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Given a system of PDEs with $n_f$ functions $f_i(x_1, x_2 ... n_v)$ each of $n_v$ real variables how many equations $n_c$ does it take before the system is unsolvable in general? I am guessing that as long as $n_c \le n_f n_v$ then these systems are usually solvable and if $n_c = n_f n_v$ then this is usually a finite dimensional infinite set and beyond that they should not be expected to be solvable. (My question is flexible in that if letting the variables be complex leads to a conceptually simpler answer than you are welcome to make that assumption).

An example would be the cauchy-riemann equations where $u,v$ are our $n_f=2$ functions $x,y$ are our $n_v =2$ variables, and the $n_c=2$ equations are given as:

$$ \frac{\partial u }{\partial x} = \frac{\partial v}{\partial y} \\ \frac{\partial u }{\partial y} = -\frac{\partial v}{\partial x} $$

We know that this system is solvable and admits an extremely large number of solutions (namely the set of all analytic functions around a point $p$). But if we add a constraint like the differential equation $y' = y$ then the solution space becomes a finite dimensional infinite set, and (I think) adding any additional PDE constraints generally speaking makes it unsolvable.

The complex differential equation $y' = y$ is actually a system of two of PDEs given as

$$ \frac{\partial u}{\partial x} = u \\ \frac{\partial v}{\partial x} = v $$

And so it seems like the sentence "what is a complex analytic function $f$ such that $f' =f$?" is really a system of $n_c=4$ PDEs of $n_f=2$ functions and $n_v=2$ variables and since $n_c = n_f * n_v$ we expect a finite dimensional infinite set of solutions (Computing the exact dimension seems to be a hard problem).

Is my intuition on this matter correct?

I guess I am asking both

  1. How can I actually state my question formally (I'm using words like usually and generally and that isn't really rigorous)
  2. Does such a simple rule hold?
Sidharth Ghoshal
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    By unsolvable do you mean no solution can exist? If so, that is not the usual meaning of unsolvable. – Paul Oct 27 '24 at 07:27
  • I'm pretty sure that a generic system of PDEs that has more equations than unknowns has no solutions. And there are simple assumptions that imply no solutions. For example, if you have a linear first order system with $n$ variables, 1 unknown function and $n + 2$ equations, then in general, there cannot be a solution. – Deane Oct 27 '24 at 18:23
  • @Paul yes your correct, my “unsolvable” means no solution exists as opposed to the algebraists unsolvable meaning “no expression in terms of fixed language such as that of algebraic radicals” – Sidharth Ghoshal Oct 27 '24 at 19:20
  • @Deane interesting! Then that would mean already that the cauchy-Riemann + diff EQ example above is a “special situation” and my intuitions about this are all wrong. Do you have any links I could consume to better see your viewpoint? – Sidharth Ghoshal Oct 27 '24 at 19:21
  • @Paul I should probably have written “infeasible”, will correct my question once I’m back at a computer – Sidharth Ghoshal Oct 27 '24 at 19:23
  • I didn't look closely at your example. Are you assuming that $u$ and $v$ are functions of $x$ and $y$ and satisfy the equations \begin{align} \partial_xu &= \partial_yv\ \partial_xv &= -\partial_yu\ \partial_xu &= u \ \partial_xv &= v \end{align} – Deane Oct 27 '24 at 19:28
  • @Deane yes that looks correct to me, the first two are Cauchy Riemann equations and second two are $y’=y$ the diff eq written as a system of PDEs – Sidharth Ghoshal Oct 27 '24 at 20:28
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    So, yes, this system is special. First, this is a homogeneous system, so $u=v=0$ is always a solution to such a system. On the other hand, if you change the system slightly, say to something like \begin{align} \partial_xu &= \partial_yv\ \partial_xv &= -\partial_yu\ \partial_xu &= u + b(x) \ \partial_xv &= v, \end{align} then for most choices of the function $b$, there will be no solution. I haven't checked this myself, though. – Deane Oct 27 '24 at 20:54

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This is a very nontrivial question and related to the Frobenius theorem. A version of it that uses PDE language can be found here. One can construct an example with two variables, three functions and two equations that has no solution, unless all three functions only depend on a single variable. That is: the inequality $$ n_c\le n_f\,n_v $$ is satisfied with $n_c=2,n_f=3,n_v=2$ but the system of PDEs is only a system of ODEs.

The idea is as follows: start with the classic socalled contact form $$ \omega=dz-y\,dx $$ which satisfies $\omega\wedge d\omega=dx\wedge dy\wedge dz$ that vanishes nowhere. Frobenius says that there is no surface in $\mathbb R^3$ on which this form vanishes. This means that there is no parametrization $$ (t_1,t_2)\mapsto \pmatrix{x(t_1,t_2)\\y(t_1,t_2)\\z(t_1,t_2)} $$ of a two-dimensional surface that satisfies the system of PDEs \begin{align} \partial_{t_i}z-y\,\partial_{t_i}x=0\,,\quad i=1,2\,. \end{align}

Kurt G.
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  • Hmm there very interesting. Let me take some time to digest your links. I guess then one might ask is there a notion of “general position” for PDEs much like “general position” for points and more generally “general position” for algebraic equations. And if your given example is itself in such general position then that tells me my intuition was very wrong. – Sidharth Ghoshal Oct 27 '24 at 20:53