Continuous differentiability of vector fields

Unparseable latex formula:

[br][br]a) What does it mean for $(f(x, y), g(x, y))$ to be differentiable and continuously differentiable? [br][br]b) Are there points where, $$F((x, y)) = (\frac{2xy}{x^2 + y^2} , \lvert 1-xy \rvert)$$ is differentiable but not continuously differentiable? [br][br]

My attempt:
a) F(x,y) = (f(x,y), g(x,y)) is differentiable at V = (a,b) if F(v + h) - F(v) = Jacobian(F) h + R(h) with the limit R(h)/norm(h) -> 0 as h -> 0
OR equivalently all partial derivates of F's components exist and are continuous in a neighbourhood of V
and F is continuously differentiable if all partial derivatives of each component exist and are continuous at V.

At (x,y) = (0,0), F isn't differentiable since approaching this point along the line y = x would yield the limit( (2h^2 / 2h^2) /h ) as h ->0 which is definitely not 0. So F isn't differentiable at (0,0)

For the rest of the domain (except this xy = 1 line and (0,0) ), F is continuously differentiable since the components of F are ratios of continously differentiable functions.

So the only place a problem could arise is this xy = 1 line:

xy - 1 < 0: the second component is 1-xy ,
d/dx(1-xy) = -y

xy -1 > 0: the second component is xy - 1
d/dx(xy-1) = y

so the partial derivatives of the 2nd component aren't continuous across the line.

I'm at a loss as to how I should show whether or not the function is differentiable across the line? It seems like it isn't since the partial derivates won't agree in any neighbourhood of a point in the line.

So would the answer to the question be no, because everywhere F isn't is differentiable it's continuously differentiable and everywhere it may not be continuously differentiable it's not even differentiable?

Reply 1

DFranklin

18

Original post by Ryanzmw

My attempt:
a) F(x,y) = (f(x,y), g(x,y)) is differentiable at V = (a,b) if F(v + h) - F(v) = Jacobian(F) h + R(h) with the limit R(h)/norm(h) -> 0 as h -> 0
OR equivalently all partial derivates of F's components exist and are continuous in a neighbourhood of V
and F is continuously differentiable if all partial derivatives of each component exist and are continuous at V.

I don't agree with the definition I've highlighted in red (I can't see how it's different from the statement I've made green).

So would the answer to the question be no, because everywhere F isn't is differentiable it's continuously differentiable and everywhere it may not be continuously differentiable it's not even differentiable?

I agree that there's no point where F it differentiable without being continuously differentiable.

(The standard example of an F that's differentiable but not continuously so is to use the 1D function g(x) = x^2 sin(1/x) (with g(0) = 0) that's diff but not cts diff at x = 0, and then define F(x, y) = g(x) + g(y)).

Reply 2

Ryanzmw

OP

16

Original post by DFranklin

I don't agree with the definition I've highlighted in red (I can't see how it's different from the statement I've made green).

I agree that there's no point where F it differentiable without being continuously differentiable.

(The standard example of an F that's differentiable but not continuously so is to use the 1D function g(x) = x^2 sin(1/x) (with g(0) = 0) that's diff but not cts diff at x = 0, and then define F(x, y) = g(x) + g(y)).

I agree that they do seem to be the same thing... I was just rattling off a theorem given that:

all continuous partial derivatives in a neighbourhood of a <=> Differentiable at a

Is there no alternative definition of differentiable in terms of partial derivatives? Or am I forced to give the first definition?
It's certainly possible I was mistaken in writing this down as a theorem though...

(edited 6 years ago)

Reply 3

DFranklin

18

Original post by Ryanzmw

I agree that they do seem to be the same thing... I was just rattling off a theorem given that:

all continuous partial derivatives in a neighbourhood of a <=> Differentiable at aThis theorem is not correct - I just gave you an example of a function differentiable at the origin but with discontinuous partial derivatives there.

Conversely, by considering

\dfrac{xy}{\sqrt{x^2+y^2}}

at the origin you can see the partial derivatives are trivially zero, but by considering the behaviour as you approach the origin on the line y = x you can see that the directional derivative in this direction is 1/2. So this function can't be differentiable.

Basically:

"partial derivatives exist and are cts" is a stronger condition than
"differentiable", which is a stronger condition than
"partial derivatives exist".

Reply 4

Ryanzmw

OP

16

Original post by DFranklin

This theorem is not correct - I just gave you an example of a function differentiable at the origin but with discontinuous partial derivatives there.

Conversely, by considering

\dfrac{xy}{\sqrt{x^2+y^2}}

at the origin you can see the partial derivatives are trivially zero, but by considering the behaviour as you approach the origin on the line y = x you can see that the directional derivative in this direction is 1/2. So this function can't be differentiable.

Basically:

"partial derivatives exist and are cts" is a stronger condition than
"differentiable", which is a stronger condition than
"partial derivatives exist".

Thank you very much!