= Solution
Away from the velocity-profile corners, $U''=0$ and an unstable <normal mode> satisfies $\phi''-k^2\phi=0$. On the upper half of the jet write
$$
\phi=A\cosh ky+B\sinh ky\quad(0<y<1),\qquad
\phi=D e^{-k(y-1)}\quad(y>1).
$$
The exterior solution is selected by decay at infinity. From the given velocity components, the perturbation <pressure> is
$$
p=-(U-c)\phi'+U'\phi.
$$
Continuity of normal velocity and <pressure> therefore gives continuity of $\phi$ and <pressure matching at a piecewise-linear shear interface>:
$$
[(U-c)\phi'-U'\phi]=0,
\qquad (U-c)[\phi']=[U']\phi.
$$
At $y=1$ the slope of $U$ jumps from $-1$ to zero. Using $\phi'(1^+)=-k\phi(1)$ yields
$$
c\,[\phi'(1^-)+k\phi(1)]=\phi(1),\qquad
ck e^k(A+B)=A\cosh k+B\sinh k.
$$
The lower interface then follows by parity.
For the even <sinuous mode of a planar jet>, the derivative can jump at the center: being an <even function> does not require the one-sided derivative to vanish. The slope jump at $y=0$ is $-2$, whereas $[\phi']=2\phi'(0^+)$. Thus
$$
(1-c)\phi'(0^+)+\phi(0)=0,
\qquad A+k(1-c)B=0.
$$
Substitute $A=-k(1-c)B$ into the upper-interface relation. With $q=e^{-2k}$, division by $e^k$ and multiplication by $2k$ give
$$
2kc(1-k+kc)+k(1-c)(1+q)-(1-q)=0.
$$
Collecting powers of the <phase velocity> gives the required <dispersion relation>:
$$
\boxed{2k^2c^2+k(1-2k-e^{-2k})c-[1-k-(1+k)e^{-2k}]=0.}
$$
These equations can also be treated as a homogeneous two-by-two system in $A,B$, avoiding division by any potentially zero <amplitude> or phase-velocity factor.
For the odd <varicose mode of a planar jet>, $\phi(0)=0$, so $A=0$; the central <pressure matching at a piecewise-linear shear interface> is automatic. The upper-interface equation becomes $ck e^k=\sinh k$, and therefore
$$
\boxed{2kc-(1-e^{-2k})=0,\qquad
c=\frac{1-e^{-2k}}{2k}.}
$$
This <phase velocity> is real for every $k>0$, so the discrete <varicose mode of a planar jet> is neutral and there is no exponentially unstable odd mode.
To determine the unstable even modes, factor the reduced <discriminant> of the quadratic:
$$
\begin{aligned}
D(k)&=(2k-3)^2-2(2k+5)e^{-2k}+e^{-4k}\\
&=(2k-3-e^{-2k}-4e^{-k})(2k-3-e^{-2k}+4e^{-k}).
\end{aligned}
$$
The second factor vanishes at zero and has derivative $2(1-e^{-k})^2>0$ for $k>0$, so it is positive there. The first factor starts at $-8$, is strictly increasing, and tends to infinity. It therefore has one positive zero $k_c$. The even <phase velocities> are
$$
c_\pm=\frac{2k+e^{-2k}-1\pm\sqrt{D(k)}}{4k}.
$$
It follows that
$$
\boxed{\text{The sinuous branch is exponentially unstable exactly for }0<k<k_c,
\quad 2k_c-3-e^{-2k_c}-4e^{-k_c}=0,
\quad k_c\simeq1.832744.}
$$
For $0<k<k_c$, one member of the <complex conjugate> pair grows at rate $k c_i=\sqrt{-D(k)}/4$ and the other decays. At $k=k_c$ the speeds merge and there is no exponential growth; for $k>k_c$ both speeds are real. This is <inviscid instability of a triangular jet>. The unstable <sinuous mode of a planar jet> also satisfies <Howard's semicircle theorem> with $U_{\min}=0$ and $U_{\max}=1$: its speed lies in the upper half-disk centered at $1/2$ with radius $1/2$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-82-jet-semicircle.png]
{title=Triangular jet profile and unstable sinuous phase velocities inside the Howard semicircle}
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