Solving Coupled Differential Equations
How do I write a system of coupled differential equations in matrix form?
 The coupled differential equations considered in this part of the course will be of the form

 are constants whose precise value will depend on the situation being modelled
 In an exam question the values of the constants will generally be given to you
 are constants whose precise value will depend on the situation being modelled
 This system of equations can also be represented in matrix form:
 It is usually more convenient, however, to use the ‘dot notation’ for the derivatives:
 This can be written even more succinctly as
 Here , , and
How do I find the exact solution for a system of coupled differential equations?
 The exact solution of the coupled system depends on the eigenvalues and eigenvectors of the matrix of coefficients
 The eigenvalues and/or eigenvectors may be given to you in an exam question
 If they are not then you will need to calculate them using the methods learned in the matrices section of the course
 On the exam you will only be asked to find exact solutions for cases where the two eigenvalues of the matrix are real, distinct, and nonzero
 Similar solution methods exist for nonreal, nondistinct and/or nonzero eigenvalues, but you don’t need to know them as part of the IB AI HL course
 Let the eigenvalues and corresponding eigenvectors of matrix be and , and and , respectively
 Remember from the definition of eigenvalues and eigenvectors that this means that and
 The exact solution to the system of coupled differential equations is then
 This solution formula is in the exam formula booklet
 are constants (they are essentially constants of integration of the sort you have when solving other forms of differential equation)
 If initial or boundary conditions have been provided you can use these to find the precise values of the constants and
 Finding the values of and will generally involve solving a set of simultaneous linear equations (see the worked example below)
Worked Example
The rates of change of two variables, and , are described by the following system of coupled differential equations:
Initially and .
Given that the matrix has eigenvalues of and with corresponding eigenvectors and , find the exact solution to the system of coupled differential equations.
Phase Portraits
What is a phase portrait for a system of coupled differential equations?
 Here we are again considering systems of coupled equations that can be represented in the matrix form , where , , and
 A phase portrait is a diagram showing how the values of x and y change over time
 On a phase portrait we will usually sketch several typical solution trajectories
 The precise trajectory that the solution for a particular system will travel along is determined by the initial conditions for the system
 Let and be the eigenvalues of the matrix
 The overall nature of the phase portrait depends in large part on the values of and
What does the phase portrait look like when and are real numbers?
 Recall that for real distinct eigenvalues the solution to a system of the above form is , where and are the eigenvalues of and and are the corresponding eigenvectors
 AI HL only considers cases where and are distinct (i.e., ) and nonzero
 A phase portrait will always include two ‘eigenvector lines’ through the origin, each one parallel to one of the eigenvectors

 So if and , for example, then these lines through the origin will have equations and , respectively
 These lines will define two sets of solution trajectories
 If the eigenvalue corresponding to a line’s eigenvector is positive, then there will be solution trajectories along the line away from the origin in both directions as t increases
 If the eigenvalue corresponding to a line’s eigenvector is negative, then there will be solution trajectories along the line towards the origin in both directions as t increases
 No solution trajectory will ever cross an eigenvector line

 If both eigenvalues are positive then all solution trajectories will be directed away from the origin as t increases
 In between the ‘eigenvector lines’ the trajectories as they move away from the origin will all curve to become approximately parallel to the line whose eigenvector corresponds to the larger eigenvalue
 If both eigenvalues are negative then all solution trajectories will be directed towards the origin as t increases
 In between the ‘eigenvector lines’ the trajectories will all curve so that at points further away from the origin they are approximately parallel to the line whose eigenvector corresponds to the more negative eigenvalue
 They will then converge on the other eigenvalue line as they move in towards the origin
 In between the ‘eigenvector lines’ the trajectories will all curve so that at points further away from the origin they are approximately parallel to the line whose eigenvector corresponds to the more negative eigenvalue
 If one eigenvalue is positive and one eigenvalue is negative then solution trajectories will generally start by heading in towards the origin before curving to head out away again from the origin as t increases
 In between the ‘eigenvector lines’ the solution trajectories will all move in towards the origin along the direction of the eigenvector line that corresponds to the negative eigenvalue, before curving away and converging on the eigenvector line that corresponds to the positive eigenvalue as they head away from the origin
What does the phase portrait look like when and are imaginary numbers?
 Here the solution trajectories will all be either circles or ellipses with their centres at the origin
 You can determine the direction (clockwise or anticlockwise) and the shape (circular or elliptical) of the trajectories by considering the values of and for points on the coordinate axes
 For example, consider the system
 The eigenvalues of are and , so the trajectories will be elliptical or circular
 When and , and
 This shows that from a point on the positive axis the solution trajectory will be moving ‘to the right and up’ in the direction of the vector
 When and , and
 This shows that from a point on the positive axis the solution trajectory will be moving ‘to the left and down’ in the direction of the vector
 The directions of the trajectories at those points tell us that the directions of the trajectories will be anticlockwise
 They also tell us that the trajectories will be ellipses
 For circular trajectories, the direction of the trajectories when they cross a coordinate axis will be perpendicular to that coordinate axis
What does the phase portrait look like when and are complex numbers?
 In this case and will be complex conjugates of the form , where and are nonzero real numbers
 If , , then we have the imaginary eigenvalues case above
 Here the solution trajectories will all be spirals
 If the real part of the eigenvalues is positive (i.e., if ), then the trajectories will spiral away from the origin
 If the real part of the eigenvalues is negative (i.e., if ), then the trajectories will spiral towards the origin
 You can determine the direction (clockwise or anticlockwise) of the trajectories by considering the values of and for points on the coordinate axes
 For example, consider the system
 The eigenvalues of are and , so the trajectories will be spirals
 For example, consider the system
 Because the real part of the eigenvalues is positive, the trajectories will spiral away from the origin
 When and , and
 This shows that from a point on the positive axis the solution trajectory will be moving ‘to the right and down’ in the direction of the vector
 When and , and
 This shows that from a point on the positive axis the solution trajectory will be moving ‘to the right and up’ in the direction of the vector
 The directions of the trajectories at those points tell us that the directions of the trajectory spirals will be clockwise
Worked Example
Consider the system of coupled differential equations
Given that and are the eigenvalues of the matrix , with corresponding eigenvectors and , draw a phase portrait for the solutions of the system.
Equilibrium Points
What is an equilibrium point?
 For a system of coupled differential equations, an equilibrium point is a point at which both and
 Because both derivatives are zero, the rates of change of both and are zero
 This means that and will not change, and therefore that if the system is ever at the point then it will remain at that point forever
 An equilibrium point can be stable or unstable
 An equilibrium point is stable if for all points close to the equilibrium point the solution trajectories move back towards the equilibrium point
 This means that if the system is perturbed away from the equilibrium point, it will tend to move back towards the state of equilibrium
 If an equilibrium point is not stable, then it is unstable
 If a system is perturbed away from an unstable equilibrium point, it will tend to continue moving further and further away from the state of equilibrium
 For a system that can be represented in the matrix form , where , , and , the origin is always an equilibrium point
 Considering the nature of the phase portrait for a particular system will tell us what sort of equilibrium point the origin is
 If both eigenvalues of the matrix are real and negative, then the origin is a stable equilibrium point
 This sort of equilibrium point is sometimes known as a sink
 If both eigenvalues of the matrix are real and positive, then the origin is an unstable equilibrium point
 This sort of equilibrium point is sometimes known as a source
 If both eigenvalues of the matrix are real, with one positive and one negative, then the origin is an unstable equilibrium point
 This sort of equilibrium point is known as a saddle point (you will be expected to identify saddle points if they occur in an AI HL exam question)
 If both eigenvalues of the matrix are imaginary, then the origin is an unstable equilibrium point
 Recall that for all points other than the origin, the solution trajectories here all ‘orbit’ around the origin along circular or elliptical paths
 If both eigenvalues of the matrix are complex with a negative real part, then the origin is an stable equilibrium point
 All solution trajectories here spiral in towards the origin
 If both eigenvalues of the matrix are complex with a positive real part, then the origin is an unstable equilibrium point
 All solution trajectories here spiral away from the origin
Worked Example
Show that is an equilibrium point for the system.
Given that and are the eigenvalues of the matrix , with corresponding eigenvectors and , determine the coordinates and nature of the equilibrium point for the system.
Sketching Solution Trajectories
How do I sketch a solution trajectory for a system of coupled differential equations?
 A phase portrait shows typical trajectories representing all the possible solutions to a system of coupled differential equations
 For a given set of initial conditions, however, the solution will only have one specific trajectory
 Sketching a particular solution trajectory will generally involve the following:
 Make sure you know what the ‘typical’ solutions for the system look like
 You don’t need to sketch a complete phase portrait unless asked, but you should know what the phase portrait for your system would look like
 If the phase portrait includes ‘eigenvector lines’, however, it is worth including these in your sketch to serve as guidelines
 Mark the starting point for your solution trajectory
 The coordinates of the starting point will be the and values when
 Usually these are given in the question as the initial conditions for the system
 Determine the initial direction of the solution trajectory
 To do this find the values of and when
 This will tell you the directions in which and are changing initially
 For example if and when , then the trajectory from the starting point will initially be ‘to the left and up’, parallel to the vector
 Use the above considerations to create your sketch
 The trajectory should begin at the starting point (be sure to mark and label the starting point on your sketch!)
 It should move away from the starting point in the correct initial direction
 As it moves further away from the starting point, the trajectory should conform to the nature of a ‘typical solution’ for the system
Worked Example
Consider the system of coupled differential equations
The initial conditions of the system are such that the exact solution is given by
Sketch the trajectory of the solution, showing the relationship between and as increases from zero.