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Let $e=\frac{1}{2}(|A{|}^2+|B{|}^2)$. The time equations and the dot product differentiation rule give the local identityThe divergence of a cross product isCombining them yields ${\partial }_{t}e+\nabla \cdot (A\times B)=0$. On a fixed bounded region with a boundary regular enough for the divergence theorem, differentiating under the integral and applying that theorem givesHere $n$ is the outward unit normal, so the sign expresses loss of the interior quadratic quantity by outward flux. No vanishing-boundary assumption is required.

For the unit-speed curve, the unit tangent vector, curvature of a space curve, principal normal vector and binormal vector areDifferentiating $t\cdot t=1$ gives $t\cdot {\gamma }^{\prime \prime }=0$, so $t,p,b$ form a positively oriented orthonormal basis. The hypothesis ${\gamma }^{\prime \prime }\ne 0$ makes the principal normal vector well-defined.

To determine the sign of the torsion of a space curve without relying on a pictorial handedness convention, take $R>0$, $h\ne 0$, $L=\sqrt{R^2+h^2}$ and use the unit-speed circular helixWriting $u=s/L$, direct differentiation gives its Frenet frame:ConsequentlyAt $s=0$ the marked point is $(R,0,0)$, with $t=(0,R/L,h/L)$, $p=(-1,0,0)$, $b=(0,-h/L,R/L)$ and $b^{\prime }=(h/L^2,0,0)$. For $h>0$, $b^{\prime }$ points opposite to $p$ and the torsion of a space curve is positive. Replacing $h$ by $-h$ reflects the circular helix across the horizontal plane and makes the torsion of a space curve negative; then $b^{\prime }$ points in the same direction as $p$. The direction of increasing $s$ is fixed in both panels.

Assume sufficient smoothness and a regular boundary for the integrations below, with $n$ outward. Put $\delta =w-\varphi$, so $\delta =0$ on $S$. Expanding the Dirichlet energy givesBy Green's first identity, the middle integral isThus the Dirichlet principle follows with an explicit remainder:Equality means $w-\varphi$ is constant on each connected component; its zero boundary value forces that constant to be zero.

Now let $w$ solve the heat equation and set $E(t)={\int }_V|\nabla w{|}^{2}dV$. Differentiation followed by Green's first identity givesUnder time-independent Dirichlet boundary conditions, $w_t=0$ on $S$, while $w_t=\Delta w$ in $V$. ThereforeIf equality holds at a particular time, continuity makes $\Delta w=0$ throughout $V$ at that time. Its boundary values are still $f$, so uniqueness of the Dirichlet problem implies $w=\varphi$. Conversely $w=\varphi$ has $\Delta w=0$ and gives equality. This proves the full equality criterion for Dirichlet energy dissipation for the heat equation.

For the dynamic boundary condition for the heat equation, substitute $w_t=-\alpha {\partial }_{n}w$ directly into the boundary term. The interior equation is unchanged, soThis is Dirichlet energy dissipation with a dynamic heat boundary condition. It remains valid when $\alpha$ vanishes, because no division by $\alpha$ is used. The last part preserves energy monotonicity; its stationary states need not have the earlier prescribed Dirichlet data. For example, with strictly positive $\alpha$ on a connected boundary, a stationary harmonic state has zero outward normal derivative and is constant, as follows by applying Green's first identity to that state itself.

For $r>0$, define the forward and backward one-sided interval averaging operatorsand set $A_0^{\pm }f=f$. Each positive-radius average is bounded in absolute value by $m_u(f)(x)$. Although $x$ is an endpoint of its averaging interval, enlarge that interval slightly so that $x$ is interior, apply the maximal bound and let the enlargement shrink to zero.

To dominate the limiting identity too without assuming the differentiation result we are proving, use the sublinear operatorThe maximal estimate and Chebyshev inequality implyThus $S$ has weak type (1,1) and dominates every $A_r^{\pm }$, including $r=0$.

Smooth compactly supported functions are dense in $L^1$, obtained by truncating and then smoothing with a compactly supported approximate identity. For each such continuous function both averages tend to its value at every point. Applying part (ii) to the two families, and intersecting their full-measure convergence sets, gives $A_r^{\pm }f(x)\to f(x)$ almost everywhere.

The forward difference quotient of the primitive is $A_r^{+}f(x)$, and the backward quotient is $A_r^{-}f(x)$:They have the same limit, proving differentiability almost everywhere, with the stronger conclusionThis proves differentiation of an indefinite Lebesgue integral directly from the maximal estimate and density, without assuming $m_u(f)\ge |f|$ as an unproved prerequisite.

The Gaussian integral over the full vertical velocity line is $\sqrt{2\pi }{\sigma }_z$. Thus the mass density of the self-gravitating isothermal slab isNeglecting radial derivatives makes the Poisson equation for Newtonian gravity ${\Phi }_{zz}=4\pi G\rho (z)$. Substituting $\Phi ={\sigma }_z^2\varphi$ and $z=z_0\zeta$, with $z_0^2={\sigma }_z^2/(8\pi G{\rho }_0)$, givesMultiply by ${\varphi }^{\prime }$ and integrate. The central value and symmetry under $z\mapsto -z$ fix the constant of integration:On the positive-$\zeta$ side, Newtonian gravity makes ${\varphi }^{\prime }\ge 0$. Put $y=e^{\varphi /2}$; then $4(y^{\prime }{)}^2=y^2-1$ with $y(0)=1$. Integration gives $2\mathrm{arcosh}y=\zeta$, and even reflection supplies the other side. ThereforeDirect differentiation verifies both the equation and central conditions. Uniqueness of the smooth initial value problem rules out other branches meeting those conditions. The surface density is $4{\rho }_0{z}_0$, and the conventional slab scale height is $2z_0$; the factor of two is only a definition of the length scale.

Assume the coefficients are sufficiently differentiable in the region considered, with no material discontinuity or caustic at the front. Let $t=T(\mathbf{x})$ describe arrival time, $\mathbf{p}=\nabla T$ the slowness vector, and $\mathbf{n}$ the unit elastic wave propagation normal. For any field, its jump is the value immediately behind the front minus that immediately ahead. If the first nonzero jump occurs in the kth time derivative, $k\ge 2$, writeA smooth background may be added without changing these leading jumps. Since $F_j^{\prime }=F_{j-1}$, differentiation expresses spatial jumps in terms of $p_j$ and time-derivative jumps. The restriction $k\ge 2$ lets displacement field and its first time derivative remain continuous and avoids an imposed impulsive source at the front.

Insert this local expansion in $\rho {\partial }_t^2{u}_i={\partial }_j(c_{ijpq}{\partial }_p{u}_q)$. The coefficient of $F_{k-2}$ isFor isotropic elasticity the slowness acoustic matrix is $\mu |\mathbf{p}{|}^{2}I+(\lambda +\mu )\mathbf{p}{\mathbf{p}}^T$. Its longitudinal eigenvalue is $(\lambda +2\mu )|\mathbf{p}{|}^2$. The P branch therefore gives the eikonal equation and polarizationThe next coefficient, $F_{k-1}$, isThe matrix on the left is symmetric and annihilates $\mathbf{n}$. Project onto $n_i$ to obtain its solvability condition. Put $a=a^{[k]}$. The two projected terms can be evaluated without dropping gradients of mass density or moduli:These identities use $|\mathbf{n}{|}^2=1$, hence $n_q{\partial }_p{n}_q=0$, and $(\lambda +2\mu )/\alpha =\rho \alpha$. Their curvature terms cancel, leavingMultiplying by $a$ proves the requested P-wavefront discontinuity transport law,The wave amplitude here is real for a real jump; a complex wave amplitude uses the corresponding modulus-squared law. An elastic ray follows $d\mathbf{x}/ds=\mathbf{n}$, with arclength $s$ and $dt/ds=1/\alpha$. If $J$ is the infinitesimal cross-sectional area of a neighboring ray tube, $d\log J/ds=\nabla \cdot \mathbf{n}$. Thus $\rho \alpha a^{2}J$ is constant along that tube, the ray-tube conservation for a P-wave jump. This defines the geometrical spreading and shows why the wave amplitude changes with material properties even before attenuation is considered. At a caustic the simple ray-tube wave amplitude is singular and a uniform local description is needed.

Put $S=u^2+v^2$. At a nonzero equilibrium point the two equations form the homogeneous linear systemIts determinant must vanish, giving $S^2/2-(\mu +\sigma )S+{\mu }^2+{\sigma }^2-1=0$. ThusEach positive real polynomial root gives an antipodal equilibrium point pair; a negative polynomial root does not describe a real equilibrium point. A direction is specified by $\sin 2\theta =S/2-\mu$, $\cos 2\theta =S/2-\sigma$, where $u=\sqrt{S}\cos \theta$, $v=\sqrt{S}\sin \theta$. This is the polar form of the cubic confinement model. The origin is always an equilibrium point.

The origin's eigenvalues are ${\lambda }_{\pm }=\mu \pm \sqrt{1-{\sigma }^2}$. A simple zero eigenvalue occurs on ${\mu }^2+{\sigma }^2=1$, except at $\mu =0$. The symmetry of an odd function $(u,v)\mapsto (-u,-v)$ makes the generic steady bifurcation a pitchfork bifurcation. At a point $(\sigma ,{\mu }_0)$ on this circle with ${\mu }_0\ne 0$, vary $\mu ={\mu }_0+\delta$. Expanding the radius equation gives $S\sim 2{\mu }_0\delta /({\mu }_0+\sigma )$. In a signed, unit-eigenvector centre manifold coordinate, the cubic coefficient therefore has sign $-({\mu }_0+\sigma )/(2{\mu }_0)$.

On the lower semicircle ${\mu }_0=-\sqrt{1-{\sigma }^2}$ the transverse eigenvalue is negative: the pitchfork bifurcation is supercritical for $\sigma <1/\sqrt{2}$ and subcritical for $\sigma >1/\sqrt{2}$. On the upper semicircle it is supercritical for $\sigma >-1/\sqrt{2}$ and subcritical for $\sigma <-1/\sqrt{2}$ in the centre manifold direction, but its transverse eigenvalue is positive, so the bifurcating equilibrium points are not fully attracting. At $(\sigma ,{\mu }_0)=(\pm 1/\sqrt{2},\mp 1/\sqrt{2})$ the cubic vanishes. The radius equation instead gives $\delta =-S^2/(4{\mu }_0)+o(S^2)$, a nonzero quintic degeneracy with amplitude proportional to $|\delta {|}^{1/4}$: these are degenerate pitchforks in the cubic confinement model.

A Hopf bifurcation occurs at $\mu =0$, $|\sigma |>1$, with frequency $\sqrt{{\sigma }^2-1}$. To determine its direction, first note that the divergence is $2\mu -2S$. The Bendixson-Dulac criterion excludes every nonconstant periodic orbit for $\mu \le 0$, since this divergence is negative except possibly at the origin. We can also calculate a nonzero saturation coefficient. Set $a=|\sigma +1|$, $b=|\sigma -1|$ and $V=a{u}^2+b{v}^2$. At $\mu =0$ the linear orbits are ellipses of constant $V$. Direct differentiation givesOn a linear ellipse, write $u=\sqrt{V/a}\cos \varphi$, $v=\sqrt{V/b}\sin \varphi$. The angular speed is constant; the averages of $u^{3}v$ and $u{v}^3$ vanish, while $⟨S⟩=V|\sigma |/({\sigma }^2-1)$. Averaging the cubic terms, or removing their oscillatory parts by a periodic change of coordinates in its normal form, gives for $\varrho =\sqrt{V}$The cubic coefficient is strictly negative. Hence a small attracting limit cycle exists for $\mu >0$, andFinally $(\sigma ,\mu )=(\pm 1,0)$ have a nonzero nilpotent matrix as their linearization with two zero eigenvalues. They are reflection-symmetric double-zero bifurcations where the Hopf bifurcation and steady thresholds meet; an ordinary nonzero-frequency Hopf bifurcation calculation does not apply there. These, the two degenerate pitchfork bifurcation points, and the generic pitchfork bifurcation/Hopf bifurcation loci above exhaust the origin's local loss-of-hyperbolicity possibilities.

The trace and determinant of the origin's linearization are $2\mu$ and ${\mu }^2+{\sigma }^2-1$. Thus its open region of linear asymptotic stability isFor $|\sigma |<1$ this means below the lower semicircle; for $|\sigma |\ge 1$ all negative $\mu$ lie in the open stable region. The origin is a saddle equilibrium inside the unit circle and a source outside it with positive $\mu$.

For nonzero equilibrium points, put $\Delta =2-(\mu -\sigma {)}^2$. Existence requires $\Delta \ge 0$ and $\mu +\sigma +\sqrt{\Delta }>0$. An equivalent region isInside the circle exactly one radius polynomial root is positive, giving two equilibrium points. Outside it, with $\mu +\sigma >0$ and $\Delta >0$, both polynomial roots are positive, giving four. On $\Delta =0$ with $\mu +\sigma >0$, the two radii coalesce in a saddle-node bifurcation at each antipodal point. On the circle a zero radius is the origin and is excluded from the nontrivial count. These distinctions are shown in the parameter sketch.

At $(\sigma ,\mu )=(1/\sqrt{2},-1/\sqrt{2})$ the linearization has one simple zero and one negative eigenvalue, while the cubic centre manifold coefficient vanishes. The quintic coefficient is negative. It is a codimension-two bifurcation with a reflection-symmetric degenerate pitchfork bifurcation: a wedge of coexistence between an attracting origin and an attracting antipodal pair opens between a subcritical pitchfork bifurcation and the nearby saddle-node bifurcation curve. The intervening smaller-radius pair consists of saddles. Indeed at a nonzero equilibrium point, direct differentiation of the polar coordinates equations givesThe smaller-radius branch has negative determinant, while in the coexistence wedge the larger-radius branch has positive determinant and negative trace.

At critical boundaries the linear stability analysis test alone is inconclusive. The origin is nonlinearly attracting at a supercritical lower pitchfork bifurcation, at its stabilizing quintic endpoint, and at the supercritical Hopf bifurcation threshold, although decay is no longer exponential. It is unstable at the subcritical lower pitchfork bifurcation. The double-zero point $(\sigma ,\mu )=(-1,0)$ is also attracting: the positive function $W=v^2+u^4/8$ hasnear the origin away from zero. Here $|u{v}^3|\le (u^2{v}^2+v^4)/2$ and $|u^{3}v|\le (u^4+u^2{v}^2)/2$ prove the bound. The other double-zero point $(1,0)$ has an unstable quartic potential, as the blow-up below shows. The shading records the open linear stability analysis regions; these boundary remarks distinguish nonlinear from exponential stability.

At $(\sigma ,\mu )=(1,0)$ the origin's matrix is $\left(\begin{array}{cc}0& 0\\ 2& 0\end{array}\right)$: it has a double zero eigenvalue and one eigenvector. Trace and determinant vary independently with $\mu$ and $\sigma$ there, so two parameters are needed. The symmetry of an odd function removes quadratic terms, giving a reflection-symmetric cubic double-zero normal form. The quadratic nonlinearity required for a generic Bogdanov–Takens bifurcation vanishes.

Use a prime for differentiation with respect to the rescaled time. Direct substitution, with all rescaled variables written without tildes, givesAt $\epsilon =0$, the equations are $u^{\prime }=-sv+v^3/2$, $v^{\prime }=2u$. Fordifferentiation gives $H^{\prime }=2u(-sv+v^3/2)+(sv-v^3/2)2u=0$. This is the Hamiltonian blow-up of a reflection-symmetric double-zero point.

For $s>0$, the equilibrium points are a centre equilibrium at $(0,0)$ and saddles at $(0,\pm \sqrt{2s})$. The centre equilibrium has eigenvalues $\pm i\sqrt{2s}$; at either saddle equilibrium they are $\pm 2\sqrt{s}$. Levels $0<H<s^2/2$ contain closed inner ovals around the centre equilibrium. The saddle equilibrium level hasThese two branches form a heteroclinic cycle connecting the saddles. The branch with $u>0$ runs from the lower saddle equilibrium to the upper one because $v^{\prime }=2u>0$; the other runs back. Outside the potential well, trajectories are open. The right panel of the preceding figure uses the coordinates $(v,u)$ to display the well and these directed connections.

Between collisions the momentum is $p_0=\sqrt{2mE}$ or $-p_0$. In phase space the cycle runs right along the upper horizontal segment from $q=0$ to $q=b$, jumps vertically to $-p_0$ at the right wall, runs left along the lower segment, and jumps back at the left wall. The jumps have $dq=0$. ThereforeDifferentiation gives $dI/dE=2bm/\sqrt{2mE}=2b/v$, exactly the out-and-back period. Under slow wall motion the action $I$ is the adiabatic invariant; a convention using $I/(2\pi )$ is equivalent. HenceCompression raises the energy and expansion lowers it, provided the wall speed is small compared with the particle speed and the change per cycle is small.

Here $dy/ds=1$, $dx/ds=u^2$, $du/ds=0$, so from initial parameter $\xi$ we get $y=s$, $x=\xi +{\xi }^{2}y$ and $u=\xi$. Solving for the branch agreeing with $u=x$ at $y=0$, and rationalizing the apparent division by $y$, givesThis formula is smooth on that open set, which contains every point of the initial line. It satisfies $x=u+y{u}^2$ and $1+2yu=\sqrt{1+4xy}>0$. Differentiation gives $u_x=1/(1+2yu)$, $u_y=-u^2/(1+2yu)$, verifying the equation and the data.

The transport field is $(u^2,1)$, with normal component one on $y=0$, so the initial line is everywhere non-characteristic. This explains why neighborhoods cover the entire initial line, although a single fixed-width strip need not work for unbounded $x$.

An entire function that is doubly periodic is bounded: its absolute value has a finite maximum on the compact square $[0,1]+i[0,1]$, and periodic translations carry every point of $\mathbb{C}$ into that square. The Liouville theorem therefore proves every such analytic function is constant.

For the zero-pole count, suppose the meromorphic function is not identically zero; otherwise isolated zero multiplicities are not defined. It has only finitely many zeros and poles modulo the lattice $\Lambda =\mathbb{Z}+i\mathbb{Z}$, because a compact square can be covered by finitely many local meromorphic neighbourhoods. Choose a translated unit square with no zeros or poles on its boundary. The logarithmic derivative $f^{\prime }/f$ is doubly periodic, so integrals on opposite sides cancel, with their opposite orientations. The argument principle givesBoth counts include multiplicities. Each half-open fundamental parallelogram contains exactly one representative of each lattice orbit, and translation by a lattice vector preserves the multiplicity of a zero or pole. Consequently the counts agree in the prescribed half-open square too, even if that particular square has boundary zeros or poles. Nonzero constants have both counts zero.

For the series, let $\Lambda =\mathbb{Z}+i\mathbb{Z}$ and write it as the Weierstrass elliptic function $\wp (z)$. On a compact set avoiding $\Lambda$, choose $R$ with $|z|\le R$. For $|w|>2R$,There are $O(j)$ lattice points with $j\le |w|<j+1$, hence ${\sum }_{w\ne 0}|w{|}^{-3}$ converges by comparison with ${\sum }_j{j}^{-2}$. The Weierstrass M-test proves absolute locally uniform convergence of the tail, and the finite remaining terms define holomorphic functions off their poles. It follows that $\wp$ is a holomorphic function off $\Lambda$. Near each $w_0\in \Lambda$, separate the term $(z-w_0{)}^{-2}$; the remaining terms converge locally uniformly to a holomorphic function. Thus $\wp$ is meromorphic on $\mathbb{C}$, with a double pole at each lattice point and no other poles.

It remains to prove the required periods, rather than merely assert them from the lattice. Termwise differentiation, justified by locally uniform convergence of these holomorphic series, givesThis series is absolutely locally uniformly convergent off $\Lambda$. For either generator $\omega =1,i$, reindexing $w\mapsto w-\omega$ proves ${\wp }^{\prime }(z+\omega )={\wp }^{\prime }(z)$. Therefore $\wp (z+\omega )-\wp (z)$ has zero derivative off $\Lambda$. Its principal parts at lattice points cancel, so it extends to an entire constant $c_{\omega }$.

The original convergent series is even: reindexing $w\mapsto -w$ gives $\wp (-z)=\wp (z)$. At $z=-\omega /2$, which is not a lattice point for either generator, evenness givesHence the series defines a meromorphic function with periods $1$ and $i$. This termwise derivative proof of Weierstrass periodicity uses an absolutely convergent derivative sum; splitting the original series into two separate inverse-square sums would not be justified.

The principal real arctangent is defined here on $x\ne 0$, so work on either of the two open half-planes. Put $r=\sqrt{x^2+y^2}$. Direct differentiation givesHence ${\varphi }_{xx}+{\varphi }_{yy}=0$: $\varphi$ is harmonic on its domain. The Cauchy-Riemann equations for $f=\varphi +i\psi$ requireIntegrating these partial derivatives gives the harmonic conjugateThe constants may be chosen independently on the two components. The harmonic angle and logarithmic radius representation supplies the holomorphic functionwhere $\mathrm{Log}$ is the branch of the complex logarithm on the right half-plane. On the left half-plane it is the angle of $-z$, rather than the principal angle of $z$, that equals $\arctan (y/x)$. This distinguishes the literal arctangent from a globally defined argument; no single-valued complex logarithm exists on the whole punctured plane.

With $C_0=0$, the level sets $\varphi =C$, for $-\pi /2<C<\pi /2$, are the two opposite rays of $y=x\tan C$ with the origin removed. On each half-plane only one ray remains. The level sets $\psi =K$ are circles $r=e^{-K}$, with their two points on $x=0$ excluded from the literal domain. The rays and circles meet at right angles, as expected for harmonic conjugates.

The contraction mapping theorem says that if $(X,d)$ is a nonempty complete metric space and $T:X\to X$ satisfies $d(Tx,Ty)\le qd(x,y)$ for a constant $0\le q<1$, then $T$ has exactly one fixed point.

For the proof, start at $x_0\in X$ and set $x_n=T^n{x}_0$. Iterating the contraction bound gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality then gives, for $m>n$,Thus $(x_n)$ is a Cauchy sequence and has a limit $x_{\ast }$ by completeness. Since the contraction is continuous, $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, then $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing distance zero. This proves existence and uniqueness.

For the integral map, the fundamental theorem of calculus shows that $Tf$ is continuous whenever $f$ is continuous. In the supremum norm,Hence $\Vert Tf-Tg{\Vert }_{\infty }\le \frac{3}{2}\max (a^2,b^2)\Vert f-g{\Vert }_{\infty }$. We can chooseThe space $C[a,b]$ with this supremum norm is complete: a uniformly Cauchy sequence has pointwise real limits, the Cauchy bound passes to those limits to give uniform convergence, and a uniform limit of continuous functions is continuous. The contraction mapping theorem therefore gives a unique continuous $y$ with $y(x)=1+{\int }_0^{x}3ty(t)dt$. The fundamental theorem of calculus makes this $y$ differentiable, with $y^{\prime }=3xy$ and $y(0)=1$. Conversely every solution of that initial-value problem obeys the integral equation, so uniqueness transfers to the ordinary differential equation. Explicitly $y(x)=e^{3x^2/2}$ on $[-1/2,1/2]$, as direct differentiation confirms.

Take the wave speed $c>0$ and introduce characteristic coordinates $\xi =x+ct$, $\eta =x-ct$. The chain rule givesThe wave equation therefore says $y_{\xi \eta }=0$. Integrating first with respect to $\eta$ gives $y_{\xi }=F(\xi )$, and integrating with respect to $\xi$ gives $y=f(x+ct)+g(x-ct)$ with twice differentiable $f,g$. Conversely substitution shows that every such sum solves the wave equation. The two terms are travelling waves of unchanged shape, moving respectively to the left and right at speed $c$.

The zero initial displacement gives $g(x)=-f(x)$, and the initial velocity gives $2c{f}^{\prime }(x)=\psi (x)$. Hence the D'Alembert formula with initial velocity isFor a classical twice differentiable solution one may take $\psi \in C^1$. Differentiation gives $y_t(x,0)=\psi (x)$ and $y(x,0)=0$ directly. The coordinate calculation also works for any nonzero signed $c$, with physical speed $|c|$. If the printed constant is allowed to be $c=0$, the equation instead gives $y=A(x)+tB(x)$ and these initial data give $y=t\psi (x)$; the displayed travelling-wave representation requires $c\ne 0$.

Let $\varphi$ denote the standard Gaussian distribution probability density function. Directly from the definitions, $S\varphi (d_1)=K{e}^{-r\tau }\varphi (d_2)$. Differentiating the price in part (ii), the terms involving derivatives of $d_1,d_2$ cancel, yielding $V_S=\Phi (d_1)$. Therefore, in units of the two traded assets, the replicating strategy isIts value is $h_t^B{B}_t+h_t^S{S}_t=V(t,S_t)$. To check self-financing explicitly, differentiation also givesHence $V_t+rS{V}_S+\frac{1}{2}{\sigma }^2{S}^2{V}_{SS}=rV$. Applying Itô formula under the physical measure givesThus the holdings form a self-financing strategy and their terminal value is the required payout. Holdings at the single maturity instant can be assigned by their limiting values; the continuous-time trading gains do not depend on that assignment. If ${\pi }_t$ denotes wealth fractions instead, the stock fraction is $S_t\Phi (d_1)/V(t,S_t)$ and the bank fraction is $K{e}^{-r\tau }\Phi (-d_2)/V(t,S_t)$. For $K\le 0$, hold one stock and no bank-account units.

Put $\xi =r/(r_0\lambda )$ and $n=n_0{\lambda }^{-2}F(\xi )$. Direct differentiation givesThus the partial differential equation reduces to an ordinary differential equation precisely when ${\lambda }^5\dot{\lambda }=C{D}_0/r_0^2$. The profile then satisfies $(\xi F^2{F}^{\prime }{)}^{\prime }=-C({\xi }^{2}F{)}^{\prime }$. Regularity and zero radial flux at the origin make the integration constant zero. Wherever $F>0$, $F{F}^{\prime }=-C\xi$, so $F^2=1-C{\xi }^2$ using $F(0)=1$. Requiring the front at $\xi =1$ selects $C=1$. Point release requires $\lambda (0)=0$, givingAlthough $F^{\prime }$ diverges at the front, $F^2{F}^{\prime }$ tends to zero, so this compactly supported profile satisfies the equation in the weak solution sense.

The compact radial cubic-diffusion profile conserves the total population:Consequently $\boxed{r_0=\sqrt{3Q/(2\pi n_0)}}$. The front radius grows as $t^{1/6}$, while the central density decays as $t^{-1/3}$. The conserved mass concentrates at the origin as $t\downarrow 0$.

Integrating the vector-field equations gives the one-parameter transformationsUnder this scaling, $v_t,v^2{v}_x,v_{xxx}$ have weights $\gamma -\beta ,3\gamma -\alpha ,\gamma -3\alpha$. Equality of all three weights givesThus the nontrivial Lie point symmetry generator can be normalized to $x{\partial }_x+3t{\partial }_t-v{\partial }_v$.

For the Miura transformation $u=v^2+v_x$, direct differentiation gives the operator identityFor example $u_x=2v{v}_x+v_{xx}$ and $u_{xxx}=6v_x{v}_{xx}+2v{v}_{xxx}+v_{xxxx}$; inserting them makes the $v_x{v}_{xx}$ terms cancel and leaves the right side. Hence $u$ solves $u_t-6u{u}_x+u_{xxx}=0$ whenever $v$ solves the mKdV equation. Both $v^2$ and $v_x$ have weight $-2\alpha$, so the corresponding Korteweg-De Vries equation symmetry is