Answers |
---|
Problem 1. Consider the system
Draw the nullclines and find all equilibrium points. Determine the fate of the solutions with initial conditions
Problem 2. Find the solution of
with the initial condition
Problem 3. Find the solution of
with the initial condition
Problem 4. Consider the system
Find the equilibrium points. Find the linearized system at these
points and discuss the behavior of the solutions at these points.
Determine if the equilibrium points are sinks, saddles, sources, and
so on...
Answers.
Answer to Problem 1. Let us find the x-nullclines and
y-nullclines.
which is equivalent to x = 0 or -4x-y+1 = 0.
which is equivalent to y = 0 or (which is the equation of a circle centered at (0,0)).
The critical points are the intersection between the x-nullclines and y-nullclines. Hence the equilibrium points are
See the figure below for more details about the nullclines and the
solutions.
Answer to Problem 2. The matrix coefficient is
The characteristic equation is
Its roots are : -3 and -3 (-3 is a double root). An eigenvector associated to -3, is
We may choose
The general solution is
where
satisfies
This gives . Hence the vector
will do (here we took ). Hence the general solution is
The initial condition Y(0) = (1,0) gives
This implies and . Therefore the solution to the initial condition is
Answer to Problem 3. The matrix coefficient is
The characteristic equation is
Its roots are
(these are complex roots). An eigenvector associated to 4 + 2i, is
We may choose
WE need the real part and imaginary part of
Easy calculations give
where
and
The general solution is
The initial condition Y(0) = (1,1) gives
which yields
This implies and . Therefore the solution to the initial condition is
Answer to Problem 4. Let us find the x-nullclines and
y-nullclines.
which is equivalent to x = 0 or . This gives
which is equivalent to y = 0 or y = 1.
The critical points are the intersection between the x-nullclines and y-nullclines. Hence the equilibrium points are
In order to find the linearized system at these points, we must find the Jacobian of the system which is
Therefore, we have
The eigenvalues are 1 and -1. Therefore, (0,0) is a saddle.
The eigenvalues are 1 (which is double). Therefore, (1,1) is a source.
The eigenvalues are -1 and -2. Therefore, is a sink.
The eigenvalues are 1 and -2. Therefore, is a saddle.
See the figure below for more details on the behavior of some
solutions.
Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.