Confusion about continuity and differentiability(In partial differential)


by athrun200
Tags: confusion, continuity, differentiabilityin, differential, partial
athrun200
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#1
Sep30-11, 03:32 AM
P: 276
Example 8 in photo 1 shows that differentiability doesn't implies continuity.
But photo 2 shows a Theorem that contradict to photo 1.
I wonder what is going on here.

Does the textbook get it wrong?
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rjvsngh
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#2
Sep30-11, 07:47 AM
P: 5
existence of partial derivatives does not imply "differentiability". in some sense, differentiability in higher dimensional spaces is a stronger condition than existence of partial derivatives. intuitively, partial derivatives only sample the function along "coordinate directions" but this is not enough to satisfy the condition of differentiability at a point because the function's behavior along coordinate directions may not represent it's behavior along other directions in higher dimensions. for functions defined on the real line, you do not encounter such problems because there is only one dimension to move around in.

look at this page for a detailed example.

i hope this is helpful.
HallsofIvy
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#3
Sep30-11, 08:09 AM
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Your example 8 shows that the existence of partial derivatives does not imply continuity. However, as rjvsngh says, the existence of partial derivatives also does not imply "differentiability" so your statement is wrong. Differentiability (whicy is much stronger than the existence of partial derivatives) does imply continuity.

athrun200
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#4
Sep30-11, 09:26 AM
P: 276

Confusion about continuity and differentiability(In partial differential)


Thx a lot!
I understand it now!
Bacle
Bacle is offline
#5
Sep30-11, 11:36 AM
P: 662
Just to add to what has been said (some nice insights, BTW), a sufficient (but not
necessary) condition for the derivative to exist is that the partials exist and that the
partials are continuous.
lugita15
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#6
Sep30-11, 02:08 PM
P: 1,583
Quote Quote by rjvsngh View Post
existence of partial derivatives does not imply "differentiability". in some sense, differentiability in higher dimensional spaces is a stronger condition than existence of partial derivatives. intuitively, partial derivatives only sample the function along "coordinate directions" but this is not enough to satisfy the condition of differentiability at a point because the function's behavior along coordinate directions may not represent it's behavior along other directions in higher dimensions.
Actually, even if the directional derivatives exist in all directions, that is still not a sufficient condition for differentiability.
Bacle
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#7
Sep30-11, 02:31 PM
P: 662
Any chance you (or any one else) have an example, lugita15 ?

How about an example of partials existing and partials continuous not

being necessary for derivative to exist?
lugita15
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#8
Sep30-11, 04:22 PM
P: 1,583
Quote Quote by Bacle View Post
Any chance you (or any one else) have an example, lugita15 ?
From this Wikipedia page, [itex]f(x,y) = \begin{cases}\frac{y^3}{x^2+y^2} \text{ if }(x,y) \ne (0,0) \\ 0 \text{ if }(x,y) = (0,0)\end{cases}[/itex].
LikeMath
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#9
Oct1-11, 08:43 AM
P: 62
Recall that if the partial derivatives exist and continuous at (a,b) then the function is differentiable at (a,b),
your example does not contradict this fact.
lugita15
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#10
Oct1-11, 08:58 AM
P: 1,583
Quote Quote by LikeMath View Post
Recall that if the partial derivatives exist and continuous at (a,b) then the function is differentiable at (a,b),
your example does not contradict this fact.
No, my example was to show that the existence of directional derivatives in all directions is still not sufficient to establish differentiability.
Omega017
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#11
Apr18-12, 12:33 AM
P: 2
Quote Quote by lugita15 View Post
From this Wikipedia page, [itex]f(x,y) = \begin{cases}\frac{y^3}{x^2+y^2} \text{ if }(x,y) \ne (0,0) \\ 0 \text{ if }(x,y) = (0,0)\end{cases}[/itex].
Can anyone explain why this function is not differentiable at (0,0)?
slider142
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#12
Apr18-12, 01:05 AM
P: 876
Quote Quote by Omega017 View Post
Can anyone explain why this function is not differentiable at (0,0)?
Consider the directional derivative in the direction of the vector (1, 1). Computed using the definition of the derivative (no shortcuts, since we do not know whether the function is differentiable here), we get 1/2.
However, the Jacobian matrix for f at the point (0, 0) is [0, 0], since both partials vanish along the axes. A standard theorem in vector calculus states that if f is differentiable at (0, 0) then the directional derivative of f at (0, 0) in the direction of the vector (1, 1) is equal to the product of the Jacobian matrix with the vector. However, that product is 0, which is not the directional derivative of f in the direction of (1, 1) at the point (0, 0), as we derived in the previous paragraph. Thus, f is not differentiable at (0, 0).
Visually, you can take directional derivatives in several directions at (0, 0); in particular, take directional derivatives in the directions (cos(t), sin(t)) for various t and you will get different derivatives. There cannot be a unique tangent plane that is tangent to all of these curves.
lugita15
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#13
Apr18-12, 01:05 AM
P: 1,583
Quote Quote by Omega017 View Post
Can anyone explain why this function is not differentiable at (0,0)?
See here (pages 4-6).
Bacle2
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#14
Apr20-12, 04:42 PM
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Check the continuity of the function at (0,0). Find different curves approaching (0,0) to show that f is not continuous there.
HallsofIvy
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#15
Apr21-12, 08:57 AM
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Quote Quote by LikeMath View Post
Recall that if the partial derivatives exist and continuous at (a,b) then the function is differentiable at (a,b),
your example does not contradict this fact.
If the partial derivatives exist and are continuous in some neighborhood of a point then the function is differentiable. Partial derivatives continuous at a point is not sufficient.


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