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For a time-independent function $F$, Hamilton's equations giveThus a constant of motion has $[F,H]=0$. For $F={\sum }_k{p}_k$ and the given Hamiltonian, $[F,H]=-{\sum }_k{V}_{q_k}$. Simultaneous translation $q_k\mapsto q_k+s$ leaves every difference $q_i-q_j$, and hence $V$, unchanged. Differentiation with respect to $s$ gives ${\sum }_k{V}_{q_k}=0$, so $[F,H]=0$ also in this example. This is conservation of total momentum from translational symmetry.

Hamilton's equations generate the phase-space vector field $(H_{p_1},\dots ,H_{p_n},-H_{q_1},\dots ,-H_{q_n})$. Its divergence vanishes:by equality of mixed derivatives. The Jacobian $J(t)$ of the flow obeys $\dot{J}=(\mathrm{div}V)J=0$ with $J(0)=1$. Consequently $d\tau =d{q}_1\cdots d{q}_{n}d{p}_1\cdots d{p}_n$ is preserved for every time on which the smooth flow exists. This proves Liouville theorem in Hamiltonian mechanics directly; explicit time dependence of the Hamiltonian would not change the divergence argument.

Choose a polynomial $P$ with $\Vert f-P{\Vert }_{\infty }<ϵ$, using part (i). For all sufficiently large $n$, its degree is at most $2n-1$, so $I_n(P)={\int }_0^{1}P$. Positive weights summing to one give $|I_n(f-P)|\le ϵ$, and $|{\int }_0^1(f-P)|\le ϵ$. HenceSince $ϵ$ is arbitrary, the quadratures converge to the integral of every continuous $f$. The positivity established in part (ii), not just polynomial exactness on its own, controls the approximation error uniformly in $n$.

For each node define the Lagrange interpolation polynomialThe nodes are distinct, so this is well defined, has degree $n-1$, and satisfies $L_j(x_{k,n})={\delta }_{jk}$. Its square has degree $2n-2$, within the assumed exactness range. ThereforeThe strict inequality follows because $L_j$ is not the zero polynomial. Exactness for the constant polynomial also gives ${\sum }_{j=1}^n{a}_{j,n}=1$. Thus the quadrature is a positive weighted average; the requested nonnegativity follows in stronger form.

For fixed $t$, let $B$ have the binomial distribution with parameters $m,t$. The Bernstein polynomial is $f_m(t)=\mathbb{E}[f(B/m)]$. Continuity on the compact interval gives uniform continuity and a finite bound $M=\Vert f{\Vert }_{\infty }$.

Given $ϵ>0$, choose $\delta >0$ so $|f(s)-f(t)|<ϵ$ whenever $|s-t|<\delta$. Since $\mathbb{E}[B/m]=t$ and $\mathrm{Var}(B/m)=t(1-t)/m\le 1/(4m)$, Chebyshev's inequality givesThe bound is independent of $t$, including both endpoints. Taking the supremum and then letting $m$ tend to infinity proves $f_m\to f$ uniformly on $[0,1]$. This provides the required polynomial-approximation theorem with its proof.

The Runge-Kutta method has $A=\left(\begin{array}{cc}1/4& 1/4-a\\ 1/4+a& 1/4\end{array}\right)$ and $b=(1/2,1/2{)}^T$. Solving the stage equations for $y^{\prime }=\lambda y$ gives the stability functionFor $a\ne 0$, the denominator roots have positive real parts: real roots are positive from their positive sum and product, while a complex conjugate pair has real part $1/(4a^2)>0$. For $a=0$ its sole root is $2$. Thus there are no poles or singular stage systems in the left half-plane.

Writing $z=x+iy$, direct expansion givesThis proves the A-stability of a symmetric two-stage implicit Runge-Kutta family:

Apply the numerical method to the test equation $y^{\prime }=\lambda y$ and put $z=h\lambda$. Its linear stability domain is the set of $z$ for which the computed amplification is stable, for a one-step method $|R(z)|\le 1$ with all stage solves well-defined. The method is A-stable if this includes the whole closed left half-plane (or its interior with boundary limits).

Balancing the time derivative, cubic nonlinearity and third spatial derivative gives $x\sim t^{1/3}$ and $u\sim t^{-1/3}$. Choose $\zeta =x/(3t{)}^{1/3}$ and $u=(3t{)}^{-1/3}w(\zeta )$. ThenThus the similarity solution isThe extra spherical term changes the coefficient of $w$. This third-order equation cannot be integrated directly into the standard second-order Painlevé II equation of the ordinary modified KdV reduction.

The initial data require $\Phi (\xi )=\exp [-(2\nu {)}^{-1}{\int }_{x_{\ast }}^{\xi }U(s)ds]$, up to a positive constant that cancels from $u$. Substituting this into the heat kernel formula gives the exponent $-G/(2\nu )$ withDifferentiate the kernel integral in $x$ and apply the Cole-Hopf transformation. Its constant prefactor cancels, leavingThe choice of $x_{\ast }$ adds an irrelevant constant to $G$; the usual convergence assumptions on the initial data justify the integrals and differentiation.

Set $u=-2\nu {\partial }_x\log \varphi$. Direct differentiation givesThus the Cole-Hopf transformation maps every nonvanishing solution of the heat equation to a solution of the viscous Burgers equation. Conversely the Burgers equation makes the ratio inside the derivative a function of time alone; a time-dependent rescaling of $\varphi$, which leaves $u$ unchanged, removes that function and gives the heat equation exactly.

False. A function of $\cos t$ is even in $t$, whereas $\sin (nt)$ is odd. If the identity held, replacing $t$ by $-t$ would imply $\sin (nt)=0$ for every $t$, contradicted by $t=\pi /(2n)$. Thus no polynomial of any degree has the required property.

True, with $R_n=T_n$. The hyperbolic cosine identity $\cosh ((n+1)t)=2\cosh t\cosh (nt)-\cosh ((n-1)t)$ and the initial values $1,\cosh t$ give $T_n(\cosh t)=\cosh (nt)$ by mathematical induction, using the same Chebyshev polynomial recurrence.

True. Define the Chebyshev polynomials by $T_0(x)=1$, $T_1(x)=x$, and $T_{n+1}(x)=2x{T}_n(x)-T_{n-1}(x)$. The cosine addition identities give the same recurrence for $\cos (nt)$, so mathematical induction proves $T_n(\cos t)=\cos (nt)$. For $n\ge 1$, the leading coefficient is $2^{n-1}$, proving that its polynomial degree is exactly $n$.

The new characteristics have invariant $x_1-x_2$. On the initial curve it is $D(s)=s$ for $s<0$ and $D(s)=s-s^2$ for $s\ge 0$. The latter branch folds at $s=1/2$, attaining its maximum $1/4$. Some characteristics meet the initial curve twice or more; generic values of $f$ at those intersections disagree, so no global solution constant along characteristics can exist.

At every initial point except $s=1/2$, the curve is non-characteristic, and the local existence and uniqueness theorem applies. At $s=1/2$, tangency requires $f^{\prime }(1/2)=0$, but this is not sufficient. The fold tangency obstruction for characteristic data requires locally $f(s)=F(D(s))$ with $F$ continuously differentiable: equal invariant values must have equal data, and $f^{\prime }(s)/(1-2s)$ must have a finite continuous limit. For example $f(s)=|s-1/2{|}^{3/2}$ satisfies $f^{\prime }(1/2)=0$ and agrees across the fold, but would force an infinite derivative of $F$ at $1/4$.

If the full compatibility condition holds, $F$ can be extended to invariant values above $1/4$ to give a local solution. Those values are not specified by the initial curve, so local uniqueness then fails.

The solution is constant on characteristics $x_1+x_2=\eta$. On the initial curve this invariant is $h(s)=s$ for $s<0$ and $h(s)=s+s^2$ for $s\ge 0$. This is a $C^1$ bijection of $\mathbb{R}$, with everywhere positive derivative. Its inverse isTherefore $\boxed{u(x_1,x_2)=f(h^{-1}(x_1+x_2)).}$ The inverse is $C^1$ even at zero, so for $C^1$ data this is a global $C^1$ solution. Every characteristic meets the initial curve exactly once.

Along the characteristics, $x_1+x_2$ is constant. Since $d{x}_2/dt=-1$, the amplitude equation gives $du/d{x}_2=-x_{2}u$. Matching the initial value at $x_2=0$ yieldsDifferentiation directly verifies both the partial differential equation and the initial data.

The characteristics solve ${\dot{x}}_j=a_j(x)$ and $\dot{u}=b(x,u)$, starting with $x$ on the initial hypersurface and $u=\phi (x)$. The non-characteristic hypersurface condition at a point is $a\cdot n\ne 0$, where $n$ is any nonzero normal there. It makes the map from initial-surface coordinates and characteristic time to $x$ locally invertible. Local uniqueness for the ordinary differential equations and the inverse function theorem then give a unique local $C^1$ solution near that point, for the stated $C^1$ coefficients and data.

Independence and the normal exponential moment give $\mathbb{E}[S_t\mid {\mathcal{F}}_s]=S_s\exp [(\mu +{\sigma }^2/2)(t-s)]$. ThusFor $\sigma \ne 0$, $B_t/t\to 0$ almost surely, so $\log S_t/t\to -{\sigma }^2/2$ and $S_t\to 0$ almost surely. Nevertheless $\mathbb{E}{S}_t=e^{x_0}$ for every finite $t$, whereas $\mathbb{E}{S}_{\infty }=0$. This Exponential martingale for Brownian motion is not uniformly integrable, so it does not converge in $L^1$. If $\sigma =0$, the process is instead constantly $e^{x_0}$.

A standard Brownian motion starts at zero, has continuous paths, and has independent increments $B_t-B_s\sim N(0,t-s)$. By the Brownian reflection principle, for $a>0$,Hence every positive integer level is reached with probability one. Intersecting these countably many probability-one events proves $\boxed{\underset{t\ge 0}{\sup }{B}_t=\infty \text{almost surely}.}$

Relative to a filtration $({\mathcal{F}}_t)$, a martingale is an adapted integrable process satisfying $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$ for $s\le t$. The martingale convergence theorem gives an almost-sure finite limit for a nonnegative martingale, with $\mathbb{E}[M_{\infty }]\le \mathbb{E}[M_0]$ by Fatou's lemma. In continuous time take the usual right-continuous version and filtration hypotheses. Convergence in $L^1$, and preservation of expectation at infinity, additionally require uniform integrability.