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I am trying to solve the following coupled system:

$$ x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-n^{2})y=z \\ x^{2}\frac{d^{2}z}{dx^{2}}+x\frac{dz}{dx}+(x^{2}-n^{2})z=a y $$

where $a\in R$. Both ODEs in their homogeneous case are the second-order Bessel differential equation. I am lost in how to interpret it as a system, especially when it is coupled. Any help would be appreciated!

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  • $\begingroup$ What is $n$, is it real, an integer...? $\endgroup$ – Moo Dec 27 '17 at 14:15
  • $\begingroup$ it's real @Moo. $\endgroup$ – kathi_h Dec 27 '17 at 16:12
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$$x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-n^{2})y=z \tag 1$$ $$\\ x^{2}\frac{d^{2}z}{dx^{2}}+x\frac{dz}{dx}+(x^{2}-n^{2})z=a y \tag 2$$

Case $\quad a>0.\quad$ (Proceed on similar manner in case $a<0$).

Multiply Eq.$(1)$ by $\sqrt{a}$ and add Eq.$(2)$ :

$$\sqrt{a}\left(x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-n^{2})y\right) +x^{2}\frac{d^{2}z}{dx^{2}}+x\frac{dz}{dx}+(x^{2}-n^{2})z=z\sqrt{a}+a y =\sqrt{a}\left(z+y\sqrt{a} \right)$$ Let $\quad u(x)=z+\sqrt{a}\:y$ $$x^{2}\frac{d^{2}u}{dx^{2}}+x\frac{du}{dx}+(x^{2}-n^{2})u=\sqrt{a}\: u$$

$$x^{2}\frac{d^{2}u}{dx^{2}}+x\frac{du}{dx}+(x^{2}-n^{2}-\sqrt{a})u=0$$

Let $\quad\nu=\sqrt{n^{2}+\sqrt{a}}$ $$x^{2}\frac{d^{2}u}{dx^{2}}+x\frac{du}{dx}+(x^{2}-\nu^{2})u=0$$ $$u(x)=c_1J_\nu(x)+c_2Y_\nu(x)$$ $z=u-\sqrt{a}\:y=c_1J_\nu(x)+c_2Y_\nu(x)-\sqrt{a}\:y \tag 3$

$$x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-n^{2})y=c_1J_\nu(x)+c_2Y_\nu(x)-\sqrt{a}\:y$$

$$x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-n^{2}+\sqrt{a})y=c_1J_\nu(x)+c_2Y_\nu(x)$$

Let $\quad\mu=\sqrt{n^{2}-\sqrt{a}}$ $$x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+(x^{2}-\mu^{2})y=c_1J_\nu(x)+c_2Y_\nu(x)$$ This is a non homogeneous Bessel equation. The particular solutions are :

$u_1(x)=c_1\frac{\pi}{2(\mu^2-\nu^2)}x\left(-J_\mu(x)J_\nu(x)Y_{\mu-1}(x) +J_{\mu-1}(x)J_\nu(x)Y_{\mu}(x)\right)$

$u_2(x)=c_2\frac{\pi}{2(\mu^2-\nu^2)}x\left(-J_\mu(x)Y_\nu(x)Y_{\mu-1}(x) +J_{\mu-1}(x)J_\nu(x)Y_{\mu}(x)\right)$

The general solution is :

$$y(x)=c_1J_\nu(x)+c_2Y_\nu(x)+c_3u_1(x)+c_4u_2(x)$$

Putting it into Eq.$(3)$ gives $z(x)$

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  • $\begingroup$ Dear @JJacquelin, could you please explain how you chose to multiply the first equation by factor $\sqrt{a}$? In my original problem there are more constant-coefficients, and I cannot figure out the right factor to multiply it with, in order to obtain the right $u(x)$. $\endgroup$ – kathi_h Dec 27 '17 at 20:59
  • $\begingroup$ @uli : I guessed a linear relationship like $u(x)=\alpha y(x)+\beta z(x)$ By luck it turns on the combination of Eq.$(1)$ and Eq.$(2)$ becomes an equation with only the function $u(x)$ in case of $a\beta^2=\alpha^2$ . $\endgroup$ – JJacquelin Dec 28 '17 at 7:46
  • $\begingroup$ @JJaquelin: that was pretty helpful, thanks! I can apply your method when both $y$,$z$ have constant-coefficients, the $u$ is then $u(x)=\sqrt{a}y+\sqrt{b}z$. Can you tell me what form $u(x)$ should have when there an additional $x^{2}$ term on the right-hand side? e.g. in (1) right-hand side is $bz+cx^{2}$? I tried $u(x)=\sqrt{a}y(x)+\sqrt{b}z(x) + c$, but it didn't work, because $c$ is multiplied with $n^{2}$ also. Alternatively, when I bring this term to the left, I have $((1-c)x^{2}+n^{2})$ in (1) and $(x^{2}+n^{2})$ in (2) as coefficient of $y$, which doesn't allow to substitute nicely.. $\endgroup$ – kathi_h Dec 29 '17 at 15:39
  • $\begingroup$ With $bz+cx^2$ it's more complicated. Try with $u(x)=\alpha y(x)+\beta z(x)+f(x)$ where $f(x)$ is unknown. The combination of the two equations will lead again to $a\beta^2=\alpha^2$ and to a second order differential equation to be solved for $f(x)$. I didn't do it , so I don't know if it is solvable or not. Good luck ! $\endgroup$ – JJacquelin Dec 29 '17 at 16:13
  • $\begingroup$ @JJaquelin: Sorry for asking so late, I am trying to imitate your way of solution for my problem and have trouble finding the particular solution. Could you please explain how you obtained it? $\endgroup$ – kathi_h Jan 22 '18 at 8:51

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