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Certain surfaces in mechanics are endowed with the fundamental forms

\begin{align} \text{I} &= \mathrm{d}u^2+\mathrm{d}v^2+2\cos\gamma\: \mathrm{d}u\: \mathrm{d}v \\ \text{II} &= \alpha\left(\gamma \right)\:\gamma_u\:\mathrm{d}u^2-\beta\left(\gamma \right)\:\gamma_v\:\mathrm{d}v^2+2\tau\:\sin\gamma\: \mathrm{d}u\: \mathrm{d}v, \\ \end{align}

with $(u,v)\in [0,a]^2$, for some $a$ (e.g. 1), and $\alpha$ and $\beta$ the following functions:

\begin{align} \alpha&=\frac{\sin\rho+\sin\sigma\cos\gamma }{\sqrt{\cos^2\sigma\sin^2\gamma-\left(\sin\rho+\sin\sigma\cos\gamma \right)^2}}\\ \beta&=-\frac{\sin\sigma+\sin\rho\cos\gamma }{\sqrt{\cos^2\rho\sin^2\gamma-\left(\sin\sigma+\sin\rho\cos\gamma \right)^2}}. \end{align}

Here $\sigma$ and $\rho$ are fixed constants (between $0$ and $\pi$). The functions $\gamma=\gamma(u,v)$ and $\tau=\tau(u,v)$ must satisfy the Gauss-Codazzi equations, which read

\begin{align} -\frac{1}{\sin\sigma}\left(\tau-\frac{\alpha}{\sin\gamma}\gamma_v \right)_u&=\frac{\gamma_u \gamma_v}{\sqrt{\cos^2\sigma\sin^2\gamma-\left(\sin\rho+\sin\sigma\cos\gamma \right)^2}} \\ -\frac{1}{\sin\rho}\left(\tau+\frac{\beta}{\sin\gamma}\gamma_u \right)_v&=\frac{\gamma_u \gamma_v}{\sqrt{\cos^2\sigma\sin^2\gamma-\left(\sin\rho+\sin\sigma\cos\gamma \right)^2}} \\ \tau^2&=\frac{\gamma_{uv}}{\sin\gamma}-\alpha\beta\:\frac{\gamma_u \gamma_v}{\sin^2\gamma}. \end{align}

The point is to characterize the space of solutions of this overdetermined PDEs system. Since the right hand sides of the first two equations are identical, we introduce some "potential" $\phi$ such that $\frac{1}{\sin\sigma}\left(\tau-\frac{\alpha}{\sin\gamma}\gamma_v \right)=\phi_v$ and $\frac{1}{\sin\rho}\left(\tau+\frac{\beta}{\sin\gamma}\gamma_u \right)=\phi_u$, thus eliminating $\tau$ the system becomes

\begin{align} \phi_{uv}&=-\frac{\gamma_u \gamma_v}{\sqrt{\cos^2\sigma\sin^2\gamma-\left(\sin\rho+\sin\sigma\cos\gamma \right)^2}} \tag{*}\\ \gamma_{uv}&=\sin\sigma\sin\rho\:\phi_u \phi_v\:\sin\gamma-\sin\sigma\:\beta \:\gamma_u\:\phi_v+\sin\rho\:\alpha\:\gamma_v\:\phi_u, \tag{**} \end{align}

with the constraint

\begin{align} \sin \sigma \:\phi_v+\frac{\alpha}{\sin\gamma}\gamma_v=\sin \rho\:\phi_u-\frac{\beta}{\sin\gamma}\gamma_u. \tag{c} \end{align}

Were it not for the constraint, $(*)$ and $(**)$ would set a well posed hyperbolic PDE system from which one can get a unique solution specifying, say, $\gamma(u,0),\:\gamma(0,v)$ and $\phi(u,0),\:\phi(0,v)$.

Is there a standard procedure to analyze systems with lower order constraints like $\{(*),\:(**),\:(\text{c})\}$, how to find particular non-trivial ($\sigma\neq \rho \neq 0$) solutions and how much freedom in boundary/initial conditions does one have?

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  • $\begingroup$ I was confused by your introduction of $\phi$ for a while, until I realized that you probably meant to use the formulas $\frac{1}{\sin\sigma}\left(\tau-\frac{\alpha}{\sin\gamma}\gamma_v \right)=\phi_v$ and $\frac{1}{\sin\rho}\left(\tau+\frac{\beta}{\sin\gamma}\gamma_u \right)=\phi_u$, without the extra derivatives on the lhs. $\endgroup$ Mar 29 at 12:33
  • $\begingroup$ @IgorKhavkine Thank you. It's exactly as you mention. $\endgroup$ Mar 29 at 13:40

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Yes, there is a standard procedure to analyze such systems, essentially, it is Cartan's method of prolongation combined with his theory of involutive systems. There are other approaches as well, but Cartan's method is the one that I have found to be the most useful in practice.

Here is an outline of the calculation that one would have to do to apply this method: It's actually easiest to go back to the original Gauss-Codazzi equations before you introduce the 'potential' function $\phi$. Up to that point, the equations essentially just involve the derivatives of $\gamma$ with respect to $u$ and $v$. The function $\tau$ is an algebraic function of the derivatives of $\gamma$ (and, of course, the constants $\rho$ and $\sigma$, but I'll regard those as fixed, so I'll supress them).

Then the Codazzi equations can expanded and rewritten in the form $$ \gamma_{uuv} - F(\gamma,\gamma_u,\gamma_v,\gamma_{uu},\gamma_{uv},\gamma_{vv})=0 \quad\text{and}\quad \gamma_{uvv} - G(\gamma,\gamma_u,\gamma_v,\gamma_{uu},\gamma_{uv},\gamma_{vv})=0 \tag1 $$
for some algebraic functions $F$ and $G$. Now, how to study this overdetermined system? Obviously, any solution $\gamma(u,v)$ of (1) would have to satisfy $$ 0 = F_v-G_u = A_0\,\gamma_{uuu} + A_1\,\gamma_{uuv} + A_2\,\gamma_{uvv} + A_3\,\gamma_{vvv} + A_4 \tag2 $$ for some algebraic functions $A_i(\gamma,\gamma_u,\gamma_v,\gamma_{uu},\gamma_{uv},\gamma_{vv})$, explicitly computable in terms of $F$ and $G$.

Now, the best thing that could happen (in the sense that the overdetermined system would be as 'compatible' as possible) would be that $A_0$ and $A_3$ vanish identically while $H=A_1\,F + A_2\,G+A_4$ also vanishes identically. These represent three conditions on the partial derivatives of $F$ and $G$. If they are satisfied, then the system (1) is involutive in Cartan's sense, and it then turns out that one can specify $\gamma(u,0)$ and $\gamma(0,v)$ arbitrarily (as long as the equations make sense), so the local solutions depend on two arbitrary functions of one variable (plus one constant because, for example, knowing $\gamma(u,0)$ and $\gamma(0,v)$ won't determine $\gamma_{uv}(0,0)$.)

Another thing that might happen is that $A_0$ and $A_3$ vanish identically, but $H=A_1\,F + A_2\,G+A_4$ does not. In this case, $$ H(\gamma,\gamma_u,\gamma_v,\gamma_{uu},\gamma_{uv},\gamma_{vv}) = 0 \tag3 $$ becomes a second order equation on $\gamma$ that is implied by (1).

The best thing that could then happen would be that the lefthand sides of the two equations in (1) turn out to be linear combinations of the $u$ and $v$ derivatives of $H$. In this case, (1) is a consequence of the determined second order equation (3), and one is now reduced to considering the solutions of a (determined) lower order system.

However, more likely would be that the third order equations $H_u=0$ and $H_v=0$ (which are linear in the third derivatives of $\gamma$) would, when combined with (1), yield 4 linearly independent equations for $\gamma_{uuu},\gamma_{uuv},\gamma_{uvv}$, and $\gamma_{vvv}$, which could then be solved (with a slight change of notation) in the form $$\begin{aligned} \gamma_{uuu} &= E_1(\gamma,\gamma_u,\gamma_v,\gamma_{uu},\gamma_{uv},\gamma_{vv}),\\ \gamma_{uuv} &= E_2(\gamma,\gamma_u,\gamma_v,\gamma_{uu},\gamma_{uv},\gamma_{vv}),\\ \gamma_{uvv} &= E_3(\gamma,\gamma_u,\gamma_v,\gamma_{uu},\gamma_{uv},\gamma_{vv}),\\ \gamma_{vvv} &= E_4(\gamma,\gamma_u,\gamma_v,\gamma_{uu},\gamma_{uv},\gamma_{vv}).\\ \end{aligned}\tag{1*} $$ Now, expand out the relations $K_i := (E_i)_v-(E_{i+1})_u=0$ for $i=1,2,3$ and use (1*) to eliminate the third derivatives of $\gamma$ from the $K_i$. This will give you up to three new second order equations for $\gamma$. The best possible case would be that these $K_i$ are all multiples of $H$ (which can happen). Then the Frobenius theorem would tell you that there is a 5-parameter family of germs of solutions at every point of $uv$-space. The worst possible case would be that the system of second order equations $H=K_1=K_2=K_3=0$ is incompatible, in which case the system has no solutions. There are intermediate cases, but basically, you keep differentiating the second order equations you have and eliminating all of the third derivatives using (1*) until you don't get any new equations of order 2 or less. This process will stop at some point either because you don't get any new equations (and then the Frobenius theorem takes over) or else because the equations you have found so far are incompatible, i.e., they have no solutions pointwise, in which case, the original system has no solutions.

Of course, for generic pairs $(F,G)$, the functions $A_0$ and $A_3$ will be nonzero, and differentiating the third order equation $$ 0 = A_0\,\gamma_{uuu} + A_1\,F + A_2\,G + A_3\,\gamma_{vvv}+A_4 \tag4 $$ with respect to u and v plus the derivatives with respect to $u$ and $v$ of the two equations in (2) will yield a system of fourth order equations that allow one to express all of the fourth derivatives of $\gamma$ as algebraic expressions in the derivatives of $\gamma$ of order at most $3$. This is now what is known as a PDE system of finite type, and it can be analyzed using the Frobenius theorem. (The best possible outcome in this case is that there turns out to be a 7-parameter family of solutions $\gamma$ to the original system.)

In the cases where one of $A_0$ and $A_3$ vanishes identically and the other does not, it either reduces to a PDE system of finite type or else it becomes an involutive system whose solutions depend on one arbitrary function of one variable plus some number of constants.

I would say that the best thing to do in the specific case being studied would be to go through the above procedure, compute $F$ and $G$ and then the $A_i$ as functions of $(\gamma,\gamma_u,\gamma_v,\gamma_{uu},\gamma_{uv},\gamma_{vv})$, and see which case your equation falls into. Since the equation is sort of symmetric with respect to $u$ and $v$, I think it's unlikely that one will encounter the case where exactly one of $A_0$ or $A_3$ vanishes identically, but "One never knows, does one?", without doing the calculations.

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  • $\begingroup$ Thanks for the guidance. A preliminary question is, shouldn't we consider a third equation? I mean, in the original system we take $\tau$ from the Gauss (third equation) and replace in the first two (Codazzi) equations, thus obtaining the two in $(1)$. But taking cross derivatives of the Codazzi's, $(\tau_u)_v=(\tau_v)_u$, puts into another equation containing $\gamma_{uvv}$ and $\gamma_{uuv}$. $\endgroup$ Mar 27 at 19:27
  • $\begingroup$ @DanielCastro: But $(\tau_u)_v = (\tau_v)_u$ is an identity, no matter what expression you plug in (as long as $\tau$ is $C^2$), so plugging in the expression for $\tau$ as above will only give you the trivial third-order equation on $\gamma$. $\endgroup$ Mar 27 at 20:18
  • $\begingroup$ Thank you. Also, in $(2)$ shouldn't we have some $A_4$ containing (only) second and lower derivatives ? $\endgroup$ Mar 28 at 9:41
  • $\begingroup$ @DanielCastro: Of course, yes you are correct. I overlooked that term when I was writing it up, but it doesn't substantially change the story. I'll fix that. $\endgroup$ Mar 29 at 9:38
  • $\begingroup$ In this case and similar ones of interest indeed $A_0=A_3=0$ identically, but $H$ does vanish and $(1)$ cannot be written as linear combinations of its derivatives. So $H=0$ is a second order determined equation for $\gamma$ but it does not necessarily imply $(1)$. $\endgroup$ Apr 9 at 13:05

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