Phase-plane analysis

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Consider the differential equations

dxdt=−x+y,dydt=log⁡(20+x)−y. \frac{dx}{dt} = -x+y, \quad \frac{dy}{dt} = \log (20+x) - y.

Let x(t)x(t) and y(t)y(t) be a solution defined for all t≥0t \geq 0 with x(0)>0x(0) > 0 and y(0)>0y(0) > 0. Prove that x(t)x(t) and y(t)y(t) are bounded.

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Consider the system of differential equations

dxdt=y+x(1−x2−y2) \frac{dx}{dt} = y + x(1-x^2-y^2) dydt=−x+y(1−x2−y2). \frac{dy}{dt} = -x + y(1-x^2-y^2).
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(1)

Show that for any x0x_0 and y0y_0, there is a unique solution (x(t),y(t))(x(t), y(t)) defined for all t∈Rt \in \mathbb {R} such that x(0)=x0x(0) = x_0, y(0)=y0y(0) = y_0.

(2)

Show that if x0≠0x_0 \neq 0 and y0≠0y_0 \neq 0, the solution referred to in Part 1 approaches the circle x2+y2=1x^2+y^2=1 as t→∞t \to \infty.

Consider the solution curve (x(t),y(t))(x(t), y(t)) to the equations

dxdt=1+12x2sin⁡y \frac{dx}{dt} = 1 + \frac{1}{2}x^2 \sin y dydt=3−x2 \frac{dy}{dt} = 3 - x^2

with initial conditions x(0)=0x(0) = 0 and y(0)=0y(0) = 0. Prove that the solution must cross the line x=1x=1 in the xyxy plane by the time t=2t=2.

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Consider the differential equation x′′+x′+x3=0x'' + x' + x^3 = 0 and the function f(x,x′)=(x+x′)2+(x′)2+x4f(x,x') = (x+x')^2 + (x')^2 + x^4.

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(1)

Show that ff decreases along trajectories of the differential equation.

(2)

Show that if x(t)x(t) is any solution, then (x(t),x′(t))(x(t), x'(t)) tends to (0,0)(0,0) as t→∞t \to \infty.

Consider the differential equation

dxdt=y,dydt=−ay−x3−x5,where a>0. \frac{dx}{dt} = y, \quad \frac{dy}{dt} = -ay - x^3 - x^5, \quad \text{where } a > 0.
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(1)

Show that

F(x,y)=y22+x44+x66 F(x,y) = \frac{y^2}{2} + \frac{x^4}{4} + \frac{x^6}{6}

decreases along solutions.

(2)

Show that for any ε>0\varepsilon > 0, there is a δ>0\delta > 0 such that whenever ∥(x(0),y(0))∥<δ\left\| (x(0),y(0))\right\| < \delta, there is a unique solution (x(t),y(t))(x(t),y(t)) of the given equations with the initial condition (x(0),y(0))(x(0),y(0)) which is defined for all t≥0t \geq 0 and satisfies ∥(x(t),y(t))∥<ε\left\| (x(t),y(t))\right\| < \varepsilon.

Transform the system of equations

dxdt=y+kx(x2+y2), \dfrac {dx}{dt}=y+kx(x^{2}+y^{2}) , dydt=−x+ky(x2+y2) \dfrac {dy}{dt}=-x+ky(x^{2}+y^{2})

into a system of equations in polar coordinates.

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Consider the Lotka-Volterra model

du1dt=f1(u1,u2)=u1−u1u2,du2dt=f2(u1,u2)=−u2+u1u2 \frac{du_{1}}{dt} = f_{1}(u_{1}, u_{2}) = u_{1} - u_{1}u_{2}, \qquad \frac{du_{2}}{dt} = f_{2}(u_{1}, u_{2}) = -u_{2} + u_{1}u_{2}

with the fixed point (u1∗=1,u2∗=1)(u_{1}^{*} = 1, u_{2}^{*} = 1). The linearized equation (variational equation) with

∂f1∂x1=1−u2,∂f1∂x2=−u1,∂f2∂x1=u2,∂f2∂x2=−1+u1 \frac{\partial f_{1}}{\partial x_{1}} = 1 - u_{2}, \quad \frac{\partial f_{1}}{\partial x_{2}} = -u_{1}, \quad \frac{\partial f_{2}}{\partial x_{1}} = u_{2}, \quad \frac{\partial f_{2}}{\partial x_{2}} = -1 + u_{1}

is given by

(dy1/dtdy2/dt)=(0−110)(y1y2). \begin{pmatrix} dy_{1}/dt \\ dy_{2}/dt \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} y_{1} \\ y_{2} \end{pmatrix} .

Solve this system of linear differential equations. First show that the eigenvalues of the 2×22 \times 2 matrix are given by ±i\pm i.

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References

[deSi04] de Souza P.N., Silva J. Berkeley Problems in Mathematics. 3rd ed., Springer, 2004.
[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[StHa17] Steeb W., Hardy Y. Problems and Solutions in Introductory and Advanced Matrix Calculus. 2nd ed., World Scientific Publishing, 2017.