As a part of my research, I am stuck in solving the analytical solution of the following 6th order differential equation:

$u_t^{VI}(\theta) + 2 u_t^{IV}(\theta) + A_3 u_t^{II}(\theta) + A_2 u_t (\theta) = 0$,

where the superscript greek letters represent the derivative order and $A_2$ and $A_3$ are the integration constants. The following boundary conditions are given:

$ u_t^V(0) + u_t^{III}(0) = A_4$,

$ u_t^V(\frac{\pi}{2}) + u_t^{III}(\frac{\pi}{2}) = A_4$,

$u_t^{II}(0) + u_t(0) = 0$

$u_t^{II}(\frac{\pi}{2}) + u_t(\frac{\pi}{2}) = 0$

$u_t^{I}(0) = 0$

$u_t^{I}(\frac{\pi}{2}) = 0$

where $A_4$ is another constant.

It should be noted that I have successfully solved this differential equation using Mathematica, however, the solution is huge and can not be shortened. I was wondering if you can suggest another method for solving the above differential equation?

Thanks in advance

  • $\begingroup$ the equation given is linear, and you can solve it $\endgroup$ Mar 12, 2018 at 19:54
  • $\begingroup$ The solution will necessarily involve solutions to the order 6 characteristic equation $r^6+2r^4+A_3r^2+A_2=0$. Since this is cubic in $r^2$, it has solutions, but their general form is intrinsically not nice, unless there are constraints on $A_2$ and $A_3$ that you're not telling us about. $\endgroup$
    – Micah
    Mar 12, 2018 at 21:01
  • $\begingroup$ Dear @Dr.SonnhardGraubner is it possible to show me the hint to solve it nicely? Beacuse as I said in the question, I can solve the equation, but the solution is huge and cannot be considered as a closed-form solution. $\endgroup$
    – KratosMath
    Mar 12, 2018 at 21:22
  • $\begingroup$ @Micah Actually there are some physical constraints on A2, A4 (the one in B.Cs). For example, both A2 and A4 cannot have negative or zero values. But regarding A3, it may have any possible value. $\endgroup$
    – KratosMath
    Mar 13, 2018 at 7:47
  • $\begingroup$ Unfortunately, that's not the kind of constraint that makes your life much easier — you'd need some specific algebraic relation between them. Best of luck! $\endgroup$
    – Micah
    Mar 13, 2018 at 17:40

1 Answer 1


$$u_t^{VI}(\theta) + 2 u_t^{IV}(\theta) + A_3 u_t^{II}(\theta) + A_2 u_t (\theta) = 0 \tag 1$$ The usual method to solve this kind of ODE is first to find particular solutions on the form : $\quad u=e^{r\:t}$. Putting it into Eq.$(1)$ leads to : $$r^6+2r^4+A_3r^2+A_2=0 \tag 2$$ Let : $\quad R=r^2$ $$R^3+2R^2+A_3R+A_2=0$$ Of course, the cubic equation is analytically solvable, but the three roots $R_1,R_2,R_3$ are not nice.

The best way to make understand what "not nice" means is to show them :

enter image description here

It's up to you to use those ugly formulas in the next equations.

The six roots of Eq.$(2)$ are :

$r_1=\sqrt{R_1}\:,\quad r_2=\sqrt{R_2}\:,\quad r_3=\sqrt{R_3}\:,\quad r_4=-\sqrt{R_1}\;,\quad r_5=-\sqrt{R_2}\:,\quad r_6=-\sqrt{R_3} $

The general solution of Eq.$(1)$ is :

$$u(t)=c_1e^{r_1t}+c_2e^{r_2t}+c_3e^{r_3t}+c_4e^{r_4t}+c_5e^{r_5t}+c_6e^{r_6t}$$ $$u(t)=\sum_{k=1}^{k=6}c_ke^{r_k t} \tag 4$$

The coefficients $c_1,c_2,c_3,c_4,c_5,c_6$ have to be determined according to the conditions :

$$ u_t^V(0) + u_t^{III}(0) = A_4 \quad\to\quad \sum_{k=1}^{k=6}(1+r_k^3)c_k=A_4$$

$$ u_t^V(\frac{\pi}{2}) + u_t^{III}(\frac{\pi}{2}) = A_4 \quad\to\quad \sum_{k=1}^{k=6}(r_k^5+r_k^3)\exp(r_k\frac{\pi}{2})c_k =A_4$$

$$u_t^{II}(0) + u_t(0) = 0 \quad\to\quad \sum_{k=1}^{k=6}(r_k^2+1)c_k =0$$

$$u_t^{II}(\frac{\pi}{2}) + u_t(\frac{\pi}{2}) = 0 \quad\to\quad \sum_{k=1}^{k=6}(r_k^2+1)\exp(r_k\frac{\pi}{2})c_k =0$$

$$u_t^{I}(0) = 0 \quad\to\quad \sum_{k=1}^{k=6}r_k c_k =0$$

$$u_t^{I}(\frac{\pi}{2}) = 0 \quad\to\quad \sum_{k=1}^{k=6}r_k\exp(r_k\frac{\pi}{2})c_k =0$$

This is a linear system of six equations that you can classically solve for the six unknown $c_k$. All the coefficients have been previously computed.

Finally, put the coefficients $c_k$ into Eq.$(4)$ for the solution $u(t)$.


Of course, all the above calculus has to be carried out in the complex domain. At end, if among the coefficients $r_k$ some are complex, transform the exponential of complex number into the product of real exponential and sinusoid functions.

  • $\begingroup$ Thanks for the answer. It appears that there are imaginary parts in the solutions of R. How to deal with them? $\endgroup$
    – KratosMath
    Mar 13, 2018 at 8:59
  • $\begingroup$ See the note added at the end of the answer. $\endgroup$
    – JJacquelin
    Mar 13, 2018 at 9:01
  • $\begingroup$ Why do you say "it is up to you to use those ugly formulas"? Is there another way to avoid them? $\endgroup$
    – KratosMath
    Mar 13, 2018 at 9:07
  • $\begingroup$ Anyway, thanks for your complete answer. However, it would be fantastic if you can show me a hint on how to transform the exponential of complex number into the product of real exponential and sinusoid functions. $\endgroup$
    – KratosMath
    Mar 13, 2018 at 9:26
  • $\begingroup$ $$\exp(\alpha +i\beta)=\exp(\alpha)\cos(\beta)+i\:\exp(\alpha)\sin(\beta)$$ If the initial parameters in the ODE and in the conditions are all real, in the calculus the complex terms are associated with the conjugates terms. At the end, The imaginary parts of the complex terms and their conjugates will simplifies one to one and disappear. The final solution will be real. $\endgroup$
    – JJacquelin
    Mar 13, 2018 at 9:49

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