Project the initial condition onto the orthogonal spatial eigenfunctions. Their squared norms in the ordinary inner product areThe latter follows from the integrals of ${\cos }^2(n\pi x)$ and ${\sin }^2(n\pi x)$ being $1/2$, with the sine-cosine cross term zero. ThereforeFor square-integrable data the expansion converges to the data in the L2 norm at $t=0$; for $t>0$ the decaying high modes are smoothed. Initial data with zero $e^x$ projection have no growing term.

Use separation of variables in the form $u=X(x)T(t)$, with $T^{\prime }/T=X^{\prime \prime }/X=-\lambda$. Then $T=e^{-\lambda t}$ and the spatial Sturm-Liouville problem isThese real separated Robin boundary conditions give a self-adjoint differential operator, so the eigenvalues are real. All three sign cases must be considered.

If $\lambda ={\mu }^2>0$, write $X=A\cos \mu x+B\sin \mu x$. The first condition gives $B=A/\mu$, while the second reduces toThus $\mu =n\pi$, and scaling the eigenfunction gives $V_n=n\pi \cos (n\pi x)+\sin (n\pi x)$.

If $\lambda =0$, then $X=A+Bx$. The first condition gives $B=A$, and the second gives $B=A+B$, so $A=B=0$; there is no zero eigenvalue.

If $\lambda =-{\kappa }^2<0$, write $X=A\cosh \kappa x+B\sinh \kappa x$. The first condition gives $B=A/\kappa$, and the second becomesHence $\kappa =1$, giving the single negative eigenvalue $-1$ and spatial eigenfunction $e^x$. Completeness of this regular Sturm-Liouville problem now givesThis is Robin diffusion with a growing mode. The growth is compatible with the stated conditions: they prescribe the same coordinate derivative at both ends, not identical dissipative outward-normal conditions.

For $G_1=⟨X⟩$, define the one-dimensional complex group representationExplicitly $W\mapsto 1$, $X\mapsto i$, $\hat{W}\mapsto -1$ and $\hat{X}\mapsto -i$. Since $i^4=1$, the map is well-defined on exponents modulo four and preserves products. Its four images are distinct, so it is also a faithful representation. The same construction applies to either of the other cyclic subgroups.

The three order-four subgroups areIn each case the generator has powers $A,A^2=\hat{W},A^3=\hat{A},A^4=W$, so all three are cyclic and mutually isomorphic to $C_4$, a cyclic group of order four. For $G_1$ the table isThe maps $X^j\mapsto Y^j$ and $X^j\mapsto Z^j$ are explicit group isomorphisms. These are all order-four subgroups: any such subgroup contains an element other than $W,\hat{W}$, and that element has order four and already generates the subgroup. They share the order-two subgroup $\{W,\hat{W}\}$ and are each normal, being of index two.

The products in (a), together with $(-A)B=A(-B)=-(AB)$ and $(-A)(-B)=AB$, give closure of the eight-element set. Associativity is inherited from matrix multiplication, $W$ is the identity element, and the inverses from (a), with $(-A{)}^{-1}=-A^{-1}$, remain in the set. Thus the group axioms hold. This is the quaternion group $Q_8$.

In the element order specified in the question, the multiplication table isInterpreting “character of the group” as the character of a representation specified by these matrices, take the trace of each matrix:A character belongs to a representation; the underlying abstract group can have other characters.

The explicit matrices areDirect matrix multiplication givesThe reverse products have the opposite signs: $YX=\hat{Z}$, $ZY=\hat{X}$, $XZ=\hat{Y}$. Since $W$ is the identity matrix and $X^2=Y^2=Z^2=-W$,These are the usual quaternion multiplication relations in the complex matrix representation of quaternions.

The trigonometric addition formula givesDividing by $\sin \theta$ and writing $x=\cos \theta$ gives the recurrence relationThe first two members are $y_1=1$ and $y_2=2x$. If $y_n$ has degree $n-1$ and leading coefficient $2^{n-1}$, then $2x{y}_n$ has degree $n$ and leading coefficient $2^n$, whereas $y_{n-1}$ has lower degree and cannot cancel that term. Mathematical induction provesFor example the sequence begins $1,2x,4x^2-1,8x^3-4x$, agreeing with the Chebyshev polynomial of the second kind.

Put $Y(\theta )=y(\cos \theta )$, with $0<\theta <\pi$. The chain rule givesConsequently the original differential expression becomes $-Y^{\prime \prime }-2\cot \theta Y^{\prime }$. If $Y=F/\sin \theta$, direct differentiation givesTaking $F=\sin n\theta$ therefore yieldsThe apparent endpoint singularities are removable. Expansion at $\theta =0,\pi$ givesThus both endpoint regularity conditions hold. This is the Chebyshev polynomial of the second kind $U_{n-1}(x)$.

Multiply the Sturm-Liouville form for $y_n$ by $y_m$ and subtract the equation for $y_m$ multiplied by $y_n$. Integration givesThe boundary expression is zero because $p\to 0$ and the functions and first derivatives remain finite. Since $n,m$ are distinct positive integers, this provesFor the derivatives, apply integration by parts once to the same Sturm-Liouville form:Thus the new weight isThis is derivative orthogonality when the Sturm-Liouville potential vanishes. For $n=1$ the derivative is identically zero, which is consistent with the identity.

Seek an integrating factor $\mu$ such that the differential expression becomes $-(p{y}^{\prime }{)}^{\prime }$. Its leading coefficient requires $p=\mu (1-x^2)$, while the first-derivative coefficient requires $-p^{\prime }=3x\mu$. ThusChoose the positive solution $\mu =\sqrt{1-x^2}$. The Sturm-Liouville form is thereforewithThe weight is positive on the open interval. The endpoints are singular because $p$ vanishes there, but the prescribed boundedness of $y_n,y_n^{\prime }$ makes the integration-by-parts boundary terms vanish.

Introduce a constant Lagrange multiplier $\mu$ for this isoperimetric constraint and make $A-\mu (C-B)$ stationary. Its integrand is $(1-\mu )y-\mu y^{\prime 2}$, so the Euler-Lagrange equation givesThe endpoint conditions therefore force a parabola $y=Kx(1-x)$. Its area and budget areSolving the quadratic and choosing positive area givesThis requires $B>0$; no positive-area barrier is possible for a nonpositive budget. The corresponding multiplier is $\mu =1/(1+4K)>0$.

To verify a global maximum rather than just stationarity, use the sharp area-cost bound for a barrier with fixed endpoints. For $w=x(1-x)$, integration by parts gives $2A={\int }_0^1{y}^{\prime }{w}^{\prime }dx$, and ${\int }_0^1{w}^{\prime 2}dx=1/3$. The Cauchy-Schwarz inequality impliesFor $A>0$ the last expression increases strictly, yielding exactly the boxed upper bound. Equality requires $y^{\prime }$ proportional to $w^{\prime }$, hence $y=Kw$ after imposing the endpoints. Thus the displayed optimum is unique and lies above the baseline throughout the interior.

The Fermat principle makes the optical path length stationary. For a graph $y=\xi (x)$ the variational integrand isThe Euler-Lagrange equation isUse the chain rule $dn(x,\xi (x))/dx=n_x+n_y{\xi }^{\prime }$ and differentiate ${\xi }^{\prime }/\sqrt{1+{\xi }^{\prime 2}}$ to obtainMultiplying through and combining the $n_y$ terms gives the required second-order ordinary differential equation with no derivatives in denominators:All partial derivatives of the refractive index are evaluated along the ray.

A finite point $x_0$ is an ordinary point when the normal-form coefficients $p,q$ are analytic at $x_0$. Then arbitrary values $y(x_0),y^{\prime }(x_0)$ determine a convergent Taylor series, and two independent choices give two linearly independent local analytic solutions.

A singular point is a regular singular point when $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ extend analytically to $x_0$. The Frobenius method then provides at least one solutionwith a locally convergent analytic factor and an exponent fixed by the indicial equation. If its two indicial roots differ by a noninteger, there are two independent Frobenius solutions. For integer-separated roots a second solution may include a logarithm; a repeated root gives a Logarithmic Frobenius solution accompanying the single pure series. Such solutions need not be finite or ordinary Taylor series at the singular point.

Set $w=z-1$. Partial fraction decomposition givesExpand both terms as geometric series:The two bracketed terms are complex conjugates. Using $1\pm i=\sqrt{2}{e}^{\pm i\pi /4}$ produces the required real coefficients:The initial terms are $1-\frac{1}{2}{w}^2+\frac{1}{2}{w}^3-\frac{1}{4}{w}^4+\cdots$. The nearest singularities are the simple poles $z=\pm i$, both at distance $\sqrt{2}$ from the center. Hence the Taylor series has