# Single line coupled 2nd order differential equation

In the paper 2111.05151 that I'm reading, there is a single coupled second-order differential equation (equation 71 in the paper),

$$$$r \ddot{z}_1 = z_1 \ddot{r}, \qquad z_1(v_0) = z_1(v_f) = 0$$$$

where the dot is derivative with respect to $$v$$, and $$v_0$$ & $$v_f$$ are the initial and final points respectively. The paper wrote that the solution that satisfies the above equation including the boundary condition on $$z_1$$ is given by,

$$$$z_1 = c r$$$$

for some constant $$c$$.

Typically, we would need two equations in order to solve for $$z_1$$ and $$r$$ as a function of $$v$$. However, in this case, I just need to get $$z_1$$ as an expression in terms of $$r$$ as is done in the paper. How is the solution obtained?

I'm applying the method in the paper to a situation that I'm studying for which I got a single coupled second-order differential equation given by,

$$$$2 r \ddot{z}_1 + 4 \dot{r} \dot{z_1} + 3 \ddot{r} z_1 = 0, \qquad z_1(v_0) = z_1(v_f) = 0$$$$

Again, how do you solve for $$z_1$$ in terms of $$r$$?

EDIT: I'm not sure if rewriting the expression in the same way as @user10354138 will be helpful, $$$$2 r \ddot{z}_1 + 4 \dot{r} \dot{z_1} + 3 \ddot{r} z_1 = 0\\ 2 r \ddot{z}_1 + 2 \dot{r} \dot{z_1} + 2 \dot{r} \dot{z_1} + 2 \ddot{r} z_1 + \ddot{r} z_1 = 0\\ 2\left( \frac{d}{dv}(r \dot{z}_1) + \frac{d}{dv}(\dot{r} z_1) \right) + \ddot{r} z_1 = 0\\ \frac{d}{dv} \left( r \dot{z}_1 + \dot{r} z_1 \right) = -\frac{1}{2} \ddot{r} z_1$$$$

Of course, we could rewrite the LHS and RHS further, but this is one version.

• The equation in the paper is a separable equation. Your equation is not. I do not think you can apply the same technique. Oct 19, 2023 at 13:15
• @Vasili I don't think the equation in paper is separable either. It was some "magic" setting the first constant (integrating LHS-RHS with respect to $v$) to zero. that come from another condition (on $r$?) Oct 19, 2023 at 13:45
• @user10354138 Based on the paper, there's no other obvious condition on $r$. It seems like it's a genuine solution that the authors have solved but did not bother to write a bit of detail, they just directly wrote the answer. Oct 19, 2023 at 14:06
• Actually that constraint comes out later... I'll post an answer for that part. Oct 19, 2023 at 14:09

## 1 Answer

For simplicity, I'll assume $$r$$ is not zero on the interval $$(v_0,v_f)$$. Note that $$0=r\ddot{z_1}-z_1\ddot{r}=(r\ddot{z_1}+\dot{r}\dot{z})-(z_1\ddot{r}+\dot{z_1}\dot{r})=\frac{\mathrm{d}}{\mathrm{d}v}\left(r\dot{z_1}-z_1\dot{r}\right)$$ Thus $$$$\label{eq1} r\dot{z_1}-z_1\dot{r}=C,\tag{1}$$$$ and so $$\frac{\mathrm{d}}{\mathrm{d}v}\left(\frac{z_1}{r}\right)=\frac{r\dot{z_1}-z_1\dot{r}}{r^2}=\frac{C}{r^2}.$$ Integrating this gives $$$$\label{eq2} \frac{z_1(v)}{r(v)}=C\int_{v_0}^v\frac{1}{r(\nu)^2}\,\mathrm{d}\nu\tag{2}$$$$ If $$r(v_f)=0$$ or $$r(v_0)=0$$ then we have $$C=0$$ from \eqref{eq1}. Otherwise \eqref{eq2} gives, at $$v=v_f$$, $$0=\frac{z_1(v_f)}{r(v_f)}=C\underbrace{\int_{v_0}^{v_f}\frac1{r^2}}_{\neq 0}$$ so you get back $$C=0$$.

Edit: Your amended equation is annoying in that the coefficient of $$\dot{r}\dot{z_1}$$ is not the sum of the other two, the reduction of order above does not work. Also $$z_1=v^\zeta$$, $$r=v^\rho$$ gives an oblique $$(\rho,\zeta)$$-ellipse that does not have any coordinate rectangles so that line of attack doesn't look promising either.

• Your answer to the paper's solution is great. For my equation, I made a small typo w/ regard to the coefficients of $\ddot{z}_1$ and $\dot{z}_1$, I corrected it above. I've tried to use your suggested solution but it seems to not give zero for the equation, I'll also try a different form. Oct 19, 2023 at 15:20
• @mathemania looks like the answer has been updated to accommodate your edit. Oct 20, 2023 at 18:19