Given the continuous real dynamical system dx/dt = Ax, we want to solve the system in the case where we get complex eigenvalues. If the eigenvalues and eigenvectors of the 2 x 2 matrix A turn out to be

        eigenvalues:  p + iq      and       p - iq

        eigenvectors: v + iw      and       v - iw

Then we showed in class that all solutions can be expressed as linear combinations of the two real solutions given below:

             e^pt cos(qt)v  -  e^pt sin(qt)w

             e^pt cos(qt) w +  e^pt cos(qt)v

Example: Again we look at salt solutions in a tank. Let me try to draw the picture:


 _________                     _______
|        |____________________|       |
|        |___________________ |       |
|        |       -----        |       |
|        |______|     |_______|       |
|        |______|     |_______|       |
|        |      |     |       |       |
|________|      |_____|       |_______|

  tank1           tank2         tank3

Now assume that the mixture in tank 1 flows into tank 2 at the rate of 10 gallons / minute, the mixture from tank 2 flows into tank 3 at the same rate and the mixture in tank 3 flows into tank 1 at the same rate. Thus the volume in each tank will remain constant over time. Assume that tanks 1 and 3 have 50 gallons of salt solution and tank 2 has 25 gallons of salt solution. Let x1, x2 and x3 denote the number of pounds of salt in the tanks. We want to understand how this changes over time.

    dx1/dt = (rate salt flows into tank 1 ) - (rate salt flows out of tank 1)

= (10 gal/min) (x3 lbs/ 50 gal) - (10 gal/min)(x1 lbs/50 gal)

    So dx1/dt = x3/5  - x1/5

Similarly since salt solution in the concentration of x1 lbs / 50 gallons flows into the second tank from the first at the rate of 10 gallons per minute and salt solution with concentration x2 lbs/ 25 gallons flows out of the second tank at the rate of 10 gallons per minute we get an expression for dx2/dt:

       dx2/dt = 10*x1/50 - 10*x2/25 =   x1/5 - 2*x2/5

Similarly,

       dx3/dt = 10*x2/25 - 10*x3/50  = 2*x2/5 - x3/5.

Thus we have a dynamical system

    [ dx1/dt ]      [ -1/5    0     1/5 ]  [ x1 ]

    [ dx2/dt ]  =   [  1/5  -2/5     0  ] *[ x2 ]

    [ dx3/dt ]      [   0    2/5   -1/5 ]  [ x3 ]

To solve this system we compute the eigenvalues and eigenvectors of this matrix and we get

      lambda = 0        lambda = -2/5 + i/5         lambda = -2/5 - i/5

       [ 2 ]              [   1  ]                       [   1   ]

       [ 1 ]              [  -i  ]                       [   i   ]       as the corresponding

       [ 2 ]              [-1 + i]                       [ -1 - i]       eigenvectors.

The first eigenvalue tells us that the solutions will contain a component of the form

            [ 2 ]       [ 2 ]

      e^0t* [ 1 ]     = [ 1 ]

            [ 2 ]       [ 2 ]

The other two solutions will be of the form

         e^(-2t/5)cos(t/5) v -e^(-2t/5)sin(t/5) w       where v = [ 1 ]  and   w = [  0 ]

[ 0 ]            [ -1 ]

 e^(-2t/5)cos(t/5) w +e^(-2t/5)sin(t/5) v                 [-1 ]            [  1 ]

The general solution will be all linear combinations of these three so we'll get solutions of the form

       x1(t) =  c1*2     +       e^(-2t/5)*c2*cos(t/5)          +           e^(-2t/5) *c3*sin(t/5)

       x2(t) =  c1*1     +       e^(-2t/5) *c2*sin(t/5)          -           e^(-2t/5) *c3*cos(t/5)

       x3(t) =  c1*2     -  e^(-2t/5) *c2(cos(t/5)  + sin(t/5))  + e^(-2t/5) *c3( cos(t/5) - sin(t/5)

Remark1: When we take linear combinations of sines and cosines of the same frequency we just change the amplitude and introduce a phase shift:

     c1*cos(qt) + c2*sin(qt)  can be rewritten as A*sin(qt + phi)  since

                                Asin(qt + phi) = A sin(qt)sin(phi) + A cos(qt) cos(phi)

      so choose A and phi to satisfy  Acos(phi) = c1 and Asin(phi)=c2.

So a function of the form exp(pt) times a linear comb of cos(qt) and sin(qt) is just a sine wave of the same frequency, but with an amplitude that is an exponential function of time.