Solving coupled linear diff. equations

In summary: I said, I'm not exactly sure how to solve the problem. In summary, the student is trying to solve a coupled system of equations.
  • #1
c0dy
15
0

Homework Statement


http://i.imgur.com/zmtJ64Z.jpg

Homework Equations


B is a column vector {{0},{0},{1}}
E is a column vector {{1},{0},{0}}
v(t) is also a column vector {{v_1},{v_2},{v_3}}

The Attempt at a Solution


I have calculated v(t) x B to get (v_2,-v_1,0) and I made a matrix,

| 0 1 0|
|-1 0 0| = A
| 0 0 0|

and now I have dv/dt = A*v + E

Am I going about this correctly? Do I integrate with respect to t? I'm not exactly sure how to solve a problem like this. The assignment is to do this in mathematica but I'd like to understand how to solve the problem first.
 
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  • #2
I'm thinking, dv/dt = (v_2, -v_1, 0) + (1, 0, 0) = (v_2 + 1, -v_1, 0)

Then integrate?

v(t) = (t*(v_2 + 1) + c1, -t*v_1 + c2, 0)

=> v(0) = (c1, c2, 0) = (1, -1, 0) => c1 = 1, c2 = -1
 
Last edited:
  • #3
It might be easier to examine the system writing down each component equation explicitly. Denoting the components of v as x, y and z:

x' = y + 1
y' = -x
z' = 0

You cannot just integrate the right hand side, because it contains unknown functions - except the third one, which you can solve right away. The first two, however, depend on each other. That's why the system is called "coupled".

What you can do, however, is differentiate the first equation and get x'' = y', and substitute y' from the second equation, getting x'' = -x.
 
  • #4
Aha, that makes sense. Thanks a lot.
 
  • #5
c0dy said:

Homework Statement


http://i.imgur.com/zmtJ64Z.jpg


Homework Equations


B is a column vector {{0},{0},{1}}
E is a column vector {{1},{0},{0}}
v(t) is also a column vector {{v_1},{v_2},{v_3}}

The Attempt at a Solution


I have calculated v(t) x B to get (v_2,-v_1,0) and I made a matrix,

| 0 1 0|
|-1 0 0| = A
| 0 0 0|

and now I have dv/dt = A*v + E

Am I going about this correctly? Do I integrate with respect to t? I'm not exactly sure how to solve a problem like this. The assignment is to do this in mathematica but I'd like to understand how to solve the problem first.

There are a couple of ways to solve such problems.
(1) Assume a form of solution, and find the parameters that "work".
(2) Use a matrix exponential.

For your system, the z-component is separate and easy, while the x and y components give x' = y, y' = -x.

Method(1): Assuming x = a*exp(r*t) and y = b*exp(r*t) [same r in both!] you have:
[tex] x' = r a e^{rt} = y = b e^{rt}\\
y' = r b e^{rt} = -x = -a e^{rt}\\
\text{so}\\
ra = b,\; rb = -a\; \Longrightarrow r^2 a = -a.[/tex]
If a = 0 then also b = 0, and so 0 is the solution. If a ≠ 0 then ##r^2 = -1##, so ##r = \pm i##. That means that we have solutions involving ##\exp(\pm i t)##, or--in real terms--##\cos(t),\; \sin(t).##

Method (2) The solution of X' = AX (A = constant) is X = C*exp(A*t). If A is a (square) matrix, you need a way to compute the matrix exponential. This can be done using an eigenvalue/eigenvector expansion of the matrix A. In a computer algebra system such as Maple or Mathematica, this can be done at the push of a button. For example, in Maple we have:
A:=Matrix(3,3,[[0,1,0],[-1,0,0],[0,0,0]]):
Et:=MatrixExponential(A,t); <---- compute the exponential of A*t
The answer, in LaTeX form, is:
[tex] Et = \pmatrix{\cos(t)&\sin(t)&0\\-\sin(t)&\cos(t)&0\\0&0&1}[/tex]
 
  • #6
Ray Vickson said:
There are a couple of ways to solve such problems.
(1) Assume a form of solution, and find the parameters that "work".
(2) Use a matrix exponential.

For your system, the z-component is separate and easy, while the x and y components give x' = y, y' = -x.

Method(1): Assuming x = a*exp(r*t) and y = b*exp(r*t) [same r in both!] you have:
[tex] x' = r a e^{rt} = y = b e^{rt}\\
y' = r b e^{rt} = -x = -a e^{rt}\\
\text{so}\\
ra = b,\; rb = -a\; \Longrightarrow r^2 a = -a.[/tex]
If a = 0 then also b = 0, and so 0 is the solution. If a ≠ 0 then ##r^2 = -1##, so ##r = \pm i##. That means that we have solutions involving ##\exp(\pm i t)##, or--in real terms--##\cos(t),\; \sin(t).##

Method (2) The solution of X' = AX (A = constant) is X = C*exp(A*t). If A is a (square) matrix, you need a way to compute the matrix exponential. This can be done using an eigenvalue/eigenvector expansion of the matrix A. In a computer algebra system such as Maple or Mathematica, this can be done at the push of a button. For example, in Maple we have:
A:=Matrix(3,3,[[0,1,0],[-1,0,0],[0,0,0]]):
Et:=MatrixExponential(A,t); <---- compute the exponential of A*t
The answer, in LaTeX form, is:
[tex] Et = \pmatrix{\cos(t)&\sin(t)&0\\-\sin(t)&\cos(t)&0\\0&0&1}[/tex]

I think method 2 is how he wanted us to solve it. We were also using a similar method for solving partial differential equations. Is there a decent textbook which focuses solving ODE or PDEs this way?
 
  • #7
For example, Arnold's book on ODEs. I am fairly sure that any book on ODEs printed within the last 30 years or so should have a section on matrix methods.
 

1. What are coupled linear differential equations?

Coupled linear differential equations are a set of two or more equations that are connected or dependent on each other. They involve the derivatives of one or more dependent variables with respect to an independent variable, and they can be solved simultaneously to find the values of the dependent variables.

2. How do you solve coupled linear differential equations?

To solve coupled linear differential equations, we use methods such as substitution, elimination, and integration. The first step is to identify the dependent and independent variables in each equation and determine the order of the derivatives. Then, we use algebraic manipulation and integration techniques to eliminate the derivatives and solve for the dependent variables.

3. What are some real-life applications of solving coupled linear differential equations?

Coupled linear differential equations are commonly used in many fields of science and engineering, such as physics, chemistry, and economics, to model and analyze dynamic systems. For example, they can be used to study the motion of objects under the influence of forces, the growth of populations, and the behavior of chemical reactions.

4. What are the challenges of solving coupled linear differential equations?

One of the main challenges of solving coupled linear differential equations is the complexity of the equations and the need for advanced mathematical techniques. It can also be difficult to identify the correct form of the equations and the initial or boundary conditions. Additionally, errors in calculations or data can affect the accuracy of the solutions.

5. Can computer software be used to solve coupled linear differential equations?

Yes, computer software can be used to solve coupled linear differential equations. There are many mathematical software programs and programming languages that have built-in functions for solving differential equations, such as Mathematica, MATLAB, and Python. These tools can handle complex equations and provide accurate solutions in a shorter amount of time compared to manual methods.

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