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I am trying to solve this third order system of homogenous ODEs.

  • $x'''=2x+y$

  • $y'''=x+2y$

Initial conditions are given as well.

Higher order systems weren't covered in the lectures hence I am a bit lost. Nor can I find any similar questions asked on this site. I have tried to solve the equations in two ways so far. Firstly express $y$ in terms of $x'''$ and $x$ from the first equation differentiate the equation three times giving: $x^{(6)}-2x=y'''$ and plugging this into the second equation finally obtaining:

$x^{(6)}-4x'''+3x=0$

A higher order homogeneous ODE with constant coefficients that I can solve with the substitution $x = e^{\lambda x}$. The solution can then be differentiated three times and plugged into $x'''=2x+y$ to obtain $y$. Using the given initial conditions we can then find the specific solution. However, this proves very tedious, especially solving the 6x6 system of equations.

Is this approach correct and applicable to other similar systems? Or is there a completely different approach to this type of problem I am not aware of? Could you try to solve the system $\vec{u}''' = A \vec{u}$ by diagonalizing A and look for a solution that way, I've tried this as well but I cannot seem to find a solution.

Any help is much appreciated.

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Because of the symmetric cyclic/circulant nature of the system matrix, you can decouple the system by Fourier transform. Set $u=x+y$, $v=x-y$ to get the system $$ u'''=3u\\ v'''=v $$ which can be both solved with standard methods for linear ODE with constant coefficients.

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  • $\begingroup$ When computing both general solutions for $v$ and $u$, the general solutions for $x$ and $y$ are $x = (u+v)/2$ and $y = (u-v)/2$. Is this correct? If so this way is much easier than the alternatives. $\endgroup$ – blaz stojanovic Jan 11 '18 at 9:32
  • $\begingroup$ Yes, that would correspond to the inverse discrete Fourier transform (which is rather trivial with sequence length 2, non-trivial in extensions to higher dimensions). $\endgroup$ – Lutz Lehmann Jan 11 '18 at 10:21
  • $\begingroup$ @blazstojanovic: Just a remark: this solution is precisely what you get if you diagonalize $A$, as you yourself suggested in the question. $\endgroup$ – Hans Lundmark Jan 11 '18 at 10:46
  • $\begingroup$ @HansLundmark This is a question coming from someone that is not familiar with Fourier transforms. So in some sense, the Fourier transform is equivalent to diagonalizing a matrix, does this mean that the Fourier transform has something to do with finding the special set of coordinate axes in which the matrix takes this canonical form? $\endgroup$ – blaz stojanovic Jan 11 '18 at 13:39
  • $\begingroup$ @blazstojanovic: In this case, I think it's actually easier to just think about diagonalization (which works in general for problems like this one), and not bother with Fourier transforms. But for circulant matrices, diagonalization and Fourier transforms are related. See this question, for example: math.stackexchange.com/questions/1427210/… $\endgroup$ – Hans Lundmark Jan 11 '18 at 13:48

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