I have reduced my problem to the following:
I have a function, $b(\theta,\phi)$, which is defined implicitly (up to integration constants) by the differential equations:
\begin{align}
\frac{\partial b}{\partial\phi} &= -\cos{\theta}\sin{\theta}\tan{\left[\phi-b(\theta,\phi)\right]}\frac{\partial b}{\partial\theta} \tag 1 \\
\frac{\partial b}{\partial\theta} &= -\cos{\left[\phi-b(\theta,\phi)\right]}\sin{\left[\phi-b(\theta,\phi)\right]}\tan{\theta} \tag 2
\end{align}
Combining, I deduce that:
$$\quad \frac{\partial b}{\partial\phi}=\sin^2{\theta}\sin^2{\left[\phi-b(\theta,\phi)\right]} \tag 3$$
My problem is to find an explicit, or implicit expression for $b(\theta,\phi)$, without derivatives. All my variables are real.
I am not sure if it is helpful, but I worked out that $(2)$ is compatible with the following
$$\cos{\theta}\tan{\left[\phi-b(\theta,\phi)\right]}=f(\phi)+\text{const} \tag 4$$
which yields $(2)$ on differentiation with respect to theta. Finally, we can add also any constant times $\sec{\theta}\cot{\left[\phi-b(\theta,\phi)\right]}$ to the LHS of $(4)$, and it is still compatible with $(2)$. But I can't find a way to develop it to make it consistent also with $(3)$. So, I am now stuck. I am not sure whether
- There is no solution – the differential equations are incompatible (the worst outcome), or
- There is no analytic solution, but the equations could be solved numerically in a computer (I could live with that), or
- There is an analytic solution (which would be best, of course).
I would very much appreciate any advice on how to proceed.
Thanks.
$$
\newcommand{\del}{\delta(\theta,\phi)}
\newcommand{\xx}{x(\theta,\phi)}
$$
Continuation
Following the suggestion of @JJacquelin continue by substituting their solution, function $b(\theta,\phi)$, into Eq. (2). First define, for brevity, two new variables:
$$\del=\phi-b(\theta,\phi)$$
$$\xx=\cos(\theta)\:\phi+g(\theta)$$
Substituting into the solution of (3) given by @JJaquelin, allows it to be re-written
$$\tan{\del}=\frac{\tan{\xx}}{\cos{\theta}} \tag 5$$
which we use below.
In order to substitute into (2), differentiate the solution of (3) w.r.t. $\theta$ and use the new variables
\begin{eqnarray}
\frac{\partial b}{\partial\theta}&=&-\frac{1}{1+\tan^2{\del}}\frac{\partial}{\partial\theta}\left(\frac{\tan \xx}{\cos{\theta}}\right)\\
&=&-\cos^2{\del}\frac{\cos{\theta}\frac{\partial}{\partial\theta}(\tan{\xx})+\sin{\theta}\tan{\xx}}{\cos^2{\theta}}
\end{eqnarray}
Now, we have from (5) that $\tan{\xx}=\cos{\theta}\tan{\del}$ but this is a function $f(\phi)$ only, from (4) (which solves(2)), and thus $\frac{\partial}{\partial\theta}(\tan{\xx})=0$ and therefore
$$\frac{\partial b}{\partial\theta}=-\cos{\del}\sin{\del}\tan{\theta}$$
which is (2). Thus all is consistent as long as
$$\cos{\theta}\tan{(\phi-b(\theta,\phi))}=f(\phi)=\tan{(\phi\cos(\theta)+g(\theta))}$$
$f(\phi)$ and $g(\theta)$ can be any functions and need to be determined from boundary conditions (which were not specified in the OP).
Best Answer
This is not a direct answer to your question but a compendium of my results.
FIRST CASE : Considering equation $(1)$ ALONE : $$\frac{\partial b}{\partial\phi} +\cos{\theta}\sin{\theta}\tan{\left[\phi-b(\theta,\phi)\right]}\frac{\partial b}{\partial\theta}=0 \tag 1 $$ This is a first order quasilinear PDE. Thanks to the Charpit-Lagrange method or of the method of characteristics, the general solution expressed on the form of implicit equation is : $$b(\theta,\phi)=F\Big(\cos\big(\phi-b(\theta,\phi)\big)\tan(\theta)\Big)\tag{S1}$$ $F$ is an arbitrary function. $$ $$
SECOND CASE : Considering equation $(2)$ ALONE : $$\frac{d b}{d\theta} +\cos{\left[\phi-b\right]}\sin{\left[\phi-b\right]}\tan{\theta}=0 \tag 2$$ This is a first order non-linear ODE. Solving it leads to : $$b(\theta,\phi)=\phi-\tan^{-1}\left(\frac{h(\phi)}{\cos(\theta)}\right) \tag{S2}$$ $h(\phi)$ is an arbitrary function. $$ $$
THIRD CASE : Considering equation $(3)$ ALONE : $$\frac{d b}{d\phi}-\sin^2{\theta}\sin^2{\left[\phi-b\right]}=0 \tag 3$$ This is a first order non-linear ODE. Solving it leads to : $$b(\theta,\phi)=\phi-\tan^{-1}\left(\frac{\tan\big(\cos(\theta)\:\phi+g(\theta)\big)}{\cos(\theta)} \right) \tag{S3}$$ $g(\theta)$ is an arbitrary function. $$ $$
CASES OF SYSTEM OF EQUATIONS :
If the equations (1),(2),(3) are not considered independently but as a system of equations, the solutions (S1) , (S2) , (S3) are equivalemt, but with not independant functions $F,h,g$, insofar they exist and can be found.
For example the system of equations (2) , (3), has a solution insofar (S2)=(S3) which obviously supposes $$h(\phi)=\tan\big(\cos(\theta)\:\phi+g(\theta)\big)$$ $h(\phi)$ is no longer a function of $\phi$ alone but is a function of $\phi$ and $\theta$. So one cannot put it into (S2) in order to find the general solution of the system of equations.
Reciprocally $g(\theta)$ is no longer a function of $\theta$ alone but is a function of $\phi$ and $\theta$. So one cannot put it into (S3) in order to find the general solution of the system of equations.
This tends to show that the system of equations $(2)$ and $(3)$ has no solution in general. One can expect a particular solution not valid for any couple $(\phi,\theta)$ but valid only on a curve which equation satisfy $h(\phi)=\tan\big(\cos(\theta)\:\phi+g(\theta)\big)$ .