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Hyperbolic functions are combinations of the exponential function analogous to trigonometric functions. The hyperbolic cosine and hyperbolic sine satisfy ${\cosh }^{2}x-{\sinh }^{2}x=1$ and parametrize a unit hyperbola.

Use a power-series solution of a differential equation, $y={\sum }_{n\ge 0}{c}_n{x}^n$. Comparing coefficients givesIn particular $c_0=0$, while $c_1$ and $c_2$ are free. The two prescribed derivative pairs select $(c_1,c_2)=(a,0)$ and $(0,b)$. The next coefficients are $c_3=c_1/2$, $c_4=c_2/6$, $c_5=c_1/24$, $c_6=c_2/120$. HenceThese are the first three nonzero terms when the respective parameter is nonzero; if $a=0$ or $b=0$, the corresponding solution vanishes identically.

For closed forms, set $y=xu$. Cancellation of the first-derivative terms reduces the equation, for $x\ne 0$, to $u^{\prime \prime }-u=0$. The analytic solutions extend across zero, and their hyperbolic cosine and hyperbolic sine expansions identifyThis is the hyperbolic reduction of a regular-singular differential equation: although the original leading coefficient vanishes at zero, both selected power-series solutions of a differential equation are entire.

Using the exponential definition of the hyperbolic cosine and the complex exponential function,ThusFor real $x$, the hyperbolic sine vanishes only at $x=0$. Also $|y|\le 1<\pi$, so the only zero of sine available to $y$ is $y=0$. The product therefore vanishes exactly when $b=0$ or $4-b^2=0$:There are three possible parameter values. At $b=0$ the roots are $\pm i$ and $\cosh (\pm i)=\cos 1$ is real; at $b=\pm 2$ the roots themselves are real.

Integrating the resistive induction equation once introduces a constant $C$:Because $b(L)=b(-L)=0$, integrating this relation again over the channel gives $C=B_{0}Q/(2L)$, where $Q$ is the volumetric flow rate per unit span. Put $m^2=B_0^2/({\mu }_0\rho \nu \eta )$ and $H=mL$, taking $B_0\ge 0$ without loss of generality. Reversing the imposed field reverses $b$ but leaves $u$ and $Q$ unchanged. Eliminating $b^{\prime }$ from the magnetohydrodynamic momentum equation givesIts two zero wall values remove the odd homogeneous solution. Write $u=U[1-\cosh (mz)/\cosh H]$. Integrating the first relation with the two magnetic wall values yieldsTherefore $U=(G{L}^2/\nu H^2)H\coth H$. Substitution and integration produce the velocity and induced field:Here $H=B_{0}L/\sqrt{{\mu }_0\rho \nu \eta }$ is the Hartmann number. The hyperbolic cosine makes $u$ even, while the hyperbolic sine makes $b$ odd. Direct differentiation verifies both coupled equations and all four wall values.

Integrating the velocity gives the fluxThe apparent singularity at $H=0$ is removable. The small-field limit is plane Poiseuille flow:The cubic coefficient of $b$ is its first-order response in $B_0$; $b$ itself vanishes at zero imposed field. For a fixed positive $G$, the flux decreases monotonically as $H>0$ increases. One exact way to see the sketch's monotonicity is the partial-fraction expansion of the hyperbolic cotangent:whose derivative is strictly negative for $H>0$. The flux has a horizontal tangent at $H=0$ and approaches zero asThe flux is even if signed $H$ is used.

For $H\gg 1$, put $\zeta =z/L$. Away from the walls,The velocity has a nearly uniform core, with thin Hartmann layers of thickness $\delta =L/H$ enforcing the no-slip boundary condition. Near the upper wall, with $s=H(1-\zeta )$ fixed, $u/U_c\sim 1-e^{-s}$ and $b/B_s\sim -1+s/H+e^{-s}$; the lower wall follows by even/odd symmetry. Thus $b$ is negative for $0<z<L$, positive for $-L<z<0$, and returns rapidly to zero at both walls. The required sketches show a decreasing flux, a flat velocity core, and an odd induced field with magnetic wall layers.

Put $t=\log (1+\sqrt{2})=\mathrm{arsinh}(1)$, so $c_K=2t/\pi$. The power series coefficients of the hyperbolic sine preprocessing areThus $f_j=|g_j|$ for every $j\ge 0$, and the series converge on the full interval. Applying coefficient-dominated entrywise positivity gives $Y⪰0$.

Every diagonal entry belongs to one of the diagonal blocks and equals $f(X_{ii})=f(1)=\sinh t$. Because $e^t=1+\sqrt{2}$ and $e^{-t}=\sqrt{2}-1$, this is $(e^t-e^{-t})/2=1$. ThereforeThe matrix lies in the elliptope and admits a Gram matrix representation by unit vectors. This preprocessing is the Krivine rounding scheme; the equality of absolute coefficients is what preserves positive semidefiniteness even though the cross-block sine coefficients alternate in sign.

The three fixed-$a$ estimates below cease to be uniform as the minimum of the phase approaches the endpoint. For the cubic endpoint-to-saddle transition, writeThe Taylor series of the hyperbolic sine gives, for bounded $s$ and fixed $l$,The cubic term controls the tail, so localization of the Laplace integral gives the uniform leading formulaHere $I$ is the cubic Laplace transition integral, equal to $\pi \mathrm{Hi}$ in terms of the Scorer Hi function.

To recover the endpoint regime, let $l\to +\infty$. Scale $s=v/(2^{1/3}l)$ in $I$; the cubic term becomes negligible and $I(-2^{1/3}l)\sim (2^{1/3}l{)}^{-1}$. Thereforewhich agrees with part (i) in the overlap $1\ll l\ll {\nu }^{2/3}$. At $l=0$, the Gamma integral gives $I(0)=3^{-2/3}\Gamma (1/3)$, recovering part (iii).

For $l\to -\infty$, set $M=-2^{1/3}l>0$. The exponent $Ms-s^3/3$ has its maximum at $s=\sqrt{M}$, with second derivative $-2\sqrt{M}$. Thus Laplace's method givesand the transition formula becomesFor $a=1-\eta$ with $\eta \downarrow 0$, part (ii) hasPutting $\eta =(-l){\nu }^{-2/3}$ reproduces both the exponential and its prefactor. For relative agreement of these leading exponentials, one may use the overlap $1\ll -l\ll {\nu }^{4/15}$, which makes $\nu {\eta }^{5/2}\to 0$. Thus the same transition integral connects all three regimes.

WriteThe first column has unit length for the indefinite quadratic form:Because $p=M_{11}>0$, there is a unique $u\in \mathbb{R}$ such thatThe two columns are orthogonal for the same form and the second has squared length $-1$:The vectors with these properties areUsing the hyperbolic cosine and hyperbolic sine, the two possible families are therefore