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The step response of a damped oscillator is the time integral of its causal Green function of a damped oscillator. Put $s=t-b$ and define $K(s)={\int }_0^s{e}^{-ku}\sin (\omega u)/\omega du$ for $s\ge 0$. Then $K(0)=K^{\prime }(0)=0$ and its oscillator equation has constant right side $1$. Integrating explicitly givesThis formula applies when $\omega \ne 0$. It is a convolution of the causal impulse response with the shifted Heaviside step function, so the response is zero before the forcing starts and has the stated zero initial data.

The integral definition also handles the degenerate parameters without division ambiguities. For $\omega =0,k\ne 0$,for $\omega =k=0$, $y_3=H(s)s^2/2$. These are the continuous zero-frequency limits of the step response.

Use the classical risk model surplus $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with $c>0$, and write $\phi$ for its ultimate survival probability. The first-claim decomposition for survival probability follows by conditioning on the first arrival time of the Poisson process. That time has exponential distribution of rate $\lambda$. A claim at time $t$ leaves capital $u+ct-x$ if $x\le u+ct$, after which the Markov property restarts the same risk model. ConsequentlyPut $C(v)={\int }_0^v\phi (v-x)f(x)dx$ and change variable $v=u+ct$:The convolution $C$ is continuous since $\phi$ is bounded and $f$ is integrable. Differentiating proves the survival integro-differential equation

For the specified claim law, expand its probability density function as $f(x)=\frac{1}{2}{e}^{-x}+\frac{1}{4}{e}^{-x/2}$. This is an equal-weight mixture distribution of exponential distributions with rates $1$ and $1/2$. In the convolution, set $z=u-x$ to obtainSince $\lambda /c=10/21$, the requested coefficients are

For the exponential-mixture differential equation for survival probability, define $A(u)=e^{-u}{I}_1(u)$, $B(u)=e^{-u/2}{I}_2(u)$ and $r=\lambda /c$. Differentiating the integrals yieldsApply $(D+1)(D+\frac{1}{2})$ to the last equation, where $D=d/du$. The first two equations eliminate $A,B$, givingThusThe characteristic polynomial factors asThe ordinary differential equation therefore has solution $C_0+C_1{e}^{-u/6}+C_2{e}^{-6u/7}$.

The claim expected value is $\frac{1}{2}\cdot 1+\frac{1}{2}\cdot 2=3/2$. The relative safety loading is $\rho =2.1/(3/2)-1=2/5$, and the given zero-capital survival probability is $\phi (0)=2/7$. The limit at infinity sets $C_0=1$. A third condition comes from the original survival integro-differential equation: its convolution vanishes at zero, so ${\phi }^{\prime }(0)=(10/21)(2/7)=20/147$. Thereforewhich gives $C_1=-20/29$ and $C_2=-5/203$. The final survival probability and ultimate ruin probability areBoth exponential coefficients of $\psi$ are positive. Hence $\phi$ increases from $2/7$ to one, and $\psi$ decreases from $5/7$ to zero. The initial slope is essential: the two stated boundary values alone do not determine all three constants of the eliminated differential equation.

Use $D_j=-i{\partial }_{x_j}$ and $⟨\xi ⟩=(1+|\xi {|}^2{)}^{1/2}$, so the Fourier transform of $P(D)u$ is $P(\xi )\hat{u}(\xi )$. Write $P_N$ for the homogeneous degree-$N$ part. An elliptic differential operator has $P_N(\xi )\ne 0$ for every real $\xi \ne 0$. By continuity on the unit sphere, $m={\min }_{|\omega |=1}|P_N(\omega )|>0$. The lower-degree terms are bounded by $C|\xi {|}^{N-1}$ for $|\xi |\ge 1$, whencefor sufficiently large $|\xi |$. This proves the high-frequency lower bound for an elliptic polynomial. Degree zero just means a nonzero constant and has no derivative gain.

With the Fourier convention in Question 3, the Sobolev space isThe condition includes that the weighted transform is represented by an $L^2$ function. For open $X$, the Local Sobolev space consists of $u\in {\mathcal{D}}^{\prime }(X)$ for which $\chi u$, extended by zero, belongs to $H^s({\mathbb{R}}^n)$ for every $\chi \in C_c^{\infty }(X)$. We use the elementary Sobolev multiplication by a smooth cutoff fact for every real $s$. It follows from Fourier convolution with the rapidly decreasing $\hat{\chi }$, the weighted inequality $⟨\xi {⟩}^s\lesssim ⟨\xi -\eta {⟩}^{|s|}⟨\eta {⟩}^s$, and the $L^1\ast L^2\to L^2$ convolution bound.

For a compactly supported distribution, continuity on test functions supported in a fixed compact neighborhood gives finite order: for some integer $M$,Insert a compact smooth cutoff function equal to one near the support, times $e^{-ix\cdot \xi }$. The resulting Fourier transform is smooth and satisfies $|\hat{u}(\xi )|\le C^{\prime }⟨\xi {⟩}^M$. The distribution is also tempered by this same finite-order bound. ThusIndeed the square of the weighted bound is integrable exactly when $2(t+M)<-n$. This is negative Sobolev regularity of a compactly supported distribution.

We next prove local regularity by a high-frequency reciprocal parametrix kernel, including its off-diagonal smoothing property. Choose $\psi \in C_c^{\infty }({\mathbb{R}}^n)$ equal to one on a ball containing all real zeros of $P$, and put $b(\xi )=(1-\psi (\xi ))/P(\xi )$, defining it smoothly as zero in the inner ball. Differentiation of the reciprocal and the elliptic lower bound giveThe Fourier multiplier $E=b(D)$ maps $H^s$ to $H^{s+N}$ by its zeroth-order bound, andThe kernel $K={\mathcal{F}}^{-1}b$ is smooth away from the origin: for $x\ne 0$, repeatedly integrate by parts using $e^{ix\cdot \xi }=(i|x{|}^2{)}^{-1}x\cdot {\partial }_{\xi }{e}^{ix\cdot \xi }$. After sufficiently many integrations, the differentiated symbol is integrable. For any desired $x$ derivative, repeat the argument with the additional polynomial ${\xi }^{\beta }$. A large-radius cutoff function justifies each step and its removal uniformly on compact sets away from zero. Thus convolution by $K$ carries a compactly supported distribution to a smooth function at points separated from its support. Also $R$ carries a compactly supported distribution to a Schwartz function, because $\psi \hat{u}$ is smooth with compact support.

For a target compact set in $X$, choose $\chi \in C_c^{\infty }(X)$ equal to one on a neighborhood of it. Then $\chi u$ and $[P(D),\chi ]u$ have compact support, and the commutator is supported where derivatives of $\chi$ occur, away from the target. The parametrix identity givesThe first term is in $H^{s+N}$ since $\chi P(D)u\in H^s$. The other two terms are smooth near the target by the proved kernel property. Since the target was arbitrary,This proof does not discard the cutoff function commutator; it places its support away from the set where regularity is sought.

For the final polynomial, select a multi-index ${\alpha }_0$ with $|{\alpha }_0|=N$ and ${\partial }^{{\alpha }_0}Q$ a nonzero constant. The derivative-ratio hypothesis immediately gives $|Q(\xi )|\ge c|\xi {|}^{\delta N}$ at large frequency. In particular $Q$ has no real zero there. For $N>0$, its degree bound also forces $\delta \le 1$. Differentiating $1/Q$ gives products of ratios ${\partial }^{\beta }Q/Q$, whose total derivative order is $|\alpha |$. Hence the corresponding high-frequency reciprocal satisfiesIts multiplier maps $H^s$ to $H^{s+\delta N}$. Its kernel is still smooth off zero: each frequency integration by parts now lowers the order by $\delta$, so more iterations may be needed, but $\delta >0$ supplies arbitrarily much decay. The same separated-support commutator identity applies without a change. The derivative-ratio Sobolev gain for a polynomial operator is thereforeSimply estimating the commutator by its differential order would lose this sharper gain; the off-diagonal kernel argument uses the full derivative-ratio hypothesis.

Use the conventional Laplace operator $\Delta ={\partial }_{x_1}^2+{\partial }_{x_2}^2$. The fundamental solution of the Laplace equation in two dimensions is $G(x)=(2\pi {)}^{-1}\log |x|$, with $\Delta G={\delta }_0$. To check its sign, integrate the outward normal derivative around a circle: ${\int }_{\partial B_{\epsilon }}{\partial }_n\log |x|ds=2\pi$. Applying integration by parts against a smooth compactly supported test function and shrinking the circle gives the distributional identity.

The printed stream function has $\varphi =-G\ast \omega$. For each multi-index $\beta$ with $|\beta |\le 2$, transfer derivatives to the compactly supported $C^2$ function:The logarithmic singularity is locally integrable. On a bounded region in $x$, the convolution uses one fixed bounded region in the integration variable; splitting off a small disk around the singularity proves continuity of these derivatives. Thus $\varphi \in C^2$, and convolution of the distributional identity givesThe asserted plus sign is false for the printed kernel and the conventional Laplacian. Any nonzero smooth compactly supported $\omega$ is a counterexample. To obtain $\Delta \varphi =\omega$, replace the kernel's minus sign by a plus sign, or explicitly adopt the negative Laplace operator. Below, retain the printed kernel and velocity convention. With that convention, $\mathrm{curl}u=\Delta \varphi =-\omega$, so $\omega$ is the negative of conventional scalar vorticity. This sign does not change the transport or flow arguments.

Use the phase-distance convention for a Bohr set:where the norm is distance to the nearest integer. This is the Bohr set in phase-distance convention; the alternative condition $|e(rx/N)-1|\le \rho$ has a different radius and corresponding constants. Assume $d\ge 1$, as $1/d$ requires.

For each residue $x$, form $t_x=(rx/N{)}_{r\in R}$ in the $d$-dimensional torus. Around these $N$ points place translates of a box of side length $ϵ$ in each coordinate. For $0<ϵ\le 1$, their volumes sum to $N{ϵ}^d$. If $ϵ>N^{-1/d}$, two boxes overlap, and their distinct indices differ by a nonzero $q$ with every phase distance less than $ϵ$. Thus the Bohr set contains a nonzero element at the stated threshold. For larger radii the entire group is already present.

Letting the box width decrease to $N^{-1/d}$ and using the finite number of possible nonzero $q$, obtain a nonzero $q$ with $\Vert rq/N\Vert \le N^{-1/d}$ for every $r\in R$. By the triangle inequality, $jq$ lies in $B(R,ϵ)$ whenever $0\le j<ϵN^{1/d}$. Since $N$ is prime, these multiples are distinct up to length $N$. Therefore, in the usual radius range $0<ϵ\le 1/2$, the arithmetic progressionhas the requested length. For literally unrestricted $ϵ>0$, the correct lower bound is $\min (N,\lceil ϵN^{1/d}\rceil )$: a progression of distinct residues cannot exceed $N$. For example, $d=1$, $ϵ=2$ makes the uncapped printed bound impossible. If $R$ is empty, the Bohr set is the whole group and has a progression of length $N$; the displayed exponent is simply inapplicable.

Now transfer the dense integer set to a larger cyclic group. Choose a prime $8N<p<16N$, using Bertrand's postulate, and regard $A$ as a subset of $G=\mathbb{Z}/p\mathbb{Z}$. Its density $\alpha =|A|/p$ is at least $1/1600$. Normalize Fourier coefficients on a finite abelian group and convolution byThe Fourier inversion and Parseval identity formulas are $f(x)={\sum }_r\hat{f}(r)e(rx/p)$ and ${\sum }_r|\hat{f}(r){|}^2=\mathbb{E}|f{|}^2$; convolution turns into multiplication of Fourier coefficients on a finite abelian group.

For $f=1_A$, set $R=\{r:|\hat{f}(r)|\ge {\alpha }^{3/2}/2\}$. Then Parseval identity boundsAlso, with $\tilde{f}(x)=f(-x)$, the nonnegative convolution $Q=f\ast f\ast \tilde{f}\ast \tilde{f}$ is supported exactly on the modular set $2A-2A$, andThe frequencies outside $R$ contribute in absolute value at most $({\alpha }^3/4){\sum }_r|\hat{f}(r){|}^2={\alpha }^4/4$. For $x\in B(R,1/6)$, the real part of each phase from $R$ is at least $1/2$; moreover $0\in R$ and $|\hat{f}(0){|}^4={\alpha }^4$. Hence $Q(x)\ge {\alpha }^4/4>0$. We have proved the cyclic Bogolyubov lemma with explicit phase radius:The previous Bohr set argument gives a modular arithmetic progression of length at least $p^{1/D}/6$. The remaining step is lifting a short modular progression to an integer progression. Every one of its elements has a unique integer lift belonging to $2A-2A\subseteq [-2N,2N]$. Successive lifted differences lie in $[-4N,4N]$ and are congruent to the same residue. Since $p>8N$, at most one integer in this difference interval represents that residue, so every successive difference is the same. The lifts form a genuine integer arithmetic progression, not merely a modular one.

For sufficiently large $N$, the constants can be absorbed into half the exponent. ThusNo attempt to optimize this absolute constant is needed.

A counterexample comes from dense integer sets with only logarithmic-length progressions. To show that $A$ itself need not have such a progression, choose a random subset $B\subseteq [1,N]$ by independent inclusion with probability $p_0=0.02$. Its size has mean $0.02N$ and variance at most $0.02N$, so Chebyshev's inequality shows $\mathbb{P}(|B|<0.01N)\le 200/N$. There are at most $N^2$ increasing arithmetic progressions of any fixed length $L\ge 2$, and each belongs to $B$ with probability $p_0^L$. Taking $L=\lceil 4\log N/\log (1/p_0)\rceil$ makes the union bound at most $N^{-2}$. For large $N$, some realization has at least $0.01N$ elements and no length-$L$ arithmetic progression. Take a subset of exactly $0.01N$ elements. It retains that avoidance. Such a set has no progression longer than $O(\log N)$, and therefore no progression of length $N^c$ for any fixed $c>0$, eventually. Interpret the prescribed exact cardinality on values of $N$ where it is integral, or use its integer part.