# A question in coupled differential equation

1. Oct 9, 2005

### uob_student

Hello

i have a question :

dx/dt=(a*x)+(b*y)+(c*z)

dy/dt=(d*x)+(e*y)+(f*z)

dz/dt=(g*x)+(h*y)+(i*z)

where a,b,c,d,e,f,g,h,i are constants

i want a proof of this and finding the values of x,y and z??

2. Oct 9, 2005

### arildno

Proof of what??.
What do you mean by finding "the values of x,y,z"?
x(t), y(t), z(t) are FUNCTIONS of the variable "t", do you want to find which set of functions satisfies your system of differential equations?

3. Oct 9, 2005

### uob_student

oh , i don't want a proof

I want the set of functions satisfies that system of differential equations

4. Oct 9, 2005

### arildno

what do you know of linear algebra?

5. Oct 9, 2005

### saltydog

So in general, determine the eigenvalues and eigenvectors of the coefficient matrix.

$$\text{det}\left(\mathbf{A}-\lambda\mathbf{I}\right)=0$$

Then calculate the eigenvectors as:

$$\mathbf{A}\mathbf{X}=\lambda_i\mathbf{X}$$

For each eigenvalue-eigenvector pair, there is a solution.

Linear combinations of the solutions create the general solution.

Edit: Oh by the way, what do the solutions look like anyway?

Last edited: Oct 9, 2005
6. Oct 11, 2005

### uob_student

my idea is :

suppose

x(t)=x(0) exp(iwt)

y(t)=y(0) exp(iwt)

z(t)=z(0) exp(iwt)

how can i solve my previous question by using this informations??

7. Oct 11, 2005

### saltydog

Well, using the eigenvalues, end up with a matrix equation for the solution in the form:

$$\mathbf{X}=c_1e^{\lambda_1 t} \left( \begin{array}{c} v_1 \\ v_2 \\ v_3 \\ \end{array} \right) + c_2 e^{\lambda_2 t} \left( \begin{array}{c} r_1 \\ r_2 \\ r_3 \\ \end{array} \right) + c_3 e^{\lambda_3 t} \left( \begin{array}{c} s_1 \\ s_2 \\ s_3 \\ \end{array} \right)$$

8. Oct 11, 2005

### saltydog

In the case of complex eigenvalues, I assume the solution is of the form:
$$\mathbf{X}=c_1e^{\lambda_1 t} \left( \begin{array}{c} v_1 \\ v_2 \\ v_3 \\ \end{array} \right) + c_2\mathbf{Y}+c_3\mathbf{Z}$$

where Y and Z are the real and imaginary parts respectively of the complex-eigenvalue contribution.

9. Oct 12, 2005

### saltydog

Well . . . suppose I should just work one and find out. Hum . . . let me see . . . Ok, how about this one:

$$\mathbf{X}^{'}= \left( \begin{array}{ccc} 1 & 0 & 3 \\ 0 & -1 & 0 \\ -3 & 0 & 1 \end{array} \right) \mathbf{X};\quad \mathbf{X}(0)= \left( \begin{array}{c} x_0 \\ y_0 \\ z_0 \end{array} \right)$$

You game Uob or what?

Last edited: Oct 12, 2005
10. Oct 12, 2005

### saltydog

$$\text{det}\left(\mathbf{A}-\lambda\mathbf{I}\right)=0$$

so for the 3x3 matrix above, that would be:

$$\text{det}\left(\mathbf{A}-\lambda\mathbf{I}\right)= (1-\lambda) \left( \begin{array}{cc} -(1+\lambda} & 0 \\ 0 & (1-\lambda) \end{array} \right)+3 \left( \begin{array}{cc} 0, &-(1+\lambda} \\ -3 & 0 \end{array}\right)=0$$

or:

$$-10-8\lambda+\lambda^2-\lambda^3=0$$

Since this is not a problem focusing on cubic equations, I simply use Mathematica to calcualte the eigenvalues:

$$\text{Eigenvalues[Matrix]}$$

They are:

$$\lambda_1=-1$$

$$\lambda_2=1+3i$$

$$\lambda_3=1-3i$$

Last edited: Oct 12, 2005
11. Oct 13, 2005

### saltydog

Did I mention Mathematica has an Eigenvector[Matrix] command? However, I'm not good at calculating eigenvectors so I really should do a few by hand:

The eigenvector equation is simple:

$$\mathbf{M}v=\lambda v$$

So for:

$$\lambda_1=-1$$

$$\left( \begin{array}{ccc} 1 & 0 & 3 \\ 0 & -1 & 0 \\ -3 & 0 & 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right)=-1 \left( \begin{array}{c} x \\ y \\ z \end{array} \right)$$

So:

$$x+3z=-x$$

$$-y=-y$$

$$-3x+z=-z$$

The middle one is easy:

$$0y=0$$

That means y can be anything so let y=1.
The other two:

$$2x+3z=0$$

$$-3x+2z=0$$

The simple thing here, since we're looking for ANY eigenvector, is to just pick the zero solution and thus:

$$v_1= \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right)$$

For:

$$\lambda_2=(1+3i)$$

$$\left( \begin{array}{ccc} 1 & 0 & 3 \\ 0 & -1 & 0 \\ -3 & 0 & 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right)=(1+3i) \left( \begin{array}{c} x \\ y \\ z \end{array} \right)$$

So that's:

$$x+3z=(1+3i)x$$

$$-y=(1+3i)y$$

$$-3x+z=(1+3i)z$$

For the middle one, the only way ay=y is if y=0. The other two yield:

$$3z=3ix$$

or:

$$z=ix$$

so let x=1 and z=i.

Thus:

$$v_2= \left( \begin{array}{c} 1 \\ 0 \\ i \end{array} \right)$$

Same dif for the other eigenvalue which yields:

$$v_3= \left( \begin{array}{c} 1 \\ 0 \\ -i \end{array} \right)$$

Mathematica returns equivalent eigenvectors.
Thus we are led to the solution in matrix form:

$$\mathbf{X}=c_1e^{-t} \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right)+ c_2e^{(1+3i)t} \left( \begin{array}{c} 1 \\ 0 \\ i \end{array} \right)+ c_3e^{(1-3i)t} \left( \begin{array}{c} 1 \\ 0 \\ -i \end{array} \right)$$

You ever work a problem and the answer is just as difficult as the question?

Last edited: Oct 13, 2005
12. Oct 14, 2005

### uob_student

thanks

but how you can find the values of Lambdas in your solution??

13. Oct 14, 2005

### saltydog

By solving the eigenvalue equation:

$$\text{det}\left(\mathbf{A}-\lambda\mathbf{I}\right)=0$$

So performing that matrix algebra gives us:

$$\text{det}\left(\mathbf{A}-\lambda\mathbf{I}\right)= (1-\lambda) \left( \begin{array}{cc} -(1+\lambda} & 0 \\ 0 & (1-\lambda) \end{array} \right)+3 \left( \begin{array}{cc} 0, &-(1+\lambda} \\ -3 & 0 \end{array}\right)=0$$
or:
$$-10-8\lambda+\lambda^2-\lambda^3=0$$

Is that what you're having problems with or just figuring the roots of that cubic equation?

14. Oct 14, 2005

### saltydog

After reviewing, I wish to clear up two points in my efforts to solve this equation:

1. There is no need to directly calculate the eigenvector of $\lambda_3$:

The complex conjugate of an eigenvector for $\lambda_2$ is an eigenvector for $\lambda_3[/tex]. So above I calculated the eigenvector for [itex](1+3i)$ to be:

$$v_2=\left(\begin{array}{c} 1 \\0 \\i\end{array}\right)$$

Therefore to calculate the eigenvector for $(1-3i)$, conjugate the eigenvector for [itex](1+3i)[/tex]:

If:

$$v_2=\left(\begin{array}{c} 1+0i \\0+0i \\0+i\end{array}\right)$$

Then:

$$\overline{v_2}=v_3=\left(\begin{array}{c} 1-0i \\0-i \\0-i\end{array}\right)=\left(\begin{array}{c} 1 \\0 \\-i\end{array}\right)$$

See how that works?

Ok that's one.

2. The complex eigenvalues both yield the same solution! That is, the solution from (1+3i) is the same solution as that from (1-3i). Go figure. I did. So that's the reason we only need calculate the Real and Complex contribution from ONE member of each complex pair. And also, it's VERY convenient to write the eigenvectors as:

$$\left(\begin{array}{c} 1 \\0 \\-i\end{array}\right)=\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+i\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)$$

Alright, so lets compute the Real part and the Complex part for:

\begin{align*} e^{(1+3i)t}\left(\begin{array}{c} 1 \\0 \\i\end{array}\right) &=e^{(1+3i)t}\left[\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+i\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right] \\ &=e^t\left[(Cos(3t)+iSin(3t))\left\{\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+ i\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right\}\right] \\ &=e^t\left[Cos(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)+iCos(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)+iSin(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)-Sin(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right] \\ &=e^t\left[\left\{Cos(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)-Sin(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right\}+i\left\{Cos(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)+Sin(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)\right\}\right] \end{align}

Note how the Real and Complex parts have been separated. Each one is a solution to the system. Therefore, this with the first solution then yields the general solution:

\begin{align*} \mathbf{X}&=C_1e^{-t}\left(\begin{array}{c} 0 \\1 \\0\end{array}\right) \\ &+e^t\left[C_2\left\{Cos(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)-Sin(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)\right\}+C_3\left\{Cos(3t)\left(\begin{array}{c} 0 \\0 \\1\end{array}\right)+Sin(3t)\left(\begin{array}{c} 1 \\0 \\0\end{array}\right)\right\}\right] \end{align}

$$x(t)=C_2e^tCos(3t)+C_3e^tSin(3t)$$

$$y(t)=C_1e^{-t}$$

$$z(t)=-C_2e^tSin(3t)+C_3e^tCos(3t)$$

So how about a 4x4, two complex-conjugate pairs. Suppose only need to calculate an eigenvalue-eigenvector for one member of each set, and then just do what I did above.

Last edited: Oct 14, 2005
15. Oct 15, 2005

### uob_student

thanks but I want just how you figuring the roots of that cubic equation?

16. Oct 15, 2005

### saltydog

You mean of course "other than Mathematica or other software" right? Well, the approach is first to just guess integer values close to 0 and see if when substituted they work. If one does, then apply synthetic division to split-out the resulting quadratic. There's always those nice formulas for cubics right? Then of course use successive approximations and Newton's method.

17. Oct 15, 2005

### uob_student

can you explain that??

18. Oct 15, 2005

### saltydog

Which part? Be more specific. I'll be leaving soon however. If you like, just start another thread in the homework section and ask about finding roots to cubics.