1
$\begingroup$

I want to solve this differential equation $$ u^{(4)}+a u^{(2)} +bu=0$$

I put $v=u^{(2)}.$ I obtain the new equation $v^{(2)}+a v+bu=0.$

What to do with $u?$ How to continue?

Thanks!

$\endgroup$
  • $\begingroup$ Are $a$ and $b$ constants? $\endgroup$ – Adrian Keister Aug 6 '19 at 14:50
  • $\begingroup$ yes @adrian keister $\endgroup$ – Vrouvrou Aug 6 '19 at 14:51
  • $\begingroup$ @complexmanifold: The notation $u^{(2)}$ means the second derivative of $u,$ not the square of $u.$ $\endgroup$ – Adrian Keister Aug 6 '19 at 15:00
  • $\begingroup$ @complexmanufold $u^{(2)}$ means the second derivative of u $\endgroup$ – Vrouvrou Aug 6 '19 at 15:01
2
$\begingroup$

I wouldn't do the substitution then, because it's unclear how to "not do" an operation like differentiation. I would go from the original DE, and substitute in the usual ansatz: $u=e^{\lambda x}$ (assuming $u=u(x).$) Then we obtain the quartic equation $\lambda^4+a\lambda^2+b=0.$ Here's where we would do the substitution $\alpha=\lambda^2,$ to obtain the quadratic $\alpha^2+a\alpha+b=0.$ The solution here is $$\alpha=\frac{-a\pm\sqrt{a^2-4b}}{2}. $$ We could get negative roots, making pairs of $\lambda$'s complex conjugates. We have $$\lambda=\pm\sqrt{\frac{-a\pm\sqrt{a^2-4b}}{2}}. $$ So the solution would be \begin{align*} u(x)= &A\exp\left(x\sqrt{\frac{-a+\sqrt{a^2-4b}}{2}}\right)+ B\exp\left(x\sqrt{\frac{-a-\sqrt{a^2-4b}}{2}}\right)+\\ &C\exp\left(-x\sqrt{\frac{-a+\sqrt{a^2-4b}}{2}}\right)+ D\exp\left(-x\sqrt{\frac{-a-\sqrt{a^2-4b}}{2}}\right) . \end{align*} You would need initial conditions to find $A, B, C,$ and $D.$ If there are complex solutions, you can translate to sine and cosine if you wish via $$e^{i\theta}=\cos(\theta)+i\sin(\theta). $$

$\endgroup$
  • $\begingroup$ thank you is there a methods when a and b are functions? $\endgroup$ – Vrouvrou Aug 6 '19 at 15:04
  • $\begingroup$ @Vrouvrou Not in general. $\endgroup$ – eyeballfrog Aug 6 '19 at 15:04
  • $\begingroup$ That's way harder! It would depend on the functions. Most often, if you need an analytical solution, series solutions are your only option. $\endgroup$ – Adrian Keister Aug 6 '19 at 15:04
  • 1
    $\begingroup$ "We're definitely going to get at least one negative root" Not if $a<0<b$. $\endgroup$ – J.G. Aug 6 '19 at 15:17
  • $\begingroup$ J.G. You're right, thanks! I'll edit. $\endgroup$ – Adrian Keister Aug 6 '19 at 15:27
1
$\begingroup$

Сharacteristic equation $t^4+at^2+b=0$. Then $t_{1,2,3,4}=\pm \sqrt{\frac{-a \pm \sqrt{a^2-4b}}{2}}$. Then $u=C_1e^{t_1x}+C_2e^{t_2x}+C_3e^{t_3x}+C_4e^{t_4x}$.

$\endgroup$
  • $\begingroup$ but $-a-\sqrt{a^2-4b}$ is negative ???? $\endgroup$ – Vrouvrou Aug 6 '19 at 20:07
  • $\begingroup$ If $a<0$ then positive. If $t_i$ is complex then there is Euler's formula. $\endgroup$ – Witold Aug 6 '19 at 21:30
  • $\begingroup$ what is euler formula? $\endgroup$ – Vrouvrou Aug 6 '19 at 22:17
  • $\begingroup$ Your problem is quite standard. Solution methods are described in textbooks ( math.psu.edu/tseng/class/Math251/Notes-HigherOrderLinEq.pdf ). $\endgroup$ – Witold Aug 6 '19 at 22:36
1
$\begingroup$

According to your idea the formulation is

$$ \begin{cases} u'' = v\\ v''=-a v-b u \end{cases} $$

You can now follow with the Laplace transform obtaining

$$ \begin{cases} s^2U(s) - su'(0)-u(0) = V(s)\\ s^2V(s)-s v'(0) - v(0) = -a V(s) - b U(s) \end{cases} $$

etc.

$\endgroup$
  • $\begingroup$ and if a and b are functions de can use this methods? $\endgroup$ – Vrouvrou Aug 6 '19 at 17:15
  • $\begingroup$ @Vrouvrou With $a(t),b(t)$ is not straightforward. $\endgroup$ – Cesareo Aug 6 '19 at 18:31
  • $\begingroup$ i don't understand what you mean? $\endgroup$ – Vrouvrou Aug 6 '19 at 18:32
  • $\begingroup$ @Vrouvrou For instance, if $a(t) = t^n$ the Laplace transform gives another differential equation now in $s$. For $a(t), b(t)$ Laplace transformable functions, after transforming we get an integral (convolutional) equation. Those diffrential or integral or integrodiffrential equations in $s$ can be quite cumbersome to solve. $\endgroup$ – Cesareo Aug 6 '19 at 18:36

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.