I have a set of four differential equations represented by the following matrix equation:

x1'[t] == (-p + I q)*x1[t] + I*m*x3[t];
x2'[t] == (-p + I q)*x2[t] + I*m*x4[t]; 
x3'[t] == (r + I s)*x3[t] + I*m*x1[t]; 
x4'[t] == (r + I s)*x4[t] + I*m*x2[t];

Here $p$, $q$, $r$, $s$ and $m$ are reals. How can I solve this system? Given that $x_1(0)=1, x_2(0)=0,x_3(0)=1,x_4(0)=0.$


closed as off-topic by corey979, MarcoB, rhermans, m_goldberg, Henrik Schumacher Jun 24 '18 at 18:24

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  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – corey979, MarcoB, rhermans, m_goldberg, Henrik Schumacher
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  • 2
    $\begingroup$ Have a look at DSolve and NDSolve. $\endgroup$ – b.gates.you.know.what Jun 24 '18 at 6:50

Use the help and type the keywords of the task you have. Then read the documentation. DSolve should become apparent as a possible avenue to explore. Notice the syntax and the examples. Reproduce the examples but using your own equations.

enter image description here

  x1'[t] == (-p + I q)*x1[t] + I*m*x3[t],
  x2'[t] == (-p + I q)*x2[t] + I*m*x4[t],
  x3'[t] == (r + I s)*x3[t] + I*m*x1[t],
  x4'[t] == (r + I s)*x4[t] + I*m*x2[t],
  x1[0] == 0, x2[0] == 0, x3[0] == 0, x4[0] == 0
 , {x1[t], x2[t], x3[t], x4[t]}
 , t
(* {{x1[t] -> 0, x3[t] -> 0, x2[t] -> 0, x4[t] -> 0}} *)

As pointed out in the comments I didn't use your boundary conditions. My system is somehow simpler. You get the idea, the rest you need to fix yourself.

  • 1
    $\begingroup$ Very good explanation. If you choose correct initial conditions MMA gives a nontrivial solution. $\endgroup$ – Ulrich Neumann Jun 24 '18 at 9:17
  • $\begingroup$ @UlrichNeumann Then the answer is too long to fit the screen! In reality I didn't bother so put the correct numbers, I'm focusing in the process. :) $\endgroup$ – rhermans Jun 24 '18 at 9:19
  • $\begingroup$ Nice video :) Just for the record: with the ICs from the OP, FullSimplify gives a not so complicated solution. $\endgroup$ – corey979 Jun 24 '18 at 11:45

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