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Write $\mathbf{x}=s\mathbf{n}+\mathbf{v}$, with $\mathbf{v}\cdot \mathbf{n}=0$. The linear map sends $s\mathbf{n}$ to $as\mathbf{n}$ and sends $\mathbf{v}$ to $a\mathbf{v}+b(\mathbf{n}\times \mathbf{v})$. On the perpendicular plane, $\mathbf{v}$ and $\mathbf{n}\times \mathbf{v}$ are perpendicular vectors of equal length. ThusThe expression in parentheses is a rotation in three dimensions by $\vartheta$ about the oriented $\mathbf{n}$ axis, with the positive direction fixed by the cross product. Therefore the map scales the axial component by $a$, scales the perpendicular component by $\sqrt{a^2+b^2}$, and rotates the perpendicular plane by $\arctan (b/a)$. Since $a,b>0$, the angle lies between zero and $\pi /2$. This is an axis-angle decomposition, with different axial and transverse scale factors.

Decompose the position vector into its components parallel and perpendicular to the unit vector $\mathbf{u}$:This is an orthogonal projection, and the Pythagorean theorem gives $|\mathbf{r}{|}^2=s^2+|\mathbf{v}{|}^2$. The two surfaces are distinguished by fixing an angle in part (a), and fixing distance from an axis in part (b).

When $p=1$ but $t^2\ne 3t-2$, the dependent coefficient rows would require an impossible equality between their right-hand sides. Thus the augmented matrix has larger matrix rank than the coefficient matrix, and the linear system is inconsistent. For $p\ne 1$ part (i) already supplies a unique solution. Consequently there is no solution exactly when $p=1$ and $t\notin \{1,2\}$.

For $p=1$, the third coefficient row equals three times the second minus twice the first. The first two rows are independent, so the coefficient matrix rank is two. The same dependence must hold for the right-hand side, givingThus infinitely many solutions occur exactly for $p=1$ and $t\in \{1,2\}$. The null space has dimension one by the rank-nullity theorem, and an explicit parameterization, with $s\in \mathbb{R}$, isSubstitution verifies all three equations in each family.

The coefficient matrix has determinantA square linear system has a unique solution for every right-hand side precisely when its coefficient matrix is an invertible matrix. Here a unique solution exists exactly when $p\ne 1$, for every real $t$.

For an explicit Gaussian elimination, subtract twice the second equation from the third, then use the second equation, obtaining $x=2z+2t-t^2$ and $y=(t^2-t)/2-3z$. The first equation determinesand the preceding formulas determine $x,y$. When $p=1$ this last coefficient vanishes; the next parts determine compatibility rather than dividing by zero.

Use suffix notation, summing repeated indices, and the Levi-Civita symbol convention $(A\times B{)}_i={ϵ}_{ijk}{A}_j{B}_k$. Contracting the outer symbol with the symbol from the second cross product gives the epsilon-delta identitywhere $\delta$ is the Kronecker delta. ThusThe second term is zero because $A_p{A}_{\ell }$ is symmetric in $p,\ell$ and the Levi-Civita symbol is antisymmetric. The cyclic rearrangement of the remaining symbol has positive sign. Its contraction is the scalar triple product $A\cdot (B\times C)$. HenceThis cross products with a common vector identity requires no independence assumption, so it also covers zero or parallel vectors.

The three spherical basis vectors form an orthonormal basis: their displayed Cartesian components have unit lengths and pairwise zero dot products. For an orthonormal basis, a vector is the sum of its projections onto the basis vectors. Taking the dot product of each Cartesian unit vector with each spherical unit vector therefore gives the inverse change of basis:Equivalently, the change-of-basis matrix is orthogonal, so its matrix inverse is its transpose. These formulas use the stated angular convention for spherical polar coordinates; at a pole the tangential basis vectors still depend on the chosen azimuth.

The PDF defines this function using $\sin (1/x)$; the converted TeX contains a conflicting duplicate $\sin (1/2)$ before its later correct version. Use the PDF function. It is bounded by one and continuous at every positive $x$, although it is discontinuous at zero. Indeed the sequences $x_n=(\pi /2+2\pi n{)}^{-1}$ and $y_n=(3\pi /2+2\pi n{)}^{-1}$ tend to zero with function values $1$ and $-1$.

That single discontinuity does not prevent Riemann integrability. Given $\epsilon >0$, choose $0<\delta <\min \{1,\epsilon /4\}$. The cell $[0,\delta ]$ contributes at most $2\delta <\epsilon /2$ to the difference between the upper Darboux sum and the lower Darboux sum. On $[\delta ,1]$ the function is uniformly continuous, so choose a partition there whose oscillation sum is less than $\epsilon /2$, by the introductory proof. Joining that partition to the cell $[0,\delta ]$ givesThus the PDF's function is Riemann integrable despite its endpoint discontinuity. This is a direct application of bounded functions continuous away from one endpoint are Riemann integrable, proved here with the actual oscillation bound rather than an improper-integral argument.

For $x>0$, the function is a composition and product of continuous functions. At zero,Thus it is continuous on the whole closed interval, including its assigned value at zero. The continuous-function proof in the introductory solution applies, so this function is Riemann integrable. The vanishing factor $x$ controls the infinitely rapid oscillation at the endpoint.

Put $q=\lambda a+(1-\lambda )b$. Since $0<\lambda <1$ and $a<b$, $a<q<b$. Apply the mean value theorem separately to $[a,q]$ and $[q,b]$. There exist $x\in (a,q)$ and $y\in (q,b)$ such thatThese are the requested points, and $x<y$. Part (i) makes $f^{\prime }(x)\le f^{\prime }(y)$, so the corresponding secant slopes satisfyMultiply by the positive denominators and collect $f(q)$ to obtainThis is precisely the defining chord inequality for a convex function. The proof uses only existence and nonnegativity of $f^{\prime \prime }$, with no additional regularity hypothesis on that derivative.

Because $f^{\prime \prime }$ exists everywhere, $f^{\prime }$ is differentiable everywhere and therefore continuous everywhere; continuity of $f^{\prime \prime }$ is not required. For $x<y$, apply the mean value theorem to $f^{\prime }$ on $[x,y]$. It gives $\xi \in (x,y)$ withFor $x=y$ equality is immediate. Hence $f^{\prime }$ is nondecreasing: $x\le y$ implies $f^{\prime }(x)\le f^{\prime }(y)$.

For the similarity solution, differentiation givesChoose ${\lambda }^2\dot{\lambda }=e^{-\mu t}$. The remaining equation is $-(\eta f{)}^{\prime }=(f{f}^{\prime }{)}^{\prime }$. Its integration constant is zero for the required decaying or compactly supported profile, so $f(f^{\prime }+\eta )=0$. Where $f>0$, this gives $f^{\prime }=-\eta$. For a nonnegative population the complete profile is thereforeNormalization gives $Q=2b^3/3$, hence $b=(3Q/2{)}^{1/3}$. Integrating the scale equation with zero initial scale yieldsThusIts support satisfiesThe profile is a weak solution at its moving edges, where $n{n}_x$ vanishes. As $t\downarrow 0$ its support shrinks to the origin and its mass tends to $Q$, so its initial distribution is $Q{\delta }_0$. The bounded limiting range assumes $\mu >0$; at zero death rate the scale instead grows as $(3t{)}^{1/3}$.

Integrating the population equation over the line and using vanishing boundary flux gives $\dot{M}=-\mu M$, where $M(t)=\int n(x,t)dx$. Since $M(0)=Q$, one has $M(t)=Q{e}^{-\mu t}$. On the other hand the similarity substitution $x=\lambda (t)\eta$ givesConsequentlyThe decay condition is on the nonlinear diffusive flux $n{n}_x$; compact support also satisfies it.

Use a Fourier mode $u_m^n=z^n{e}^{im\theta }$. Put $a=4\mu {\sin }^2(\theta /2)\ge 0$. Its amplification polynomial of a multistep method isThe two roots are real, one nonnegative and one nonpositive. For $a>0$, evaluation at $1$ gives $a>0$ and at zero gives $-a/2<0$, so the positive root lies below one. The negative root is at least $-1$ exactly when the value at $-1$, namely $2-2a$, is nonnegative. Thus both roots have modulus at most one exactly for $0\le a\le 1$. At $a=0$ they are $1,0$, and at $a=1$ they are $1/2,-1$; all unit-modulus roots are simple, satisfying the multistep root condition for a multistep method. Since the largest available $a$ is $4\mu$, the Adams-Bashforth stability for centered diffusion condition isFor $\mu >1/4$, the highest-frequency mode has a root less than $-1$ and grows, proving necessity as well as sufficiency.

Write $h=\Delta x$, $k=\Delta t$ and apply the recurrence to an exact smooth solution. The centred second difference is $h^2{u}_{xx}+h^4{u}_{xxxx}/12+O(h^6)$. Expanding the previous-time second difference and using $u_t=u_{xx}$ gives the one-step residualwhere $h^2=k/\mu$ with fixed positive $\mu$. Thus the local one-step error is $O((\Delta t{)}^2)$ as stated. If local truncation error is defined after division by the step, it is instead $O(\Delta t)$ in this coupled refinement; the spatial error is only second order in $h$. This distinguishes the printed local-error convention from the second-order time accuracy of the underlying Adams-Bashforth method at fixed spatial discretization.

Under the usual smoothness assumptions for WKB approximation, the strict positive lower bound on $g$ excludes turning point. Put $S(x)={\int }_0^x\sqrt{g(s)}ds$. The oscillatory approximation is $g(x{)}^{-1/4}[A\sin (\lambda S(x))+B\cos (\lambda S(x))]$. The zero boundary value at the origin forces $B=0$, and the one at the right end gives the quantization conditionHere $n$ is large and $C$ is an arbitrary normalization constant. Nonzero solutions exist only at the eigenvalues, not for every arbitrary large $\lambda$. The approximation concerns the leading phase and envelope.

Choose an analytic square-root branch of $f$ near infinity in a sector. The leading Liouville–Green approximation iswith successive inverse-power corrections; the two signs give an asymptotic basis in suitable sectors. Since $\sqrt{f}=\sqrt{f_0}+f_1/(2\sqrt{f_0}z)+\cdots$, the exponent contains a linear term and a logarithm. For the specified $f$, $\sqrt{f}=1+1/(2z)+7/(8z^2)+\cdots$, explaining $e^z{z}^{1/2}$ and $e^{-z}{z}^{-1/2}$ on the positive real axis.

To determine the requested coefficient from the differential equation itself, put $w_1=e^z{z}^{1/2}h(z)$. Substitution givesFor $h=1+a_1/z+\cdots$, the order-$z^{-2}$ coefficient is $-2a_1-9/4$. It must vanish, giving

Insert the uniformly convergent Fourier series of $g$ into the interval formula and integrate termwise. The constant mode contributes $t{g}_0$, while for $k\ne 0$,The sum is exactly the expression from part (b), proving agreement of the transform and Fourier-series formulas.

A smooth $2\pi$-periodic function has coefficients $g_k=(2\pi {)}^{-1}{\int }_0^{2\pi }g(x)e^{-ikx}dx$ decaying faster than any power, by repeated integration by parts. Thus its Fourier series and every differentiated series converge absolutely and uniformly to the corresponding function and derivatives. Solving each spatial Fourier mode givesThe zero-frequency term is linear in time, not obtained by division by zero. Uniform convergence on compact time intervals permits termwise differentiation and verifies the initial conditions and the wave equation.

Use $\hat{f}(\xi )=\int e^{-ix\xi }f(x)dx$. For a Schwartz function, Fourier inversion is $f(x)=(2\pi {)}^{-1}\int e^{ix\xi }\hat{f}(\xi )d\xi$. For a tempered distribution define $⟨\hat{T},\varphi ⟩=⟨T,\hat{\varphi }⟩$ for Schwartz functions; this agrees with the function transform in the stated convention. The half-indicator kernel has transformwith value $t$ at zero. Transforming the wave equation gives ${\hat{u}}_{tt}+{\xi }^2\hat{u}=0$, zero initial value and initial derivative $\hat{g}$. Hence $\hat{u}=\hat{g}\sin (t\xi )/\xi$, and convolution yieldsFinite propagation speed means that the solution at $(t,x)$ depends only on initial data in this interval, so disturbances cannot travel faster than one. The formula also makes sense for any smooth $g$, regardless of its growth at infinity, because the integration interval is compact. Direct differentiation verifies the equation and both initial conditions. Equivalently, locally replace $g$ by a compactly supported smooth function agreeing throughout the relevant domain of dependence.