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

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

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,

\begin{equation} z_1 = c r \end{equation}

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,

\begin{equation} 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 \end{equation}

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, \begin{equation} 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 \end{equation}

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

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    $\begingroup$ The equation in the paper is a separable equation. Your equation is not. I do not think you can apply the same technique. $\endgroup$
    – Vasili
    Oct 19, 2023 at 13:15
  • $\begingroup$ @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 $r$?) $\endgroup$ Oct 19, 2023 at 13:45
  • $\begingroup$ @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. $\endgroup$
    – mathemania
    Oct 19, 2023 at 14:06
  • $\begingroup$ Actually that constraint comes out later... I'll post an answer for that part. $\endgroup$ Oct 19, 2023 at 14:09

1 Answer 1

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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 \begin{equation} \label{eq1} r\dot{z_1}-z_1\dot{r}=C,\tag{1} \end{equation} 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 \begin{equation} \label{eq2} \frac{z_1(v)}{r(v)}=C\int_{v_0}^v\frac{1}{r(\nu)^2}\,\mathrm{d}\nu\tag{2} \end{equation} 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.

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  • $\begingroup$ 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. $\endgroup$
    – mathemania
    Oct 19, 2023 at 15:20
  • $\begingroup$ @mathemania looks like the answer has been updated to accommodate your edit. $\endgroup$ Oct 20, 2023 at 18:19

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