Let $a,b$ two smooth functions from the open square $I^{2}$ in $\mathbb{R}^{2}$ to $\mathbb{R}^{4}$. In particular, assume $a(t,u)$ and $b(t,u)$ be linearly independent for all $(t,u) \in I^{2}$.

I need to study the PDE problem in $ x \in C^{\infty}(I^{2},\mathbb{R}^{4})$ given by the underdetermined system $$ \begin{cases} \displaystyle \frac{\partial x}{\partial t} \cdot a = 0 \\ \displaystyle \frac{\partial x}{\partial u} \cdot b = 0 \\ \displaystyle \frac{\partial x}{\partial t} \cdot b = \frac{\partial x}{\partial u} \cdot a \end{cases} $$ and the auxiliary condition $$x(\cdot,0)= \frac{\partial x}{\partial u}(\cdot,0)=0\,.$$

Question: how would you go about solving this problem, i.e. representing its solution set?

Before somebody legitimately asks for motivation, let me finally say that this problem is linked to the isometric flexibility of a surface in $\mathbb{R}^{4}$.

  • 1
    $\begingroup$ Have you looked at Gromov's book Partial Differential Relations? He studies underdetermined systems of PDEs and applies them to prove existence and perhaps flexibility of isometric embeddings. $\endgroup$ – Deane Yang Dec 12 '18 at 14:33
  • 3
    $\begingroup$ An example, where a sufficiently nondegenerate $1$-by-$2$ system can be solved using Gromov's approach, can be found here: deaneyang.com/papers/underdetermined_ode.pdf. $\endgroup$ – Deane Yang Dec 12 '18 at 14:41
  • $\begingroup$ Are you assuming anything about $a(t,u)$ and $b(t,u)$? For example, do you assume that they are linearly independent at every point $(t,u)$ in the unit square? $\endgroup$ – Robert Bryant Dec 12 '18 at 14:42
  • $\begingroup$ @DeaneYang Thanks for the book suggestion and for the paper. I will look at it carefully $\endgroup$ – MK7 Dec 12 '18 at 14:49
  • $\begingroup$ @RobertBryant Yes, I am assuming linear independence $\endgroup$ – MK7 Dec 12 '18 at 14:50

Is there anything else that you are not telling us about $a$ and $b$? The particulars of these two vector-valued functions have a great influence on what the general solution of the system $$ a\cdot x_t = b\cdot x_u = a\cdot x_u - b\cdot x_t = 0\tag 1 $$ looks like.

For example, take the very special case in which $a$ and $b$ are constant. Then the above equations become $$ (a\cdot x)_t = (b\cdot x)_u = (a\cdot x)_u - (b\cdot x)_t = 0\tag 2 $$

Let $a^*$ and $b^*$ be the two linear combinations of $a$ and $b$ that satisfy $$ a\cdot a^* = b\cdot b^* = 1\qquad\text{while}\qquad a\cdot b^* = b\cdot a^* = 0, $$ and write $x = v\,a^* + w\,b^* + c$ where $c:I^2\to\mathbb{R}^4$ satisfies $a\cdot c = b\cdot c = 0$. Then the above equations become $v_t = w_u = v_u-w_t=0$, which implies that there are constants $v_0$, $w_0$, and $r_0$ such that $$ x = (v_0 + r_0\, u)\,a^* + (w_0 + r_0\, t)\,b^* + c.\tag 3 $$ Moreover, $c$ is arbitrary subject to the equations $a\cdot c = b\cdot c = 0$. This is the general solution of the system in the case that $a$ and $b$ are (linearly independent) constants. Imposing the given 'auxilliary conditions' forces $v_0=r_0=w_0=0$ and $c(t,0)=c_u(t,0)=0$, so the solution reduces to $x(t,u) = u^2w(t,u)$ where $w:I^2\to\mathbb{R}^4$ is any mapping satisfying the two linear constraints $a\cdot w = b\cdot w = 0$. Thus, the general solution subject to the auxilliary conditions still depends on two arbitrary functions of two variables. (A similar story holds whenever $a$ and $b$ take values in a fixed $2$-dimensional subspace $S\subset\mathbb{R}^4$.)

At the other extreme, consider the case in which the $5$ vector-valued functions $a$, $b$, $a_t$, $b_u$, and $a_u{-}b_t$ span $\mathbb{R}^4$ at every $(t,u)\in I^2$. Let us write $$ p\,a_t + q\,b_u + r\,(a_u{-}b_t) + f\,a + g\,b = 0\tag 4 $$ for some functions $p$, $q$, $r$, $f$ and $g$, where $p$, $q$, and $r$ do not simultaneously vanish. Up to a common nonzero multiple, these $5$ functions are determined by $a$ and $b$, so they can be regarded as known.

Set $$x\cdot a = v\qquad \text{and} \qquad x\cdot b = w.\tag 5 $$
Then, with the given equations, we have the three identities $$ x\cdot a_t = v_t\qquad x\cdot b_u = w_u\qquad x\cdot(a_u-b_t) = v_u-w_t\,.\tag 6 $$ Given the relation (4), we then get a single linear first order PDE for the pair $(v,w)$: $$ p\,v_t + q\,w_u + r\,(v_u-w_t) + f v + g\,w = 0.\tag 7 $$ Given a solution $(v,w)$ to (7), the five inhomogeneous linear equations (5)$+$(6) for $x$ are consistent and uniquely determine a solution $x$ to the original system of $3$ equations. Thus, the solutions of the original overdetermined system of three equations for four unknowns have now been expressed in terms of the solutions to a single linear PDE for two unknowns. Imposing the 'auxilliary conditions' is equivalent to requiring that the five functions $v$, $w$, $v_t$, $w_u$, and $v_u{-}w_t$ and their first derivatives with respect to $u$ should all vanish along the edge $u=0$ in $I^2$.

Methods for writing down the solutions of an equation such as (7) can be found in the 19th century PDE literature. I can say more about this if you are interested. The most elementary thing to do is to simply take one of the pair $(v,w)$ to be an arbitrary function on $I^2$ and then use the integrating factor method to solve the resulting linear PDE for the other.

There are, of course, more sophisticated things one can do to reduce the problem further to an 'explicit solution' without any integration involved. In particular, it can be shown that, for 'generic' $a$ and $b$ (i.e., for $(a,b)$ in a dense, open set in the $C^\infty$-topology on pairs of smooth maps $I^2\to\mathbb{R}^4$), there exist second-order linear differential operators $A$ and $B$ on $C^\infty(I^2)$ such that $(v,w) = \bigl(A(z),B(z)\bigr)$ solves (7) for every smooth function $z$ on $I^2$ and, moreover, every smooth solution $(v,w)$ of (7) is of the form $(v,w) = \bigl(A(z),B(z)\bigr)$ for some smooth function $z$ on $I^2$. The auxilliary conditions then translate into a pair of linear conditions on $z$ and its partials up to fourth order along the edge $u=0$.

In between these two extreme cases of hypotheses on $a$ and $b$, there are some exceptional cases that require separate handling. Also, I can say more about this, if you want.

  • $\begingroup$ Thanks a lot for the help. I have now included in the question the assumption that $a$ and $b$ are always linearly independent. Could you add something about that specific case in the answer? $\endgroup$ – MK7 Dec 14 '18 at 16:11
  • $\begingroup$ Also, I am not sure I understand how equation 4 follows from the assumption that the 5 vector-valued functions span $\mathbb{R}^{4}$ everywhere. $\endgroup$ – MK7 Dec 14 '18 at 16:13
  • 1
    $\begingroup$ @MK7: Given 5 vectors spanning a 4-dimensional vector space, there has to be a linear relation between them, and that linear relation is unique up to a nonzero multiple. (I'm not sure what you are asking if it's not this.) That's all that I'm using to get the existence of the scalar functions $p$, $q$, $r$, $f$ and $g$ on $I^2$ that do not simultaneously vanish. The functions $p$, $q$, and $r$ cannot simultaneously vanish either because, at such a point in $I^2$, we'd then have a linear relation between $a$ and $b. $\endgroup$ – Robert Bryant Dec 14 '18 at 22:37
  • $\begingroup$ Of course, sorry for the trivial question $\endgroup$ – MK7 Dec 15 '18 at 21:48
  • 1
    $\begingroup$ @MK7: Don't worry about that; sometimes, when you are in unfamiliar territory, even familiar facts can seem mysterious. I was wondering whether you had any other questions. Since you haven't accepted my answer, I'm assuming that either there is still something more you'd like to know or that this answer is not helpful for what you were trying to do. I'm happy to think more about it, but it would be good to know what the issues are. $\endgroup$ – Robert Bryant Jan 5 at 22:47

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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