Small System of First Order Coupled PDEs

Question

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).

Answer 1

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

Answer 2

$$\frac{d b}{d\phi}=\sin^2(\theta)\sin^2(b-\phi) \tag 3$$ $b(\phi)=\psi(\phi)+\phi$ $$\quad \frac{d \psi}{d\phi}+1=\sin^2(\theta)\sin^2(\psi)$$ $y(\phi)=\tan\left(\psi(\phi) \right)$

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

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

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

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

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

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

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

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.