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Consider a nonreal phase velocity $c$, so $U-c$ never vanishes. Substitute $\varphi =(U-c)\psi$ into the Rayleigh equation for inviscid shear flow. Differentiating cancels the two terms involving $U^{\prime \prime }$, leavingFor $k>0$, the wall condition on normal velocity gives $\varphi =0$, hence $\psi =0$, at both walls. Multiply by $\overline{\psi }$ and use integration by parts to obtain the weighted identity for Howard's semicircle theorem:For a nonzero normal mode, $N=\int Wdy>0$. The imaginary and real parts of this identity give, respectively,Since $c_i\ne 0$, these implyThe first equation says that $c_r$ is a positive weighted mean of $U$, and consequently $U_{\min }\le c_r\le U_{\max }$. The division by $c_i$ is essential: the weighted-mean assertion concerns nonreal modes. A neutral mode may have a critical level where $U=c$ and the substitution for $\psi$ is singular.

Pointwise, $(U-U_{\max })(U-U_{\min })\le 0$. Integrating against the nonnegative weight $W$ and substituting both identities yieldsCompleting the square proves Howard's semicircle theorem:Thus all nonreal phase velocities lie in the closed disk with the real interval $[U_{\min },U_{\max }]$ as diameter. For the convention $e^{ik(x-ct)}$ and $k>0$, exponentially growing normal modes occupy its upper semicircle, while decaying modes occupy its lower semicircle.

Away from the velocity-profile corners, $U^{\prime \prime }=0$ and an unstable normal mode satisfies ${\varphi }^{\prime \prime }-k^2\varphi =0$. On the upper half of the jet writeThe exterior solution is selected by decay at infinity. From the given velocity components, the perturbation pressure isContinuity of normal velocity and pressure therefore gives continuity of $\varphi$ and pressure matching at a piecewise-linear shear interface:At $y=1$ the slope of $U$ jumps from $-1$ to zero. Using ${\varphi }^{\prime }(1^{+})=-k\varphi (1)$ yieldsThe 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 $[{\varphi }^{\prime }]=2{\varphi }^{\prime }(0^{+})$. ThusSubstitute $A=-k(1-c)B$ into the upper-interface relation. With $q=e^{-2k}$, division by $e^k$ and multiplication by $2k$ giveCollecting powers of the phase velocity gives the required dispersion relation: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, $\varphi (0)=0$, 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 thereforeThis 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: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 areIt follows thatFor $0<k<k_c$, 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 $1/2$.

First suppose $c_i\ne 0$ and put $F=\hat{w}/(U-c)$. The wall condition gives $F(\pm L)=0$. Substitution into the Rayleigh equation for inviscid shear flow givesMultiply by $\overline{F}$ and use integration by parts. The resulting weighted identity for Howard's semicircle theorem isIts imaginary part gives $\int UQ=c_{r}I$, and its real part then gives $\int U^{2}Q=(c_r^2+c_i^2)I$. Set $a=U_{\min }$ and $b=U_{\max }$. Since $(U-a)(b-U)\ge 0$,Completing the square proves Howard's semicircle theorem:In particular every growing normal mode lies in the upper semicircle for $k>0$. If $c$ is real and outside $[a,b]$, $F$ is again regular, but the same identity has a strictly positive integrand unless the mode vanishes. Such a neutral mode is impossible. Consequently neutral regular modes also satisfy the bound; no division by $c_i$ is needed for this last conclusion. The nonreal proof avoids critical-level singularities altogether.

Put $\varphi =(U-c)F$. The Rayleigh equation for inviscid shear flow becomesMultiplication by $F^{\ast }$ and integration givesFor $c_i>0$, its real and imaginary parts implyThus $c_r$ is a weighted mean of $U$ and $c_i^2$ is its weighted variance. If $U_{-}\le U\le U_{+}$, the sharp bounded-variable variance estimate givesCompleting the square proves Howard's semicircle theorem:

For a nonreal regular Rayleigh equation for inviscid shear flow mode vanishing at the walls, put $F=\hat{w}/(U-c)$. The equation becomes $[(U-c{)}^2{F}^{\prime }{]}^{\prime }-k^2(U-c{)}^{2}F=0$. Multiplication by $\overline{F}$ and integration by parts gives the displayed identity. With $Q=|F^{\prime }{|}^2+k^2|F{|}^2$, its imaginary and real parts give $\int UQ=c_r\int Q$ and $\int U^{2}Q=|c{|}^2\int Q$. Since $(U-U_{\min })(U_{\max }-U)\ge 0$, these imply Howard's semicircle theorem.