Given two PDE(s): $F(x,y,z,p,q)=0$
and $G(x,y,z,p,q)=0$
In I.A.N Sneddon's "Elements of Partial Differential Equations",If every solution of $F=0$ is a solution of $G=0$,then $F=0$ and $G=0$ are said to be compatible. But according to another textbook if $F=0$ and $G=0$ have atleast one common solution then they are said to be compatible.So here my question is:
What is the actual "definition of compatibility of two pde(s)"?

Now if I assume that $\left|\frac{\partial(F,G)}{\partial(p,q)}\right|\ne0$ and solve the system $F=0$ and $G=0$ for $p$ and $q$,suppose that I get $p=\phi(x,y,z)$ and $q=\psi(x,y,z)$ so writing complete differential for $dz$,
$dz=pdx +qdy$ ,so this Pfaffian equation is integrable?(This condition has been used in textbook to deduce the condition for two partial differential equations to be compatible).
My second question is:
How does compatibility of "$F=0$ and $G=0$" plus the condition $\left|\frac{\partial(F,G)}{\partial(p,q)}\right|\ne0$ imply integrability of Pfaffian equation "$dz=pdx +qdy$" ?
Please help me know answers to my questions.
Your thoughts are always appreciated.


Actually, you have described three entirely different notions of 'compatibility' of a pair of first order PDE for a single function of two variables. The first two that you have listed are not the standard ones (and, in fact, are not very useful), and I am surprised that you were able to find them stated that way in textbooks.

Your third condition, which is that, when one solves for $p$ and $q$ as functions of $x$, $y$, and $z$ and substitutes these into the $1$-form $\theta = \mathrm{d}z - p\,\mathrm{d}x - q\,\mathrm{d}y$, one obtains an integrable $1$-form on $xyz$-space, i.e., one that satisfies that $\theta \wedge \mathrm{d}\theta$ vanishes identically, is the standard definition of compatibility of the pair of equations. What it ensures is that there is a nontrivial $1$-parameter family of (local) functions $z(x,y,c)$ (where $c$ is the nontrivial parameter) that simultaneously satisfy both equations.

Note that this is properly stronger than the condition that there exists a single function $z(x,y)$ that satisfies both equations, and it does not imply that every (local) solution of one of the equations is a (local) solution of the other.

It is not possible to prove that this third notion of compatibility either implies or is implied by either of the first two, because that is not the case, as easy examples show.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.