- 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 $α$.
- Write \eqref{1} as a plane autonomous system of two first-order differential equations. Explain what is meant by autonomous.
- Show that there is only one critical point of this system, and classify it.
- 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.
- 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 $|x|<α$.
- 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 $t$.
- \begin{cases}\dot x=y\\\dot y=F(x,y)\end{cases}autonomous – time $t$ does not appear explicitly in the governing equations
- 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. - starts at rest$⇔x'(0)=0$
- 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 linesthe closed trajectory that starts at rest from $x=\gamma$ where $\gamma>\sqrt{2} α$
- 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 $|y-b| \leqslant k\}$.
- State the conditions that must hold for \eqref{2} to have a unique solution in $R$.
- Show that solutions to \eqref{2} in $R$ depend continuously on initial data. [Gronwall's inequality may be stated without proof.]
- Now consider the case where $f(x, y)=\sqrt{\left|x\left(1-y(x)^2\right)\right|}$, and restrict attention to $x≥0$.
- Show that $y$ is an increasing function of $x$ for $y≠1$.
- Suppose $y(0)=0$. Show that the conditions of (a) hold for suitable choices of $h$ and $k$. Find the solution in this case.
- 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).
- $f$ is continuous in $R$, with bound $M$ and $Mh≤k$.
- $f$ satisfies Lipschitz condition in $R$: $∃L>0$ such that $|f(x,u)-f(x,v)|≤L|u-v|$ for $(x,u)∈R,(x,v)∈R$.
- 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. - For $y≠1$, $f(x,y)>0⇒y$ is an increasing function of $x$.
- 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 $h≤k^{2/3}$.
$\frac{dy}{dx}=x^{1/2}(1-y(x)^2)^{1/2}⇒\int\frac1{\sqrt{1-y^2}}dy=\int x^{1/2}dx⇒y=\sin\left(\frac23x^{3/2}+C\right)$
$y(0)=0⇒C=0⇒y(x)=\sin\left(\frac23x^{3/2}\right)$ - 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.
- 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 $1/2≤x≤1$.
- Explain what is meant by a characteristic curve and a characteristic projection. Write down the characteristic equations for \eqref{3}.
- 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.
- 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.
- 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.
- 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.
- $\cases{x\frac{dx}{dt}=-xy\\y\frac{dy}{dt}=xy}⇒x^2+y^2=\text{const}$ 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. - 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. - $ze^{x^2}=s^2$ 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.