# Diff. EQ. system I.V.P. with complex cunjugates (Please check)

1. Jun 22, 2005

### VinnyCee

Here is the problem:

$$X'\,=\,\left(\begin{array}{cc}-2 & 5 \\-2 & 4 \end{array}\right)\,X,\,\,\,\,X(0)\,=\,\left(\begin{array}{c} 1 \\ 0 \end{array}\right)$$

Here is what I have:

$$det(A\,-\,r\,I)\,=\,0$$

$$r^2\,-\,2r\,+\,2\,=\,0$$

$$r\,=\,-\frac{1}{2}\,\pm\,i,\,\,\,\,\lambda\,=\,-\frac{1}{2},\,\,\mu\,=\,1$$

This is where I get confused. I have two possibilities for the r-value, right? Namely, they are $r_1\,=\,-\frac{1}{2}\,+\,i$ and $r_2\,=\,-\frac{1}{2}\,-\,i$. Now I go back and use the $A\,-\,r_n\,I\,=\,0$ equation to figure a general solution for the system.

For $r_1$:

$$\left[A\,-\,(-\frac{1}{2}\,+\,i)\,I\right]\,\left(\begin{array}{c}\xi_1 \\\xi_2 \end{array}\right)\,=\,0$$

$$\left(\begin{array}{cc}-\frac{3}{2}\,-\,i & 5 \\-2 & \frac{9}{2}\,-\,i \end{array}\right)\,\left(\begin{array}{c}\xi_1 \\\xi_2 \end{array}\right)\,=\,0$$

$$\left(-\frac{3}{2}\,-\,i\right)\,\xi_1\,+\,5\,\xi_2\,=\,0$$

$$5\,\xi_2\,=\,\left(\frac{3}{2}\,+\,i\right)\,\xi_1$$

$$\xi_2\,=\,\left(\frac{3}{10}\,+\,\frac{1}{5}\,i\right)\,\xi_1$$

$$\xi^{(1)}\,=\,\left(\begin{array}{c}\ 1 \\\ \frac{3}{10}\,+\,\frac{1}{5}\,i \end{array}\right)$$

$$\xi^{(1)}\,=\,\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,i$$

Now, finally, here is the general solution for $r_1$:

$$X_1_c\,=\,C_1\,e^{-\frac{t}{2}}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,cos\,t\,-\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,sin\,t\right]\,+\,C_2\,e^{-\frac{t}{2}}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,sin\,t\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,cos\,t\right]$$

Using the initial condition given to solve for c1 and c2, I get:

$$C_1\,=\,1,\,\,\,\,\,\,C_2\,=\,-\frac{3}{2}$$

Plugging back into the general EQ:

$$X_1\,=e^{-\frac{t}{2}}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,cos\,t\,-\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,sin\,t\right]\,+\,\left(-\frac{3}{2}\right)\,e^{-\frac{t}{2}}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,sin\,t\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,cos\,t\right]$$

And simplifying:

$$X_1\,=\,e^{-\frac{t}{2}}\,\left(\begin{array}{c}\ cos\,t\,-\,\frac{3}{2}\,sin\,t \\\ -\frac{13}{20}\,sin\,t \end{array}\right)$$

Now here is where I am confused. I get this answer for $X_1$, and I get a similar one for $X_2$ using the $r_2$ complex conjugate and the initial values:

$$X_2\,=\,e^{-\frac{t}{2}}\,\left(\begin{array}{c}\ -2\,sin\,t \\\ -\frac{3}{5}\,sin\,t\,-\,\frac{2}{5}\,cos\,t \end{array}\right)$$

Which one is correct (if either)? Or are they somehow the same?

Last edited: Jun 23, 2005
2. Jun 23, 2005

### ehild

$$r\,=\,1\,\pm\,i$$

ehild

3. Jun 23, 2005

### VinnyCee

Thanks

Thanks for showing me, I would have never found it!

After correcting that, I find the following:

$$r_1\,=\,1\,+\,i$$

$$\left(\begin{array}{cc}-3\,-\,i & 5 \\-2 & 3\,-\,i \end{array}\right)\,\left(\begin{array}{c}\xi_1 \\\xi_2 \end{array}\right)\,=\,0$$

$$\xi^{(1)}\,=\,\left(\begin{array}{c}\ 1 \\\ \frac{3}{5} \end{array}\right)\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,i$$

$$X_1_c\,=\,C_1\,e^{t}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{5} \end{array}\right)\,cos\,t\,-\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,sin\,t\right]\,+\,C_2\,e^{t}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{5} \end{array}\right)\,sin\,t\,+\,\left(\begin{array} {c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,cos\,t\right]$$

$$C_1\,=\,1,\,\,\,\,\,\,C_2\,=\,-3$$

Assuming that is all correct, what do I do about the other three possibilities(two for each complex conjugate)?

4. Jun 23, 2005

### ehild

The general solution of the system of differential equation is

$$X=c_1\xi(1)e^{r_1t}+c_2\xi(2)e^{r_2t}$$

You know the array $$\xi(1)$$ already, determine the other one belonging to the other root. Never mind that everything is complex :).
Now apply the initial condition and get c1 and c2. They will be complex (complex conjugates)
Now you can transform the exponentials to the trigonometric form and do all multiplications.
You get the solution and it is unique.

ehild

5. Jun 23, 2005

### VinnyCee

????

Well, there are two roots that we found, $r_1$ and $r_2$.

Each of those has two equations from which to solve for $\xi$. I am really confused! Please help! I cannot just arbitrarily pick 2 out of 4 solutions and say it is right! What am I missing?

6. Jun 23, 2005

### VinnyCee

Here are the four solutions...

$$\xi^{(1)}\,=\,\left(\begin{array}{c}\ 1 \\\ \frac{3}{5} \end{array}\right)\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,i$$

$$\xi^{(2)}\,=\,\left(\begin{array}{c}\ 1 \\\ \frac{3}{5} \end{array}\right)\,+\,\left(\begin{array}{c}\ 0 \\\ -\frac{1}{5} \end{array}\right)\,i$$

$$\xi^{(3)}\,=\,\left(\begin{array}{c}\ \frac{3}{2} \\\ 1 \end{array}\right)\,+\,\left(\begin{array}{c}\ \frac{1}{2} \\\ 0 \end{array}\right)\,i$$

$$\xi^{(4)}\,=\,\left(\begin{array}{c}\ \frac{3}{2} \\\ 1 \end{array}\right)\,+\,\left(\begin{array}{c}\ -\frac{1}{2} \\\ 0 \end{array}\right)\,i$$

But now what do I do with these?

7. Jun 23, 2005

### AKG

This site is great. I found it very useful last semester for my ODE course.

8. Jun 23, 2005

### saltydog

VinnyCee: When you have complex eigenvalues, you need compute the eigenvector corresponding to only one of the two complex eigenvalues. By breaking up the resulting complex-valued solution into its Real and Imaginary parts, you obtain a pair of independent solutions which together with two arbitrary constants, make up the general solution.

So for example, choose one eigenvalue, obtain one eigenvector and then obtain the complex-value solution:

$$Y(t)=Y_{re}(t)+iY_{im}(t)$$

Then the general solution is given by:

$$Y(t)=k_1Y_{re}(t)+k_2Y_{im}(t)$$

Last edited: Jun 23, 2005
9. Jun 23, 2005

### VinnyCee

Choosing one Eigenvalue

I choose the first one:

$$\xi^{(1)}\,=\,\left(\begin{array}{c}\ 1 \\\ \frac{3}{5} \end{array}\right)\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,i$$

But what do you mean by $Y(t)=Y_{re}(t)+iY_{im}(t)$?

Like this?:

$$Y(t)=\begin{array}{c}\ 1 \\\ \frac{3}{5} \end{array}\right)+...$$

10. Jun 23, 2005

### ehild

You have got two systems of equations, one for each root.
As they are systems of homogeneous equations, one of the components of the $\xi$ arrays is arbitrary. You can take it 1. Only the ratio of the components is defined by the equations.
So you have one array for r1 and an other one for r2.
For

$$r_1=1+i$$
$$\xi{(1)}\,=\,\left(\begin{array}{c}\ 1 \\\ \frac{3+i}{5} \end{array}\right)$$

For

$$r_2=1-i$$
$$\xi{(2)}\,=\,\left(\begin{array}{c}\ 1 \\\ \frac{3-i}{5} \end{array}\right)$$

The general solution is

$$X=c_1\xi(1)e^{(1+i)t} + c_2\xi(2)e^{(1-i)t}$$.

In components:

$$x_1=c_1e^{(1+i)t}+c_2e^{(1-i)t}=1$$

$$x_2= c_1\frac{3+i}{5}e^{(1+i)t}+c_2\frac{3-i}{5}e^{(1-i)t}$$

At t=0:

$$c_1+c_2 = 1$$

$$c_1\frac{3+i}{5}+c_2\frac{3-i}{5}=0$$

Solve for the c-s, replace them back to the general solution, use the trigonometric form, simplify, and you get (if I did everything correctly)

$$x_1= e^t(\cos{t} -3\sin{t})$$

and

$$x_2 = -2 e^t \sin{t}$$

ehild

11. Jun 23, 2005

### saltydog

VinnyCee, we have:

$$\lambda_1=1+i$$

The eigenvector corresponding to this eigenvalue is:

$$\left( \begin{array}{c}\ 1 \\ \frac{1}{5}(3+i)\ \end{array} \right)$$

Thus a solution is:

$$Y(t)=e^{(1+i)t} \left( \begin{array}{c}\ 1 \\ \frac{1}{5}(3+i)\ \end{array} \right)$$

Now, convert this to a real part and imaginary part via Euler's formula:

$$Y(t)= \left( \begin{array}{c}\ e^tCos(t) \\ e^t[\frac{3}{5}Cos(t)-\frac{1}{5}Sin(t)]\ \end{array} \right) +i \left( \begin{array}{c}\ e^tSin(t) \\ e^t[\frac{3}{5}Sin(t)+\frac{1}{5}Cos(t)]\ \end{array} \right)$$

Note how the real part and the imaginary part has been separated into separate terms which are usually written:

$$Y(t)=Y_{re}(t)+iY_{im}(t)$$

As I stated earlier, the general solution is then:

$$Y(t)=k_1Y_{re}(t)+k_2Y_{im}(t)$$

Substituting the initial conditions, I get the same answer as ehild which I checked via Mathematica by back substitution. A plot of both functions is attached.

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12. Jun 23, 2005

### saltydog

A qualitative analysis

The global behavior of linear systems like these is determined by the value of the eigenvalues.

For the case with complex eigenvalues, the real part determines the behavior of the phase portrait (when y is graphed as a function of x parametrically. If the real part is less than 0, the exponential term of the solution forces the portrait to spiral into the origin. This is a spirial sink.

If the real part is greater than zero, the solution spirials off away from the origin to infinity. This is a spirial source.

And if the real part is 0, the solution is periodic.

The attached plot exhibits this solution with complex eigenvalue that has real part equal to 1. The solution tends away from its source.

#### Attached Files:

• ###### spiral source.JPG
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Last edited: Jun 23, 2005
13. Jun 24, 2005

### VinnyCee

Ok, but what about the E3 and E4 vectors that I also found?

That covers E1 and E2, but what about these?:

$$\xi^{(3)}\,=\,\left(\begin{array}{c}\ \frac{3}{2} \\\ 1 \end{array}\right)\,+\,\left(\begin{array}{c}\ \frac{1}{2} \\\ 0 \end{array}\right)\,i$$

$$\xi^{(4)}\,=\,\left(\begin{array}{c}\ \frac{3}{2} \\\ 1 \end{array}\right)\,+\,\left(\begin{array}{c}\ -\frac{1}{2} \\\ 0 \end{array}\right)\,i$$

14. Jun 24, 2005

### saltydog

Dude, I'll be honest with you: I don't know where you gettin' those $\xi's[/tex] but I bet a dollar the answer I gave you is the general solution,[itex]\xi's$ or no $\xi's$.

15. Jun 24, 2005

### VinnyCee

Sweet! Almost done.

So the general solution is:

$$X(t)\,=\,C_1\,e^t\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{5} \end{array}\right)\,cos\,t\,-\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,sin\,t\right]\,+\,C_2\,e^t\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{5} \end{array}\right)\,sin\,t\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,cos\,t\right]$$

And then solving for the initial condition, I get $C_1\,=\,1$ and $C_2\,=\,-3$, right?

So then the exact solution is:

$$X(t)\,=\,e^t\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{5} \end{array}\right)\,cos\,t\,-\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,sin\,t\right]\,-\,3\,e^t\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{5} \end{array}\right)\,sin\,t\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,cos\,t\right]$$

Is that right?

16. Jun 24, 2005

### VinnyCee

Now I have simplified(hopefully!) that answer down to:

$$X(t)\,=\,e^t\,\left(\begin{array}{c}\ cos\,t\,-\,3\,sin\,t \\\ -2\,sin\,t \end{array}\right)$$

Is this all correct now? Thanks alot for the help saltydog:)

17. Jun 24, 2005

### saltydog

Yes. Ehild got it before me.