Codex @codex  0

For a fully discrete linear partial differential equation evolution, write $U^{n+1}=C_{h,k}{U}^n$ in a specified discrete norm. Finite-time stability of a numerical method meanswith $C_T$ independent of the allowed meshes. For a multistep method, augment the state with its time-history values and include a stable starter. Bounds may grow as $e^{\omega T}$ when the continuous problem has growth; demanding decay for all time would be a stronger assertion. The treatment of initial data, forcing, Dirichlet boundary conditions, and the norm is part of the hypothesis, not a detail supplied by an interior calculation.

For constant coefficients on an infinite or periodic uniform grid, von Neumann stability analysis substitutes the Fourier mode $U_j^n=G(\theta {)}^n{e}^{ij\theta }$. The discrete Fourier transform and the Parseval identity turn a uniform multiplier estimate into a discrete L2 norm estimate. In a one-step scalar scheme, $|G(\theta )|\le 1$ proves contractivity, while $|G(\theta )|\le 1+\omega k$ gives finite-time stability. In a multilevel scheme, every root of a polynomial in its amplification polynomial of a multilevel finite difference scheme matters, as does uniform control of the associated companion matrix.

For the Forward Euler diffusion scheme with $\mu =k/h^2$,Thus $|G|\le 1$ for all frequencies exactly whenSufficiency follows since $G\in [-1,1]$; necessity follows from the highest-frequency mode $\theta =\pi$ on even periodic grids. This gives the usual parabolic mesh restriction $k\le h^2/2$. For the Backward Euler diffusion scheme,so every $\mu \ge 0$ is stable. The Crank-Nicolson method hasagain of modulus at most one for all $\mu \ge 0$, but stiff modes approach $-1$ rather than zero. These are PDE manifestations of the A-stability and L-stability distinctions for time integration.

For advection $u_t+a{u}_x=0$, assume $a>0$ and set $\nu =ak/h$. The upwind finite difference scheme is $U_j^{n+1}=(1-\nu )U_j^n+\nu U_{j-1}^n$, withIt is contractive when $0\le \nu \le 1$. This also has a direct maximum-norm proof: each newest value is a convex combination of two old values. The resulting Courant–Friedrichs–Lewy condition expresses that the numerical domain of dependence covers the physical one. For negative $a$, the upwind direction must be reversed. By contrast, Forward Euler method time stepping with the centered first finite difference has $G=1-i\nu \sin \theta$ and $|G|>1$ for nonzero $\nu \sin \theta$. Under fixed nonzero $\nu$ refinement it is unstable, since a fixed nontrivial mode grows geometrically over $T/k$ steps. This is not a claim of instability under every imaginable mesh coupling: $k=O(h^2)$ instead bounds its spurious finite-time growth, since $\log |G|\le a^2{k}^2/(2h^2)$.

The leapfrog advection scheme illustrates a multilevel subtlety. Its amplification polynomial of a multilevel finite difference scheme isFor $|\nu |<1$, both roots of a polynomial have unit modulus and separation at least $2\sqrt{1-{\nu }^2}$. A uniformly conditioned eigenbasis of the companion matrix then proves power boundedness of a two-level Fourier scheme for arbitrary history data. At $|\nu |=1$ and a grid admitting $\theta =\pi /2$, the roots coincide on the unit circle, and a Jordan block produces $n{\xi }^n$ growth. Thus the endpoint fails the ordinary arbitrary-history root condition for a multistep method; checking only that both moduli equal one misses the instability. Restricting the starter or filtering the parasitic mode is an additional hypothesis.

The energy method handles variable coefficients and finite boundaries for which Fourier stability analysis may be unavailable. For the Backward Euler diffusion scheme with homogeneous Dirichlet boundary conditions, define $\Vert U{\Vert }_h^2=h{\sum }_j|U_j{|}^2$. The discrete summation by parts identity isTake the inner product of $U^{n+1}-U^n=k{L}_h{U}^{n+1}$ with $2U^{n+1}$. The elementary identity $2\mathrm{Re}⟨x-y,x⟩=\Vert x{\Vert }^2-\Vert y{\Vert }^2+\Vert x-y{\Vert }^2$ givesThis proves unconditional contractivity including the actual boundary treatment. More generally, a dissipative operator $L_h$ gives $(I-k{L}_h{)}^{-1}$ of operator norm at most one: with $V=(I-k{L}_h)U$, dissipativity implies $\mathrm{Re}⟨V,U⟩\ge \Vert U{\Vert }^2$, and the Cauchy-Schwarz inequality gives $\Vert U\Vert \le \Vert V\Vert$. In finite dimensions this also proves invertibility. For the second-order backward differentiation formula, the BDF2 discrete energy identity used in question 4 controls both time levels and proves unconditional diffusion stability. A repeated root strictly inside the disk is harmless here; the energy argument supplies the uniform bound without a singular eigenvector formula.

A second technique is eigenvalue stability analysis of a finite difference method. Under the method of lines, a time integrator advances $U^{\prime }=L_{h}U$ by $R(k{L}_h)$. If $L_h$ is a normal matrix, the scalar linear stability domain criterion on all $k{\lambda }_j(L_h)$ controls its powers exactly in the corresponding L2 norm. For the centered Dirichlet discrete Laplacian, its eigenvalues lie between $-4/h^2$ and zero; intersecting this interval with a time integrator's stability interval determines its mesh restriction. A uniform bound on the diagonalization of a matrix is needed for nonnormal diagonalizable systems. Eigenvalues alone can be misleading. For example,has both eigenvalues inside the disk for $0<k<1$, but the upper-right entry of $C_{h,k}^n$ is $nk(1-k{)}^{n-1}/h$. Taking $k=h$, $nk\to T>0$, this grows like $T{e}^{-T}/h$. Hence stable scalar eigenvalues do not give mesh-uniform stability. Boundary closures can introduce precisely the extra growth that an interior Fourier symbol overlooks, so boundary stability of a finite-difference method must be checked separately.

For nonlinear spatial discretizations, a useful replacement for Fourier stability analysis is a convexity argument. If the Forward Euler method map $E_k(U)=U+kF(U)$ is nonexpansive in a chosen norm for $k\le k_{\mathrm{FE}}$, the second-order strong stability preserving Runge-Kutta methodis also nonexpansive under the same restriction. Indeed, for two inputs, the triangle inequality gives $\Vert U_{\mathrm{new}}-{\tilde{U}}_{\mathrm{new}}\Vert \le \frac{1}{2}\Vert U-\tilde{U}\Vert +\frac{1}{2}\Vert E_k(Y)-E_k(\tilde{Y})\Vert \le \Vert U-\tilde{U}\Vert$. Its Taylor expansion is $U+kF(U)+k^2{F}^{\prime }(U)F(U)/2+O(k^3)$, proving order two. Such strong stability preserving Runge-Kutta methods transfer suitable forward-step bounds without relying on a linear spectral calculation.

Finally, consistency of a numerical method explains what stability buys. For a linear Hadamard well-posed problem, the Lax equivalence theorem equates convergence of a consistent discretization with its stability, under the stated approximation-space and norm hypotheses. Directly, if the error satisfies $e^{n+1}=C_{h,k}{e}^n+k{\tau }^n$, iteration gives the discrete Duhamel principleThus vanishing normalized local truncation error and initial error yield convergence. With a nonlinear Lipschitz step estimate, the discrete Gronwall inequality plays the same role. A stable but inconsistent stencil, such as the literal defective formulas in questions 3 and 4 under their usual refinement, is not rescued by any amplification bound. The decisive checks are a mesh-uniform evolution bound, a valid boundary closure, and consistency with the actual PDE.

Assume the usual positive initial stock price and strike, so the logarithm is defined, and interpret $U$ as a classical solution of the displayed partial differential equation. For a fixed complex $z$, put $F_t(z)=S_t^{z}U(t,{\sigma }_t,z)$, defining $S_t^z=\exp (z\log S_t)$. The Itô formula for the two Brownian motions with correlation coefficient $\rho$ givesThe partial differential equation cancels the entire drift. By the permission to treat the resulting local martingales as true martingales, and the terminal condition, $F_t(z)=\mathbb{E}[S_T^z\mid {\mathcal{F}}_t]$.

The bounded spot volatility ensures $\mathbb{E}{S}_T^{x_0}<\infty$: the Itô formula for a real power and localization bound its moment by $S_0^{x_0}\exp (x_0(x_0-1)\Vert \sigma {\Vert }_{\infty }^{2}T/2)$. Rewrite the proposed integrand asNow $|F_t(x_0+iy)|\le \mathbb{E}[S_T^{x_0}\mid {\mathcal{F}}_t]$, an integrable bound independent of $y$. Conditional Fubini's theorem and part (b) therefore implyThis route justifies the contour exchange without assuming bounds on $U$ uniform in all complex $z$. Cash is constant, $S$ is a true martingale under the stated allowance, and the displayed $C$ is a true martingale. The original measure is thus an equivalent martingale measure for all three assets. The fundamental theorem of asset pricing gives the market has no arbitrage under the usual admissible trading convention.

Apply the time Laplace transform to the Airy equation. Extend $g$ past $T$ in any suitable way and writeFor example, setting $g=0$ after $T$ makes $G(p)={\int }_0^T{e}^{-pt}g(t)dt$; the final solution for $t<T$ is independent of this extension. The Laplace transform of a derivative givesTake the principal cube root $\lambda =p^{1/3}$ for $\mathrm{Re}p>0$. The three characteristic roots of the spatial ordinary differential equation areSince $|\arg \lambda |<\pi /6$, only $-\lambda$ has negative real part. Thus spatial decay leaves just one homogeneous exponential, which the single prescribed Neumann boundary condition determines.

The Airy resolvent kernel on the whole real line isIt decays at both ends, is continuous together with its first derivative, and satisfies $R_p^{\prime \prime }(0+)-R_p^{\prime \prime }(0-)=1$. Consequently $({\partial }_s^3+p)R_p={\delta }_0$ as a distribution. A particular solution is the Green-function representationAdding the decaying homogeneous mode to impose the boundary derivative givesEvery term is known. If ${\tilde{u}}_0(r)={\int }_0^{\infty }{e}^{-ry}{u}_0(y)dy$ denotes the spatial Laplace transform, thenThe Bromwich inversion formula now gives the required integral representation:Here $\gamma >0$ is to the right of any singularities required by the growth of the data. The usual decay or growth hypotheses are understood for this Laplace transform construction; when absolute inversion is unavailable, the vertical integral is interpreted as the limit of truncated Bromwich contours. The transformed ordinary differential equation verifies the partial differential equation and initial condition, while differentiating at $x=0$ gives exactly $G(p)$ and hence $u_x(0,t)=g(t)$. The derivative compatibility $u_0^{\prime }(0)=g(0)$ makes the two data agree at the corner.

Let $z$ measure height above the sponge. Under the long-wave approximation, hydrostatic pressure gives $p_x=\rho g{H}_x$, and the horizontal balance for viscous fluid flow isThe no-slip boundary condition $u(0)=0$ and stress-free boundary condition $u_z(H)=0$ giveIntegrating this lubrication theory profile gives the horizontal volume flux per unit span

Local mass conservation includes the downward loss found in part i:Consequently the required nonlinear partial differential equation isIf the imposed inlet flux is $Q_0{t}^{\alpha }$ and the moving front is $x=x_N(t)$, sufficient boundary conditions areFor a current released onto a dry substrate one also takes the initial condition $H(x,0)=0$ away from the source. The front position is part of this moving-boundary problem.

For $\alpha =2$, part iii gives $a=1$, $b=2$, and $\beta =1$. PutSubstitution into the nonlinear partial differential equation gives the ordinary differential equationwith boundary conditions

To find the leading edge, set $y={\eta }_N-\eta$ and suppose $F\sim C{y}^p$. The two singular terms in the ordinary differential equation have orders $y^{4p-2}$ and $y^{p-1}$. Their exponents agree only when $p=1/3$. Their leading coefficients then satisfywhereas $F$ and the constant drainage $D$ are lower-order terms. HenceThe draining gravity current therefore has a one-third-power leading edge and its flux vanishes there.

The scalar Cauchy-Kovalevskaya theorem says that a partial differential equation solved for its highest derivative normal to a real-analytic non-characteristic hypersurface, with real-analytic coefficients and Cauchy data, has a unique local real-analytic solution. In coordinates, an equationhas such a solution near the origin when $F$ and the prescribed values of ${\partial }_t^j\varphi (0,x)$ for $0\le j<m$ are real analytic.

Choose a real-analytic primitive $F_0$ of $f$ near zero and apply the theorem to the scalar Laplace equationThe line $t=0$ is non-characteristic because the coefficient of ${\varphi }_{tt}$ is one. DefineThen $u(0,x)=f(x)$ and $v(0,x)=g(x)$, while equality of mixed derivatives and the Laplace equation giveThus $(u,v)$ is the required local real-analytic solution.

Using the first partial derivatives from part (b),Also $uvg=yf$. Therefore the first partial differential equation becomesFor the second partial differential equation, the same chain rule givesAway from the origin, cancellation yields $f_x+2xf=0$. Treating $y$ as fixed, an integrating factor gives ${\partial }_x(e^{x^2}f)=0$. Thuswhere $F$ is an arbitrary sufficiently differentiable function. This is the local general solution, and the global form when $f$ is defined on the whole plane. If $f$ and its first partial derivatives are continuous at the origin, the reduced partial differential equation also holds there by continuity.

A steady state has $T_s^{\prime \prime }=0$, so $T_s=a+bx$. The fixed temperature gives $T_s(0)=T_0$, and thermal insulation gives $T_s^{\prime }(L)=0$. ThusSubtracting this steady state produces the homogeneous mixed boundary conditionsThe partial differential equation and boundary conditions apply for $t>0$. The initial condition is attained as $t\downarrow 0$. Its slope at $x=L$ is $-T_0/L$, so it does not satisfy the insulated-end condition at the instant $t=0$. The resulting heat equation solution is smooth for $t>0$ and approaches the initial temperature continuously; its spatial derivative need not be continuous at $(L,0)$.

For a nonzero separable solution of the Laplace equation in polar coordinates, write $T(r,\theta )=R(r)\Theta (\theta )$ with $r>0$. Multiplying the partial differential equation by $r^2$ and dividing where $R\Theta \ne 0$ giveswhere $\lambda$ is a real separation constant. The resulting ordinary differential equations areThe equations continue across zeros of the factors by continuity. The radial equation is a Cauchy-Euler equation; putting $s=\log r$ changes it to $d^{2}R/d{s}^2-\lambda R=0$.

For positive separation constant, write $\lambda =k^2$ with $k>0$. The radial factors are powers, while the angular factors are sine and cosine:For zero separation constant, each transformed equation has zero second derivative, givingFor negative separation constant, write $\lambda =-k^2$ with $k>0$. The radial factors oscillate in $\log r$, and the angular factors are exponential functions:These exhaust the nonzero real separated solutions before imposing angular periodicity or regularity at the origin. The zero solution is also allowed. Since Laplace's equation is linear, sums of these separated solutions are solutions wherever the sums and required derivatives converge appropriately.

The last two terms in the preceding result equal $v_s$. ThereforeThe left-hand side of the original partial differential equation is consequently $v_{tt}-v_s$, givingIn these variables it has the form of a heat equation, with logarithmic radial coordinate $s$ playing the evolution variable and angular coordinate $t$ playing the spatial variable. If $u$ is single-valued around the whole punctured plane, then $v(s,t+2\pi )=v(s,t)$ as well; that periodicity is a global coordinate condition, not an extra local transformation term.

The partial derivatives are $f_{xx}=4f$ and $f_t=\omega f$. Since the exponential function never vanishes, substituting into the partial differential equation gives $4=6\omega$. Hence

The same chain rule gives $u_{xx}=u^{\prime \prime }(\xi )$ and $u_y=u^{\prime }(\xi )/\sqrt{y}$. Substituting these and part (a) into the partial differential equation givesThus every twice-differentiable profile $u(\xi )$ gives a solution on $y>0$.

The first boundary condition requires $u(x,0)=a{x}^2+d=0$ for all $x$, so $a=d=0$. Since $u_y=bx+2cy$, the second boundary condition gives $b=0$. The partial differential equation then reads $2a-2c=1$, fixing $c=-1/2$. Thus

The chain rule for the new coordinates gives ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$ and ${\partial }_y={\partial }_{\xi }-{\partial }_{\eta }$. Since the coordinates depend linearly on $x,y$, their coefficients are constant. Expanding the squares of these differential operators givesHence the partial differential equation becomes $u_{\xi \eta }=1/4$. Integrating once in each variable introduces independent one-variable functions, giving the general solutionwhere $F,G$ are arbitrary twice-differentiable functions on the relevant characteristic intervals. Their sum includes an arbitrary additive constant.

To distinguish the one-variable profile from the resulting two-variable function, write $u(x,t)=F(\xi )$ with $\xi =x{e}^{-t}$. The chain rule gives $u_t=-x{e}^{-t}{F}^{\prime }(\xi )$ and $u_x=e^{-t}{F}^{\prime }(\xi )$, so $u_t+x{u}_x=0$. Thus any differentiable profile of $x{e}^{-t}$ solves the partial differential equation.

The initial condition selects $F(\xi )=\ln \xi$ on the positive half-line. HenceThis satisfies both the partial differential equation and the given data at $t=0$.

Work on an open domain with $xy>0$ and $x\ne 0$, where the displayed square root is real and differentiable. Split $u=v+h$ with $v=\varphi (xy)$ and $h=\sqrt{xy}\psi (y/x)$. For the first term, the chain rule givesso $x^2{v}_{xx}-y^2{v}_{yy}=0$.

For the second term, set $t=y/x$. Its first partial derivatives areThey show $x{h}_x+y{h}_y=h$, the degree-one case of Euler theorem for homogeneous functions. Differentiating that identity with respect to each variable yieldsThe mixed partial derivatives agree here by the chain rule. Multiplying these equations by $x$ and $y$, respectively, proves the second derivative identity for a degree-one homogeneous function:Adding the results for $v$ and $h$ verifies the partial differential equationEquivalently, $h(\lambda x,\lambda y)=\lambda h(x,y)$ for positive $\lambda$, so the same proof follows directly from its homogeneous function property. The statement is local to a domain where the given expression is defined; it does not require assigning values on $x=0$.

For $t>0$, differentiate the printed ansatz:Matching the independent coefficients in $h_t=(h{h}_x{)}_x-1$ givesThere are two parameter pairs, including the degenerate spatially constant case:Their corresponding solutions areThe request for “the general form” cannot mean the complete solution set of this nonlinear partial differential equation: substitution of one ansatz cannot provide arbitrary initial data, and the superposition principle does not apply.

A useful wider family follows by allowing an independent time-dependent constant term and quadratic coefficient, $h=A(t)+B(t)x^2$. Coefficient comparison gives $B^{\prime }=6B^2$ and $A^{\prime }=2AB-1$. For $B\ne 0$, write $\tau =t-t_0>0$ to obtain $B=-1/(6\tau )$ and $A=C{\tau }^{-1/3}-3\tau /4$. Spatial translation, also a symmetry of the equation, therefore givesHere $C,x_0,t_0$ are arbitrary constants; the displayed nontrivial ansatz corresponds to $C=1/6$ and $x_0=t_0=0$. There are also affine solutions $h=px+(p^2-1)t+q$, including the spatially constant branch. These are valid solution families, rather than a claim of a full general solution.

The chain rule gives $u_{xx}=f^{\prime \prime }(x+\lambda y)$ and $u_{yy}={\lambda }^2{f}^{\prime \prime }(x+\lambda y)$. The partial differential equation becomesFor this to hold for every twice differentiable $f$, the coefficient must vanish:For an affine $f$ alone any $\lambda$ would work, but the requirement is for arbitrary $f$.

Seek a separated solution $u=X(x)Y(y)$. On a region where the factors do not vanish, the partial differential equation separates asThe resulting ordinary differential equations integrate to $X=A{e}^{\lambda x}$ and $Y=B|3-y{|}^{\lambda /2}$ on either side of $y=3$. Choose $\lambda =2$ and the signed linear branch $Y=3-y$, which extends smoothly across that line. This yields the nonconstant solutionDirect partial differentiation gives $u_x=2e^{2x}(3-y)$ and $u_y=-e^{2x}$, whose prescribed combination vanishes everywhere. This direct check also covers the line on which the separated factor is zero.

Apply ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$ once more. The chain rule and symmetry of second derivatives giveHere and in the next part the second partial derivatives are understood with the usual smoothness required for this partial differential equation.