How to symbolically solve this transcendental equation

Question

The equation below shows up when dealing with geodesic curves on elliptic paraboloids:

\begin{equation} u-c^2=u(1+4c^2)\sin^2\left(v-2c \ln\left(k(2\sqrt{u-c^2}+\sqrt{4u+1})\right)\right) \tag{1} \end{equation}

Obtaining a solution for $v(u)$ is rather straightforward using Solve, like so:

Solve[u-c^2==u(1+4 c^2) Sin[v-2c Log[k (2 Sqrt[u-c^2]+Sqrt[4u+1])]]^2,v,Reals]

[output omitted]

It would be much more convenient to find the inverse function $u(v)$, so I tried solving for $u$, but MMA gave up with "... cannot be solve with the methods available to Solve". That's when I gave Reduce a shot:

Reduce[u-c^2==u(1+4c^2)Sin[v-2c Log[k\left(2 Sqrt[u-c^2]+Sqrt[4u+1]\right)]]^2,u,Reals]

Same message as before.

Is there a way to make this equation more tractable for Mathematica, to let it arrive at a symbolic solution for $u(v)$?

NB: $c$ and $k$ are constants, more precisely $c,k \in \mathbb{R}$; $u \geq 0$ ($u$ denotes the radius)

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Edit: How can we obtain the paraboloid geodesic $(1)$ above?

First, define the paraboloid surface as follows:

\begin{equation}(u,v)\mapsto\begin{pmatrix}\sqrt{u}\cos{v} \\ \sqrt{u}\sin{v} \\ u \end{pmatrix} \tag{2}\end{equation}

In order to trace out a complete paraboloid, let $u\geq 0$ and $v \in [0,2\pi)$. I would rather use $r$ and $\varphi$ to make their meaning a little more clear, but let's stick to the usual naming scheme w.r.t. differential geometry.

Next, calculate the general solution to the geodesics using the Euler-Lagrange equation:

\begin{equation} \underbrace{\frac{\frac{\partial P}{\partial v}+2v\,'\frac{\partial Q}{\partial v}+v\,'^2\frac{\partial R}{\partial v}}{2\sqrt{P+2Qv\,'+Rv\,'^2}}}_{\displaystyle{=0}}-\frac{d}{du}\left(\frac{Q+Rv'}{\sqrt{P+2Qv\,'+Rv\,'^2}}\right)=0 \end{equation}

$P, Q, R$ are the coefficients of the first fundamental form; I managed to verify that the solution indeed matches equation $(1)$ given in the literature, see here.

Rearrange $(1)$ so we have $v(u)$, and use that inside $(2)$ to describe geodesics that travel across the paraboloid surface.

I was able to come up with a Notebook which computes the integration constants $c,k$ to yield the specific geodesic passing through two points $(u_1,v_1),(u_2,v_2)$.

However, if these points have the same "height" on the paraboloid, that is $u_1=u_2$, Mathematica won't find a solution. I assume knowing the inverse function $u(v)$ would be helpful here.

Answer 1

In situations like this I use InverseFunction. Although it does not generate an analytical expression, you still end up with something which returns an exact result rather than an approximate one:

lhsMinusRhs[c_, k_][u_, v_] := u-c^2 - u(1+4 c^2) Sin[v-2c Log[k (2 Sqrt[u-c^2]+Sqrt[4u+1])]]^2
assumptions[c_, k_][u_, v_] := u >= 0;
uFunction[c_, k_][v_] :=
  InverseFunction[
    Function[{u},
      ConditionalExpression[
        lhsMinusRhs[c, k][u, v],
        assumptions[c, k][u, v]
      ] // Evaluate
    ]
  ][0] // Evaluate;

For example, uFunction[3, 4][5] will return an exact Root object corresponding to $(c, k) = (3, 4)$ and $v = 5$.

uFunction[3, 4][5]
(* Root[{..., 16.33...}] *)
lhsMinusRhs[3, 4][%, 5] // FullSimplify
(* 0 *)

Note that InverseFunction makes no guarantees on which branch of the inverse will be chosen. If I am after a specific branch, the approach I take is to make assumptions specific enough to pin down the unique value corresponding to that branch.