Lecture Notes in Differential Equations - Bruce E. Shapiro

375
Since both f and g are differentiable, their quotient is differentiable (away
from regions wheref = 0) and hence Lipshitz; consequently the initial value
problem
dy g(x, y)
=
dx f(x, y) , y(x 0) = y 0 (34.10)
has a unique solution, which corresponds to the trajectory of equation
(34.8) through the point (x 0 , y 0 ). Since the solution is unique, we conclude
that there is only one trajectory through any point, except possibly
wheref(x, y) = 0. A plot showing the one-parameter family of solutions to
(34.9), annotated with arrows to indicate the direction of motion in time,
is called a phase portrait of the system.
Example 34.2. Let {x 1 , y 1 } and {x 2 , y 2 } be the solutions of
and
respectively.
x ′ = −y, x(0) = 1 (34.11)
y ′ = x, y(0) = 0 (34.12)
x ′ = −y, x(π/4) = 1 (34.13)
y ′ = x, y(π/4) = 0 (34.14)
The solutions are different, but both solutions follow the same trajectory.
To see this we solve the system by forming the second order differential
equation representing this system: differentiate the second equation to obtain
y ′′ = x ′ ; then substitute the first equation to obtain
y ′′ = −y (34.15)
The characteristic equation is r 2 +1 = 0 with roots of ±i; hence the solutions
are linear combinations of sines and cosines. As we have seen, the general
solution to this system is therefore
y = A cos t + B sin t (34.16)
x = −A sin t + B cos t (34.17)
where the second equation is obtained by differentiating the first (because
x = y ′ ).