Solve PDE on a subspace

Question

Let us assume that we are given some partial differential equation. Just for example, let me consider the following.

$$ a(x,y,z) \partial_x F(x,y,z) + b(x,y,z) \partial_y F(x,y,z) + c(x,y,z) \partial_z F(x,y,z) = 0$$

I know how to solve it with the usual methods. However, what if we know that the previous partial differential equation is satisfied only on some subspace? Let us say, we know that the previous equation is true only on $F(x,y,z) = 0$. How to approach such case? And how to generalize this to an arbitrary partial differential equation?

My first idea would be to solve differential equation in general for the whole space, and then put the constraint in the end. Can someone (at least intuitively) show that in that case, we obtain the most general solution? Is it possible that previous equation is satisfied on $F(x,y,z) = 0$ but solution cannot be obtained by such method? Is there a criterion for that?

Answer 1

I had the exact same question come up in my work recently so I will provide the answer I came to. I have no references for this though so maybe you'll come across a better answer. I will speak specifically to the following domain:

• Smooth Category
• Linear PDEs

Let me restate the problem in a slightly different way. Let $M$ be an $m$-dimensional smooth manifold with coordinates $x^i$ and let $N \subset M$ be an embedded submanifold with inclusion $\iota: N \to M$. Let $X\in \mathcal{T}N$ be a smooth vector field. The question is:

Find a smooth function $f:M \to \mathbb{R}$ that solves $$X\left(\iota^* f\right) = 0.$$ That is, find the functions that, on $N,$ solve the PDE induced by the smooth vector field $X.$

To me, this is the most natural interpretation of your statement. Solutions can be found by first solving the PDE on $N$ for some solution $g: N \to \mathbb{R}.$ Any smooth extension of $g$ to a function $f: M \to \mathbb{R}$ solves the PDE locally on $N$ and answers the question. This is important, since it does imply that solving the PDE globally and restricting doesn't include the entire class of solutions. Let me illustrate this by way of example.

Let $M = \mathbb{R}^3$ with coordinates $x = (x^1, x^2, x^3)$ and let $N = \{ x \in M \;\colon\; x^3 = 0 \}.$ Consider the smooth vector field $$X = -x^2 \frac{\partial}{\partial x^1}+ x^1 \frac{\partial}{\partial x^2}$$ on $N.$ Pick $$g(x^1, x^2) = (x^1)^2 + (x^2)^2 - 1$$ which satisfies the PDE $X(g)=0$ on $N.$ Now let $\lambda: M \to \mathbb{R}$ be a smooth bump function satisfying $\left.\lambda\right|_N = 0$ but non-zero everywhere else. Extend $g$ in the natural way for simplicity and define $$ f_1(x) = \lambda(x) x^2 + (1 - \lambda(x)) g(x). $$ Another example is $$ f_2(x) = g(x) + (x^1)^4 (x^3)^2. $$ I think you can convince yourself that neither $f_1$ nor $f_2$ solve the PDE off of $N.$ I'm willing to bet that the problem doesn't disappear in the real analytic category for a similar reason (how we choose to extend the solution matters).

So I would solve the PDE in a coordinate system for the subset first then consider the family of extensions that satisfy whatever other assumptions you wish to enforce (smoothness, analyticity, etc.)