i was wondering if somebody could help me solving the following two coupled differential equations...

$$\Delta u_{xx}(x)=dw_{xxx}(x)+\frac{K}{EA}\Delta u(x) \\$$ $$-EI_\infty w_{xxxx}(x)+2d\,EA\;\Delta u_{xxx}(x)+q(x)=0$$

where $q(x)$ is a given function e.g. $q(x)=q_0\sin(\frac{\pi x}{l})$

I tried the following step in Mathematica

DSolve[{-EIinf w''''[x] + 2 d EA du'''[x] + q[x] == 0, du''[x] == d w'''[x] + 
K/EA du[x]}, {du, w}, x]

i also tried adding the boundary conditions to the DSolve function but it did not work either. The evaluation is going on "forever" that i have to abort it after some time.

FYI: i have following boundary conditions:

w[0]==0 | w[l]==0 | w''[0]==0 | w''[l]==0 | w''''[0]==q[0]/(EI0) | w''''[l]==q[l]/(EI0) | du'[0]==0 | du'[l]==0 

If i transfer the two differential equations to one DE 6th order i get a analytical solution. Why isn't that the case if i solve both equations simultaneously? Thank you in advance!

  • 2
    $\begingroup$ It is very unlikely that there will be a simple symbolic solution. Use NDSolve to get a numerical solution. (And of course, you have then to include you initial and boundary conditions in the call to NDSolve.) $\endgroup$ Feb 3, 2020 at 10:28
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    $\begingroup$ You have $3$ unknown functions $\Delta u(x), w(x), u(x)$ ( why such a very misleading notation?) and only two equations and so you cannot sove the system. I'm voting to close this question as impossible to answer without more detailed information. $\endgroup$
    – Artes
    Feb 3, 2020 at 12:46
  • 1
    $\begingroup$ I would guess $q$ is a known function, but hard to say without any specification... Anyway, I would start by eliminating $w$ from the second equation using the first equation. That should give an linear second order ODE in $\Delta u,x$. $\endgroup$
    – anderstood
    Feb 3, 2020 at 12:59
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    $\begingroup$ @anderstood You cannot solve the system if you don't know $q(x)$, that's all $\endgroup$
    – Artes
    Feb 3, 2020 at 13:12
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    $\begingroup$ Also, please address issues in the question directly rather than in the comments; in particular, you should specify that q is a known function. I voted to reopen because this question seems clear enough. $\endgroup$
    – anderstood
    Feb 4, 2020 at 8:42

1 Answer 1


The reopening of the question allows me to turn my comment into an answer and expand a bit:

You can solve for $\Delta u$ by eliminating $w$ in the second equation, using the first equation, without knowing $q$, but it gives a long expression:

  eq1 = du''[x] == d w'''[x] + K/EA du[x];
  eq2 = -EIinf w''''[x] + 2 d EA du'''[x] + q[x] == 0;
  sol = First@First@Solve[eq1, w'''[x]]
  eq2b = eq2 /. w''''[x] -> D[w'''[x] /. sol, x] // FullSimplify;
  dusol = DSolve[{eq2b, du'[0] == 0, du'[l] == 0}, du[x], x]
  dusol = Assuming[Reals, Simplify@dusol]

You can simplify the expression a bit if the variables are real (see last line above).

If you prescribe $q(x)=q_0\sin(\pi x/l)$ it becomes much much simpler:

  dusol = DSolve[{eq2b /. q[x] -> q0*Sin[Pi*x/l], du'[0] == 0, du'[l] == 0}, du[x], x]
  (* {{du[x] -> (-EIinf K l^2 π C[3] + 2 d^2 EA^2 π^3 C[3] - 
   EA EIinf π^3 C[3] - d EA l^3 q0 Cos[(π x)/l])/(π (-EIinf K l^2 + 
   2 d^2 EA^2 π^2 - EA EIinf π^2))}} *)

You can then inject the solution $\Delta u$ in the first equation and solve for $w$:

  eq1b = eq1 /. dusol /. du''[x] -> D[dusol[[1, 1, 2]], {x, 2}] // FullSimplify // First
  wsol = DSolve[{eq1b, w[0] == 0, w''[0] == 0}, w[x], x] // Simplify

  (* {{w[x] -> (π^4 (EIinf K l^2 - 2 d^2 EA^2 π^2 + 
   EA EIinf π^2) x (K x^2 C[3] - 6 d EA C[5]) - 
    6 d EA l^4 (K l^2 + EA π^2) q0 Sin[(π x)/l])/(
    12 d^3 EA^3 π^6 - 6 d EA EIinf π^4 (K l^2 + EA π^2))}}  *)

You should now be able to find C[3] and C[5] using the boundary conditions in l. By the way it seems you have two many boundary conditions (overdetermined problem).

  • 1
    $\begingroup$ I want to thank you so much for your effort helping me to solve this problem! $\endgroup$
    – WM_noob5
    Feb 6, 2020 at 11:16

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