I would like to solve the following partial differential equation
$$\begin{cases} u_{x}^{2}+u_{y}^{2} & =\dfrac{(1+\sin u)^{2}}{4a^{2}},\\ u|\varGamma & =0, \end{cases}$$
where $a\in\mathbb{R}-\{0\}$. The equation can be rewritten to
$$F(x,y,z,p,q)=p^{2}+q^{2}-\frac{(1+\sin z)^{2}}{4a^{2}}=0.$$
The characteristic equation is given by
\begin{align*} \frac{dx}{ds}(r,s) & =F_{p}=2p,\\ \frac{dy}{ds}(r,s) & =F_{q}=2q,\\ \frac{dz}{ds}(r,s) & =pF_{p}+qF_{q}=2p^{2}+2q^{2},\\ \frac{dp}{ds}(r,s) & =-F_{x}-pF_{z}=\frac{p}{2a^{2}}\cos z(1+\sin z),\\ \frac{dq}{ds}(r,s) & =-F_{y}-qF_{z}=\frac{q}{2a^{2}}\cos z(1+\sin z). \end{align*}
The parametrization $\Gamma$ is
\begin{align*} x(r,0) & =c\cos r,\\ x(r,0) & =c\sin r,\\ z(r,0) & =c^{2}(\cos^{2}r+\sin^{2}r)-\frac{(1+\sin z)^{2}}{4a^{2}}=0, \end{align*}
where $c=(1+\sin z)/(2a)$. We search for functions $\psi_{1}(r)$, $\psi_{2}(r)$ satisfying
\begin{align*} F(\gamma_{1}(r),\gamma_{2}(r),\phi(r),\psi_{1}(r),\psi_{2}(r)) & =0,\\ \phi^{\prime}(r) & =\psi_{1}(r)\gamma_{1}^{\prime}(r)+\psi_{2}(r)\gamma_{2}^{\prime}(r). \end{align*}
Hence
$$ \psi_{1}^{2}+\psi_{2}^{2}=c^{2},$$
and
$$0=\psi_{1}(r)c(-\sin r)c+\psi_{2}(r)c\cos r,$$
with the solutions
\begin{align*} (1)\psi_{1}(r) & =c\cos r\qquad\psi_{2}(r)=c\sin r,\\ (2)\psi_{1}(r) & =-c\cos r\qquad\psi_{2}(r)=-c\sin r. \end{align*}
For the first solution are
\begin{align*} p(r,s) & =c\cos r,\\ q(r,s) & =c\sin r, \end{align*}
and
\begin{align*} x(r,s) & =2cs\cos r+c\cos r=c\cos r(2s+1),\\ y(r,s) & =2cs\sin r+c\sin r=c\sin r(2s+1),\\ z(r,s) & =2cs(\cos^{2}r+\sin^{2}r)=2cs. \end{align*}
Then,
$$x^{2}+y^{2}=c^{2}(2s+1)=(z+c)^{2}.$$
Because $u(x,y)=z(r,s),$ then
\begin{align*} \left(u+c\right)^{2} & =x^{2}+y^{2},\\ u+\frac{1+\sin u}{2a} & =\sqrt{x^{2}+y^{2}}, \end{align*}
which represents the transcendental equation for $u$, but the solution is wrong. The correct result is
$$ u=\frac{\pi}{2}-2\arctan\frac{\sqrt{x^{2}+y^{2}}}{2a}.$$
Could I ask for your help with the solution of this PDE? Thank you very much...
Updated question:
In my opinion, the problem of the last solution was the substitution, because $c=f(z)$. Let us put
$$ c(z)=\sqrt{\frac{(1+\sin z)^{2}}{4a^{2}}}=\pm(1+\sin z)/(2a).$$
We parametrize curve $\Gamma$ by $r$ such that $\Gamma=\{c(r)\cos r,c(r)\sin r\}$, on $\Gamma$ is $u|\varGamma=0$.
\begin{align*} x(r,0) & =c(r)\cos r=\frac{1+\sin r}{2a}\cos r,\\ x(r,0) & =c(r)\sin r=\frac{1+\sin r}{2a}\sin r,\\ z(r,0) & =\frac{(1+\sin r)^{2}}{4a^{2}}(\cos^{2}r+\sin^{2}r)-\frac{(1+\sin r)^{2}}{4a^{2}}=0, \end{align*}
We search for functions $\psi_{1}(r)$, $\psi_{2}(r)$ satisfying
\begin{align*} F(\gamma_{1}(r),\gamma_{2}(r),\phi(r),\psi_{1}(r),\psi_{2}(r)) & =0,\\ \phi^{\prime}(r) & =\psi_{1}(r)\gamma_{1}^{\prime}(r)+\psi_{2}(r)\gamma_{2}^{\prime}(r). \end{align*}
Hence $$ \psi_{1}^{2}+\psi_{2}^{2}=c^{2},$$ and $$0=-\psi_{1}(r)\frac{\sin r+2\sin^{2}r-1}{2a}+\psi_{2}(r)\frac{\sin2r+\cos r}{2a}.$$
However, I am not sure, if I am not completely wrong with my assumptions and substitutions.
Thank for your help.
They are many typos in your calculus ($x$ instead of $y$, $\:p$ instead of $q$, in several places and a power $2$ missing elsewhere) which makes difficult to follow what is correct or not.
HINT : it seems that the initial mistake is here: