I try to solve a system of six couple linear PDF. I applied the Laplace transform method and with some simplifications I retain with the following three coupled ODE

   eq1=s^2*(1-w)*e[x] + C1*D[e[x],{x,2}]+ C2*D[e[x],{x,4}]- C3*D[θ[x], 
        {x,4}] + C4 D[θ[x],{x,6}];
   eq2=-s*a (s^2 β + ϵ1*ϵ2)*e[x] - s*a*θ[x] + 
       b1*g[x] + (1 +s*a*β)*(w*D[e[x], {x, 2}]+β*D[θ[x], {x, 4}]);
   eq3=b3*s^2*e[x]-(b2 + b*s)*g[x]-b3*(s^2*β+ w)*D[e[x], {x, 2}] + D[g[x], 
      {x, 2}] +  b3*β*w*D[e[x], {x, 4}] - b3*β*D[θ[x], {x, 4}] + 
      b3*β^2*D[θ[x], {x, 6}];

It is suppose to obtain a six order ODE in $θ[x]$ but I could not simplify this system or to solve even with the following DSolve command:


it gives no response. I tried to use the command Reduce, but Mathematica gives no response too. If there is no any syntax mistakes, would some one suggests what do can I do? Thank you. In this system the only functions are $θ[x]$, $e[x]$ and $g[x]$, and all other characters are non zero constants.

  • $\begingroup$ Bad syntax. The equations must go in a list. However, even with that fixed, I highly doubt that a solution is possible with so many symbolic parameters. $\endgroup$
    – Szabolcs
    Sep 23, 2018 at 12:11
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    $\begingroup$ Also, these are ordinary differential equations—there's only one variable. $\endgroup$
    – Szabolcs
    Sep 23, 2018 at 12:12
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    $\begingroup$ Because the equations are linear and homogeneous, the only solution is that all dependent variables are zero unless the constants happen to satisfy specific relationships. $\endgroup$
    – bbgodfrey
    Sep 23, 2018 at 13:02
  • $\begingroup$ A list like: sys={eq1==0,eq2==0,eq3==0}? The original system becomes ordinary differential equation in the variable x after application of the Laplace transform. $\endgroup$
    – Essam
    Sep 23, 2018 at 13:54
  • $\begingroup$ A list like:` sys={eq1==0,eq2==0,eq3==0}?` The original system becomes ordinary differential equation in the variable x after application of the Laplace transform.These constant are the material constants. I put it in the form D[θ[x], {x,2}]=A1 θ[x]+A2 e[x]-A3 g[x]; D[e[x], {x,2}]=B1 θ[x]+B2 e[x]-B3 g[x]; D[g[x], {x,2}]=C1 θ[x]+C2 e[x]-C3 G[x] where A1,...,C3 are constants. it is not slove also with DSolve. $\endgroup$
    – Essam
    Sep 23, 2018 at 14:19

1 Answer 1


As I noted in an a comment above, the general solution of the coupled ODEs is zero for all three dependent variables, because the equations are linear and homogeneous. However, because the coefficients all are constants, the equations can be Fourier-decomposed, after which the values of k that cause the determinant of the equations to vanish can be determined. Only for these eigenvalues are non-zero solutions possible.

det = Simplify[CoefficientArrays[Exp[-I k x] {eq1, eq2, eq3} /. 
    {θ -> Function[x, θ0 Exp[I k x]], e -> Function[x, e0 Exp[I k x]], 
     g -> Function[x, g0 Exp[I k x]]}, {θ0, e0, g0}] 
    // Normal // Last // Det] /. k^n_ -> ksq^(n/2)
(* b1 b3 ksq^2 (1 + ksq β) (C3 (s^2 + ksq w) + C4 ksq (s^2 + ksq w) + (C1 ksq - 
   C2 ksq^2 + s^2 (-1 + w)) β) + (-b2 - ksq - b s) (-(-C1 ksq + C2 ksq^2 - s^2 
   (-1 + w)) (-a s + ksq^2 β (1 + a s β)) + ksq^2 (C3 + C4 ksq) 
   (ksq (w + a s w β) + a s (s^2 β + ϵ1 ϵ2)))*)

Thus, we have a fifth order polynomial in k^2, which can be solved by

Solve[det == 0, ksq] // Flatten

Not unexpectedly, the results are five Root functions, which are rather long to reproduce here.

  • 1
    $\begingroup$ If I copy the three equations from the question and the code from my answer into a fresh notebook, I obtain the result shown in the answer with no error messages. Try to do the same. $\endgroup$
    – bbgodfrey
    Sep 23, 2018 at 17:02
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    $\begingroup$ I believe that you are mistyping something. $\endgroup$
    – bbgodfrey
    Sep 24, 2018 at 12:13
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    $\begingroup$ k is the wavenumber. q does not appear in the equations. By the way, I noticed that you have four questions with answers but have accepted none of them.. Eventually, this will discourage readers from answering your questions. $\endgroup$
    – bbgodfrey
    Sep 24, 2018 at 14:31
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    $\begingroup$ I do not know what you mean by, the answer "contains many # slots". I see none. Whether you accept this, or any, answer, is your choice. It does not accomplish much to write comment that you accept a solution. Again, though, that is your choice. $\endgroup$
    – bbgodfrey
    Sep 25, 2018 at 18:07
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    $\begingroup$ I urge you to read the documentation for Root to understand the usage of # in this context. By the way, the message to "Please avoid extended discussions in comments. Would you like to automatically move this discussion to chat?" was from the web site, not from me. $\endgroup$
    – bbgodfrey
    Sep 25, 2018 at 19:30

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