Differential equations 1 paper 2016

 
  1. The displacement $x(t)$ of a particle at time $t$ satisfies \begin{equation}\label1 \ddot{x}=F(x, \dot{x}) \quad \text { where } \quad F(x, \dot{x})= \begin{cases}-x & |x|>α, \\ -x-\dot{x} & |x| \leqslant α,|\dot{x}|<α, \\ 0 & \text { otherwise }\end{cases} \end{equation} for positive real constant $α$.
    1. Write \eqref{1} as a plane autonomous system of two first-order differential equations. Explain what is meant by autonomous.
    2. Show that there is only one critical point of this system, and classify it.
    3. Find and sketch the nullclines for (i) $|x|>α$ and (ii) $|x|<α,|\dot{x}|<α$. Indicate the direction of the trajectories across the nullclines.
    4. Sketch a trajectory in $|x|<α,|\dot{x}|<α$. Sketch $x(t)$ as a function of $t$ for a trajectory that starts at rest from $|x|<α$.
    5. Find the trajectories in $|x|>α$ and $|x|<α,|\dot{x}|>α$ and hence sketch on the phase plane the closed trajectory that starts at rest from $x=\gamma$ where $\gamma>\sqrt{2} α$. For this trajectory, sketch $x(t)$ as a function of $t$.
    Solution.
    1. \begin{cases}\dot x=y\\\dot y=F(x,y)\end{cases}autonomous – time $t$ does not appear explicitly in the governing equations
    2. critical point $\dot x=\dot y=0$.
      $\dot x=0⇒y=0$
      $F(x,0)=0⇒x=0$
      critical point at (0,0)
      eigenvalues of $\pmatrix{0&1\\-1&-1}$ are $\frac{-1\pm\sqrt3i}2⇒$critical point is a stable spiral.
    3. starts at rest$⇔x'(0)=0$
    4. For $|x|>α$ we have $\cases{\dot x=y\\\dot y=-x}⇒x^2+y^2=\text{const}⇒$trajectories are circular arcs
      For $|x|<α,|\dot{x}|>α$ we have $\cases{\dot x=y\\\dot y=0}⇒y=\text{const}⇒$trajectories are straight horizontal lines
      the closed trajectory that starts at rest from $x=\gamma$ where $\gamma>\sqrt{2} α$
  2. Consider the initial value problem \begin{equation}\label2 y'(x)=f(x, y(x)), y(a)=b, \end{equation} and define the rectangle $R=\{(x, y):|x-a| \leqslant h$ and $|y-b| \leqslant k\}$.
    1. State the conditions that must hold for \eqref{2} to have a unique solution in $R$.
    2. Show that solutions to \eqref{2} in $R$ depend continuously on initial data. [Gronwall's inequality may be stated without proof.]
    3. Now consider the case where $f(x, y)=\sqrt{\left|x\left(1-y(x)^2\right)\right|}$, and restrict attention to $x≥0$.
      1. Show that $y$ is an increasing function of $x$ for $y≠1$.
      2. Suppose $y(0)=0$. Show that the conditions of (a) hold for suitable choices of $h$ and $k$. Find the solution in this case.
      3. Now suppose that $y(0)=1$. Show that there are infinitely many solutions of \eqref{2} subject to $y(0)=1$. Explain why this does not contradict the uniqueness result in part (a).
    Solution.
      1. $f$ is continuous in $R$, with bound $M$ and $Mh≤k$.
      2. $f$ satisfies Lipschitz condition in $R$: $∃L>0$ such that $|f(x,u)-f(x,v)|≤L|u-v|$ for $(x,u)∈R,(x,v)∈R$.
    1. Gronwall's inequality: Suppose $A≥0$ and $b≥0$ are constants and $v$ is a non-negative continuous function satisfying $v(x)≤b+A\left|\int_a^xv(s)ds\right|$. Then $v(x)≤be^{A|x-a|}$.
      Suppose now that $y$ and $z$ are solutions of $y'(x)=f(x,y(x))$ with $y(a)=b$ and $z(a)=c$, where $f$ satisfies P(i) and P(ii). Then $y(x)-z(x)=b-c+\int_a^xf(s,y(s))-f(s,z(s))ds$ so that $|y(s)-z(s)|≤|b-c|+\left|\int_a^x|f(s,y(s))-f(s,z(s))|ds\right|≤|b-c|+\left|\int_a^xL|y(s)-z(s)|ds\right|$. By Gronwall's inequality, $|y(x)-z(x)|≤|b-c|e^{L|x-a|}≤|b-c|e^{Lh}$
      Solution is continuously dependent on the initial data if we can make $|y(x)-z(x)|$ as small as we like by taking $|b-c|$ small enough.
      i.e. $∀x∈[a-h,a+h]∀ε>0∃δ>0$ s.t. $|b-c|<δ⇒|y(x)-z(x)|≤ε$
      Take $δ=e^{-Lh}ε$ the result follows.
      1. For $y≠1$, $f(x,y)>0⇒y$ is an increasing function of $x$.
      2. We need $y$ bounded away from 1, so $k<1$, then $f(x,y)$ is continuous on $R$ and $M=\sup_R\left|x(1-y(x)^2)\right|^{1/2}≤h^{1/2}$. We need $Mh≤k⇐h^{3/2}≤k$. So P(i) holds for $h≤k^{2/3}$.
        Also, by MVT there exists ξ between $y$ and $z$ such that $|f(x,y)-f(x,z)|=|f_y(x,ξ)||y-z|=|x|^{1/2}\left|\frac12(1-ξ^2)^{-1/2}⋅-2ξ\right||y-z|=\frac{|x|^{1/2}}{(1-ξ^2)^{1/2}}|ξ||y-z|≤\frac{h^{1/2}k}{(1-k^2)^{1/2}}|y-z|$
        So P(ii) holds with $L=\frac{h^{1/2}k}{(1-k^2)^{1/2}}$. Thus there is a unique solution in $R$ for $h≤k^{2/3}$.
        $\frac{dy}{dx}=x^{1/2}(1-y(x)^2)^{1/2}⇒\int\frac1{\sqrt{1-y^2}}dy=\int x^{1/2}dx⇒y=\sin\left(\frac23x^{3/2}+C\right)$
        $y(0)=0⇒C=0⇒y(x)=\sin\left(\frac23x^{3/2}\right)$
      3. Clearly $y(x)=1$ is a solution. $y$ increasing, so $\frac{dy}{dx}=\sqrt{x(y^2-1)}⇒\int\frac1{\sqrt{y^2-1}}dy=\int x^{1/2}dx⇒y=\cosh^{-1}\left(\frac23x^{3/2}+\text{const}\right)$
        Suppose $y=1$ at $x=a$, then $y=\cosh^{-1}\left(\frac23(x^{3/2}-a^{3/2})+\cosh1\right)$

        Infinitely many solutions as $f(x,y)$ not Lipschitz at $y=1$. Hence does not contradict uniqueness result.
  3. Consider the differential equation \begin{equation}\label3 -y z_x+x z_y=2 x y z, \end{equation} with $z=x^2 \exp \left(-x^2\right)$ on $y=1-x^2$ for $1/2≤x≤1$.
    1. Explain what is meant by a characteristic curve and a characteristic projection. Write down the characteristic equations for \eqref{3}.
    2. Describe the characteristic projections in the $(x, y)$-plane. By differentiating along the characteristic curve, show that $z \exp \left(x^2\right)$ is constant on each characteristic curve.
    3. Identify the two separate segments of the data curve where the data is Cauchy. Comment on why it is necessary to consider the two segments separately.
    4. For each segment, find the explicit solution for $z(x, y)$, and identify the domain on which the solution is defined and uniquely determined by the given initial conditions.
    Solution.
    1. Suppose $Γ=(x(t),y(t),z(t))$ in terms of a parameter $t$. The characteristic equations are\[\frac{dx}{dt}=-y; \frac{dy}{dt}=x; \frac{dz}{dt}=2xyz\]The curve Γ is a characteristic curve. The curve $(x(t),y(t),0)$ which lies below the characteristic curve in the xy-plane is called a characteristic projection.
    2. $\cases{x\frac{dx}{dt}=-xy\\y\frac{dy}{dt}=xy}⇒x^2+y^2=\text{const}$ are characteristic projections. These are circles with centre the origin in the xy-plane.
      Consider $\frac{d}{dt}(ze^{x^2})=e^{x^2}\frac{dz}{dt}+2xze^{x^2}\frac{dx}{dt}=e^{x^2}2xyz+2xze^{x^2}⋅-y=0⇒ze^{x^2}$ is constant on each characteristic curve.
    3. Consider the initial data, and parametrise by $s$: $(s,1-s^2,s^2e^{-s^2})$.
      So characteristic projections are $x^2+y^2=s^2+(1-s^2)^2,\frac12≤s≤1$
      $\cases{x(t)=\sqrt{s^2+(1-s^2)^2}\cos t\\y(t)=\sqrt{s^2+(1-s^2)^2}\sin t}$
      $J=\begin{vmatrix}x_s&x_t\\y_s&y_t\end{vmatrix}=s-2s(1-s^2)$
      $J=0$ when $s=0$ or $s=±\frac1{\sqrt2}$
      So data is Cauchy for $\frac12≤s<\frac1{\sqrt2}$ and $\frac1{\sqrt2}<s≤1$
      At $s=\frac1{\sqrt2}$, characteristic projections and data curves touch.
    4. $ze^{x^2}=s^2$ constant on characteristics.
      For $\frac12≤s<\frac1{\sqrt2}$, characteristics are circles of radius between $\sqrt{\frac12+\left(1-\frac12\right)^2}=\sqrt{3\over4}$ and $\sqrt{\frac14+\left(1-\frac14\right)^2}=\sqrt{13\over16}$
      For $\frac1{\sqrt2}<s≤1$ circles of radius between $\sqrt{3\over4}$ and 1.
      $x^2+y^2=s^2+(1-s^2)^2⇒s^2=\frac{1±\sqrt{4(x^2+y^2)-3}}2$⋯(*)
      $x^2+y^2≥\frac{13}{16}⇒4(x^2+y^2)-3≥\frac{13}4-3>0$ So (*) has real roots.
      $x^2+y^2≤1⇒4(x^2+y^2)-3≤1$ so both $s^2$ positive. So $s^2=\frac{1±\sqrt{4(x^2+y^2)-3}}2$ (Two positive roots)
      So $z=\frac{1-\sqrt{4(x^2+y^2)-3}}2⋅e^{-x^2}$ for $\frac34<x^2+y^2≤\frac{13}{16}$ and $\frac{1+\sqrt{4(x^2+y^2)-3}}2⋅e^{-x^2}$ for $\frac34<x^2+y^2≤1$
      So $\frac{13}{16}<x^2+y^2≤1$ is the domain on which the solution is unique.