Coupled ODE with missing connecting derivatives

[FrankST]

Hi,

I have a coupled system of ODE like:

w1'' = A w2'' + B w1 + C w2
w2'' = D w1'' + E w1 + F w2

I need to solve it analytically but it seems it cannot be solved using eigenvalue method. My concern is first that if this system have sufficient equations and if so how it can be solved analytically. If there is no analytical solution, is there any numerical solution for that?


Thanks for your assistance,

Frank

 

[tiny-tim]

Welcome to PF!

Hi Frank! Welcome to PF!

Isn't that just PW'' = QW, with P and Q matrices?

 

[JJacquelin]

Solving is possible by substitution.
w1'' = A w2'' + B w1 + C w2
w2'' = D w1'' + E w1 + F w2
w2''= D(A w2'' + B w1 + C w2) + E w1 + F w2
w1 = (w2'' - D A w2''- D C w2 - F w2 ) / ( D A + E )
w1'' = (w2'''' - D A w2''''- D C w2'' - F w2'' ) / ( D A + E )
Then, bringing back w1 and w1'' into w2'' = D w1'' + E w1 + F w2 leads to a fourth order linear ODE with only one unknown function w2

 

[jackmell]

I personally prefer the more elegant operator approach. First, I'll use lower-case constants and subscripts to make it easier to read. The operator approach is more methodical and easy to apply:

$$w_1^{''}=aw_2^{''}+bw_1+cw_2$$

$$w_2^{''}=dw_1^{''}+ew_1+fw_2$$

Now move everything over:

$$w_1^{''}-bw_1-aw_2^{''}-cw_2=0$$

$$w_2^{''}-fw_2-dw_1^{''}-ew_1=0$$

Now convert to operators and line-up everything:

$$(D^2-b)w_1-(aD^2+c)w_2=0$$

$$(dD^2+e)w_1-(D^2-f)w_2=0$$

Now, to eliminate [itex]w_2[/tex], operate on the first by $(D^2-f)$ and on the second by $-(aD^2+c)$ and add. I get then:<br /> <br /> $$(D^2-f)(D^2-b)w_1-(aD^2+c)(dD^2+e)w_1=0$$<br /> <br /> Do something similar to get the equation in $w_2$[/itex]

 

[FrankST]

Thank you guys for you valuable advices.

Your answers helped me a lot.