Math 4791/5791 Solutions 3
Spring 2000
1.
Detailed analysis of a linear system. (Problem 5.1.9)
$\textstyle \parbox{3.5in}{ \item To plot the direction field note that $x'(t)=... ...igin, so the arrows should increase in length as they recede from the origin.}$
(a)
It is easiest to eliminate t from the problem by dividing one ODE into the other to get y'(x)=x/y. This equation is separable and can be integrated directly to give x2-y2=C where C is an arbitrary constant. This says that the solutions are hyperbolas symmetric about both the x- and y- axes (depending on the sign of C).
(b)
The characteristic polynomial of the matrix of coefficients is $\lambda^2-1=0$ which has roots $\lambda=\pm 1$. Thus the origin is a saddle point. The eigenvector associated with $\lambda=1$ is (1,-1)T which means that the unstable manifold (just a fancy way of saying the direction along which trajectories escape) is y=-x. The eigenvector associated with $\lambda=-1$ is (1,1)T which means that the stable manifold (the direction along which trajectories approach the origin) is y=x.
(c)
The change of variables u=x+y and v=x-y is a rotation of the coordinate axes by $\pi/4$ radians which aligns the (u,v) axes with the lines $y=\pm x$. This change of variables also means that $x=\frac 12 (u+v)$ and $y=\frac 12 (u-v)$. Adding and subtracting the two ODEs, we find a new system for u and v:
x'(t)+y'(t) = $\displaystyle -x(t)-y(t) \quad \mbox{or} \quad u'(t)=-u(t)$ (1)
x'(t)-y'(t) = $\displaystyle x(t)-y(t) \quad \mbox{or} \quad v'(t)=-v(t)$ (2)

These decoupled equations can be solved immediately u(t)=u0e-t and v(t)=v0et where u0 and v0 are arbitrary initial conditions. The change of variables rotates the axes so they are aligned with the eigenvectors.
(d)
In the (u,v) coordinate system, trajectories approach the origin along the line u=0 (the stable manifold) and recede from the origin along the line v=0 (the unstable manifold).
(e)
Using the change of coordinate equations given above, it is now possible to write the solution in terms of the original variables. We find that

\begin{displaymath}x(t)=c_1 e^{-t} +c_2 e^t \quad \mbox{and} \quad y(t)=c_1 e^{-t} -c_2 e^t \end{displaymath}

where we have gathered all of the constants in the two arbitrary constants c1 and c2.
2.
Complex eigenvalues. (Extension of Problem 5.2.2) There are at least three approaches to this problem.
(a)
As the book suggests, it is possible to find the eigenvalues and eigenvectors of the system. The characteristic polynomial is $(a-\lambda)^2+b^2=0$ which can be solved directly (avoid the quadratic formula) to give us the eigenvalues $\lambda=a \pm ib$. The eigenvector associated with $\lambda=a+ib$ is ${\bf v} = (i,1)^T$ or ${\bf v} = (-1i)^T$ . Recall from the discussion in class that the eigenvectors and eigenvalues appear in complex conjugate pairs. We can go directly from here to the general solution with the arbitrary constant $c=\alpha+i\beta$:

\begin{eqnarray*}{\bf x}(t) = \left( \begin{array}{c} x(t) \\ y(t) \end{array... ...bt) \\ \beta \sin(bt)+ \alpha \cos(bt) \end{array} \right). \end{eqnarray*}


In this business, constants can be folded into the arbitrary constants $\alpha$ and $\beta$ as the calculation proceeds.
(b)
Adding i times the second ODE to the first gives us

(x(t)+iy(t))'=ax-by+i(bx+ay) = a(x+iy)+ib(x+iy).

Letting z=x+iy and $\lambda=a+ib$ we have $z'(t)=\lambda z(t)$ which has solutions $z(t)=ce^{\lambda t}$. Letting $c=\alpha+i\beta$, it follows that

\begin{eqnarray*}x(t) = \mbox{Real} \;\; z(t)& = & e^{at}(\alpha \cos(bt)-\beta ... ...{Imag} \;\; z(t) & = & e^{at}(\alpha \sin(bt)+\beta \cos(bt)). \end{eqnarray*}


This is the same general solution found in part(a) up to the arbitrary constants.
(c)
We see that the parameter a determines whether the trajectories approach or recede from the origin. If a>0 then the trajectories spiral outward with an angular frequency of b. If a<0 then the trajectories spiral inward with an angular frequency of b. And if a=0 the trajectories are circles about the origin.




$\textstyle \parbox{3.0in}{ \item {\bf Classification of systems.} (Problem 5.2... ...ifold is the line $y=x$\space and whose stable manifold is the line $y=x/2$ .}$



3.
The pig problem. Note that all quantities have constant absolutes rate of change, which implies linear change. Let t=0 correspond to the day that the pig weighs 200 pounds. The pig's weight is given by w(t)=200+5t. The price per pound is given by p(t)= 0.65-0.01t. So the market value of the pig is m(t) = w(t)p(t) = (200+5t)( 0.65-0.01t). However, the accumulated cost for feeding the pig is c(t) = 0.45t. So the net value of the pig is V(t)=m(t)-c(t)= (200+5t)( 0.65-0.01t)-0.45t. The pig should be sold when the net value is a maximum. From here, it's straightforward: the graph of V(t) is a parabola with a single maximum located at the point where V'(t)=0. The net value of the pig is a maximum at t=8 days.
4.
The soda can problem. In order to clarify ideas, I will solve a slightly simpler problem: let's imagine the soda confined within a weightless can, which allows us to neglect the can. The method of solution with the can exactly the same with slightly different numbers. Also the final result is the same: the center of gravity reaches the lowest point when the height of the soda equals the height of the center of gravity. The main fact we need is the old teeter-totter principle: if a person of mass m1 is located at a coordinate x1 and another person of mass m2 is located at a coordinate x2, the balance point (center of gravity) has the coordinate $\overline{x}$ where $m_1(x_1-\overline{x}) = m_2(x_2-\overline{x})$. For example, a 100 kg person sitting 3 meters from the pivot balances a 150 kg person sitting 2 meters from the pivot (in which case, $\overline{x}=0$). Solving this equation for the position of the center of gravity we have

\begin{displaymath}\overline{x}= \frac{m_1x_1+m_2x_2}{m_1+m_2}. \end{displaymath}

$\textstyle \parbox{3.0in}{ As shown in the figure to the right, assume the bot... ...ocalized at its midpoint which is $x=(L+h)/2$ . Thus the center of gravity is}$

\begin{displaymath}\overline{x}= \frac{ \pi r^2 h \rho_\ell\cdot \frac h2 + \pi ... ...ell h^2 + \frac 12 \rho_a(L^2-h^2)}{\rho_\ell h+\rho_a(L-h)}. \end{displaymath}

We now have the center of gravity, $\overline{x}$, as a function of the soda level, h. What value of h minimizes $\overline{x}$? With a bit of work, we can compute $\overline{x}'(h)$ and set it equal to zero. This leads to the quadratic

\begin{displaymath}h^2 + \frac{2\rho_aL}{\rho_\ell-\rho_a}h-\frac{\rho_aL^2}{\rho_\ell-\rho_a}=0. \end{displaymath}

The positive root that corresponds to the physical solution is

\begin{displaymath}h^\ast =\frac{\rho_aL}{\rho_\ell-\rho_a} \left(\sqrt{\frac{\rho_\ell}{\rho_a}}-1 \right). \end{displaymath}

Using the given values of the constants ( $\rho_a=0.001, \rho_\ell=1,L=12$), we find that h= 0.368. (In cases like ours for which $\rho_a <<\rho_\ell$, we can make the good approximation $h^\ast \approx L\sqrt{\rho_a/\rho_\ell}=0.379$.) Recall that these results assume no can. The graph of $\overline{x}$ as a function of h is show below for the given constants and for the ``heavy air'' case $\rho_a=0.1, \rho_\ell=1,L=12$. The identity line $\overline{x}=h$ is also shown. In both cases, it appears that the minimum point on the curve occurs at the point of intersection with the identity line. This is the case: the point at which the two curves intersect satisfies $\overline{x}(h)=h$ which reduces to the quadratic found above. Thus we have the somewhat unexpected result that the lowest level of the center of gravity occurs when the height of the soda equals the height of the center of gravity.
\includegraphics[height=2.0in,width=6.0in]{sol3_1.eps}