Find all the stationary solutions to the following system of differential equations.
\(\frac{\dx}{\dt} = 1 + 2y, \hspace{.3in} \frac{\dy}{\dt} = 1 - 3x^2 \)
\(\frac{\dx}{\dt} = x - y, \hspace{.3in} \frac{\dy}{\dt} = y + 2x \)
\(\frac{\dx}{\dt} = y, \hspace{.3in} \frac{\dy}{\dt} = x(x-2)(y-2) \)
\(\frac{\dx}{\dt} = -(2+y)y, \hspace{.3in} \frac{\dy}{\dt} = x(1-y)\)
\(\frac{\dx}{\dt} = x(1 - x - y), \hspace{.3in} \frac{\dy}{\dt} = (1-y)(2+x) \)
\(\frac{\dx}{\dt} = (2+x)(y-x), \hspace{.3in} \frac{\dy}{\dt} = y(2 + x - x^2) \)
For problems \(7-10\), find the semistationary solutions. If they do not exist, state as such.
\(\frac{\dx}{\dt} = y, \hspace{.3in} \frac{\dy}{\dt} = x^2 \)
\(\frac{\dx}{\dt} = y(x-1), \hspace{.3in} \frac{\dy}{\dt} = (x-2)(y-2)\)
\(\frac{\dx}{\dt} = (x^2 - 2x)(y+1), \hspace{.3in} \frac{\dy}{\dt} = y(2 + x^2)\)
\(\frac{\dx}{\dt} = x(1 + y), \hspace{.3in} \frac{\dy}{\dt} = (y^2 - y)(x+1) \)
Show that if \((x(t),y(t))\) is a solution to a Hamiltonian system, then \(H(x(t),y(t))\) is some constant.
Consider the Hamiltonian system \[x^\prime = \partial_yH(x,y), \quad y^\prime = -\partial_xH(x,y)\] and some function \(g(x,y)\) which satisfies \[\partial_xg(x,y)\partial_yH(x,y) = \partial_xH(x,y)\partial_yg(x,y).\] Show that \(g(x(t),y(t))\) is constant.
Show the system is Hamiltonian and find its Hamiltonian \(H(x,y)\)
\(\frac{\dx}{\dt} = y, \hspace{.3in} \frac{\dy}{\dt} = -x + \frac{x^3}{6}\)
\(\frac{\dx}{\dt} = (y-x), \hspace{.3in} \frac{\dy}{\dt} = (y-x)(2+x) \)
\(\frac{\dx}{\dt} = 1 + 2y, \hspace{.3in} \frac{\dy}{\dt} = 1 - 3x^2 \)
\(\frac{\dx}{\dt} = x-xy, \hspace{.3in} \frac{\dy}{\dt} = -y + \frac{y^2}{2} \)
\(\frac{\dx}{\dt} = -x-x^2, \hspace{.3in} \frac{\dy}{\dt} = y + 2xy \)
\(\frac{\dx}{\dt} = x(-2 + \frac{x}{2} + \frac{x^2}{3}), \hspace{.3in} \frac{\dy}{\dt} = y(2-x-x^2) \)
Draw the phase portraits for the following system of equations
\(\frac{\dx}{\dt} = y, \hspace{.3in} \frac{\dy}{\dt} = 2x-x^3 \)
\(\frac{\dx}{\dt} = 2y, \hspace{.3in} \frac{\dy}{\dt} = -3x\)
\(\frac{\dx}{\dt} = -x + y + x^2, \hspace{.3in} \frac{\dy}{\dt} = y - 2xy \)
\(\frac{\dx}{\dt} = 2 - y, \hspace{.3in} \frac{\dy}{\dt} = 4 - x^2\)
\(\frac{\dx}{\dt} = y, \hspace{.3in} \frac{\dy}{\dt} = -x + \frac{x^2}{4}\)
\(x^\prime = x^2 + y - 1, \hspace{.3in} y^\prime = 2xy\)
\(\frac{\dx}{\dt} = y-2, \hspace{.3in} \frac{\dy}{\dt} = \sin(x) \)
Consider a body falling through the air with drag as described in the text. Recall that the equations of motion are \[\frac{\dee}{\dt}u = -ku\sqrt{u^2 + v^2}, \qquad \frac{\dee}{\dt}v = -kv\sqrt{u^2 + v^2} - a.\] where \(u\) and \(v\) are the horizonal and vertical components of the velocity \(\uBld\), \(a>0\) is the accelleration due to gravity and \(k = \rho_{\mathrm{air}}A/m\), where \(\rho_{\mathrm{air}}\) is the density of air, \(A\) is the crosssectional area of the falling body, and \(m\) is the mass of the body.
Often times a good candidate for a Hamiltonian for a physical system is the total mechanical energy. In this case it is given by \[E(t) = \frac{1}{2}m|\uBld(t)|^2 + ma\int_0^t v(s)\;\ds.\] Show that the system is always losing energy, i.e. \[\frac{\dee}{\dt}E(t) = - \rho_{\mathrm{air}}A|\uBld|^3 \leq 0.\] What is the rate of energy loss at the stationary solution (i.e. at terminal velocity?).
Suppose a particle of mass \(m\) is the moving in the presence of an attractive central potential \(\displaystyle V(r) = -\frac{1}{\alpha r^\alpha}\), here \(\alpha\) is a positive integer. It is well known from Physics that the distance \(r\) of the particle from the origin is govered by the equation \[\frac{\dee^2r}{\dt^2} - \frac{L^2}{mr^3} + \frac{1}{mr^{\alpha+1}}=0,\quad r>0,\] where \(L\) is the angular momentum.
Write this second order differential equation as a first order planar system and show that it is Hamiltonian. Give its Hamiltonian \(H\).
Solve the differential equation for \(r\) in the case \(\alpha = 2\), \(r(0) = r_0 >0\), and \(r^\prime(0) = 0\) by using the Hamiltonian to reduce the equations of motion for \(r\) to a first order seperable differential equation.
In your solution for part (b) suppose the angular momentum \(L > 1\). What is the behavior of \(r\) as \(t\to \infty\).
In your solution for part (b), suppose \(L<1\). After what time \(t_0\) does the solution become undefined? What does \(r(t)\) approach as \(t\to t_0\). What happens to \(p(t)\) as \(t\to t_0\)?
In your solution for part (b), what happens when \(L=1\).
Consider Newton’s equations for a one-dimensional particle of mass \(m\) in a potential \(V(x)\), \[m\frac{\dee^2}{\dt^2} x = - V^\prime(x).\]
Write this equation as a planar system. Show that this system is Hamiltonian and find its’ Hamiltonian \(H\).
Using the Hamiltonian to reduce the system, show that the evolution of \(x\) can be given implicitly in terms of the formula, \[t = \pm\int \frac{1}{\sqrt{c - \frac{2}{m}V(x)}}\;\dx.\]
Consider a Hamiltonian of the form \(H(x,y) = \frac{1}{2}y^2 + V(x)\), and the corresponding Hamiltonian system \[x^\prime = \partial_yH(x,y), \quad y^\prime = -\partial_xH(x,y).\] We will assume that \(V(x)\) is a smooth function. Show the following
Equilibrium points of are of the form \((x_0,0)\), where \(x_0\) is a critical point of \(V(x)\).
If \(x_0\) is a (strict) local maximum of \(V(x)\), then \((x_0,0)\) is a saddle point for the system.
If \(x_0\) is a (strict) local minimum of \(V(x)\), then \((x_0,0)\) is a locally a center
(Liouville’s Theorem) Consider a Hamiltonian system \(x^\prime = \partial_yH(x,y)\), \(y^\prime = -\partial_xH(x,y)\) and define \[\PhiBld_t(x_0,y_0) = \begin{pmatrix} \varphi_t(x_0,y_0)\\ \psi_t(x_0,y_0) \end{pmatrix} := \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}\] to be the unique solution \((x(t),y(t))\) to the Hamiltonian system with initial conditions \((x(0),y(0)) = (x_0,y_0)\). Let \(A\) be a set of initial points in \({\mathbb R}^2\), then \[\PhiBld_t(A) = \{\PhiBld(x,y) : (x,y) \text{ is in } A\}\] is the evolution of that set of initial points under the dynamics of the system (think of a blob of ink moving in a fluid). Answer the following:
Show that \[{\boldsymbol \partial}\PhiBld_t(x,y) = \begin{pmatrix} \partial_x\varphi_t(x,y) & \partial_y\varphi_t(x,y)\\ \partial_x\psi_t(x,y) & \partial_y\psi_t(x,y) \end{pmatrix}\] satisfies \[\frac{\dee}{\dt}{\boldsymbol \partial}\PhiBld_t(x,y) = \begin{pmatrix} \partial_{yx}H(\PhiBld_t(x,y)) & \partial_{yy}H(\PhiBld_t(x,y))\\ -\partial_{xx}H(\PhiBld_t(x,y)) & - \partial_{xy}H(\PhiBld_t(x,y)) \end{pmatrix} {\boldsymbol \partial}\PhiBld_t(x,y).\]
Use Liouville’s Wronskian Theorem to conclude that \[\frac{\dee}{\dt} \det\left({\boldsymbol \partial}\PhiBld_t(x,y)\right) = 0.\]
Finally, use the following multivariable calculus formula for the volume of a set \[\text{Vol}\left(\PhiBld_t(A)\right) = \iint_A\det\left({\boldsymbol \partial}\PhiBld_t(x,y)\right)\;\dx\dy\] to show that \(\text{Vol}\left(\PhiBld_t(A)\right) = \text{Vol}\left(A\right)\) (i.e. Hamiltonian evolution preserves the volume of sets of initial points).