Closed curves of the form $F(x)+G(y)=0$

Question

For certain problem in mechanics, it is useful to assume that a simple (smooth probably, but not strictly necessary) closed curve can be expressed in implicit form as

\begin{align} F\left (x\right)+G\left (y\right)=0, \end{align}

being $F(x)$ and $G(y)$ respectively $C^1$ functions of the Cartesian coordinates $x$ and $y$. For instance, in the case of a circle this is true with $F(x)=x^2,\:G(y)=y^2-c^2$, and similarly for an ellipse. The question is

Which (simple closed) curves can be expressed in such a Cartesian separable way ?

It is likely that not any closed curve can be put in such form, but probably the family is much bigger than simply ellipses. I'm looking for some criteria that a curve may or may not satisfy, like convexity.

Answer 1

Suppose a curve is defined by $f(x,y)=c$. Then $f(x,y)=F(x)+G(y)$ for some $F,G$ if and only if $f_{xy}=0$. So one way to characterize your curves is that they are level sets of a function whose Hessian matrix is diagonal.

Answer 2

You are looking for conditions whether the equation $f(x,y)=0$ can be expressed as $F(x)+G(y)=0$. For $f(x,y)\in C^1(D)$ where $D\subseteq \Bbb R^2$, the following deduction can be made: $$ f(x,y)=F(x)+G(y)\implies \begin{cases} \frac{\partial f(x,y)}{\partial x}=F'(x)\\ \frac{\partial f(x,y)}{\partial y}=G'(y) \end{cases}. $$ To prove the converse, i.e. $$ \begin{cases} \frac{\partial f(x,y)}{\partial x}=F'(x)\\ \frac{\partial f(x,y)}{\partial y}=G'(y) \end{cases} \implies \exists H(x),J(y)\ \ \ \text{such that}\ \ \ f(x,y)=H(x)+J(y), $$ note that one can apply integration w.r.t. $x$ to $\frac{\partial f(x,y)}{\partial x}=F'(x)$ to achieve $f(x,y)=F(x)+\alpha(y)$. Likewise, $\frac{\partial f(x,y)}{\partial x}=G'(y)$ implies $f(x,y)=G(y)+\beta(x)$. Comparing the last two results yields $ H(x)=\alpha(x)=F(x) $ and $ J(y)=\beta(y)=G(y) $. Hence, the following criteria is obtained:

Let $f(x,y)\in C^1(D)$ where $D\subseteq \Bbb R^2$. Then, $$ \exists F(x),G(y)\ \ \ \text{such that}\ \ \ f(x,y)=F(x)+G(y) \iff \begin{cases} \frac{\partial f(x,y)}{\partial x}\text{ is a function of $x$}\\ \frac{\partial f(x,y)}{\partial y}\text{ is a function of $y$} \end{cases}. $$

Answer 3

Consider any parametric curve consisting of points $(x,y)=(u(t),v(t)), t\in T \subset \mathbb R$.

Ignoring the restriction that the curve is closed, we can observe that:

• If both $u$ and $v$ are invertible functions, we are done, as we can put $$ F(x) = u^{-1}(x), \quad G(y) = -v^{-1}(y), $$ so that $$ F(u(t)) + G(v(t)) = t - t = 0. $$

• If at least one of the functions $u$ and $v$ is invertible, say $u$ (e.g. $u(t)=t, v(t) = \sin(t)$), then we can put $$ F(x)= v(u^{-1}(x)), \quad G(y) = y,\tag{*} \label{vu} $$ so that $$ F(u(t)) + G(v(t)) = v(t) - v(t) = 0. $$

If neither $u$ not $v$ is invertible, it does not have to be the case that all is lost yet, as illustrated by the example of a circle provided in the question. However, based on the condition given by @DavidESpeyer in the comments, if one of the quadrants of the circle was missing, then such a curve couldn't be characterised as in the desired way.

To be able to analyse the case when $u$ is not invertible, define $u^{-1}$ as the multi-functions $$ u^{-1}(x) = \{t\in T: u(t) = x\}, \quad x\in \mathbb R. $$

Claim: The curve $\{(u(t),v(t)),t\in T \}$ can be characterised by the equation $F(x)+G(y)=0$ if and only if $$ v(t_1) = v(t_2) \quad \Longrightarrow \quad u^{-1}(u(t_1)) = u^{-1}(u(t_2)) \quad \forall t_1,t_2 \in T. \tag{**} \label{cond} $$

Why the condition \eqref{cond} is necessary should be obvious from @DavidESpeyer's comment. It is sufficient because once it is satisfied $F$ and $G$ can be defined as in \eqref{vu}. If we want $F$ and $G$ to be $C^1$, we need to impose some restrictions on $u$ and $v$, e.g. that both of them are $C^1$ and additional conditions at the points where the derivative of one of them is zero.

Additional: Taking into account that the curve is closed.

I closed curve has the nice property that it splits the space into two connected regions: it's interior and the exterior. However, that the interior of the curve does not need to be convex can be seen for example from the curve given by $$ 0=11-\left|x\right|-2\sin\left(x\right)- y^{2}, $$ as can be seen in the graph:

Let $S$ be the set of the point of the curve and let $$ S_x = \{x\in \mathbb R: (x,y)\in S\} $$ be the projection of $S$ onto the $x$-axis. Analogously we define $S_y$. For any $x\in S_x$, define $$ \bar{y}(x) = \sup \{ y: (x,y)\in S \}, $$ and analogously we define $\underline{y}(x)$ and $\bar{x}(y)$\underline{y}(x)$$

The following lemma is a consequence of the condition described by
@ChristopheLeuridan in the question comments: $$ \text{If if three of the points } (x_1,x_2),(x_1,y_2),(y_1,x_2),(y_1,y_2) \text{ are on the the curve, then the four points are on the curve.} \tag{$3\to 4$} $$

Lemma: The set of local (global) minima (maxima) of $\underline{y}(x)$ is equal to the set of local (global) maxima (minima) of $\bar{y}(x)$. An analogous relation holds between the functions $\underline{x}$ and $\bar{x}$.