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For isotropic incompressible flow of a Newtonian fluid, the stress tensor is the displayed expression, with pressure $p$, dynamic viscosity $\mu$, velocity $\mathbf{u}$, and identity matrix $I$. In Couette flow, its off-diagonal component is the shear stress.

No slip requires fluid velocity to equal solid-boundary velocity.

At a fluid interface, normal and tangential tractions satisfy the imposed stress balance.

A steady film down a wall has a parabolic velocity profile set by gravity, ambient pressure gradient, and surface shear.

Couette flow is viscous shear flow driven by relative tangential motion of parallel boundaries. In the steady no-pressure-gradient case its velocity profile is linear.

Momentum diffuses across a distance $L$ in the characteristic time $L^2/\nu$, where $\nu$ is the kinematic viscosity.

For liquid occupying $a(t)<r<b(t)$, radial symmetry and incompressible flow implyThe kinematic boundary condition at the outer interface giveswhich expresses conservation of the liquid's area.

Uniform irrotational flow of speed $U$ past a circular cylinder of radius $a$, with zero circulation, has velocity potentialOn the cylinder, the tangential speed is $-2U\sin \theta$, so the Bernoulli equation gives

In a two-level system, let the Hamiltonian operator have matrix $H=\mathrm{diag}(0,2)$ in the orthonormal basis $|0⟩,|1⟩$. The normalized real trial quantum states and their Rayleigh quotient areThe excited-state probability amplitude is $\sin \theta$, so its probability is ${\sin }^2\theta$. Thus $E(\theta )\ge 0=E_0$, with equality if and only if the excited-state probability amplitude vanishes: $\theta \in \pi \mathbb{Z}$. The remaining quantum state is $|0⟩$ up to a global phase. At $\theta =\pi /4$ the trial energy is $1$; at $\theta =\pi /2$ it is $2$. These are upper bounds on the ground-state energy, even though neither is a minimizing trial.

The real trial family contains the ground state, so it attains the exact minimum despite not covering every complex quantum state. More generally, a family covering all normalized vectors in a two-dimensional Hilbert space gives the exact lowest eigenvalue of any Hermitian matrix by minimizing its Rayleigh quotient. For a complex trial $c_0|0⟩+c_1|1⟩$ with $|c_0{|}^2+|c_1{|}^2=1$, the same calculation gives $E=2|c_1{|}^2$: a relative quantum phase cannot change this diagonal Hamiltonian energy. This finite example is the weighted mean proof of the variational method with just two energy eigenstates.

For a normalized dilation ${\psi }_{\lambda }(x)={\lambda }^{d/2}\psi (\lambda x)$ and a homogeneous potential of degree $-n$, the energy is ${\lambda }^{2}T+{\lambda }^{n}V$. Stationarity at an eigenstate gives $2T+nV=0$.

If $V_1\ge V_2$ pointwise for Hamiltonians with the same kinetic term, using the ground state of $H_1$ as a trial state for $H_2$ gives $E_1\ge E_2$.

For the trial state $e^{-\alpha x^2/2}$, the dimensionless Rayleigh quotient isIts unique minimizer isand at the minimum $\epsilon ({\alpha }_{\ast })=2+2{\alpha }_{\ast }/3-1/{\alpha }_{\ast }$.

Forthe nodeless wavefunction $e^{-x^4/4}$ is an exact eigenstate with energy ${\hslash }^2/m$, and hence is the ground state.

For a regular one-dimensional confining potential, the bound-state eigenfunctions can be ordered by their number of nodes. The unique nodeless normalizable eigenfunction is the ground state.

The spectral arc length $\theta (U)$ is the length of the shortest closed unit-circle arc containing every eigenvalue of $U$.

If $Uv=\lambda v$ and $U$ is unitary, norm preservation gives $|\lambda |=1$.

The numerical range of an operator $A$ is

Perfect discrimination is possible exactly when some unit vector satisfies

The units of a ring $R$ form a group $R^{\times }$ under multiplication.

Every height-one prime ideal in a unique factorization domain is principal and generated by an irreducible element. Consequently, two elements contained in the same height-one prime have a common irreducible factor.