A conformal map preserves oriented angles between intersecting smooth curves at each point, with a nonzero local scale factor. A sufficient criterion is that $f$ be a holomorphic function on the domain and that $f^{\prime }(z)\ne 0$ everywhere there. Its first-order action then multiplies every tangent direction by the same nonzero complex number, combining a rotation and a dilation. A conformal bijection additionally requires injectivity and surjectivity onto the target.

For $z=x+iy$ in the given half-strip,Thus $0<|w|<1$ and $0<\arg w<\pi$. Conversely every such $w$ has a unique argument in $(0,\pi )$ and unique radius in $(0,1)$, giving $x=\arg w$, $y=-\log |w|$. HenceThe complex exponential is injective on $S$, and its derivative $i{e}^{iz}$ never vanishes.

On $D_{+}$, use the Joukowski map with a sign change,Sinceit maps into the upper half-plane. To prove injectivity, $J(w_1)=J(w_2)$ givesBoth moduli are below one, so $w_1{w}_2\ne 1$ and $w_1=w_2$.

For surjectivity, take any $\zeta$ in the upper half-plane and solve $w^2+\zeta w+1=0$. The two roots have product one. Neither can lie on the unit circle, since there $-(w+w^{-1})$ is real. Exactly one root therefore has modulus below one. For this root, the displayed imaginary-part identity forces $\mathrm{Im}w>0$, so it belongs to $D_{+}$. Finally $J^{\prime }(w)=w^{-2}-1$ has no zero in $D_{+}$. This proves the Joukowski map from the upper half-disc to the upper half-plane is a conformal bijection.

Composing gives the required conformal mapIts derivative $2\sin z$ is nonzero on $S$, since the zeros of the complex sine are the real integer multiples of $\pi$.

Introduce a characteristic parameter $s$ and label each point of the initial segment by $r\in [0,1]$. The method of characteristics givesIntegrating gives the parametric solutionThe projected characteristic curves satisfy $y=r(2-x)$, so each lies on a straight line through $(2,0)$. Its branch from the data always has $x<2$; the common vertex is approached only as $s\to -\infty$.

Eliminating $s,r$ using $e^s=2-x$ and $r=y/(2-x)$ yieldsIts derivatives give $(x-2)z_x+y{z}_y=y$, and substitution of $x=1$ gives the prescribed value one.

The Jacobian determinant of $(s,r)\mapsto (x,y)$ is $-e^{2s}\ne 0$, and the inverse just displayed is unique. Hence the data determine the solution exactly onThis is the domain determined by a finite segment under a radial characteristic flow. Its interior is $x<2$, $0<y<2-x$; the two boundary rays come from the endpoints of the initial segment. Every point there lies on exactly one characteristic from the data. Outside this sector, a characteristic either misses the initial segment or lies on the branch $x>2$, which cannot reach it. Thus the formula may be extended beyond this sector, but such an extension is not uniquely specified by the given data.

Both components vanish atThe Jacobian matrix of the vector field isIts eigenvalues classify these equilibria by linearization of a dynamical system:All real parts are nonzero, so the linear classifications determine the local stability of the nonlinear autonomous system.

At the origin, the stable direction is the $y$-axis and the unstable direction is the $x$-axis. At $(1,0)$ the eigendirections are $(1,0)$ for $-1$ and $(1,-2)$ for $-3$. At $(0,1)$,has only one eigendirection, the $y$-axis; its linearization of a dynamical system has solutions $\delta x=A{e}^{2t}$, $\delta y=(B-At)e^{2t}$, explaining the improper node. At $(4,3)$, stable and unstable tangent vectors are respectively $(4,5-\sqrt{13})$ and $(4,5+\sqrt{13})$.

For the phase portrait, the axes are invariant, and the remaining nullclines are $y=x-1$ and $y=x/2+1$. Off these lines, the complete directional information isIn the positive quadrant, arrows point right above $y=x-1$ and left below it, and point up above $y=x/2+1$ and down below it. On the positive $x$-axis, trajectories tend to $(1,0)$. On the $y$-axis, $0<y<1$ flows toward the origin and $y>1$ flows upward. These signs and the saddle directions give the following sketch; the plotted curves illustrate the phase portrait, with the marked saddle curves tangent to the calculated eigenvectors.

With $y=xv$, direct differentiation givesThe homogeneous ordinary differential equation therefore has general solution $y=x(A\cos x+B\sin x)$. Chooseso $y_1(\pi /2)=0$ and $y_2(\pi )=0$. Their Wronskian in the convention $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$ is $W(x)=x^2$. There is no nonzero homogeneous solution satisfying both boundary conditions, since the first forces $B=0$ and the second forces $A=0$.

For the operator as printed, whose leading coefficient is $x^2$, the Green's function isIt satisfies the endpoint boundary conditions, is continuous across $x=s$, and its first derivative has jump $1/s^2$. Multiplying that jump by the leading coefficient $s^2$ produces the unit source required by a Green's function. This factor matters because the operator was not normalized to leading coefficient one; it is an instance of a Green function under dependent-variable rescaling.

Thus $y(x)={\int }_{\pi /2}^{\pi }G(x,s)f(s)ds$. For $f(s)=s^3$, split the integral at $s=x$ to obtainThereforeThe substitution above gives $v^{\prime \prime }+v=1$, so the forcing is indeed $x^3$, and both endpoint values vanish.

The contour integral along a piecewise smooth path is defined by parametrization:with a sum over the smooth pieces. The residue theorem states that for a positively oriented simple closed contour and a function holomorphic on and inside it apart from finitely many isolated singularities $a_j$ inside,More generally a closed contour uses the winding number of each singularity as its weight.

Set $z=e^{it}$, so $dt=dz/(iz)$ and $\cos t=(z+z^{-1})/2$. The required real integral becomesThe denominator factors as $(2z^2+1)(z^2+2)$. Only the simple poles $z=\pm i/\sqrt{2}$ lie inside the unit circle. At either one,The residue theorem now gives