({\xi}_{i})<x$. It follows from the definition of

*y* in (4.2a) that

$\mu ((-\mathrm{\infty},x))+x\ge {\xi}_{i}$. Passing to the limit, we obtain

$\mu ((-\mathrm{\infty},x))+x\ge g(x)$ which, together with (4.8), yields

$\mu ((-\mathrm{\infty},x))+x=g(x).$

(4.9)

We obtain that $\overline{\mu}=\mu $ by comparing (4.9) and (4.7). It is clear from the definitions that $\overline{z}=z$. Hence, $(\overline{z},\overline{\mu})=(z,\mu )$ and $M\circ L={Id}_{D}$. □

Our next task is to transport the topology defined in $F/G$ into *D*, which is guaranteed by the fact that we have established a bijection between the two equivalent systems and then obtained a continuous semigroup of solutions for the original system.

Let us define the distance

${d}_{D}$ on

*D* as

${d}_{D}((z,\mu ),(\overline{z},\overline{\mu}))={d}_{F/G}(L(z,\mu ),L(\overline{z},\overline{\mu})),$

which makes the bijection

*L* between

*D* and

$F/G$ into an isometry. Since

$F/G$ equipped with

${d}_{F/G}$ is a complete metric space, it is not hard to know that

*D* equipped with the metric

${d}_{D}$ is also a complete metric space. For each

$t\in R$, we define the mapping

${T}_{t}:D\to D$ as

${T}_{t}=M{\tilde{S}}_{t}L.$

**Theorem 4.2** *Given* $(\overline{z},\overline{\mu})\in D$,

*if we denote* $t\to (z,\mu )(t)={T}_{t}(\overline{z},\overline{\mu})$ *the corresponding trajectory*,

*then* $z=(u,v)$ *is a weak solution of the two*-

*component Camassa*-

*Holm equations* (2.1),

*which constructs a continuous semigroup*.

*Moreover*,

*μ* *is a weak solution of the following transport equation*:

${\mu}_{t}+{(u\mu )}_{x}={({u}^{3}-2Pu)}_{x}.$

(4.10)

*Furthermore*,

*we have* $\mu (t)(R)=\mu (0)(R)\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.25em}{0ex}}t$

(4.11)

*and*
$\begin{array}{rl}\mu (t)(R)& ={\mu}_{\mathrm{ac}}(t)(R)={\parallel z(t)\parallel}_{{H}^{1}}^{2}\\ ={\parallel u(t)\parallel}_{{H}^{1}}^{2}+{\parallel v(t)\parallel}_{{L}^{2}}^{2}=\mu (0)(R)\phantom{\rule{1em}{0ex}}\mathit{\text{for almost all}}\phantom{\rule{0.25em}{0ex}}t.\end{array}$

(4.12)

*Thus*, *the unique solution described here is a conservative weak solution of system* (2.1).

*Proof* To prove that

$z=(u,v)$ is a weak solution of the original system (2.1), it suffices to show that, for all

$\varphi \in {C}^{\mathrm{\infty}}({R}^{+}\times R)$ with compact support,

$\begin{array}{r}{\int}_{{R}^{+}\times R}(-u{\varphi}_{t}+u{u}_{x}\varphi )(t,x)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt=-{\int}_{{R}^{+}\times R}({P}_{x}\varphi )(t,x)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt,\\ {\int}_{{R}^{+}\times R}(-v{\varphi}_{t}+u{v}_{x}\varphi )(t,x)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt=-{\int}_{{R}^{+}\times R}[(v+1){u}_{x}\varphi ](t,x)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt,\end{array}$

(4.13)

where

${P}_{x}$ is given by (2.1). We denote by the solution

$(y,U,V,N,H)(t)$ of (2.10) a representative of

$L(z(t),\mu (t))$. On the one hand, since

$y(t,\xi )$ is Lipschitz and invertible w.r.t.

*ξ* for almost all

*t*, we can use the change of variables

$x=y(t,\xi )$, then we get

$\begin{array}{l}{\int}_{{R}^{+}\times R}(-u{\varphi}_{t}+u{u}_{x}\varphi )(t,x)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}={\int}_{{R}^{+}\times R}[-(U{y}_{\xi})(t,\xi ){\varphi}_{t}(t,y(t,\xi ))+(U{U}_{\xi})(t,\xi )\varphi (t,y(t,\xi ))]\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}dt.\end{array}$

(4.14)

Since

${y}_{t}=U$ and

${y}_{\xi t}={U}_{\xi}$, it then follows from (2.10) that

$\begin{array}{l}{\int}_{{R}^{+}\times R}[-U{y}_{\xi}{\varphi}_{t}(t,y)+U{U}_{\xi}\varphi (t,y)]\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}=\frac{1}{4}{\int}_{{R}^{+}\times {R}^{2}}\{sgn(\xi -{\xi}^{\prime}){e}^{-|y(\xi )-y({\xi}^{\prime})|}[{H}_{\xi}+({U}^{2}+2V){y}_{\xi}]\}\left({\xi}^{\prime}\right)\\ \phantom{\rule{2em}{0ex}}\cdot \varphi (t,y(\xi )){y}_{\xi}(\xi )\phantom{\rule{0.2em}{0ex}}d{\xi}^{\prime}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}dt.\end{array}$

(4.15)

On the other hand, using the change of variables

$x=y(t,\xi )$ and

${x}^{\prime}=y(t,{\xi}^{\prime})$, and since

*y* is an increasing function, we have

$\begin{array}{c}-{\int}_{{R}^{+}\times R}({P}_{x}\varphi )(t,x)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}{\int}_{{R}^{+}\times {R}^{2}}[sgn(\xi -{\xi}^{\prime}){e}^{-|y(\xi )-y({\xi}^{\prime})|}({u}^{2}+\frac{1}{2}{u}_{x}^{2}+\frac{1}{2}{v}^{2}+v)](t,y\left({\xi}^{\prime}\right))\hfill \\ \phantom{\rule{2em}{0ex}}\cdot \varphi (t,y(\xi )){y}_{\xi}\left({\xi}^{\prime}\right){y}_{\xi}(\xi )\phantom{\rule{0.2em}{0ex}}d{\xi}^{\prime}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}dt.\hfill \end{array}$

It follows from the identity (3.5c) that

$\begin{array}{r}-{\int}_{{R}^{+}\times R}({P}_{x}\varphi )(t,x)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}=\frac{1}{4}{\int}_{{R}^{+}\times {R}^{2}}\{sgn(\xi -{\xi}^{\prime}){e}^{-|y(\xi )-y({\xi}^{\prime})|}[{H}_{\xi}+({U}^{2}+2V){y}_{\xi}]\}\left({\xi}^{\prime}\right)\\ \phantom{\rule{2em}{0ex}}\cdot \varphi (t,y(\xi )){y}_{\xi}(\xi )\phantom{\rule{0.2em}{0ex}}d{\xi}^{\prime}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}dt.\end{array}$

(4.16)

By comparing (4.15) and (4.16), we know that

${\int}_{{R}^{+}\times R}[-U{y}_{\xi}{\varphi}_{t}(t,y)+U{U}_{\xi}\varphi (t,y)]\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}dt=-{\int}_{{R}^{+}\times R}({P}_{x}\varphi )(t,x)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt.$

Thus, the first identity in (4.13) follows directly from (4.14) and the second identity in (4.13) follows in the same way. It is not hard to check that

$\mu (t)$ is the solution of (4.10). From the definition

*μ* in (4.1b), we can get that

$\mu (t)(R)={\int}_{R}{H}_{\xi}\phantom{\rule{0.2em}{0ex}}d\xi =H(t,\mathrm{\infty}),$

which is constant in time from Lemma 3.3(iii). Thus, we have proved (4.11).

Since

${y}_{\xi}(t,\xi )>0$ a.e. for almost every

$\xi \in R$, it then follows from (3.5c) that

$\mu (t)(B)={\int}_{{y}^{-1}(B)}{H}_{\xi}\phantom{\rule{0.2em}{0ex}}d\xi ={\int}_{{y}^{-1}(B)}({U}^{2}+{U}_{\xi}^{2}/{y}_{\xi}^{2}+{V}^{2}){y}_{\xi}\phantom{\rule{0.2em}{0ex}}d\xi $

(4.17)

for any Borel set

*B*. Since

*y* is one-to-one and

${u}_{x}\circ y{y}_{\xi}={U}_{\xi}$ almost everywhere, then (4.17) implies that

$\mu (t)(B)={\int}_{B}({u}^{2}+{u}_{x}^{2}+{v}^{2})(t,x)\phantom{\rule{0.2em}{0ex}}dx.$

Thus, (4.12) holds and the proof is completed. □