Write $z(x,y)=Z(r,s)$ in the new coordinates. Sincethe partial differential equation becomesFor a twice continuously differentiable solution of a differential equation, $Z_r$ is independent of $s$, so integration first in $r$ and then in $s$ givesThis is the decomposition of the one-dimensional wave equation into waves traveling in opposite directions.

The first initial condition gives $\varphi (y)+\psi (y)=\cos y$. The second initial condition gives ${\varphi }^{\prime }(y)-{\psi }^{\prime }(y)=0$. Differentiating the first relation and combining the two relations yields ${\varphi }^{\prime }=ps{i}^{\prime }=-\frac{1}{2}\sin y$. Thus $\varphi (y)=\frac{1}{2}\cos y+C$ and $\psi (y)=\frac{1}{2}\cos y-C$. The constants cancel, and the cosine addition formula givesBoth initial conditions and the partial differential equation follow directly by differentiation.

Multiplying the ordinary differential equation by two suggests using the dependent variable $Y=y^2$, whose derivative is $Y^{\prime }=2y{y}^{\prime }$. On the interval through $x=1$, this gives the linear ordinary differential equationAfter integration, $Y=x^2+Cx$. The initial condition $Y(1)=4$ gives $C=3$. As $y(1)>0$, continuity selects the positive branch:The interval $x>0$ is the maximal real interval through the given initial condition: at $x=0$ the formula approaches $0$, but its derivative is unbounded, so it does not extend through zero as a differentiable solution of a differential equation.

The quotient is a Klein bottle, equivalently a connected sum of two real projective planes.

Use $z\in S^1$ for the angular coordinate and $r\in [1/2,1]$ for the radius. Parametrize the first annulus by $\tau =2r-1\in [0,1]$, running from its inner to its outer boundary. Parametrize the second by $\tau =3-2r\in [1,2]$, running from its outer to its inner boundary. The outer-boundary identity gluing puts them together at $\tau =1$. The remaining inner-boundary identification reverses the angle, so the entire quotient isThis is the mapping torus of reflection of a circle, after rescaling the interval.

To verify the homeomorphism, send $rz$ in the first annulus to $[z,2r-1]$ in $M$, and $rz$ in the second to $[z,3-2r]$. Both formulas are continuous, agree under the outer gluing, and agree under the reflected inner gluing. They therefore descend by the universal property of the quotient topology. Conversely, on the lower half of the cylinder send $(z,\tau )$ to the class of $((1+\tau )/2)z$ in the first annulus; on the upper half send it to the class of $((3-\tau )/2)z$ in the second. The formulas agree at $\tau =1$, and at the identified endpoints. The pasting lemma and the quotient property give a continuous inverse. Thus the two quotient descriptions are homeomorphic.

The quotient $M$ is compact. Its equivalence relation on the compact Hausdorff cylinder is closed: it is the union of the diagonal and the two graphs of the boundary reflection. Therefore $M$ is Hausdorff by compact Hausdorff quotient by a closed equivalence relation. Away from the seam its local charts are circle arcs times open intervals. At the seam, an arc near $z$ at $\tau =0$ glues to its conjugate arc at $\tau =2$; using the reflection as the transition gives another open rectangle. These are disc neighbourhoods, and a finite choice of such charts gives a countable basis. Hence $M$ is a closed surface.

Cutting the circle coordinate gives a rectangle with its vertical sides matched and its horizontal sides reflected. The reflection changes the sign of the angular coordinate while leaving the interval direction unchanged, so an orientation transported around the interval loop comes back reversed. Thus $M$ is a nonorientable surface. The rectangle quotient has one vertex, two edges and one face, giving Euler characteristic zero. By the classification theorem for closed combinatorial surfaces, the nonorientable genus is $k=2-\chi (M)=2$, so $M\cong {\mathbb{RP}}^2\#{\mathbb{RP}}^2$. This is the gluing annuli with one reflected boundary construction.