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Writing the velocity as $(u,v)=(y,ax)$ givesso the flow is incompressible. With the conventiona stream function isThe streamlines are its level sets. For $a>0$, they are the hyperbolaswith separatrices $y=\pm \sqrt{a}x$ and a saddle at the origin. For $a<0$, they are concentric ellipsestraversed clockwise. This is the streamline classification of a planar linear saddle or centre.

The scalar vorticity isHence the flow is irrotational exactly whenThen $(u,v)=(y,x)=\nabla \varphi$, and a velocity potential is

Vanishing net charge density does not require vanishing current. Positive and negative charge carriers can cancel in charge density while their oppositely directed motions add to a nonzero current. Charge conservation only requiresfor magnetostatics this becomes $\nabla \cdot J=0$.

Because $\nabla \cdot B=0$, one may introduce a magnetic vector potential withIt is not unique: $A+\nabla \chi$ gives the same field for any scalar $\chi$.

For the stated current,so it is consistent with stationary charge conservation. Direct calculation givesThus $J$ is a Beltrami field. For $\lambda \ne 0$, chooseThen $\nabla \cdot B=0$ and $\nabla \times B={\mu }_{0}J$. A convenient Coulomb-gauge potential issince $\nabla \times A=B$ and $\nabla \cdot A=0$. Gradient gauge terms may of course be added.

If $\lambda =0$, the current is the constant field $J=(0,J_0,0)$. The formulas involving $1/\lambda$ do not apply; one valid choice is

Multiplication by $(1-x^2{)}^{-1/2}$ puts the equation in Sturm-Liouville theory form:Multiply the equations for $T_n,T_m$ by $T_m,T_n$, subtract and integrate. The boundary term vanishes because $\sqrt{1-x^2}=0$ at both endpoints and the polynomial derivatives are bounded. Therefore, for $n^2\ne m^2$,This is the usual Chebyshev polynomial orthogonality.

Differentiating the original equation and writing $U_n=T_n^{\prime }$ givesIts self-adjoint form isThe same subtraction argument now yields, for $n^2\ne m^2$,These two relations form the Chebyshev derivative Sturm-Liouville pair.

A map $h:X\to X$ is a contraction if there is a constant $q<1$ such thatfor every $x,y\in X$.

The contraction mapping theorem states that a contraction of a nonempty complete metric space has a unique fixed point, and that the iterates from every starting point converge to it. To prove this, choose $x_0\in X$ and put $x_{n+1}=h(x_n)$. Thenso, for $m>n$,Thus $(x_n)$ is Cauchy and converges, by completeness, to some $x_{\ast }$. A contraction is continuous, soIf $y_{\ast }$ is another fixed point, thenforcing $x_{\ast }=y_{\ast }$.

For the Newton mapone has $g(r)=r$ andOn the given neighbourhood,Since $f(r)=0$, choose a closed interval $U^{\prime }$ centred at $r$ and contained in $U$ so small that $M|f(x)|/{\delta }^2\le 1/2$ there. Thenso $g(U^{\prime })\subseteq U^{\prime }$ and $g$ is a contraction on the complete interval $U^{\prime }$. The local contraction proof for Newton iteration therefore shows that $r$ is the unique fixed point of $g$ on $U^{\prime }$.

The Eisenstein criterion says that a primitive polynomialis irreducible if there is a prime $p$ such thatBy the Gauss lemma for polynomials, irreducibility over $\mathbb{Q}$ and over $\mathbb{Z}$ agree for primitive polynomials.