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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

  1. There is no solution - the differential equations are incompatible (the worst outcome), or
  2. There is no analytic solution, but the equations could be solved numerically in a computer (I could live with that), or
  3. 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).

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  • $\begingroup$ Hi Paul ! You wrote : All is consistent as long as $f(\phi)=\tan{(\phi\cos(\theta)+g(\theta))}\quad$ . $f(\phi)$ and $g(\theta)$ can be any functions. I cannot understand how a function of only one variable $f(\phi)$ can be equal to a function which includes a second variable $\theta$ not constant. $\endgroup$
    – JJacquelin
    May 13, 2020 at 21:40
  • $\begingroup$ @JJaquelin As answer, I see it like a constraint. To take a more familiar analogy, if we have some function, f(x), then we can set f(x)=7, and this fixes x to a point. Similarly, if we have some other function g(x,y) and we set g(x,y)=h(x), then this defines some locus within the x-y plane (which otherwise was unconstrained). Thus, setting, e.g. $\cos(\theta)\:\phi+g(\theta)=f(\phi)$ defines a locus in the $\theta-\phi$ plane. Is that reasonable in your view? Perhaps it is then a mistake to say (as I did) that $\frac{\partial}{\partial \theta}(\cos(\theta)\:\phi+g(\theta))=0$ in that case? $\endgroup$ May 13, 2020 at 23:22
  • $\begingroup$ Well I understand that. This means that the variables $\phi$ and $\theta$ are not independant but are related defining a curve on the $\phi,\theta$ plane. Then the solution we are looking for is not a general solution but a particular solution valid only on this curve. Do you agree with this wording of the problem ? $\endgroup$
    – JJacquelin
    May 14, 2020 at 4:59
  • $\begingroup$ I noticed that I made some typos in my original question (for which, my apologies that I only noticed them just now). All the derivatives in my question are supposed to be partial, although I see that I did not include the correct "$\partial$" in my denominators. I will correct that now. But perhaps it was implicit anyway that all should be partial, since $b(\theta,\phi)$ is a function of two variables. $\endgroup$ May 14, 2020 at 10:51
  • $\begingroup$ I have fixed the partial derivative typos. I wonder if this changes the discussion? $\endgroup$ May 14, 2020 at 10:55

2 Answers 2

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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 $h(\phi)=\tan\big(\cos(\theta)\:\phi+g(\theta)\big)$ .

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  • $\begingroup$ Thank-you for this latest compendium. There are quite a lot of details there for me to assimilate. But first, a question: I noticed in your Case 2 and 3, you notated full derivatives, not partial. Is the difference significant? For my original question, they should all be partial (though I notice, with apologies, that as typed, they are ambiguous - which I will fix now). $\endgroup$ May 14, 2020 at 10:50
  • $\begingroup$ That is only a matter of habit of notations. This changes nothing in my answer; $\endgroup$
    – JJacquelin
    May 14, 2020 at 11:09
  • $\begingroup$ I confirm by direct differentiation the quoted solutions for the three cases considering the equations as independent. $\endgroup$ May 14, 2020 at 14:24
  • $\begingroup$ I will go back to my original problem (which is a practical physics problem) to check the correspondence between the variables there with those of the derived mathematical question asked here, in order to see if there are any inconsistencies in interpretation etc. This is likely to take me some time, since I will do a "deep" check. $\endgroup$ May 14, 2020 at 15:22
  • $\begingroup$ I have discovered issues with the physical "set-up" of this problem, and I cannot now be confident that the equations, as stated, are really valid for my problem. The discussion here was very helpful in getting me to this "conclusion", and I thank you for your help. I will uptick the compendium as the best answer to the mathematical problem I posted, and I consider this post as answered. Thanks again. $\endgroup$ May 20, 2020 at 7:45
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$$\frac{d b}{d\phi}=\sin^2(\theta)\sin^2(b-\phi) \tag 3$$ $b(\phi)=\psi(\phi)+\phi$ $$\quad \frac{d \psi}{d\phi}+1=\sin^2(\theta)\sin^2(\psi)$$ $y(\phi)=\tan\left(\psi(\phi) \right)$

$\sin^2(\psi)=\frac{\tan^2(\psi)}{1+\tan^2(\psi)}=\frac{y^2}{1+y^2}$

$\frac{dy}{d\phi}=(1+y^2)\frac{d\psi}{d\phi}\quad\implies\quad \frac{d\psi}{d\phi}=\frac{1}{1+y^2}\frac{dy}{d\phi}=\sin^2(\theta)\sin^2(\psi)-1=\sin^2(\theta)\frac{y^2}{1+y^2}-1$

$$\frac{dy}{d\phi}=\sin^2(\theta)\:y^2-(1+y^2)=-\cos^2(\theta)\:y^2-1$$

$$\int d\phi=-\int\frac{dy}{\cos^2(\theta)\:y^2+1}$$

$$y=-\frac{\tan\left(\cos(\theta)\:\phi+c\right)}{\cos(\theta)}$$

$$\psi=-\tan^{-1}\left(\frac{\tan\left(\cos(\theta)\:\phi+c\right)}{\cos(\theta)} \right)$$

$$b=\phi-\tan^{-1}\left(\frac{\tan\left(\cos(\theta)\:\phi+c\right)}{\cos(\theta)} \right)$$

Important : $c$ is a constant parameter with respect to $\phi$. But it can be a function of $\theta$.

$$b(\theta,\phi)=\phi-\tan^{-1}\left(\frac{\tan\big(\cos(\theta)\:\phi+g(\theta)\big)}{\cos(\theta)} \right)$$

This solves Eq.$(3)$. The end of the task is for you. This will not be an easy task: Putting the above function $b(\theta,\phi)$ into Eqs.$(1)$ and $(2)$ and checking if a function $g(\theta)$ exists or not.

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  • $\begingroup$ Thank-you for your helpful answer. I have followed through your working. I have suggested a small edit to fix four minus signs in the last four lines of your solution. Otherwise it seems perfect. I will proceed to attempt to finish the problem in the way you suggest. Thanks again. $\endgroup$ May 12, 2020 at 20:22
  • $\begingroup$ You are right, there was a sign mistake. I accepted your suggested edit. $\endgroup$
    – JJacquelin
    May 12, 2020 at 21:40
  • $\begingroup$ In notifications I read this : Leucippus reviewed this 7 hours ago: Reject This edit was intended to address the author of the post and makes no sense as an edit. It should have been written as a comment or an answer. Harish Chandra Rajpoot reviewed this 7 hours ago: Reject This edit deviates from the original intent of the post. Even edits that must make drastic changes should strive to preserve the goals of the post's owner. $\endgroup$
    – JJacquelin
    May 13, 2020 at 7:56
  • $\begingroup$ As far as I can understand from the rules of the forum : One is allowed to correct my own answer when they are mistakes in it. But one is not allowed to add something to my answer. You are allowed do edit your question and add what you want into it. $\endgroup$
    – JJacquelin
    May 13, 2020 at 8:02
  • $\begingroup$ I don't think that it is a good idea. I didn't reject your suggested addition. But this is not my own addition, so editing it in my answer is not the right place. Better edit it as an addition into your own question. $\endgroup$
    – JJacquelin
    May 13, 2020 at 8:07

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