The tangent half-angle substitution $t=\tan (x/2)$ gives $\sin x=2t/(1+t^2)$ and $dx=2dt/(1+t^2)$. The endpoints $x=0,\pi /2$ become $t=0,1$, respectively. HenceCompleting the square, $t^2+t+1=(t+1/2{)}^2+3/4$, then gives

Substitute $u(y,t)=Y(y)T(t)$ into the diffusion equation and divide on a region where both factors are nonzero. This gives $Y^{\prime \prime }/Y=T^{\prime }/(\nu T)$. The two sides depend on different variables, so they equal a separation constant $a$:These identities then extend through the zeros of the separated factors. The zero Dirichlet boundary conditions require $Y(0)=Y(1)=0$. For a nontrivial solution, $a$ must be negative: $a=0$ gives only a vanishing linear function and $a>0$ gives only a vanishing hyperbolic-sine solution. Writing $a=-k^2$ gives $Y=A\sin (ky)$, and $k=n\pi$, $n=1,2,\dots$. Consequently the separated modes of free viscous decay between parallel plates are $\sin (n\pi y)e^{-\nu n^2{\pi }^{2}t}$.

The first initial condition selects the lowest mode:The superposition principle for a homogeneous linear differential equation states that any linear combination of solutions is again a solution, with the same combination of homogeneous boundary conditions and initial data. It therefore givesThe higher spatial mode decays nine times as fast, so the velocity profile approaches a multiple of the first sine mode before vanishing altogether.

Substituting $x=r\cos \theta$, $y=r\sin \theta$ into the circle equation gives $r^2-2r\cos \theta =0$. Thus its nonzero radial intersection is $r_Q=2\cos \theta$, with the origin also on the circle. The tangent to a circle at the rightmost point is $x=2$, or $r_R\cos \theta =2$.

Take rays with $-\pi /2<\theta <\pi /2$. The difference of the two radial distances gives the cissoid of Diocles:Multiplying by $\cos \theta$ gives $x=2{\sin }^2\theta =2y^2/(x^2+y^2)$, hence the Cartesian equation $y^2(2-x)=x^3$. A convenient parametric curve, with $t=\tan \theta$, isThe real curve has $0\le x<2$, a cusp at the origin, symmetry about the $x$ axis, and the vertical asymptote $x=2$. Near the origin, $x\sim 2t^2$ and $y\sim 2t^3$, so its cusp has horizontal tangent direction. As $|t|\to \infty$, $x\to 2^{-}$ while $|y|\to \infty$. The circle and curve meet away from the origin at $(1,\pm 1)$.

For $x_0>0$, separation of variables gives $1/x=kt+1/x_0$, hence $x(t)=x_0/(1+k{x}_{0}t)$. Setting $x(t)=x_0/2$ gives the half-lifeThis is the same reciprocal decay law as a second-order reaction with the specified disappearance rate coefficient. A positive initial value is needed for a positive half-time; $x_0=0$ instead gives the identically zero solution.

The logarithmic series gives the first two nonzero terms:

The two factors have moduli $\sqrt{2}$ and $2$, and complex arguments $\pi /4$ and $\pi /6$. Multiplication multiplies the moduli and adds the complex arguments, givingThe displayed angle is the principal argument in radians; equivalently the argument is $75^{\circ }$ modulo $360^{\circ }$.