- #1

- 8

- 0

f ''

_{i}=

**C**

_{ij}f

_{j}

C is a matrix, f

_{j}(z) are functions dependent of z.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter qetuol
- Start date

- #1

- 8

- 0

f ''

C is a matrix, f

- #2

- 160

- 0

f ''

===>

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

- #3

- 8

- 0

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

- 240

- 2

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

- 8

- 0

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.. c

if so, one question remains... how are the indexes assigned to eigenvalues? i mean which eigenvalue will be c

Last edited:

- #6

- 8

- 0

anyone can answer me? please?

- #7

- 64

- 0

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]

[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

- 8

- 0

ross, thank you very much, i will look into it

- #9

- 64

- 0

Share:

- Replies
- 1

- Views
- 3K

- Replies
- 1

- Views
- 887

- Replies
- 1

- Views
- 3K