1. Systems of Differential Equations Phase Plane Analysis
Math 4B
Systems of Differential Equations
Phase Plane Analysis
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2. 𝐴= 𝑎 11 𝑎 12 𝑎 21 𝑎 22 𝑥 ′ 𝑡 =𝐴∙ 𝑥 (𝑡) 𝑥 𝑡 = 𝑥 1 (𝑡) 𝑥 2 (𝑡)
A 2x2 system of 1st order linear differential equations will have the form
𝐴= 𝑎 11 𝑎 12 𝑎 21 𝑎 22
𝑥 ′ 𝑡 =𝐴∙ 𝑥 (𝑡)
𝑥 𝑡 = 𝑥 1 (𝑡) 𝑥 2 (𝑡)
The solution vector x(t) can be visualized by a phase plane diagram.
At each point on a solution curve, the tangent is given by x’. The phase plane diagram looks like a vector field, where at each point the slope of the vector is just x2/x1 (i.e. rise/run).
Let’s look at an example first, then get a more general description.
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3. Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 6 − 𝑥 𝑡 ; 𝑥 0 = 1 3
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4. 𝑥 ′ 𝑡 = 6 −3 2 1 𝑥 𝑡 ; 𝑥 0 = 1 3 𝑟 1 =3 ; 𝑟 2 =4 Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 6 − 𝑥 𝑡 ; 𝑥 0 = 1 3
𝑟 1 =3 ; 𝑟 2 =4
Unstable Node
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5. Here are the possibilities for equilibrium solutions to a 2x2 system.
The outcome depends on the values of the eigenvalues.
Real Eigenvalues r1 and r2:
If both are negative the result is an asymptotically stable node.
If both are positive the result is an unstable node.
If they have opposite signs, there is an unstable saddle point.
Complex Eigenvalues λ±iμ:
If λ=0, a stable center point.
If λ<0 the result is a stable spiral.
If λ>0 the result is an unstable spiral.
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6. Same information in a graph.
By calculating p(trace) and q(determinant), we can find the eigenvalues and categorize the equilibrium point.
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7. 𝑥 ′ 𝑡 = 2 1 −1 −2 𝑥 𝑡 ; 𝑥 0 = −2 2 Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 2 1 −1 −2 𝑥 𝑡 ; 𝑥 0 = −2 2
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8. 𝑥 ′ 𝑡 = 2 1 −1 −2 𝑥 𝑡 ; 𝑥 0 = −2 2 𝑟 1 =+ 3 ; 𝑟 2 =− 3 Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 2 1 −1 −2 𝑥 𝑡 ; 𝑥 0 = −2 2
𝑟 1 =+ 3 ; 𝑟 2 =− 3
Unstable Saddle
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9. Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2
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10. 𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2 𝑟=2±3𝑖 Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2
𝑟=2±3𝑖
Unstable Spiral
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11. 𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2 𝑟=2±3𝑖 Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2
𝑟=2±3𝑖
Let’s work through the solution procedure for this one as well.
Find eigenvectors for the eigenvalues:
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12. 𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2 𝑟=2±3𝑖 Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2
𝑟=2±3𝑖
Let’s work through the solution procedure for this one as well.
Find eigenvectors for the eigenvalues:
𝑣 = 1 𝑖 = 𝑖
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13. 𝑋 𝑡 = 𝑒 2𝑡 cos⁡(3𝑡) 𝑒 2𝑡 sin⁡(3𝑡) − 𝑒 2𝑡 sin⁡(3𝑡) 𝑒 2𝑡 cos⁡(3𝑡)
Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2
𝑟=2±3𝑖
Let’s work through the solution procedure for this one as well.
Find eigenvectors for the eigenvalues:
𝑣 = 1 𝑖 = 𝑖
Here is the corresponding fundamental matrix:
𝑋 𝑡 = 𝑒 2𝑡 cos⁡(3𝑡) 𝑒 2𝑡 sin⁡(3𝑡) − 𝑒 2𝑡 sin⁡(3𝑡) 𝑒 2𝑡 cos⁡(3𝑡)
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14. 𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2 𝑟=2±3𝑖 𝑥 𝑡 =𝑋 𝑡 𝑐 ; 𝑐 = 𝑐 1 𝑐 2
Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2
𝑟=2±3𝑖
The general solution can now be written in terms of this fundamental matrix:
𝑥 𝑡 =𝑋 𝑡 𝑐 ; 𝑐 = 𝑐 1 𝑐 2
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15. 𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2 𝑟=2±3𝑖 𝑥 𝑡 =𝑋 𝑡 𝑐 ; 𝑐 = 𝑐 1 𝑐 2
Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2
𝑟=2±3𝑖
The general solution can now be written in terms of this fundamental matrix:
𝑥 𝑡 =𝑋 𝑡 𝑐 ; 𝑐 = 𝑐 1 𝑐 2
To solve the initial value problem, we can incorporate the initial value by writing the solution this way:
𝑥 𝑡 =𝑋(𝑡)∙ 𝑋(0) −1 ∙ 𝑥 (0)
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16. 𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2 𝑟=2±3𝑖 𝑥 𝑡 =𝑋 𝑡 𝑐 ; 𝑐 = 𝑐 1 𝑐 2
Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2
𝑟=2±3𝑖
The general solution can now be written in terms of this fundamental matrix:
𝑥 𝑡 =𝑋 𝑡 𝑐 ; 𝑐 = 𝑐 1 𝑐 2
To solve the initial value problem, we can incorporate the initial value by writing the solution this way:
𝑥 𝑡 =𝑋(𝑡)∙ 𝑋(0) −1 ∙ 𝑥 (0)
𝑋 0 =
By coincidence this matrix is already the identity, so we don’t need to find the inverse.
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17. 𝑥 𝑡 = 𝑋(𝑡)∙ 𝑋(0) −1 𝑌(𝑡) ∙ 𝑥 (0)
Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2
𝑟=2±3𝑖
𝑥 𝑡 = 𝑋(𝑡)∙ 𝑋(0) −1 𝑌(𝑡) ∙ 𝑥 (0)
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18. 𝑥 𝑡 = 𝑋(𝑡)∙ 𝑋(0) −1 𝑌(𝑡) ∙ 𝑥 (0)
Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 2 3 −3 2 𝑥 𝑡 ; 𝑥 0 = 0 2
𝑟=2±3𝑖
𝑥 𝑡 = 𝑋(𝑡)∙ 𝑋(0) −1 𝑌(𝑡) ∙ 𝑥 (0)
𝑥 𝑡 = 𝑒 2𝑡 cos⁡(3𝑡) 𝑒 2𝑡 sin⁡(3𝑡) − 𝑒 2𝑡 sin⁡(3𝑡) 𝑒 2𝑡 cos⁡(3𝑡) ∙ = 2 𝑒 2𝑡 sin⁡(3𝑡) 2 𝑒 2𝑡 cos⁡(3𝑡)
Notice that with this formulation, we can quickly change the initial value to obtain a new solution, without recalculating the fundamental matrix.
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19. Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 1 −5 1 −1 𝑥 𝑡 ; 𝑥 0 = 1 0
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20. 𝑥 ′ 𝑡 = 1 −5 1 −1 𝑥 𝑡 ; 𝑥 0 = 1 0 𝑟=0±2𝑖 Example:
Draw the Phase Plane diagram, and sketch the solution curve.
𝑥 ′ 𝑡 = 1 −5 1 −1 𝑥 𝑡 ; 𝑥 0 = 1 0
𝑟=0±2𝑖
Stable Center
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21. Example:
Find the solution to this initial value problem.
𝑥 ′ 𝑡 = 1 −5 1 −1 𝑥 𝑡 ; 𝑥 0 = 1 0
Find eigenvectors for the eigenvalues:
𝑣 = 1+2𝑖 1 = 𝑖
Eigenvector is split into real and imaginary parts.
Solutions can be written in terms of complex exponentials, or sine/cosines.
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22. Example:
Find the solution to this initial value problem.
𝑥 ′ 𝑡 = 1 −5 1 −1 𝑥 𝑡 ; 𝑥 0 = 1 0
Here is the eigenvector for r=2i. The other one is the conjugate.
𝑣 = 𝑖
Eigenvector is split into real and imaginary parts.
Solutions can be written in terms of complex exponentials, or sine/cosines.
After some rearranging two independent solutions are:
𝑥 (1) = cos 2𝑡 − sin 2𝑡 = cos 2𝑡 −2sin⁡(2𝑡) cos⁡(2𝑡)
𝑥 (2) = sin 2𝑡 cos 2𝑡 = sin 2𝑡 +2cos⁡(2𝑡) sin⁡(2𝑡)
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23. Example:
Find the solution to this initial value problem.
𝑥 ′ 𝑡 = 1 −5 1 −1 𝑥 𝑡 ; 𝑥 0 = 1 0
Solution can be written as a Fundamental Matrix:
𝑋 𝑡 = cos 2𝑡 −2sin⁡(2𝑡) sin 2𝑡 +2cos⁡(2𝑡) cos⁡(2𝑡) sin⁡(2𝑡)
𝑥 𝑡 =𝑋(𝑡)∙ 𝑐
The columns of this fundamental matrix X(t) are the solutions x(1) and x(2).
This is the general solution. The next step is to incorporate the initial value.
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24. 𝑥 ′ 𝑡 = 1 −5 1 −1 𝑥 𝑡 ; 𝑥 0 = 1 0 𝑥 𝑡 = 𝑋(𝑡)∙ 𝑋(0) −1 𝑌(𝑡) ∙ 𝑥 (0)
Example:
Find the solution to this initial value problem.
𝑥 ′ 𝑡 = 1 −5 1 −1 𝑥 𝑡 ; 𝑥 0 = 1 0
Here is the format for the solution with the initial value:
The matrix Y(t) is another version of the fundamental matrix that also has the property that Y(0)=I.
𝑥 𝑡 = 𝑋(𝑡)∙ 𝑋(0) −1 𝑌(𝑡) ∙ 𝑥 (0)
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25. Example:
Find the solution to this initial value problem.
𝑥 ′ 𝑡 = 1 −5 1 −1 𝑥 𝑡 ; 𝑥 0 = 1 0
Multiplying matrices, we find that Y(t) is:
𝑌 𝑡 = 1 −2 − sin 2𝑡 −2cos⁡(2𝑡) 5sin⁡(2𝑡) −sin⁡(2𝑡) −2 cos 2𝑡 +sin⁡(2𝑡)
Now we can multiply this by the initial value vector to get the solution.
𝑥 𝑡 =𝑌 𝑡 ∙ 𝑥 0 = 1 −2 − sin 2𝑡 −2cos⁡(2𝑡) 5sin⁡(2𝑡) −sin⁡(2𝑡) −2 cos 2𝑡 +sin⁡(2𝑡)
𝑥 𝑡 = sin 2𝑡 +cos⁡(2𝑡) sin 2𝑡
this is x1(t)
this is x2(t)
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