Q13RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Solve the related phase plane equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows) and describe the stability of the critical points (i.e., compare with Figure \({\bf{5}}{\bf{.12}}\), page \({\bf{267}}\)).

\(\begin{array}{c}{\bf{x' = 4 - 4y}}\\{\bf{y' =  - 4x}}\end{array}\)

 

Step-by-Step Solution

Answer

The related phase plane equation for the given system is\({{\bf{x}}^{\bf{2}}}{\bf{ - (y - 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = c}}.\)

The critical point \(\left( {{\bf{0,1}}} \right)\) is the saddle and it is unstable.

 

Step 1: Finding the critical point

Let's first find the critical point. One has to solve the system\({\bf{x' = 0,y' = 0}}\), so, one has:

 

\(\begin{array}{c}{\bf{0 = 4 - 4y}} \Rightarrow {\bf{y = 1,}}\\{\bf{0 =   - 4x}} \Rightarrow {\bf{x = 0}}\end{array}\)

 

So, the critical point is\(\left( {{\bf{x, y}}} \right){\bf{ = }}\left( {{\bf{0,1}}} \right)\). Now one can solve the related phase plane equation which is:

 

\(\begin{array}{c}\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = }}\frac{{\frac{{{\bf{dy}}}}{{{\bf{dt}}}}}}{{\frac{{{\bf{dx}}}}{{{\bf{dt}}}}}}{\bf{ = }}\frac{{{\bf{ - 4x}}}}{{{\bf{4 - 4y}}}}\\{\bf{ = }}\frac{{{\bf{4x}}}}{{{\bf{4y - 4}}}}\\{\bf{(4y - 4)dy = 4xdx}}\end{array}\)

 

Step 2: Integrate the given equation

Integrating the previous equation, one will get:

\(\begin{array}{c}\int {{\bf{(4y - 4)}}} {\bf{dy = }}\int {\bf{4}} {\bf{xdx}}\\{\bf{4}}\frac{{{{\bf{y}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - 4y = 4}}\frac{{{{\bf{x}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ + C}}\\{{\bf{y}}^{\bf{2}}}{\bf{ - 2y = }}{{\bf{x}}^{\bf{2}}}{\bf{ + }}\frac{{\bf{C}}}{{\bf{2}}}\\{{\bf{x}}^{\bf{2}}}{\bf{ - (y - 1}}{{\bf{)}}^{\bf{2}}}{\bf{ =  - 1 - }}\frac{{\bf{C}}}{{\bf{2}}}\end{array}\)

Introducing a new constant \({\bf{c =  - 1 - }}\frac{{\bf{C}}}{{\bf{2}}}\)one has that the solution is\({{\bf{x}}^{\bf{2}}}{\bf{ - (y - 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = c}}\)which in \({\bf{xy - }}\)plane represents hyperbole with a focus on the critical point. 

 

The critical point is the saddle and it is unstable. In the figure below are some trajectories with their flow arrows, the critical point, and one particular solution for \({\bf{c = 1}}\) (red line).