# Finding a relation between functions according to known constraints

I am solving a problem on geodesics with ideas from General Relativity and got stuck with one step. The simplified version is the following:

With notations $$\dot{x}\equiv \frac{dx}{dt}, \quad \mathring{x} \equiv \frac{dx}{d\lambda}, \quad x' \equiv \frac{dx}{dr}$$ $$A=A(r), B=B(r), t=t(\lambda), r=r(\lambda), \varphi=\varphi(\lambda), v= \text{ constant}$$

Given the constraints $$\mathring{t}=\frac{1}{A},\quad \mathring{\varphi}=\frac{1}{B\cdot r^2},\quad \dot{r}^2 +r^2\dot{\varphi}^2 = v^2$$

Can we find a relation between $$A,B,A',B',r$$?

I can convert $$\dot{\varphi}$$ in terms of functions of $$r$$. But for $$\dot{r}$$, I can't see any way to do it so that I can use the last constraint.

$$\mathring{\varphi}=\frac{dt}{d\lambda}\frac{d\varphi}{dt}=\frac{1}{A}\dot{\varphi}$$ $$\dot{\varphi}=A \mathring{\varphi}=\frac{A}{B}\cdot r^{-2}$$

Any hints are appreciated! Thanks.

I do not think it is possible. You have 3 equations and 3 unknowns. The first two equations are used to eliminate $$t$$ and $$\varphi$$ from the 3rd equation. There is only one way to do this - by solving them and plugging in the result, so there is no flexibility there. You get a first order ODE for $$r$$ and $$\lambda$$. This ODE can be solved by integration, and produce and integration constant. This means that $$r$$ that satisfies above equation has an extra degree of freedom, and thus there can't exist an equation you request that does not include this extra degree of freedom.