Coupled Differential Equations (DP IB Maths: AI HL)

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
• 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 non-zero
  • Similar solution methods exist for non-real, non-distinct and/or non-zero 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)

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 non-zero
• 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

• 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 non-zero 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

• 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

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

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