The trace is invariant under a similarity transformation, by cyclicity of the trace:Hence equivalent representations have the same character of a representation.

If $f$ has no explicit $y$ dependence, the Euler-Lagrange equation becomes $d(f_{y^{\prime }})/dx=0$. ThusThe variable $y$ is then a cyclic coordinate. This is the case used for the azimuth in the ramp calculation.

If the integrand has no explicit $x$ dependence, differentiate $f-y^{\prime }{f}_{y^{\prime }}$ along a stationary path:The last equality uses $f_x=0$ and the Euler-Lagrange equation. This proves the Beltrami identity, the first integral $\boxed{f-y^{\prime }{f}_{y^{\prime }}=C}$.

For the given weighted inner product, a self-adjoint operator $L$ obeys $L=L^{†}$, including equality of its domain with the adjoint domain. In the differential-equation calculation this requiresfor all functions in a specified common domain, with endpoint conditions that remove the boundary form. Equality of these inner products by itself describes symmetry; the domain conditions complete the self-adjoint realization.

A Hermitian matrix obeys $H^{†}=H$, where the dagger is conjugate transpose. If $Hv=\lambda v$ with $v\ne 0$, then $v^{†}Hv=\lambda v^{†}v$. This scalar is real because its complex conjugate equals $v^{†}{H}^{†}v=v^{†}Hv$. Since $v^{†}v>0$, the eigenvalue $\lambda$ is real.

For the Gram matrix, conjugate symmetry of the inner product givesThus $G$ is Hermitian. The assumption that the vectors form a basis additionally makes $G$ positive definite, though only Hermiticity is needed here.

Use the Fourier transform conventionThese hold directly under suitable integrability and inversion hypotheses, and extend to square-integrable functions through the Plancherel theorem. At a jump, ordinary Fourier inversion uses the mean of the one-sided limits under the usual piecewise smooth hypotheses.

By Fourier's law, insulation means zero outward heat flux. The Neumann boundary conditions are ${\theta }_x(0,t)={\theta }_x(L,t)=0$. The constant mode equals the initial integral average $L/4$. For $n\ge 1$, the cosine coefficients areOdd modes and modes divisible by four vanish; write the remaining indices as $n=2(2j+1)$. The heat flow from a triangular initial temperature with insulated endpoints givesThe nonconstant modes decay, so $\boxed{{\theta }_N\to L/4}$ uniformly as $t\to \infty$. Integrating the heat equation gives $d({\int }_0^L\theta dx)/dt=\nu [{\theta }_x{]}_0^L=0$, explaining the conserved mean. The cooled case tends instead to zero because heat leaves through its ends.

The initial triangle has nonzero endpoint slopes, so it is not compatible with the insulated condition at the corner $t=0$. The series nevertheless recovers the continuous initial temperature and satisfies the insulated condition for every $t>0$; one should not assert continuity of its spatial derivative at that initial corner.

The cooled endpoints impose Dirichlet boundary conditions $\theta (0,t)=\theta (L,t)=0$. The spatial modes are $\sin (n\pi x/L)$. Integrating the initial profile on its two straight segments givesThe heat flow from a triangular initial temperature with height $L/2$ is thereforeEvery nonzero mode decays, so ${\theta }_D\to 0$ as $t\to \infty$. There is no nonzero constant mode compatible with these endpoint temperatures.

Under an orthogonal Cartesian change of axes $Q$, the vector components transform as $v_i^{\prime }=Q_{ia}{v}_a$, while $v_k^{\prime }{v}_k^{\prime }=v_a{v}_a$ and ${\delta }_{ij}=Q_{ia}{Q}_{ja}$. Substitution givesThis is the rank-two Cartesian tensor transformation law. The nonzero assumption ensures that the denominator is positive. In matrix form $P=I-v{v}^T/(v^{T}v)$ is the orthogonal projection matrix onto the orthogonal complement of $v$.

For Cartesian tensors, let $R$ be the orthogonal matrix taking old vector components to new ones, $v_i^{\prime }=R_{ij}{v}_j$. A rank-$n$ tensor transforms by one copy of $R$ on each index:Repeated indices are summed according to Einstein notation. In general coordinates a contravariant tensor index instead carries the corresponding Jacobian matrix $\partial x^{\prime i}/\partial x^j$, and a covariant tensor index carries its inverse; in orthonormal bases of Cartesian axes their component laws coincide with the displayed form.

An isotropic tensor has identical components after every proper orthogonal transformation, so $T^{\prime }=T$ for every $R$ with $R^{T}R=I$ and $\det R=1$. Requiring reflection invariance as well is a stronger convention for odd tensor ranks. The Kronecker delta is invariant under all orthogonal transformations, while the Levi-Civita symbol is invariant under proper rotations and acquires a determinant sign under reflections.

Multiply the differential equation by $-p(x)$ and compare the coefficients of $y^{\prime \prime }$ and $y^{\prime }$ with $-(p{y}^{\prime }{)}^{\prime }=-p{y}^{\prime \prime }-p^{\prime }{y}^{\prime }$. The required integrating factor satisfies $p^{\prime }=-bp$. Thus one may takeThen $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$, the desired Sturm-Liouville form, with the same homogeneous Dirichlet boundary conditions. A common positive constant multiplying $p,q,w$ is immaterial. For the usual regular Sturm-Liouville problem one also assumes real sufficiently regular coefficients and $d>0$, ensuring $p,w>0$.

Length functional. On the unit sphere, write the position vector in spherical coordinates asIts two coordinate tangent vectors satisfy $r_{\theta }\cdot r_{\varphi }=0$, $|r_{\theta }|=1$ and $|r_{\varphi }|=\sin \theta$. Consequently the induced Riemannian metric is $d{s}^2=d{\theta }^2+{\sin }^2\theta d{\varphi }^2$. On a segment for which $\theta$ increases and $\varphi =\varphi (\theta )$,Segments where $\theta$ is not monotone require another parameter or a subdivision. Endpoint turning points can be handled by a limit.

First integral. With $F=\sqrt{1+{\sin }^2\theta ({\varphi }^{\prime }{)}^2}$, there is no explicit $\varphi$ dependence. The Euler-Lagrange equation therefore says thatis constant. For $c\ne 0$, put $k=1/|c|$ and solve:The radicand must be nonnegative along the chosen segment. The printed expression with both limits fixed at ${\theta }_1,{\theta }_2$ describes a longitude difference between endpoints; a formula for $\varphi (\theta )$ needs the variable upper limit and the additive constant shown here. The case $c=0$ gives a meridian with constant longitude. The equator is also a stationary route, although it cannot be parameterized using changing $\theta$.

Integration and geometric meaning. Put $t=\cot \theta$, so $d\theta =-dt/(1+t^2)$ andFor a nonconstant segment $k>1$, and writing $K=\sqrt{k^2-1}$, the integral becomesRearranging, and absorbing phase and sign into two constants, givesMultiplying by $\sin \theta$ shows that the corresponding Cartesian coordinates satisfy $z=Ax+By$. Thus every such stationary route lies in a plane through the sphere's center: it is a great circle, the geodesic of the sphere. The meridians and equator are the limiting or separately parameterized cases of the same description.

Rome to Singapore. Use exactly the approximate coordinates supplied in the question: $(\theta ,\varphi )=(\pi /4,\pi /12)$ at departure and $(\pi /2,7\pi /12)$ at arrival. Inserting them into the preceding relation determinesIndeed the right side is one at Rome and zero at Singapore. This corresponds to $K=1$, $k=\sqrt{2}$. In latitude $\ell =\pi /2-\theta$, the route is $\ell =\arctan [\cos (\varphi -\pi /12)]$: it starts at $45^{\circ }$ north with an initially eastward heading and then curves southward to the equator. The first-integral expression in $\theta$ has an integrable endpoint singularity at departure, where $\theta$ has a turning point; the longitude parameter is regular there.

For an explicit regular arc-length parametrization, let $a=r(\pi /4,\pi /12)$ and $b=r(\pi /2,7\pi /12)$. Their dot product is $\sin (\pi /4)\cos (\pi /2)=0$. HenceBoth $|r(s)|$ and $|r^{\prime }(s)|$ equal one. The shorter great circle arc is the global shortest path: for any unit-speed curve from $a$, its central angle $\alpha =\arccos (a\cdot r)$ satisfies $|d\alpha /ds|\le 1$ by the Cauchy-Schwarz inequality, since the tangential projection of $a$ has length $\sin \alpha$. Integrating gives length at least the endpoint angle $\pi /2$, which this route attains. For a sphere of radius $R$, the corresponding shortest distance is $\pi R/2$.

Let $y$ be a stationary function and perturb it to $y_{\epsilon }=y+\epsilon \eta$, where $\eta$ is smooth and $\eta (a)=\eta (b)=0$. These perturbations preserve the prescribed endpoint values. Differentiate the functional at $\epsilon =0$:By integration by parts,The boundary term vanishes. The remaining integral must vanish for every such $\eta$, so the fundamental lemma of the calculus of variations gives the Euler-Lagrange equationIt is a necessary stationarity condition; deciding whether the stationary path is a minimum requires additional information.

Write $v=(x+1{)}^l(x-1{)}^l$. In the Leibniz rule for its $l$th derivative at one, any term with fewer than $l$ derivatives on $(x-1{)}^l$ vanishes. The sole surviving term leaves $(x+1{)}^l$ undifferentiated and differentiates the second factor $l$ times, givingThe normalization in Rodrigues' formula consequently givesThe same argument covers $l=0$, with both zeroth powers equal to one.

The polynomial $v$ has leading term $x^{2l}$. Its $l$th derivative has leading term $(2l)!x^l/l!$, and all other terms have lower degree. Thus the Rodrigues formula for Legendre polynomials defines a polynomial with nonzero leading coefficientIt therefore has degree exactly $l$, including the constant case $P_0=1$.

Direct differentiation of $v=(x^2-1{)}^l$ gives $(x^2-1)v^{\prime }=2lxv$; the identity also holds for $l=0$. Differentiate it $l+1$ times using the Leibniz rule. The quadratic factor has only its first two derivatives nonzero, soWriting $y=v^{(l)}$ and collecting terms gives $(x^2-1)y^{\prime \prime }+2x{y}^{\prime }-l(l+1)y=0$. Multiplication by $-1$ gives the Legendre differential equation:

Choose the upward orientation. The boundary is the circle $x^2+y^2=4$ at $z=1$, oriented counterclockwise from above. By Stokes theorem, the surface integral is ${\oint }_{\partial S}G\cdot dr$. On this boundary $dz=0$ and $z^2=1$, soThe first term integrates to zero around a closed curve. By Green's theorem, the second equals the disk integral ${\int }_{x^2+y^2\le 4}(-3y)dxdy=0$, by oddness in $y$. ThusHere reversing orientation leaves the zero value unchanged.

The divergence is $\nabla \cdot F=-1+0+1=0$. Take the hemisphere's outward orientation, and close it with the unit disk $D$ at $z=0$, whose outward normal is $-e_z$. The divergence theorem then gives ${\int }_{S}F\cdot dS+{\int }_{D}F\cdot dS=0$. On the disk, $F\cdot (-e_z)=-x^2$, and polar coordinates giveConsequently the curved hemisphere has fluxThe reversed hemisphere orientation would reverse the sign.