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Write the second-order ODE
\[
\ddot{x}-\dot{x}^2+x^2-x=0
\]
as a plane autonomous system. Find and classify the critical points of this system. Analyse each critical point in detail and hence sketch the phase plane including the arrows for the trajectories.
Show that there are periodic orbits of the original ODE.
[Hint: Either make the transformation $u=x+y, v=x-y$ and solve the resulting equation for $d u / d v$ to write the characteristics as level sets of a function and then argue graphically that these level sets are closed curves.
Or exploit the symmetry of the phase plane about the $x$ axis to prove the claim.]
Solution. \begin{cases} \dot x=y\\ \dot y=y^2-x^2+x \end{cases} Nullclines: $y=0$ and $y^2-x^2+x=0$
Critical points: $(0,0)(1,0)$. \[M=\pmatrix{0&1\\1-2x&2y}\] At (0,0) eigenvalues ±1 ⇒ saddle. Eigenvectors $(±1,1)$. Two solutions $x=e^t,y=e^t;x=e^{-t},y=-e^{-t}⇒y=±x$ are trajectories.
At (1,0) eigenvalues ±i ⇒ center. $\frac{∂X}{∂y}=1>0⇒$clockwise.
Let $u=x+y, v=x-y,$ \begin{cases} \dot u=u(1-v)\\ \dot v=(u-1)v \end{cases} Dividing, $\frac{du}{dv}=\frac{u(1-v)}{(u-1)v}⇒\frac{u-1}u\,du=\frac{1-v}v\,dv$.
Integrating, $u+v-\ln uv$ is constant ⇒ $2x-\ln(x^2-y^2)$ is constant ⇒ $e^{-2x}(x^2-y^2)$ is constant.
For $0< c< e^{-2}$, the equation $e^{-2x}x^2=c$ have 2 solutions, so the trajectory intersects x-axis twice.
Time reversal symmetry: with $y↦-y,t↦-t$ we get exactly the same equations! The backward trajectory is precisely the vertical reflection of the forward trajectory, and they start and end at the same points.
phase plane:
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Show that the system
\[
\frac{d x}{d t}=-y-x\left(x^2+y^2\right), \quad \frac{d y}{d t}=x-y\left(x^2+y^2\right)
\]
has no closed trajectories. Show that in polar coordinates the system becomes
\[
\frac{d r}{d t}=-r^3, \quad \frac{d θ}{d t}=1,
\]
and sketch the trajectories (you can solve this system explicitly).
Show that the only critical point is the origin. Consider a linear approximation to the $(x, y)$ system in a neighbourhood of the origin and determine its type. Comment on any differences between the approximation and the exact behaviour.
Solution. \[\frac{∂X}{∂x}+\frac{∂Y}{∂y}=-4(x^2+y^2)< 0 ∀(x,y)≠(0,0)\]By Bendixson–Dulac Theorem, the system has no closed trajectories. \begin{array}l r^2=x^2+y^2⇒r\dot r=x\dot x+y\dot y=-(x^2+y^2)^2=-r^4⇒\dot r=-r^3\\ y=r\sinθ⇒\dot y=\dot r\sinθ+r\dotθ\cosθ⇒x-yr^2=-r^3\sinθ+r\dotθ\cosθ⇒\dotθ=1 \end{array} Solving this system explicitly, \begin{cases} r=\left(2t+c_1\right)^{-1/2}\\ θ=t+c_2 \end{cases} ∴ the trajectory has polar equation $r=\left(2θ+c\right)^{-1/2}$
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Find $z(x, y)$ explicitly if
\[
-y z_x+x z_y=\frac{x z}{\sqrt{x^2+y^2}}
\]
and
\[
z(x, 0)=1 \quad \text { for } \quad x \geq 1 .
\]
Describe the projections of the characteristic curves upon the $(x, y)$-plane. In what region of the plane is $z(x, y)$ determined uniquely?
Solution.
We introduce a parameter $s$ for the data, say $γ(s)=(s,0,1)$, for $s≥1$, and then solve the characteristic equations with this as data at $t=0$. \begin{align*} \frac{dx}{dt}&=-y&x&=-A\sin t+B\cos t&x(0, s)&=B=s \\ \frac{dy}{dt}&=x&y&=A\cos t+B\sin t&y(0, s)&=A=0 \\ \frac{dz}{dt}&=\frac{xz}{\sqrt{x^2+y^2}}&z&=C\exp\left(\frac{A\cos t+B\sin t}{\sqrt{A^2+B^2}}\right)&z(0, s)&=C\exp\left(A\over\sqrt{A^2+B^2}\right)=1 \end{align*} So, $A=0,B=s,C=1$ and the parametric form of the solution is \begin{cases}x=s\cos t\\y=s\sin t\\z=\exp\left(\sin t\right)\end{cases} The characteristic projections are the circle with center $(0,0)$ and radius $s$. Given $s≥1$, the characteristic projections sweep out the exterior of unit circle.
$z(x,y)$ is 2π periodic, so $z(x,y)$ is determined uniquely on $\{(x,y)∈ℝ^2:x^2+y^2≥1\}$. $$z(x, y)=\exp\left(\frac{y}{\sqrt{x^2+y^2}}\right)$$ -
Find in parametric form the characteristics of the differential equation
\[
x z_x+y z_y=2 z,
\]
and describe the characteristic projections into the $(x, y)$-plane.
Find in explicit form the solution satisfying $z=x^3$ on $x+y=1$.
In what region of the plane is the solution uniquely determined by the data?
Solution.
We introduce a parameter $s$ for the data, say $γ(s)=(s,1-s,s^3)$, and then solve the characteristic equations with this as data at $t=0$. \begin{align*} \frac{dx}{dt}&=x&x&=A\exp t&x(0,s)&=A=s\\ \frac{dy}{dt}&=y&y&=B\exp t&y(0,s)&=B=1-s\\ \frac{dz}{dt}&=2z&z&=C\exp(2t)&z(0,s)&=C=s^3 \end{align*} The characteristic projections are halflines starting at $(0,0)$. The solution is uniquely determined by the data if $x+y>0$. $$z=\frac{A^3}{A+B}\exp(2t)=\frac{x^3}{x+y}$$
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Consider the differential equation
\[
z_x-y z_y=-z
\]
with data $z(0, y)=\min(1, y)$ for $-∞< y< ∞$.
(Here $\min (1, y)$ means the lesser of 1 and $y$, so the data is not smooth; draw its graph.)
Show that the solution of this problem is the following: \begin{aligned} z &=\mathrm{e}^{-x} &\text{ for }y≥\mathrm{e}^{-x}\\ &=y &\text{ for } y<\mathrm{e}^{-x} \end{aligned} Try to sketch the solution surface.
Solution.
We introduce a parameter $s$ for the data, say $γ(s)=(0,s,\min(1,s))$, and then solve the characteristic equations with this as data at $t=0$. \begin{align*} \frac{dx}{dt}&=1&x&=t+A&x(0,s)&=A=0\\ \frac{dy}{dt}&=-y&y&=B\exp(-t)&y(0,s)&=B=s\\ \frac{dz}{dt}&=-z&z&=C\exp(-t)&z(0,s)&=C=\min(1,s) \end{align*} The characteristic projections are the curves $y=s\exp(-x)$.For $y≥\mathrm{e}^{-x},s≥1$, we have $z=\exp(-t)=\exp(-x)$.
For $y<\mathrm{e}^{-x},s< 1$, we have $z=s\exp(-t)=y$.
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Consider the differential equation
\[
x z_x+y z_y=(x+y) z
\]
with $z=1$ on $y=x^2+1 / 4$ for $x ≥ 0$. Parametrise the initial curve as $\left(s, s^2+1 / 4,1\right), s≥0$, and find the characteristics of the differential equation, $x=x(s, t), y=y(s, t), z=z(s, t)$.
Identify the two separate segments of the data curve where the data is Cauchy. Find the domain of definition for each segment. On a sketch show the initial curve with these domains. For each segment solve the first two equations for $s$, and find $z$ explicitly for each segment. Comment on why we need to consider the segments separately.
Solution. \begin{align*} \frac{dx}{dt}&=x&x&=A\exp t&x(0,s)&=A=s\\ \frac{dy}{dt}&=y&y&=B\exp t&y(0,s)&=B=s^2+1/4\\ \frac{dz}{dt}&=(x+y)z&z&=C\exp((A+B)\exp t)&z(0,s)&=C=\exp\left(-\left(s+\frac12\right)^2\right) \end{align*} The data is Cauchy when $0≠P(x, y)y_s-Q(x, y)x_s=s⋅2s-(s^2+1/4)⋅1=s^2-1/4⇒s≠1/2$.
The two separate segments of the data curve are $0≤s< 1/2$ and $1/2< s≤1$. The domains of definition of both segments are $\{(x,y):y≥x≥0\}$ \[z=\exp\left(x+y-\left(s+\frac12\right)^2\right)=\exp\left(x+y-\left(\frac{x+y±\sqrt{y^2-x^2}}{2x}\right)^2\right)\] We get different solutions for the 2 segments.
