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Named after Charles Hermite.

A monic orthogonal polynomial has leading coefficient one. For a fixed positive weight, there is a unique monic orthogonal polynomial of each degree.

A matrix is skew-symmetric when $A^T=-A$.

The special orthogonal group isEvery element of $SO(2)$ is a rotation in the plane. Every element of $SO(3)$ fixes an axis and restricts to a planar rotation on its perpendicular plane.

A real square matrix is orthogonal when $R^{T}R=I$, equivalently when its columns form an orthonormal basis.

An orientation of a surface induces the boundary direction used in Stokes theorem. Walking in the positive boundary direction keeps an upward-oriented surface on the left.

A multiplicative subset $S$ is left Ore when, for every $s\in S$ and $r\in R$, there are $s^{\prime }\in S$ and $r^{\prime }\in R$ such that $s^{\prime }r=r^{\prime }s$.

The sequence $\lfloor n/2\rfloor$ has generating function

A first integral is a function of time, state, and derivatives that remains constant along every solution.

A linear ordinary differential equation is linear in the unknown function and its derivatives.

A differentiable real function satisfyingon an interval has $f(0)=0$ and $f^{\prime }=C(1+f^2)$, hence $f(x)=\tan (Cx)$ wherever the tangent remains finite.

A Bernoulli differential equation is a first-order ordinary differential equation of the displayed form, with $P$ and $Q$ continuous functions on an interval and a fixed real exponent $n$. A nonlinear power can hide a first-order linear differential equation: on a suitable branch, the Bernoulli power substitution $z=y^{1-n}$ removes it exactly. The transformed solution of a differential equation can be completely regular even when recovering $y$ produces finite-time blowup.

For example, the reciprocal Bernoulli equation with finite-time blowup $y^{\prime }+y=e^x{y}^2$ becomes $z^{\prime }-z=-e^x$ under $z=1/y$. The initial conditions $y(0)=1/2$ and $y(0)=-1$ select a positive trajectory blowing up at $x=2$ and a negative trajectory tending to zero as $x\to \infty$, respectively. The integrating factor exposes both behaviors with the same short calculation.

The cases $n=0$ and $n=1$ are already first-order linear differential equations: respectively $y^{\prime }+Py=Q$, with the zeroth-power term interpreted as the constant $1$, and $y^{\prime }+(P-Q)y=0$. There is no need for a nonlinear change of variable; for $n=1$, $y^{1-n}=1$ would discard the unknown entirely. For other exponents, the substitution must respect the domain of exponentiation and may lose the zero solution of a differential equation.^[ref]

A differential inequality bounds derivatives rather than specifying them exactly; comparison and integration turn it into bounds on the function.

Differentiating a parameter-dependent initial-value problem with respect to its parameter gives a linear inhomogeneous equation for the solution sensitivity, with initial data obtained by differentiating the original initial condition.

For $y_0>0$, the initial-value problemhas the continuous piecewise differentiable solutionThe vector field is not Lipschitz at zero, but the direction on each side prevents any solution from leaving zero, so this solution is unique in the piecewise differentiable class.

A boundary condition prescribes values of an unknown function or its derivatives at the boundary of the domain of a differential equation.

A boundary value problem asks for a differential-equation solution satisfying conditions at more than one point or boundary component.

A Lie point symmetry is a local one-parameter transformation of the independent and dependent variables that maps solution graphs of a differential equation to solution graphs.

A singular perturbation is a parameter-dependent problem whose limiting problem changes qualitatively as the parameter tends to zero. In an ordinary differential equation, a small coefficient of the highest derivative can reduce the order of the limiting differential equation, preventing it from satisfying all the original boundary conditions. Different length or time scales often require different asymptotic expansions.^[ref]

Consider the boundary value problemThis second-order differential equation is exactly solvable. Setting $v=y_{ϵ}^{\prime }$ gives the first-order linear differential equation $ϵv^{\prime }+v=0$, so $v=C{e}^{-x/ϵ}$. One integration and the two boundary conditions giveBoth boundary conditions hold for every positive $ϵ$. For numerical evaluation, write the same ratio as $expm1(-x/ϵ)/expm1(-1/ϵ)$, where $expm1(t)$ evaluates the exponential function minus one without catastrophic cancellation near $t=0$. Keep the denominator for finite $ϵ$: at $ϵ=0.2$ it is approximately $0.993262$, rather than exactly one.

Setting $ϵ=0$ gives $y_0^{\prime }=0$. The outer solution selected by the right boundary condition is therefore $y_0=1$, which fails the left boundary condition. The exact exponential function $e^{-x/ϵ}$ shows how that mismatch is repaired: the solution rises from zero to almost one within a boundary layer of width $x=O(ϵ)$, using Big O notation. For each fixed $x>0$, $y_{ϵ}(x)\to 1$, while $y_{ϵ}(0)=0$ always. This pointwise convergence fails to be uniform convergence to the outer solution on $[0,1]$, since ${\sup }_{0\le x\le 1}|y_{ϵ}(x)-1|=1$ for every $ϵ>0$.

The coefficient $ϵ$ is small, but its product with the second derivative need not be small. Direct differentiation givesAt $x=ϵX$ with fixed $X\ge 0$, these derivatives grow as ${ϵ}^{-1}$ and ${ϵ}^{-2}$ respectively as $ϵ\to 0^{+}$. Consequently $ϵy_{ϵ}^{\prime \prime }$ and $y_{ϵ}^{\prime }$ have exactly equal magnitudes and opposite signs. Even away from the boundary layer, their magnitudes remain equal in this example; both become exponentially small at each fixed positive $x$. The outer expansion neglects an exponentially small transient there, rather than establishing that one derivative term is smaller than the other everywhere.

To predict the width without knowing the exact solution, suppose an order-one change occurs over distance $\delta$. The derivative scales are $y^{\prime }=O({\delta }^{-1})$ and $y^{\prime \prime }=O({\delta }^{-2})$. The dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives $\delta \sim ϵ$. The resulting inner variable $X=x/ϵ$ makes the boundary layer visible at a fixed scale. This is the useful feature of singular perturbation: an apparently disappearing term identifies the missing structure when one measures its derivatives on the appropriate scale.

For continuous $\varphi :\mathbb{R}\to \mathbb{R}$, the initial-value problemhas a local continuously differentiable solution. If $\varphi (0)=0$, use the constant solution; otherwise locally invert $F(x)={\int }_0^{x}du/\varphi (u)$.