a question in coupled differential equation


by uob_student
Tags: coupled, differential, equation
uob_student
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#1
Oct9-05, 08:41 AM
P: 22
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??

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arildno
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#2
Oct9-05, 08:55 AM
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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?
uob_student
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#3
Oct9-05, 10:30 AM
P: 22
Quote Quote by 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?
oh , i don't want a proof

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

arildno
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#4
Oct9-05, 11:31 AM
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a question in coupled differential equation


what do you know of linear algebra?
saltydog
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#5
Oct9-05, 03:29 PM
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So in general, determine the eigenvalues and eigenvectors of the coefficient matrix.

First start with finding the eigenvalues via:

[tex]\text{det}\left(\mathbf{A}-\lambda\mathbf{I}\right)=0[/tex]

Then calculate the eigenvectors as:

[tex]\mathbf{A}\mathbf{X}=\lambda_i\mathbf{X}[/tex]

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?
uob_student
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#6
Oct11-05, 05:25 AM
P: 22
thank you for your helping

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??
saltydog
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#7
Oct11-05, 07:24 AM
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Well, using the eigenvalues, end up with a matrix equation for the solution in the form:

[tex]
\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)
[/tex]
saltydog
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#8
Oct11-05, 08:05 AM
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Well, I have a question about this. That means what I'm about to say I'm not sure of:

In the case of complex eigenvalues, I assume the solution is of the form:
[tex]
\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}
[/tex]

where Y and Z are the real and imaginary parts respectively of the complex-eigenvalue contribution.
saltydog
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#9
Oct12-05, 07:42 AM
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Well . . . suppose I should just work one and find out. Hum . . . let me see . . . Ok, how about this one:

[tex]
\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)
[/tex]

You game Uob or what?
saltydog
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#10
Oct12-05, 06:01 PM
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Suppose I should start with calculating eigenvalues:

[tex]
\text{det}\left(\mathbf{A}-\lambda\mathbf{I}\right)=0
[/tex]

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

[tex]
\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
[/tex]

or:

[tex]
-10-8\lambda+\lambda^2-\lambda^3=0
[/tex]

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

[tex]\text{Eigenvalues[Matrix]}[/tex]

They are:

[tex]\lambda_1=-1[/tex]

[tex]\lambda_2=1+3i[/tex]

[tex]\lambda_3=1-3i[/tex]
saltydog
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#11
Oct13-05, 03:09 PM
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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:

[tex]
\mathbf{M}v=\lambda v
[/tex]

So for:

[tex]
\lambda_1=-1
[/tex]

[tex]
\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)
[/tex]

So:

[tex]x+3z=-x[/tex]

[tex]-y=-y[/tex]

[tex]-3x+z=-z[/tex]


The middle one is easy:

[tex]
0y=0
[/tex]

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

[tex]2x+3z=0 [/tex]

[tex]-3x+2z=0[/tex]


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

[tex]
v_1=
\left(
\begin{array}{c} 0 \\
1 \\
0
\end{array}
\right)
[/tex]

For:

[tex]
\lambda_2=(1+3i)
[/tex]

[tex]
\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)
[/tex]

So that's:

[tex]x+3z=(1+3i)x[/tex]

[tex]-y=(1+3i)y[/tex]

[tex]-3x+z=(1+3i)z[/tex]

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

[tex]
3z=3ix
[/tex]

or:

[tex]z=ix[/tex]

so let x=1 and z=i.

Thus:

[tex]
v_2=
\left(
\begin{array}{c} 1 \\
0 \\
i
\end{array}
\right)
[/tex]

Same dif for the other eigenvalue which yields:

[tex]
v_3=
\left(
\begin{array}{c} 1 \\
0 \\
-i
\end{array}
\right)
[/tex]

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

[tex]
\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)
[/tex]

You ever work a problem and the answer is just as difficult as the question?
uob_student
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#12
Oct14-05, 06:40 AM
P: 22
Quote Quote by saltydog
Suppose I should start with calculating eigenvalues:
[tex]
\text{det}\left(\mathbf{A}-\lambda\mathbf{I}\right)=0
[/tex]
so for the 3x3 matrix above, that would be:
[tex]
\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
[/tex]
or:
[tex]
-10-8\lambda+\lambda^2-\lambda^3=0
[/tex]
Since this is not a problem focusing on cubic equations, I simply use Mathematica to calcualte the eigenvalues:
[tex]\text{Eigenvalues[Matrix]}[/tex]
They are:
[tex]\lambda_1=-1[/tex]
[tex]\lambda_2=1+3i[/tex]
[tex]\lambda_3=1-3i[/tex]

thanks

but how you can find the values of Lambdas in your solution??
saltydog
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#13
Oct14-05, 05:57 PM
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Quote Quote by uob_student
thanks
but how you can find the values of Lambdas in your solution??
By solving the eigenvalue equation:

[tex]
\text{det}\left(\mathbf{A}-\lambda\mathbf{I}\right)=0
[/tex]

So performing that matrix algebra gives us:

[tex]
\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
[/tex]
or:
[tex]
-10-8\lambda+\lambda^2-\lambda^3=0
[/tex]

Is that what you're having problems with or just figuring the roots of that cubic equation?
saltydog
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#14
Oct14-05, 07:26 PM
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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 [itex] \lambda_3[/itex]:

The complex conjugate of an eigenvector for [itex] \lambda_2[/itex] is an eigenvector for [itex]\lambda_3[/tex].

So above I calculated the eigenvector for [itex](1+3i)[/itex] to be:

[tex]v_2=\left(\begin{array}{c} 1 \\0 \\i\end{array}\right)[/tex]

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

If:

[tex]v_2=\left(\begin{array}{c} 1+0i \\0+0i \\0+i\end{array}\right)[/tex]

Then:

[tex]\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)[/tex]

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:

[tex]\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)[/tex]

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

[tex]
\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}
[/tex]

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:

[tex]
\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}
[/tex]

That's read as:

[tex]x(t)=C_2e^tCos(3t)+C_3e^tSin(3t)[/tex]

[tex]y(t)=C_1e^{-t}[/tex]

[tex]z(t)=-C_2e^tSin(3t)+C_3e^tCos(3t)[/tex]

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.
uob_student
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#15
Oct15-05, 01:11 AM
P: 22
Quote Quote by saltydog
By solving the eigenvalue equation:


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

thanks but I want just how you figuring the roots of that cubic equation?
saltydog
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#16
Oct15-05, 04:58 AM
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Quote Quote by uob_student
thanks but I want just how you figuring the roots of that cubic equation?
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.
uob_student
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#17
Oct15-05, 05:06 AM
P: 22
Quote Quote by 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.

can you explain that??

saltydog
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#18
Oct15-05, 05:26 AM
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Quote Quote by uob_student
can you explain that??
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.


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