Graduate Functions with "antisymmetric partial"

[economicsnerd]

Sorry for the terribly vague title; I just can't think of a better name for the thread.

I'm interested in functions $f:[0,1]^2\to\mathbb{R}$ which solve the DE, $\tfrac{\partial}{\partial y} f(y, x) = -\tfrac{\partial}{\partial x} f(x,y)$.

I know this is a huge collection of functions, amounting to everything of the form $\left\{ \int g(x,y) dx: \ g \text{ antisymmetric} \right\}$, but I'm wondering whether there's a nice description of the family of functions $f$ satisfying the equation. Ideally, I'd like a description which doesn't use derivatives or integrals.

 

[epenguin]

-1 = (∂ƒ/∂x)/(∂ƒ/∂y) = - (dy/dx)_ƒ.
which is not hard to integrate to
y - x = K at constant ƒ.

So ƒ(y - x) = C. , any constant C and any differentiable function ƒ is a solution.

IMHO not a bad starting point for study of solution of pde's (which I must start one of these days. )

 

[andrewkirk]

solve the DE, $\tfrac{\partial}{\partial y} f(y, x) = -\tfrac{\partial}{\partial x} f(x,y)$.

Are you sure you mean that and not $\tfrac{\partial}{\partial y} f(x,y) = -\tfrac{\partial}{\partial x} f(x,y)$?

On my calculations, every differentiable anti-symmetric function is a solution of the former (the one you wrote) because both partial derivs are taken wrt the first argument of $f$. So if $g:[0,1]^2\to\mathbb{R}$ is the partial diff of $f$ wrt its first argument then the DE is just $g(y,x)=-g(x,y)$, and the problem becomes uninteresting.

However, if the DE is really $\tfrac{\partial}{\partial y} f(x,y) = -\tfrac{\partial}{\partial x} f(x,y)$, or to make it even clearer:

$$D_2 f(x,y) =-D_1f(x,y) $$
where $D_n$ indicates partial differentiation wrt the $n$th argument, the problem becomes interesting and penguin's solution comes into play.

So ƒ(y - x) = C. , any constant C and any differentiable function ƒ is a solution.

I wasn't quite sure what you meant by this last part. Do you mean that, for any differentiable function $g:\mathbb{R}\to\mathbb{R}$, the function $f:[0,1]^2\to\mathbb{R}$ defined by $f(x,y)=g(y-x)$ is a solution of the DE?

That certainly seems to work, and is a nice solution [of what I think the problem was intended to be, per the above].

Do you think we can conclude that all solutions are of that form?

 

[epenguin]

penguin's solution comes into play.

Aaargh I misread.

 

[economicsnerd]

I did mean $\tfrac{\partial}{\partial y} f(y, x) = -\tfrac{\partial}{\partial x} f(x,y)$, not $\tfrac{\partial}{\partial y} f(x, y) = -\tfrac{\partial}{\partial x} f(x,y)$.

So I'm after the set of functions whose first partial is an antisymmetric function. I'm wondering if there's a way to specify that class of functions without referencing derivatives or integrals.

 

[stevendaryl]

It seems to me that every such f(x,y) can be written in the form f(x,y) = g(x-y), where g is an arbitrary differentiable function.

 

[andrewkirk]

I'm after the set of functions whose first partial is an antisymmetric function.

A large class of solutions will be of the form

$$f(u,v)=g(u-v)+h(v)$$

where
$g:[-1,1]\to\mathbb{r}$ and $h:[0,1]\to\mathbb{r}$ are both once-differentiable and $g$ is an odd function.
Note that the $h$ term is a function only of the second argument, and hence disappears upon partial differentiation wrt the first argument.
The extra requirement that the function $g$ be odd appears necessary. A function that is even, such as $g(x)=x^2$, or neither even nor odd, such as $g(x)=e^x$, would not work.

I can't prove that all solutions are of this form. Perhaps there may be other classes of solutions, but I can't think of any examples.