System of second order linear homogenous differential coupled equations

In summary, the system of coupled differential equations has a general solution where all the functions are linear multiples of each other and can be represented by trigonometric functions. The Einstein summation convention is used and the eigenvalues of the matrix C determine the indexes of the functions. A method is provided to reduce the equation into a system of first-order differential equations, which can be solved using matrix exponential.
  • #1
qetuol
8
0
my question is: what is the general solution of this system of coupled diff. equations:

f ''i = Cijfj

C is a matrix, fj(z) are functions dependent of z.
 
Physics news on Phys.org
  • #2
f ''i = Cijfj
f ''i = Cikfk

===> Cijfj = Cikfk

Which means all the functions are just linear multiples of each other, and each is just a trigonometric function.

Are you using the Einstein summation convention? If so, you should specify that you are! The summation convention is not something that is commonly used in non-physics-related mathematics.
 
  • #3
oh my bad.
yes, i am using Einsteins's summarizing convention. By the way this is a physics related problem, i am analyzing waves propagating a periodic media with RCWA method.
 
  • #4
It will be almost the same as a first-order equation.
Guess

[tex]\vec{f}(t)=\vec{f}_{s}e^{st}[/tex]

Plug in and get:

[tex]s^{2}\vec{f}_{s}e^{st}=C \vec{f}_{s}e^{st}[/tex]

[tex](C-s^{2}I)\vec{f}=0[/tex]

So here s^2 are the eigenvalues of C. And each eigenvalue has two corresponding solutions with s and -s. I haven't seen treatment of different cases (complex s, multiplicity and such) but I guess it'll be the same as with the first-order analysis.
 
  • #5
elibj123 said:
It will be almost the same as a first-order equation.
Guess

[tex]\vec{f}(t)=\vec{f}_{s}e^{st}[/tex]

Plug in and get:

[tex]s^{2}\vec{f}_{s}e^{st}=C \vec{f}_{s}e^{st}[/tex]

[tex](C-s^{2}I)\vec{f}=0[/tex]

So here s^2 are the eigenvalues of C. And each eigenvalue has two corresponding solutions with s and -s. I haven't seen treatment of different cases (complex s, multiplicity and such) but I guess it'll be the same as with the first-order analysis.

thank you for your answer, so is this correct?:

[tex]f_j=G_je^{i\sqrt{c_j}z}+H_je^{-i\sqrt{c_j}z}[/tex] where [tex]G_j and H_j[/tex] are integrating constants.. cj are eigenvalues of C and are complex..
if so, one question remains... how are the indexes assigned to eigenvalues? i mean which eigenvalue will be c1...?
 
Last edited:
  • #6
anyone can answer me? please?
 
  • #7
I am now providing a method to reduce the equation into system of 1st-order differential equations, so that you can use matrix exponential to solve it easily.

[tex]\mathbf{x}'= C \mathbf{y}[/tex]
[tex]\mathbf{y}'=\mathbf{x}[/tex]
[tex]\frac{d}{\text{dt}}\left(
\begin{array}{c}
\mathbf{x} \\
\mathbf{y}
\end{array}
\right)=\left(
\begin{array}{cc}
0 & C \\
I & 0
\end{array}
\right)\left(
\begin{array}{c}
\mathbf{x} \\
\mathbf{y}
\end{array}
\right)[/tex]

After writing it into system of 1st-order differential equations, you can use matrix exponential to solve it. If you are not sure how to do it, please refer to the tutorial I have written here.

http://www.voofie.com/content/18/solving-system-of-first-order-linear-differential-equations-with-matrix-exponential-method/" [Broken]
 
Last edited by a moderator:
  • #8
ross, thank you very much, i will look into it
 
  • #9
Welcome. Hope it can help you. Please inform us if you can solve your problem, or you have more question. I am willing to help.
 

1. What does "system of second order linear homogenous differential coupled equations" mean?

A system of second order linear homogenous differential coupled equations refers to a set of two or more equations that involve derivatives of the same dependent variable, with no constant term, and where the coefficients of the derivatives are constant. This type of system is called "homogenous" because all the equations have the same dependent variable and "coupled" because the equations are connected through the dependent variable.

2. What makes these types of equations different from other differential equations?

The main difference is that these equations involve more than one dependent variable and they are connected through derivatives. This means that the solutions to these equations will be a combination of functions for each dependent variable, rather than a single function for a single dependent variable. It also means that the solutions can be more complex and require more advanced mathematical techniques to solve.

3. What are some common applications of systems of second order linear homogenous differential coupled equations?

These types of equations are commonly used in physics, engineering, and other scientific fields to model complex systems. They can be used to study the motion of particles, the behavior of circuits, and the dynamics of chemical reactions, among other things.

4. What is the general method for solving these types of equations?

The general method involves finding the characteristic equation of the system, which is a polynomial equation that relates the coefficients of the derivatives to the dependent variables. This equation can then be solved using various techniques, such as factoring, substitution, or using the quadratic formula. The solutions to the characteristic equation will give the general form of the solutions to the system of equations.

5. Are there any special cases or techniques for solving these equations?

One special case is when the coefficients of the derivatives are constant and the equations are uncoupled, meaning they do not involve derivatives of the same dependent variable. In this case, the solutions can be found by solving each equation separately. Another technique is using Laplace transforms, which can simplify the solutions to these equations and make them easier to solve in certain cases.

Similar threads

  • Differential Equations
Replies
2
Views
1K
Replies
2
Views
2K
  • Differential Equations
Replies
5
Views
1K
  • Differential Equations
Replies
3
Views
1K
  • Differential Equations
Replies
7
Views
1K
  • Differential Equations
Replies
7
Views
2K
  • Differential Equations
Replies
1
Views
1K
Replies
6
Views
1K
Replies
1
Views
1K
Replies
1
Views
2K
Back
Top