Integrate real and imaginary parts from the lip value $z(-\pi )=i$:The identities $\sin s\cot (s/2)=1+\cos s$ andgive the explicit parametrizationAs $\theta \uparrow 0$, $x\to +\infty$ and $y\to 1-a$. This must be the downstream height $a$, soThe surface initially moves slightly left of the lip before turning right. Its leftmost point occurs at $\theta =-\pi /2$, with $x_{\min }=-(1-\log 2)/(2\pi )$. This is the half-contraction free streamline at a horizontal canal lip.

On the fluid free boundary write $q=e^{-i\theta }$, with $-\pi <\theta <0$: the lip has $q=-1$ and the downstream limit has $q=1$. ThenSubstitution into the relation from part (ii) givesThe surface lies on the top of the velocity potential strip, so its logarithm has imaginary part $\pi$, givingFinally $dz/dw=1/q=e^{i\theta }$, and thereforeThe source's typeset $\cot \theta /2$ is interpreted as $\cot (\theta /2)$, as the derivation requires.

For the velocity potential strip, $e^{\pi w/a}$ maps to the upper half-plane. The real Möbius transformationpreserves that half-plane and sends its marked endpoints to $-1,\infty ,0$ in the order $A/E,D,B/C$.

For the hodograph plane, first put $r=(q-1)/(q+1)$. Its image is the second quadrant, bounded by the negative real ray and positive imaginary ray. Squaring takes that quadrant to the lower half-plane, and negating restores the upper half-plane. Thushas exactly the same three marked values. Uniqueness of a conformal map to a half-plane with three assigned boundary values makes these maps equal. Cancel the minus signs to obtainThe sign in the intermediate map is important for identifying the upper, rather than lower, half-plane.

The lower wall has streamfunction zero. The streamline through the lip and along the upper fluid free boundary has streamfunction equal to the downstream volumetric flow rate, namely $a$ because the downstream speed is one and its depth is $a$. The complex potential of a two-dimensional potential flow therefore fillsThe chosen velocity potential origin gives $w(D)=ia$. The remote reservoir endpoints $A,E$ have velocity potential tending to $-\infty$; the downstream endpoints $B,C$ have velocity potential tending to $+\infty$.

Put $q=dw/dz=u-iv$. On the floor its values run from zero to one along the positive real axis. Above the plate the flow goes left toward the lip, so the plate velocities run along the negative real axis from zero at $E$ to $-1$ at $D$. On the fluid free boundary Bernoulli's equation and its constant pressure give $|q|=1$. Its downward velocity has $v<0$, hence $\mathrm{Im}q>0$. Consequently the hodograph plane domain isthe upper unit semicircle. Its marked boundary values are $q(A)=q(E)=0$, $q(D)=-1$, and $q(B)=q(C)=1$.

The finite corners have interior angle $3\pi /2$, hence exterior-angle parameters $-1/2$. The channel end at $Z=0$ is logarithmic. Their Schwarz-Christoffel mapping is consequentlyChoose $R(Z)=\sqrt{Z^2-1}$ analytic in the upper half-plane with $R(Z)\sim Z$ at upper infinity. In particular $R(iy)=i\sqrt{1+y^2}$ for $y>0$. A direct differentiation givesHence $f=A+CH$ with $H=R+{\tan }^{-1}(1/R)$.

The inverse trigonometric functions here is the branch obtained by analytic continuation from $Z>1$ through the upper half-plane, not an independent Cauchy principal value evaluation at each point. Its endpoint values give $H(1)=\pi /2$, $H(-1)=-\pi /2$. Imposing $f(1)=1$ and $f(-1)=-1$ therefore gives $A=0$, $C=2/\pi$:Along the positive imaginary axis, $H$ is purely imaginary and tends to $-i\infty$ as $Z\to 0$; at upper infinity $H\sim Z$. The real intervals $(0,1)$ and $(-1,0)$ trace the channel walls $x=1$ and $x=-1$ downward, while the outer intervals trace the horizontal boundary. This verifies the stated end assignments as well as the constants. The map is the Schwarz-Christoffel map of a half-plane joined to a strip.

Near a prevertex of a Schwarz-Christoffel map $X_j$, a conformal map to an interior angle $(1-{\beta }_j)\pi$ has local formIts derivative therefore has the factor $(Z-X_j{)}^{-{\beta }_j}$. On each real interval between prevertex, the argument of the product of these factors is constant, so its integral traces a straight polygon side. Passing a prevertex of a Schwarz-Christoffel map changes the argument by the prescribed exterior angle ${\beta }_j\pi$. Thushas exactly the required straight sides and corner angles, with the branches chosen in the upper half-plane. Equivalently $f^{\prime \prime }/f^{\prime }$ has simple poles $-{\beta }_j/(Z-X_j)$; removing them leaves the constant normalization determined by behavior at infinity. The additive integration constant translates the polygon and $C$ rotates and scales it. This is the Schwarz-Christoffel formula.

The automorphisms of the upper half-plane are real Möbius transformations with positive determinant, modulo a common nonzero scalar. They have three real degrees of freedom, so three prevertex of a Schwarz-Christoffel map positions may be assigned independently. The remaining positions are constrained by the side-length ratios.

If $X_n=\infty$, omit its finite factor:Since ${\sum }_j{\beta }_j=2$, the derivative behaves like $Z^{{\beta }_n-2}$ at infinity; in the coordinate $U=1/Z$, this is exactly the required corner exponent there. Thus the missing angle is encoded by behavior at infinity.

The order-${ϵ}^2$ equation isUsing $\cos t\cos 2t=(\cos 3t+\cos t)/2$ and $\cos t\sin 2t=(\sin 3t+\sin t)/2$, its resonance coefficients aresecular term are removed by setting these to zero:The eigenvalues of this two-by-two system satisfyThey are real with opposite signs for $-1/12<a_2<5/12$. Generic initial amplitudes then contain an exponentially growing slow-time component. At either leading endpoint the matrix is nonzero nilpotent matrix, so suitable initial amplitudes give linear slow-time growth; higher-order detuning decides the precise corrected boundary. Outside this interval the leading eigenvalues are purely imaginary and the slow amplitudes remain oscillatory.

Substitution into $a=1+a_2{ϵ}^2+\cdots$ gives the requested leading instability interval“Can become unbounded” permits special initial amplitudes without growth. The calculation identifies the second instability tongue of a weak Mathieu oscillator and its slow amplitudes for the second Mathieu instability tongue; it is not an assertion that every solution grows throughout the closed interval.

At order one, $({\partial }_t^2+1)y_0=0$, soAt order $ϵ$,The operator sends a constant to itself and sends $\cos 2t,\sin 2t$ to minus three times themselves. HenceAn additional homogeneous $\cos t,\sin t$ contribution can be absorbed into the definitions of $A,B$. There is no resonance forcing at this order; the first amplitude evolution therefore occurs on the chosen slower scale $T={ϵ}^{2}t$.

Treat $t,T$ as independent variables, with $T={ϵ}^{2}t$. The chain rule givesTherefore the method of multiple scales isThe unspecified $O({ϵ}^4)$ term includes the higher correction in $a$. Only the displayed terms through order ${ϵ}^2$ are needed to determine the leading slow-amplitude system.

The only nondegenerate stationary point of a phase of the phase $t^2$ is the endpoint zero. Its usual leading stationary phase method term is proportional to the amplitude there, but ${\tan }^2(0)=0$. Thus the nominal $x^{-1/2}$ stationary contribution vanishes.

Expand one order further in the local amplitude: ${\tan }^{2}t=t^2+O(t^4)$. The rescaling $t=u/\sqrt{x}$ then gives a first nonzero stationary contribution of order $x^{-3/2}$, with coefficient $-\sqrt{2\pi }/8$ in the real integral, as in part (i). Meanwhile the upper endpoint is nonstationary and has nonzero amplitude; its integration by parts contribution is $2\sin ({\pi }^{2}x/16)/(\pi x)$, of order $x^{-1}$.

Consequently the nonstationary endpoint dominates the first nonzero saddle contribution. A stationary phase method formula retaining only the usual leading saddle term would miss the answer. The full stationary phase method, with the vanishing-amplitude correction and the endpoint term, agrees with method of steepest descent. This is quadratic stationary endpoint with quadratically vanishing amplitude.

Let $b=\pi /4$ and introduceDeform the contour upward to the steepest descent contour $t=e^{i\pi /4}r$ from zero and the endpoint path $t=(b^2+is{)}^{1/2}$ from $b$. More precisely the original integral is the first path traversed forward minus the endpoint path traversed forward. Connect the truncated paths where $\mathrm{Im}(t^2)=R$ and let $R\to \infty$. That connector has exponentially small factor $e^{-xR}$, and no pole of $\tan t$ is crossed: all its poles are real and outside the original segment.

On the first path ${\tan }^{2}t=t^2+O(t^4)$, so its leading contribution isThe last equality follows by $u=x{r}^2$, $\Gamma (3/2)=\sqrt{\pi }/2$.

On the endpoint path $dt=ids/(2\sqrt{b^2+is})$, and ${\tan }^{2}b=1$. Its negative orientation gives the leading endpoint termTaking real parts and inserting $b=\pi /4$ givesThe oscillatory first coefficient can vanish at particular values of $x$; the stated remainder still holds as an additive asymptotic expansion.

Dropping the small highest-derivative term gives $y_o^{\prime }+x{y}_o=0$. The outer solution satisfying the right endpoint condition isIt fails the left endpoint condition, so a boundary layer is needed at $x=0$. If $X=x/{ϵ}^{\gamma }$, diffusion and first derivative balance when ${ϵ}^{1-2\gamma }={ϵ}^{-\gamma }$, giving $\gamma =1$.

With $y(x)=Y(X)$, the exact inner expansion after multiplication by $ϵ$ isIts leading solution is $Y=A+B{e}^{-X}$. The zero boundary value gives $B=-A$, and asymptotic matching condition as $X\to \infty$, $x\to 0$ fixes $A=e^{1/2}$. ThusThe common overlap region value is $e^{1/2}$. Add the outer and inner approximations and subtract that overlap region:The decaying inner exponential function can accommodate the left condition, whereas a right-end inner coordinate would give a mode that grows into the interval. There is no right boundary layer. The composite asymptotic expansion satisfies the left condition exactly and the right one up to the exponentially small term $e^{1/2-1/ϵ}$; the remaining approximation error is algebraic and vanishes as $ϵ\downarrow 0$.

Write $y=e^{S/\delta }$ and divide the equation by $y$:Balancing the first and third terms gives $\delta =ϵ$. With $S=S_0+ϵS_1+\cdots$, the terms of order ${ϵ}^{-1}$ and one giveFor the slow branch $S_0^{\prime }=0$, choose $S_0=0$ and obtain $S_1=-x^2/2$. For the fast branch $S_0^{\prime }=-1$, choose $S_0=-x$ and obtain $S_1=x^2/2$. Additive integration constants enter the two independent amplitudes. The two-term WKB approximation is consequentlyImposing the boundary conditions on these two approximating modes yieldsThis uses the first two exponent functions $S_0,S_1$ for both branches. The next transport equations, if desired to identify the algebraic error, give $S_2=x-x^3/3$ for the slow mode and $S_2=x+x^3/3$ for the fast mode. Thus each displayed mode has relative amplitude correction $1+O(ϵ)$ uniformly on the fixed interval; no turning point occurs because the two leading exponent roots remain distinct.

The function $e^{-x^2}/x$ is strictly decreasing from infinity to zero on $x>0$, so there is a unique positive root, which tends to infinity. Let $L=\log (1/ϵ)$. Taking logarithms givesFirst $x^2\sim L$. Put $x=\sqrt{L}+d$ with $d=o(\sqrt{L})$, expand, and cancel $L$:The leading balance gives $d=-(\log L)/(4\sqrt{L})$. Substitution leaves a residual $O((\log L{)}^2/L)$; the derivative of $x^2+\log x$ is of order $\sqrt{L}$, so the root error is $O((\log L{)}^2/L^{3/2})$. ThusThe second term is logarithmic rather than a constant multiple of $L^{-1/2}$. This is logarithmic inversion of a Gaussian divided by its argument.

First take the usual positive-parameter limit $ϵ\downarrow 0$. For real $x$, the equation implies $|x^2-1|\le ϵ$, so every root lies near $1$ or $-1$. Near the positive root,and hence $x^2=1-ϵ$ up to an exponentially small correction. Near the negative root, the corresponding limit is minus one, giving $x^2=1+ϵ$ to all algebraic orders. Expanding the square roots givesThere is one root in each neighborhood: the derivative $2x+{\mathrm{sech}}^2(x/ϵ)$ has the sign of $x$ there for sufficiently small positive $ϵ$. The bound above excludes further real roots.

If the source's unqualified $ϵ\to 0$ is read as a two-sided limit, use the exact identityThe two-sided versions therefore replace $ϵ$ by $|ϵ|$ in both displayed first corrections. The two one-sided algebraic expansions otherwise have the same reasoning.

The ratio can be evaluated directly:The assumption $|f-g|\to 0$ gives $f-g\to 0$, so continuity of the exponential function givesThe additional premise $f\sim g$ is not needed for this implication. The useful requirement is additive control of the exponents, as explained by asymptotic comparison of exponential functions.

Take $g(x)=x$ and $f(x)=x+1$. Then $f/g=1+1/x\to 1$, so $f\sim g$. However,for every $x$, and this ratio does not approach one. Thus asymptotic equivalence need not survive exponentiation. A small relative error between large exponents can leave a nonvanishing additive error. This is the distinction in asymptotic comparison of exponential functions.

For every fixed truncation index $N$, the remainder must be smaller than the last retained term:This is an asymptotic expansion relative to the sequence in part (i). The number of retained terms is fixed before taking the limit. The associated series need not converge for any nonzero parameter; its role is to give successively smaller truncation errors. If $a_n$ is an asymptotic sequence, the same condition can be stated as a remainder $O(a_{N+1})$ when the next expansion condition is available, but the displayed little o notation formulation is the defining one.