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Hartmann flow is a fully developed viscous conducting-fluid flow driven along a planar channel while a uniform magnetic field crosses the walls. Induction and the Lorentz force density couple its velocity and tangential induced field. For no-slip, normal-field walls separated by $2L$, the velocity is a hyperbolic cosine profile that approaches plane Poiseuille flow at zero imposed field and develops a uniform core with Hartmann layers at large Hartmann number.

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.

For $x\ge 1$, the inverse hyperbolic cosine on its nonnegative real branch is $\mathrm{arcosh}x=\log (x+\sqrt{x^2-1})$.

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.

For the simple symmetric random walk, independence and the Rademacher distribution give the moment-generating functionThe function ${\theta }^2/2-\log \cosh \theta$ is even; for $\theta \ge 0$ its derivative is $\theta -\tanh \theta \ge 0$, since the derivative of $\tanh \theta$ is at most one and $\tanh 0=0$. Here $\cosh$ and $\tanh$ are the hyperbolic cosine and hyperbolic tangent. Hence the exponential-moment estimate is

The supplied exponential maximal bound for a symmetric random walk now gives $\mathbb{P}({\max }_{k\le n}{S}_k\ge t\sqrt{n})\le \exp (-\theta t\sqrt{n}+n{\theta }^2/2)$. For $t>0$, choose $\theta =t/\sqrt{n}$; for $t=0$, use the elementary bound by one. Thus

Fix $a>1$, choose $0<\delta <a^2-1$, and set $n_m=\lfloor (1+\delta {)}^m\rfloor$. For large $m$ these form an increasing sequence. Apply the preceding exponential maximal bound for a symmetric random walk at $n_{m+1}$ with threshold $a\sqrt{2n_m\log \log n_m}$. The resulting probability is at mostWe have $n_m/n_{m+1}\to (1+\delta {)}^{-1}$ and $\log \log n_m=\log m+O(1)$. Choose $b$ strictly between $1$ and $a^2/(1+\delta )$; the displayed probabilities are $O(m^{-b})$. By the Borel-Cantelli first lemma, almost surely, eventually ${\max }_{k\le n_{m+1}}{S}_k<a\sqrt{2n_m\log \log n_m}$. For every $n_m\le n\le n_{m+1}$, monotonicity of $n\log \log n$ for large $n$ therefore givesIntersect the probability-one events for $a=1+1/r$ to obtain the upper law of the iterated logarithm:

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.

Write $S=x_1+\cdots +x_n$. The Taylor series of the cosine isAfter expanding each power by the multinomial theorem, the coefficient of $x^{\alpha }$ has absolute value $1/\alpha !$ when $|\alpha |$ is even and is zero when $|\alpha |$ is odd. The hyperbolic cosinetherefore has exactly the absolute values of the coefficients of $f$. Thus $g$ is an entire majorant series for $f$.

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

Write $s=\log r$, so the two circular boundaries are $s=0$ and $s=2$. The zero angular mode must interpolate between $-1$ and $1$, giving $s-1=\log r-1$.

For the $n$th cosine mode, the radial factor has equal value $A_n$ at both boundaries. The unique harmonic function with those data isHence the solution of the annular Dirichlet problem isAt $r=1$ and $r=e^2$, the hyperbolic cosine quotient equals one, so the boundary conditions are satisfied term by term.