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An addendum to Deanne Yang's answer. Actually, if a linear hyperbolic equation is degenerate, meaning that the characteristic roots are real but not distinct, then the inverse operator presents a loss of derivatives, meaning that it takes $H^s$ to $H^{s-m}$ for some $m>0$. This is an actual phenomenon and not a defect of the estimates. Indeed, the solution of the 1D equation $u_{tt} − t^2 u_{xx} = (4k+1) u_x $ can be written explicitly for integer $k$ and loses exactly $k$ derivatives (this was proved by Qi Min-you in 1958).

Thus the inverse of a degenerate hyperbolic equation is a perfect example of a smooth tame map, and the Nash-Moser theorem can be used to handle the nonlinear equations. I used myself this idea in a couple of papers.

EDIT: I lost track of the hardcopy of QMY's paper, which I had at some point, anyway from my notes the solution with initial data 0 and $\phi(x)$ should be $$u(t,x)=\sum_{j=1}^{k} \frac{\sqrt{\pi}t^{2j}}{j!(k-j)!\Gamma(j+1/2)} \phi^{(j)}(x+t^{2}/2).$$

An addendum to Deanne Yang's answer. Actually, if a linear hyperbolic equation is degenerate, meaning that the characteristic roots are real but not distinct, then the inverse operator presents a loss of derivatives, meaning that it takes $H^s$ to $H^{s-m}$ for some $m>0$. This is an actual phenomenon and not a defect of the estimates. Indeed, the solution of the 1D equation $u_{tt} − t^2 u_{xx} = (4k+1) u_x $ can be written explicitly for integer $k$ and loses exactly $k$ derivatives (this was proved by Qi Min-you in 1958).

Thus the inverse of a degenerate hyperbolic equation is a perfect example of a smooth tame map, and the Nash-Moser theorem can be used to handle the nonlinear equations. I used myself this idea in a couple of papers.

EDIT: I lost track of the hardcopy of QMY's paper, which I had at some point, anyway from my notes the solution with initial data 0 and $\phi(x)$ should be $$u(t,x)=\sum_{j=1}^{k} \frac{\sqrt{\pi}t^{2j}}{j!(k-j)!\Gamma(j+1/2)} \phi^{(j)}(x+t^{2}/2).$$

EDIT 2: Wait: in section 6 of this paper the example is extended to a more general class of equations.

An addendum to Deanne Yang's answer. Actually, if a linear hyperbolic equation is degenerate, meaning that the characteristic roots are real but not distinct, then the inverse operator presents a loss of derivatives, meaning that it takes $H^s$ to $H^{s-m}$ for some $m>0$. This is an actual phenomenon and not a defect of the estimates. Indeed, the solution of the 1D equation $u_{tt} − t^2 u_{xx} = (2k+1) u_x $ can be written explicitly for integer $k$ and loses exactly $k$ derivatives (this was proved by Qi Min-you in 1958).

Thus the inverse of a degenerate hyperbolic equation is a perfect example of a smooth tame map, and the Nash-Moser theorem can be used to handle the nonlinear equations. I used myself this idea in a couple of papers.

EDIT: I lost track of the hardcopy of QMY's paper, which I had at some point, anyway from my notes the solution with initial data 0 and $\phi(x)$ should be $$u(t,x)=\sum_{j=1}^{k} \frac{\sqrt{\pi}t^{2j}}{j!(k-j)!\Gamma(j+1/2)} \phi^{(j)}(x+t^{2}/2).$$

An addendum to Deanne Yang's answer. Actually, if a linear hyperbolic equation is degenerate, meaning that the characteristic roots are real but not distinct, then the inverse operator presents a loss of derivatives, meaning that it takes $H^s$ to $H^{s-m}$ for some $m>0$. This is an actual phenomenon and not a defect of the estimates. Indeed, the solution of the 1D equation $u_{tt} − t^2 u_{xx} = (4k+1) u_x $ can be written explicitly for integer $k$ and loses exactly $k$ derivatives (this was proved by Qi Min-you in 1958).

Thus the inverse of a degenerate hyperbolic equation is a perfect example of a smooth tame map, and the Nash-Moser theorem can be used to handle the nonlinear equations. I used myself this idea in a couple of papers.

EDIT: I lost track of the hardcopy of QMY's paper, which I had at some point, anyway from my notes the solution with initial data 0 and $\phi(x)$ should be $$u(t,x)=\sum_{j=1}^{k} \frac{\sqrt{\pi}t^{2j}}{j!(k-j)!\Gamma(j+1/2)} \phi^{(j)}(x+t^{2}/2).$$

An addendum to Deanne Yang's answer. Actually, if a linear hyperbolic equation is degenerate, meaning that the characteristic roots are real but not distinct, then the inverse operator presents a loss of derivatives, meaning that it takes $H^s$ to $H^{s-m}$ for some $m>0$. This is an actual phenomenon and not a defect of the estimates. Indeed, the solution of the 1D equation $u_{tt} − t^2 u_{xx} = (2k+1) u_x $ can be written explicitly for integer $k$ and loses exactly $k$ derivatives (this was proved by Qi Min-you in 1958).

Thus the inverse of a degenerate hyperbolic equation is a perfect example of a smooth tame map, and the Nash-Moser theorem can be used to handle the nonlinear equations. I used myself this idea in a couple of papers.

An addendum to Deanne Yang's answer. Actually, if a linear hyperbolic equation is degenerate, meaning that the characteristic roots are real but not distinct, then the inverse operator presents a loss of derivatives, meaning that it takes $H^s$ to $H^{s-m}$ for some $m>0$. This is an actual phenomenon and not a defect of the estimates. Indeed, the solution of the 1D equation $u_{tt} − t^2 u_{xx} = (2k+1) u_x $ can be written explicitly for integer $k$ and loses exactly $k$ derivatives (this was proved by Qi Min-you in 1958).

Thus the inverse of a degenerate hyperbolic equation is a perfect example of a smooth tame map, and the Nash-Moser theorem can be used to handle the nonlinear equations. I used myself this idea in a couple of papers.

EDIT: I lost track of the hardcopy of QMY's paper, which I had at some point, anyway from my notes the solution with initial data 0 and $\phi(x)$ should be $$u(t,x)=\sum_{j=1}^{k} \frac{\sqrt{\pi}t^{2j}}{j!(k-j)!\Gamma(j+1/2)} \phi^{(j)}(x+t^{2}/2).$$