# Phase portraits

1. Sep 5, 2006

### Benny

Hi, I'm unsure about how to do the following question.

I am given the following system for which I first need to find the general solution.

$$\left[ {\begin{array}{*{20}c} {\mathop x\limits^ \bullet } \\ {\mathop y\limits^ \bullet } \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} 3 & { - 2} \\ 6 & { - 5} \\ \end{array}} \right]\left[ {\begin{array}{*{20}c} x \\ y \\ \end{array}} \right]$$

I found the general solution to be:

$$\left[ {\begin{array}{*{20}c} x \\ y \\ \end{array}} \right] = c_1 \left[ {\begin{array}{*{20}c} 1 \\ 1 \\ \end{array}} \right]e^t + c_2 \left[ {\begin{array}{*{20}c} 1 \\ 3 \\ \end{array}} \right]e^{ - 3t}$$

I then sketched the phase portrait which in this case is a saddle. The question asks me to use this phase portrait to sketch x(t) and y(t) on the same graph for the case of x = 1, y = 2.9 when t = 0.

I can't think of a way to do this. The phase portrait I've sketched is for the general solution. I've tried a few things with the point (x,y) = (1,2.9) including shifting the 'centre' of the saddle to that point but none of the things I've tried have any reasoning behind them. They're just random things I've tried which haven't lead me anywhere. Can someone please help me out? Thanks.

2. Sep 5, 2006

### HallsofIvy

If you have the general solution
$$\left[ {\begin{array}{*{20}c} x \\ y \\\end{array}} \right] = c_1 \left[ {\begin{array}{*{20}c} 1 \\ 1 \\\end{array}} \right]e^t + c_2 \left[ {\begin{array}{*{20}c} 1 \\ 3 \\\end{array}} \right]e^{ - 3t}$$
then surely
$$\left[ {\begin{array}{*{20}c} x(0) \\ y(0) \\\end{array}} \right] = c_1 \left[ {\begin{array}{*{20}c} 1 \\ 1 \\\end{array}} \right] + c_2 \left[ {\begin{array}{*{20}c} 1 \\ 3 \\\end{array}} \right]= \left[\begin{array}{c}{c_1+ c_2}\\{c_1+ 3c_2}\end{array}\right]= \left[\begin{array}{c}1 \\ 2.9\end{array}\right]$$
so that $c_1+ c_2= 1$ and $c_1+ 3c_2= 2.9$. Once you've found the actual solution, it should be easy to graph it.

But that isn't really "using the phase portrait". I presume you calculated that the eigenvalues of the coefficient matrix are -3 and 1 and that y= 3x and y= x give eigenvalues of each respectively. Are you clear on exactly what the "phase portrait is? You make it sound as if you have just drawn those two lines. Of course, those lines are part of the phase portrait with "flow" along the line y= 3x directed inward (since it corresponds to the negative eigenvalue) and "flow" along the line y= x directed outward. But the phase diagram also consists of all hyperbolas having those lines as asymptotes and the same "flow".

No, the graph of the solution satisfying x(0)= 1, y(0)= 2.9 is not the phase portrait "shifted". It is the particular hyperbola, out of all those having asymptotes y= 3x, y= x. that passes through (1, 2.9)

3. Sep 6, 2006

### Benny

The eigenvectors are the axes of the hyperbolas. Their directions, along with the general solution allow for a sketch of the family of solutions if the appropriate values of t are considered. I've drawn the saddle (ie. the hyperbolas).

If I locate the point I'll get one of the hyperbolas - there are an infinite number of trajectories but the sketch will only have the key hyperbolas for clarity. Having found the hyperbola which corresponds to the point (x,y) = (1, 2.9), I still only have a curve of y against x and not y(t) and x(t) separately - I just don't get how I can sketch the curves of x(t) and y(t) using only the phase portrait. Nor do I understand the relevance of "t=0" to the required sketches. Can you please offer further assistance?