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For $x^2/a^2-y^2/b^2=1$, let $c=\sqrt{a^2+b^2}$ and place the foci of a conic at $(\pm c,0)$. On the right branch write $x=a\cosh t$, $y=b\sinh t$. Direct expansion givesSince $c>a>0$, both square roots are positive and $d_{-}-d_{+}=2a$. On the reflected left branch the signed difference reverses, while its absolute value stays $2a$. This proves the constant-difference characterization of a hyperbola and keeps track of which branch a hyperbolic function parametrization covers.

A hyperbola is a conic section with two unbounded branches. In suitable coordinates its equation is $x^2/a^2-y^2/b^2=1$ with $a,b>0$. The unit hyperbola is parametrized by $x=\cosh t$, $y=\sinh t$ on its right branch, relating it to the hyperbolic functions.

For example, $x^2/9-y^2/16=1$ has foci of a conic at $(\pm 5,0)$, because $5^2=3^2+4^2$. Its focal-distance difference has absolute value $6$, while its eccentricity of a conic is $5/3>1$. At the point $(3,0)$, the focal distances are $8$ and $2$, verifying that difference. The asymptotes are $y=\pm 4x/3$; a distant point approaches those directions while remaining on one of the two branches. The focal-distance difference, rather than a constant sum, distinguishes this hyperbola from an ellipse.

Hyperbolic functions combine the exponential function to describe a hyperbola much as trigonometric functions describe a circle. Their two basic members are the hyperbolic cosine and hyperbolic sine, defined for every real number $t$ byTheir derivatives exchange the two functions:These formulas follow by applying differentiation to the exponential function definitions. ^[ref]

Adding and subtracting gives $\cosh t+\sinh t=e^t$ and $\cosh t-\sinh t=e^{-t}$. Multiplying these equalities proves the hyperbolic Pythagorean identity:Consequently the parametric equation $(x,y)=(\cosh t,\sinh t)$ traverses the entire right branch of the unit hyperbola $x^2-y^2=1$, where $x\ge 1$. For example, $t=0$ gives $(1,0)$, while $t=\pm 1$ gives approximately $(1.543,\pm 1.175)$. Compare $(\cos \theta ,\sin \theta )$ on the unit circle, where the squared coordinates add to one. The minus sign changes the geometry from a closed circle to an unbounded hyperbola.

The other members are ratios and reciprocals: hyperbolic tangent is $\tanh t=\sinh t/\cosh t$, hyperbolic cotangent is $\coth t=\cosh t/\sinh t$, and hyperbolic secant is $\mathrm{sech}t=1/\cosh t$; denominators must be nonzero. The hyperbolic sine and hyperbolic cosine both solve the second-order linear differential equation $f^{\prime \prime }=f$. In fact $f(t)=A\cosh t+B\sinh t$ has $f(0)=A$ and $f^{\prime }(0)=B$, making this basis convenient for an initial value problem. Their geometry also makes hyperbolic substitution useful for integrals containing $\sqrt{1+u^2}$, and the hyperbolic cosine gives the physical catenary of a hanging chain.

The identity ${\cosh }^{2}x-{\sinh }^{2}x=1$ implies

The identity follows directly from the exponential definitions of the hyperbolic functions.

The inverse hyperbolic functions invert the hyperbolic functions, with real branches on their standard ranges and complex branches obtained through analytic continuation. For example, $\mathrm{arsinh}z=\log (z+\sqrt{z^2+1})$ on a chosen branch; the multivalued inverse hyperbolic sine records all inverse values.

Choose an inertia unit $I_{\ast }$ and an angular speed unit ${\Omega }_{\ast }>0$, so physical principal moments of inertia are $I_{\ast }(1,2,3)$, physical angular velocity is ${\Omega }_{\ast }\mathbf{\omega }$, and normalized time is $t={\Omega }_{\ast }{t}_{\mathrm{physical}}$. The Euler equations for a torque-free rigid body becomeThis is a physically admissible ideal planar mass distribution: $I_3=I_1+I_2$ is the perpendicular axis theorem. A book of finite thickness need not have these exact ratios.

For an original comparison, start with angular velocity $(1,\epsilon ,0)$, $(\epsilon ,1,0)$ or $(\epsilon ,0,1)$, respectively. With $\epsilon =0.01$, the transverse Euclidean norm stays small about axes one and three, but becomes order one about axis two. For the intermediate-axis start, its linearization givesThe hyperbolic functions contain a growing exponential function with rate ${\lambda }_2=1/\sqrt{3}$. Reducing $\epsilon$ tenfold delays reaching the same small transverse amplitude by approximately $\sqrt{3}\log 10\simeq 3.99$ normalized time units, once the growing term dominates. That delays the instability. Once the transverse angular velocity is comparable to the main component, the linearization ceases to describe the motion.

The hyperbolic-sine infinite product gives, by taking its logarithmic derivative,The identity holds away from the poles, with locally convergent sums. For real $z>0$, it makes $(z\coth z-1)/z^2$ positive and strictly decreasing. It is useful for Hartmann flow flux monotonicity and small-argument expansions of hyperbolic functions.

Introduce the positive dimensionless quantity $x=(1-\Omega )a/\Omega$ and the time scale $A=\Omega /[H_0(1-\Omega {)}^{3/2}]$. The Friedmann equation gives the separable relationFor $x=(\cosh \theta -1)/2={\sinh }^2(\theta /2)$, $\theta \ge 0$, the hyperbolic function identities giveChoose the constant of integration so that $t=0$ at $a=0$. The resulting parametric equation for the open matter-dominated Friedmann solution isFor $\theta \ll 1$, the Taylor series giveEliminating $\theta$ yields $a^{3/2}\simeq (3/2)H_0\sqrt{\Omega }t$, exactly the early matter-dominated universe result. For $\theta \gg 1$, $\sinh \theta \sim \cosh \theta \sim e^{\theta }/2$, whereas $\theta /e^{\theta }\to 0$. The ratio of the two parametric expressions tends to $a/t=H_0\sqrt{1-\Omega }$, recovering the curvature-dominated universe result including its coefficient.

In the integral from part (d), set $y={\sinh }^{2}u$. Its upper endpoint is $u_x=\mathrm{arsinh}\sqrt{x}$, equal to $\theta /2$ in part (b). The hyperbolic function identitygives the antiderivativeMultiplication by ${\delta }_1$ therefore yieldsAt small $x$, cancellation between the terms leaves the Taylor series ${\delta }_2=2x/5-8x^2/35+16x^3/105+O(x^4)$. At large $x$, $\mathrm{arsinh}\sqrt{x}=\frac{1}{2}\ln (4x)+O(x^{-1})$, so the correction to one is $[3-\frac{3}{2}\ln (4x)]/x+O(\ln x/x^2)$. Both checks agree with part (d); the integral or the small-$x$ Taylor series is preferable to subtracting large nearly equal terms in a numerical evaluation near zero.

For a positive eigenvalue write $\mu =\sqrt{\lambda }>0$. The general solution of the limiting constant-coefficient differential equation is $y=A\cos (\mu x)+B\sin (\mu x)$. The endpoint at zero gives $A=0$. A nonzero eigenfunction then requires $B\ne 0$, and the endpoint at $\pi$ gives $\sin (\mu \pi )=0$. ConsequentlyFor real eigenfunctions the constants may be any nonzero real amplitudes. Requiring unit L2 norm would set $|C_k|=\sqrt{2/\pi }$, but that normalization is not required by the question. The lowest eigenvalue is ${\lambda }_1=1$ and its eigenspace is spanned by $\sin x$. For completeness, zero yields only an affine solution annihilated by both endpoints, and negative eigenvalues yield hyperbolic functions whose two zero endpoint values also force the zero solution.

The addition formulas for the hyperbolic functions implySince $M=B(y)/2$, its fourth matrix power isAlternatively, the orthonormal eigenvectors $(1,\pm 1)/\sqrt{2}$ have eigenvalues $e^{\pm y}/2$. Raising these eigenvalues to the fourth power gives the same result by spectral decomposition.

The hyperbolic functions satisfy ${\cosh }^2\theta -{\sinh }^2\theta =1$, so a useful family isFor real $\theta$ this gives a one-real-parameter family of complex orthogonal matrices, which is enough to exhibit the requested hyperbolic parametrization. It does not parametrize every matrix in part (ii). For example $a=b=1/\sqrt{2}$ obeys the constraint but cannot have $a=\cosh \theta$ with real $\theta$, since then $a\ge 1$.

Allowing complex $\theta$ gives the full parametrization: $(a-ib)(a+ib)=1$, so choose a complex logarithm $\theta$ with $e^{\theta }=a-ib$. Then $a=\cosh \theta$ and $b=i\sinh \theta$. A general complex pair therefore needs a complex parameter; the real-parameter restriction in the question should not be mistaken for a completeness claim.

Expand the sine of the complex number using hyperbolic functions:Its squared complex modulus isTaking nonnegative square roots proves $\boxed{|\sin z|\ge |\sin x|}$. Equality holds exactly when $y=0$.

Using the hyperbolic functions identities $\sin (ix)=i\sinh x$ and $\cos (ix)=\cosh x$, divide by $i$ and put $u=e^x\ne 0$. The equation becomes $-u/2-5/(2u)=3$, henceTaking all branches of the complex logarithm givesIf $x$ is required to be real, there are no real solutions, since $e^x$ cannot be negative. The complex branches are essential to the unrestricted answer.

For two solutions $y_1,y_2$, define the WronskianDifferentiate and substitute the second-order linear differential equation:Integration proves the Abel identity, including the identically zero case:For $p=-2/x$, this is $W=C{x}^2$ on either interval excluding zero, since the integral uses $\log |x|$. The coefficient singularity at zero must not be ignored.

Multiply the particular equation by $x^2$ and substitute the Frobenius solution $y=x^{\lambda }{\sum }_{n\ge 0}{a}_n{x}^n$. With $a_{-1}=a_{-2}=0$, equating coefficients givesThe leading coefficient is nonzero, so the indicial equation is $\lambda (\lambda -3)=0$. Thus only $\lambda =0,3$ are possible, and the recurrences below construct both.

For $\lambda =0$, $a_1=0$, $a_2=-a_0/2$, and the $n=3$ recurrence reads $0=a_1=0$, leaving $a_3$ free. For all later indices,This is an undetermined coefficient at Frobenius resonance; choosing $a_3=0$ selects a convenient even basis solution but is not forced by the differential equation.

For $\lambda =3$, all odd-indexed coefficients vanish andThe factorial denominators show convergence for every finite $x$. Expanding the given hyperbolic functions directly yieldsConsequently the exponent-zero family is $a_0(\cosh x-x\sinh x)-3a_3(\sinh x-x\cosh x)$, and the normalized exponent-three solution is $-3a_0(\sinh x-x\cosh x)$. This verifies the coefficient formulas and supplies two independent basis functions.

Finally $y_1^{\prime }=-x\cosh x$ and $y_2^{\prime }=-x\sinh x$, soThis agrees with the Abel calculation. Its vanishing at zero does not contradict independence on regular intervals: the original equation is singular there.

For the Minkowski metric $J=\mathrm{diag}(1,-1)$, preservation means $(Ax{)}^{T}J(Ay)=x^{T}Jy$ for all real vectors $x,y$. This is equivalent to $A^{T}JA=J$, or, entrywise,The restriction $a>0$ puts the first column on the right branch of the unit hyperbola, so uniquely $a=\cosh t$, $c=\sinh t$ for a real parameter $t$. The third equation makes the second column a scalar multiple of $(\sinh t,\cosh t{)}^T$: this also covers $t=0$, when $b=0$. The second equation makes that multiple $\epsilon =\pm 1$. Thus all the required matrices areThe hyperbolic function addition formulas give $H(t)H(s)=H(t+s)$, while $D_{\epsilon }H(s)=H(\epsilon s)D_{\epsilon }$. ConsequentlyThe product remains of the required form and its upper-left entry is positive. The identity is $H(0)D_1$, and the inverse is $H(-\epsilon t)D_{\epsilon }$. Together with associativity of matrix multiplication, these identities establish the group property. This is the time-orientation-preserving Lorentz group in one spatial dimension.

Use the Bromwich inversion formula in the real variable $\omega$, with $p=\gamma +i\omega$:When the relevant Laplace transforms and spatial integral transforms are explicit, every sample of this integrand can be evaluated directly, without a time-stepping approximation to the partial differential equation. Split the initial condition integral at $y=x$: the two pieces of $R_q$ are linear combinations of $e^{\pm qy}$, so truncated spatial exponential transforms suffice. For real data the negative-frequency half is the complex conjugate of the positive-frequency half.

A practical numerical integration is to truncate to $|\omega |\le \Omega$, apply an adaptive quadrature rule, and independently increase $\Omega$ and refine the quadrature. Choose $\gamma >0$ large enough to remain to the right of all singularities, but avoid an unnecessarily large $e^{\gamma t}$. Evaluate hyperbolic function ratios in scaled form; for example,This prevents overflow when $\mathrm{Re}qL$ is large. The boundary contributions at an interior point are damped by $e^{-x\mathrm{Re}q}$ or $e^{-(L-x)\mathrm{Re}q}$, with $\mathrm{Re}q\sim \sqrt{|\omega |/2}$. Near an endpoint more frequencies are needed; at the endpoint itself use the prescribed Dirichlet boundary condition. The initial condition resolvent has a generally only $O(1/\omega )$ tail. Treat its oscillatory tail accurately, subtract a known transform with the same leading term, or evaluate that contribution separately with the heat kernel.

An equally useful check is the eigenfunction expansion. DefineThe causal modal formula isIt follows either from the heat kernel formula or directly from integration by parts against a sine eigenfunction. Exponential-transform formulas evaluate the time integral explicitly when available. The initial condition contribution has a Gaussian function cutoff in $n$ for each positive time. Nonzero endpoint forcing produces a more slowly convergent sine series tail. A boundary lifting $\ell (x,t)=(1-x/L)g_0(t)+(x/L)e^{aL}h(t)$ gives $w=\ell +z$, where $z$ has zero Dirichlet boundary conditions and forcing $-{\ell }_t-a^2\ell$; expanding $z$ gives better convergence and imposes the endpoint values exactly. At very small times the method of images is often more efficient than retaining many modes.

Check convergence by increasing both the frequency cutoff and quadrature resolution, and compare with a separately truncated causal modal or image-kernel evaluation. Causality in finite-time Laplace contour inversion means a leftward contour deformation must respect both resolvent operator poles and growth of the transformed data. In particular, the transform of data extended by zero after $T$ can grow like $e^{-pT}$ in the left half-plane: for $t<T$ one cannot close that contour indiscriminately and discard its large arc. The modal forcing integral with upper limit $t$ avoids this causality error.

The boundary ray has $y/x=3/5$, and its intersection with the hyperbola is $(5/4,3/4)$. Under the given change of variables, $x^2-y^2=r^2$ and $y/x=\tanh \theta$. Thus the region becomesThe hyperbolic functions satisfy ${\cosh }^2\theta -{\sinh }^2\theta =1$, so the Jacobian determinant isIt is positive in the interior. The coordinate degeneracy at $r=0$ is a boundary set of area zero and does not affect the change of variables formula. ThereforeHere $\cosh (\log 2)=5/4$. The PDF specifies two line segments and one hyperbolic arc; the duplicated $y=0$ line in the TeX is not an extra boundary condition.

The wall removes the far-field decay condition and instead imposes ${\psi }_{1x}(d)=0$, ${\psi }_{1y}(d)=U_1$. The force-free condition eliminates any first-order mean shear; since mean sheet tangential velocity is zero, $U_1=0$. Thus ${\psi }_1=f_d(y)\sin \theta$, whereStart with the hyperbolic function form $f_d=(a+by)\cosh y+(c+ey)\sinh y$. The lower conditions give $a=1$, $c=-b$. Defining $\Delta ={\sinh }^{2}d-d^2$, the wall conditions giveConsequently the first-order flow for Taylor-sheet swimming next to a rigid wall isFor $d>0$, $\sinh d>d$ ensures a nonzero denominator; as $d\to \infty$, this reduces to the unbounded first-order field on every fixed height interval.

For the graph of a differentiable function, the arc length is $L=\int \sqrt{1+(y^{\prime }{)}^2}dx$. Since $y^{\prime }=2x$,Use the hyperbolic substitution $2x=\sinh t$: then $dx=\frac{1}{2}\cosh tdt$ and $\sqrt{1+4x^2}=\cosh t$. The hyperbolic functions ${\cosh }^{2}t=(1+\cosh 2t)/2$ yieldsHere $\mathrm{arsinh}t=\ln (t+\sqrt{1+t^2})$ is the inverse hyperbolic sine.