0
$\begingroup$

I have a problem with homogenous system of two differential equations (fourth order). Only thing which I know is that solutions of characteristic equation (polynomial of 8th order) are complex in the following form

$$s_1=m_1+n_1*i,\; s_2=m_1-n_1*i,\; s_3=-m_1+n_1*i,\; s_4=-m_1-n_1*i,$$ and $$s_5=m_2+n_2*i,\; s_6=m_2-n_2*i,\; s_7=-m_2+n_2*i,\; s_8=-m_2-n_2*i.$$

Is there anyway to obtain analytically solutions for $y_1(x)$ and $y_2(x)$ with 8 unknown constants (constants can be solved using initial conditions). The system is in the following form where $a, b, c$ and $d$ are known constants in the equation.

$$y_1''''(x) + a*y_1''(x) + b*(y_1(x) - y_2(x))=0,$$ $$y_2''''(x) + c*y_2''(x) + d*y_2(x) - b*(y_1(x) - y_2(x))=0.$$

$\endgroup$
0
$\begingroup$

Symplectic systems typically have such an eigenvalue structure. But on the other hand, they are of order $1$, or as Newton equation, of order $2$, not $4$ as in this case.

The eigenvalues tell you that the homogeneous solutions are linear combinations of $\cosh(m_kt)\cos(n_kt)$, $\cosh(m_kt)\sin(n_kt)$, $\sinh(m_kt)\cos(n_kt)$, $\sinh(m_kt)\sin(n_kt)$ where $k=1,2$.

Insertion with vector valued coefficients should result in the eigen equation for the corresponding eigenvectors.


These functions have the advantage to be real, but the decided disadvantage that they are not eigensolutions, since they combine eigenvectors of different eigenvalues. While one can make an ansatz with each group ($k=1,2$) of real functions at once, it is easier to do the complex calculations and set for the first eigenvalue $y_1(t)=c_1e^{(m_1+n_1i)t}$ and $y_2(t)=c_2e^{(m_1+n_1i)t}$, etc., insert into the homogeneous system and after factoring out the exponential, the resulting $2x2$ system should be singular and give the solution subspace. Then, the real and imaginary part of the eigensolutions are real solutions.

$\endgroup$
  • $\begingroup$ Can You give me some instructions in detail or just start point that I can continue $\endgroup$ – Pipe Mar 11 '15 at 20:28
  • $\begingroup$ how many constants? just present us the solution for each function $\endgroup$ – Pipe Mar 11 '15 at 21:37
  • $\begingroup$ Done. Alternatively, use $y_1(t)=e^{m_1t}(c_1\cos(n_1t)+d_1\sin(n_1t))$ and $y_2(t)=e^{m_1t}(c_2\cos(n_1t)+d_2\sin(n_1t))$ for the first pair of eigenvalues, etc. $\endgroup$ – LutzL Mar 11 '15 at 22:00
  • $\begingroup$ so, as I understood, firstly I should insert this first solution into homogenous system and determine first pair of constants c1 and c2, then I should insert second solution and determine c3 and c4. Finally when I will obtain c1,c2, c3, c4... c8, what will be the function y1(x) and what y2(x) in the last step? $\endgroup$ – Pipe Mar 12 '15 at 11:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.