Math 4791/5791 Solutions 4
Fall 1998

FIGURES HAVE BEEN OMITTED.

2. Detailed analysis of a linear system. (Problem 5.1.9)

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  1. 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 tex2html_wrap_inline157 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).
  2. The characteristic polynomial of the matrix of coefficients is tex2html_wrap_inline167 which has roots tex2html_wrap_inline169 . Thus the origin is a saddle point. The eigenvector associated with tex2html_wrap_inline171 is tex2html_wrap_inline173 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 tex2html_wrap_inline177 is tex2html_wrap_inline179 which means that the stable manifold (the direction along which trajectories approach the origin) is y=x.
  3. The change of variables u=x+y and v=x-y is a rotation of the coordinate axes by tex2html_wrap_inline187 radians which aligns the (u,v) axes with the lines tex2html_wrap_inline191 . This change of variables also means that tex2html_wrap_inline193 and tex2html_wrap_inline195 . Now do the change of variables in the system of ODEs:

    displaymath111

    A little more simplification gives the decoupled equations u'(t)=-u(t) and v'(t)=v(t) which have solutions tex2html_wrap_inline201 and tex2html_wrap_inline203 where tex2html_wrap_inline205 and tex2html_wrap_inline207 are arbitrary initial conditions.

  4. 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).
  5. 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

    displaymath112

    where we have gathered all of the constants in the two arbitrary constants tex2html_wrap_inline215 and tex2html_wrap_inline217 .

4. Complex eigenvalues. (Extension of Problem 5.2.2) There are at least three approaches to this problem.

  1. As the book suggests, it is possible to find the eigenvalues and eigenvectors of the system. The characteristic polynomial is tex2html_wrap_inline219 which can be solved directly (avoid the quadratic formula) to give us the eigenvalues tex2html_wrap_inline221 . The eigenvector associated with tex2html_wrap_inline223 is tex2html_wrap_inline225 or tex2html_wrap_inline227 . 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 tex2html_wrap_inline229 :

    eqnarray25

    In this business, constants can be folded into the arbitrary constants tex2html_wrap_inline231 and tex2html_wrap_inline233 as the calculation proceeds.

  2. Adding i times the second ODE to the first gives us

    displaymath113

    Letting z=x+iy and tex2html_wrap_inline223 we have tex2html_wrap_inline241 which has solutions tex2html_wrap_inline243 . Letting tex2html_wrap_inline229 , it follows that

    eqnarray42

    This is the same general solution found in part(a) up to the arbitrary constants.

  3. 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.

tex2html_wrap371

6. Pursuit problem. The diagram for this problem is shown at the end of the problem. The goal is to find the function f such that y=f(x) describes the path of the master. As the pursuit is taking place both x and y are functions of t; in fact, x'(t) and y'(t) are the east-west and north-south components of the master's velocity and y'(x)=y'(t)/x'(t). The fact that the master's speed s (given) is constant throughout the pursuit is given by

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From this relation we find that

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where the minus sign insures that x is decreasing as it should. Noting that the position of the dog moving at one mile per hour northward is tex2html_wrap_inline313 and tex2html_wrap_inline315 , we can formulate the condition that the master always walks with the tangent to his/her path pointing at the dog as

displaymath116

Now the task is to reduce these two conditions in three variable to one condition in two variables by eliminating t. There are undoubtedly several ways to proceed. The shortest path I can find is to differentiate condition 2 with respect to t carefully! Note that tex2html_wrap_inline323 . It also helps to multiply across by x first:

displaymath117

Dividing through by x'(t) and substituting for x'(t) from condition 1a, we have

displaymath118

At last we have an ODE subject to the conditions y(1)=y'(1)=0 (when x=1, y and the slope of the pursuit curve are zero). The ODE is second order in y, but first order in y'. Therefore let v=y' and after separating variables we have

displaymath119

Integrating once gives us

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The condition y'(1)=v(1)=0 implies that C=1. Now solving for v leads us to

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which can be integrated directly to give

displaymath122

Using the condition y(1)=0 tells us that tex2html_wrap_inline351 . Finally we have the path of the master given by

displaymath123

What does it mean? Is it correct? First, check the initial conditions: indeed y(1)=y'(1)=0. Second we see that when x=0, tex2html_wrap_inline357 which gives us the location of the encounter (in miles) of the master and the dog along the y-axis; it also gives the time of the encounter in hours. This makes sense since as the speed of the master approaches the speed of the dog, the time to meeting gets arbitrarily large. If tex2html_wrap_inline361 there is no solution (no meeting time and place). If you plot the y=f(x) curves for various s>1, you see the paths of the master tex2html_wrap_inline367 .


Wed Sep 30 17:07:03 MDT 1998