- #1

qetuol

- 8

- 0

f ''

_{i}=

**C**

_{ij}f

_{j}

C is a matrix, f

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

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- Thread starter qetuol
- Start date

- #1

qetuol

- 8

- 0

f ''

C is a matrix, f

- #2

IttyBittyBit

- 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

qetuol

- 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

elibj123

- 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

qetuol

- 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

qetuol

- 8

- 0

anyone can answer me? please?

- #7

ross_tang

- 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

qetuol

- 8

- 0

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

- #9

ross_tang

- 64

- 0

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