What is the meaning by a function with continuous 1st and 2nd derivatives?

The 3rd derivative of y = 2x is zero, and so is the 4th derivative, and the 5th, and so on. What's the point of this? If you graph y = 2x, you get a slanting line. The slope of the tangent line is constant at 2. The graph of y'' = 0 is a horizontal line, the slope of any tangent line is zero. And the graph of y''' = 0 is a horizontal line, the slope of any tangent line is zero. And the graph of y'''' = 0 is a horizontal line, the slope of any tangent line is zero. And the graph of y''
  • #1
yungman
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The definitions of a harmonic function u are: It has continuous 1st and 2nd derivatives and it satisfies [itex]\nabla^2 u = 0[/itex].

Is the second derivative equal zero consider continuous?

Example: [tex] u=x^2+y^2 ,\; \hbox{ 1st derivative }=u_x + u_y = 2x+2y,\; \hbox{ 2nd derivative }=u_{xx} + u_{yy} + u_{xy} + u_{yx} = 2+2=4[/tex] is continuous.

How about [tex] u=x+y,\; \hbox{ 1st derivative }= 1+1=2,\; \hbox{ 2nd derivative }= 0[/tex],

Is the 2nd derivative continuous if equal to zero?

For [itex]\; u=x+y,\; \nabla^2u=0[/itex]!
 
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  • #2
yungman said:
The definitions of a harmonic function u are: It has continuous 1st and 2nd derivatives and it satisfies [itex]\nabla^2 u = 0[/itex].

Is the second derivative equal zero consider continuous?
Yes, and why would it not be considered continuous?
yungman said:
Example: [itex] u=x^2+y^2 ,\;u' = 2x+2y,\; u''= 2+2=4[/itex] is continuous.
Your work in both examples is incorrect. In both examples, u is a function of both x and y, so you need to be working with partial derivatives, not ordinary derivatives. The notation u' makes no sense with a function of two or more variables.

If u = f(x, y) = x2 + y2,
then ux = 2x and uy = 2y
and uxx = 2 and u yy = 2, and uxy = uyx = 0

I have used subscript notation to indicate partial derivatives, with [tex]u_x = \frac{\partial u}{\partial x}[/tex]

yungman said:
How about [itex] u=x+y,\; u' = 1+1=2,\; u''= 0[/itex], is the 2nd derivative continuous?

For [itex]\; u=x+y,\; \nabla^2u=0[/itex]!
 
  • #3
Mark44 said:
Yes, and why would it not be considered continuous?
Your work in both examples is incorrect. In both examples, u is a function of both x and y, so you need to be working with partial derivatives, not ordinary derivatives. The notation u' makes no sense with a function of two or more variables.

If u = f(x, y) = x2 + y2,
then ux = 2x and uy = 2y
and uxx = 2 and u yy = 2, and uxy = uyx = 0

I have used subscript notation to indicate partial derivatives, with [tex]u_x = \frac{\partial u}{\partial x}[/tex]

I change the symbol on the original post already using the full PDE form. But my question remain the same.
 
  • #4
In your examples, unless x and y are functions of a single variable (t, for example), there is no single first derivative. There are two first partial derivatives, ux and uy. These can also be written as
[tex]\frac{\partial u}{\partial x}[/tex]
and
[tex]\frac{\partial u}{\partial x}[/tex].

If it turns out that x and y are functions of, say, t, then you can talk about the total derivative, du/dt, which is given by
[tex]\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt}[/tex].

So if u = x2 + y2, ux = 2x and uy = 2y, but there is no "first derivative" that equals 2x + 2y, as you have.

For more on harmonic functions, see this wiki article.
 
  • #5
Thanks for you time.

I looked at the Wiki link before. Notice they talked about "twice continuously differentiable function satisfies Laplace's equation". And they show multi-variables. Laplace's eq. by it self is a multiple variable equation.

They even specifically give example of harmonic functions of three independent variables of [itex]r^2=x^2+y^2+z^2[/itex].

So what is the meaning of Harmonic function having continuous 1st and 2nd derivatives. This is being talked in the PDE textbook by Asmar.

Everything I've seen in this topic has multi independent variables, not just x(t), y(t) and z(t) that really has single variable t.
 
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  • #6
I read the book again. I left out the word partial. The book said Harmonic function have continuous 1st and 2nd partial derivatives. My mistake.

But the rest of my post still hold. The question is the same.

The book use parametric representation of the boundary with parameter t as angle in polar coordinate. The function u itself is not necessary parametric.
 
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  • #7
For your example function, u = x2 + y2,
ux = 2x
uy = 2y

uxx = 2
uyy = 2
uxy = uxy = 0

The first two partials above, the first partials, are continuous.
The next four partials, including the two mixed partials, are the second partials. All of them are continuous as well.

I didn't want to answer your question until you realized that we were talking about partial derivatives, and until your misconceptions about derivatives were straightened out.
 
  • #8
Mark44 said:
For your example function, u = x2 + y2,
ux = 2x
uy = 2y

uxx = 2
uyy = 2
uxy = uxy = 0

The first two partials above, the first partials, are continuous.
The next four partials, including the two mixed partials, are the second partials. All of them are continuous as well.

I didn't want to answer your question until you realized that we were talking about partial derivatives, and until your misconceptions about derivatives were straightened out.

Thanks for you reply. It is very helpful.

So I want to clear up. In case of u=x+y, even the second partial derivative equal to zero, it's second partial derivative is still continuous. Also [itex]\nabla^2 u=0[/itex], so I can conclude u=x+y is a Harmonic function.

How about 3rd partial derivative of u=x+y? The 3rd partial derivative will be taking the partial derivative of zero! That should not be consider continuous, should it?

You got me thinking about the parametric equation. I am going to post a thread to ask that question.

Thanks

Alan.
 
  • #9
yungman said:
Thanks for you reply. It is very helpful.

So I want to clear up. In case of u=x+y, even the second partial derivative equal to zero, it's second partial derivative is still continuous.
Any function that is identically equal to a constant, including zero, is continuous. You seem to be laboring under the misconception that if a function is always zero, then it isn't continuous.
yungman said:
Also [itex]\nabla^2 u=0[/itex], so I can conclude u=x+y is a Harmonic function.

How about 3rd partial derivative of u=x+y? The 3rd partial derivative will be taking the partial derivative of zero!
So what? Let's step back from multivariate functions for a minute and consider a function of one variable, y = 2x
y' = 2
y'' = 0
y''' = ?
y''' is just the derivative (with respect to x) of y''. Remember that the derivative gives the slope of the tangent line. What's the slope of the tangent line to the graph of y'' = 0?

yungman said:
That should not be consider continuous, should it?

You got me thinking about the parametric equation. I am going to post a thread to ask that question.

Thanks

Alan.

You should probably post it in the Homework section, in Calculus & Beyond.
 
  • #10
Mark44 said:
Any function that is identically equal to a constant, including zero, is continuous. You seem to be laboring under the misconception that if a function is always zero, then it isn't continuous.
So what? Let's step back from multivariate functions for a minute and consider a function of one variable, y = 2x
y' = 2
y'' = 0
y''' = ?
y''' is just the derivative (with respect to x) of y''. Remember that the derivative gives the slope of the tangent line. What's the slope of the tangent line to the graph of y'' = 0?
y''' being the slope of y'' is 0.


You should probably post it in the Homework section, in Calculus & Beyond.

It has been almost 30 year between my first one and half class of calculus ( at the time, we broke up the first two calculus into 4 classes and I only have 3). I was very lazy on top and get barely passing grades on the first two! I have very little interest in math during high school.

I started getting into Calculus the pass 4 years, re-study the calculus and continue onto multi-variables on my own. I took to ODE class and I did very well ( first in class). I just finish studying the PDE according to SJSU requirement on my own.

As you can see, I mostly study on my own, there are holes in my knowledge. Some very fundamental stuffs that people take for granted and I don't know. Like what you are referring to the characteristic of function. I don't know that if a function is identity to zero, it is still continuous! Yes, I have more than one occasion when I asked a question in the ODE class, the instructor and the students turn and look at me like...you should know this to be here!

I just need to keep plugging those holes!
 
  • #11
Well, keep "plugging" away!
 

1. What is the definition of a function with continuous 1st and 2nd derivatives?

A function with continuous 1st and 2nd derivatives is a mathematical function that is smooth and does not have any abrupt changes or sharp corners. This means that the function is differentiable at every point and its first and second derivatives exist and are continuous.

2. How can you tell if a function has continuous 1st and 2nd derivatives?

A function has continuous 1st and 2nd derivatives if it is differentiable at every point in its domain and its first and second derivatives are continuous at those points. This can be checked by graphing the function and looking for any sharp corners or discontinuities, or by using the limit definition of derivative.

3. What is the significance of a function having continuous 1st and 2nd derivatives?

A function with continuous 1st and 2nd derivatives is considered to be a well-behaved function. It is used in many mathematical applications, such as optimization problems and curve fitting. It also allows us to apply theorems and techniques from calculus, such as the Mean Value Theorem and Taylor series expansions.

4. Can a function have continuous 1st and 2nd derivatives everywhere?

No, a function cannot have continuous 1st and 2nd derivatives everywhere. For example, functions with sharp corners or discontinuities will not have continuous derivatives at those points. Additionally, some functions may have points where the derivatives do not exist, such as singularities.

5. How does the continuity of 1st and 2nd derivatives relate to the smoothness of a function?

The continuity of 1st and 2nd derivatives is directly related to the smoothness of a function. A function with continuous 1st and 2nd derivatives is considered to be a smooth function, meaning it has no abrupt changes or sharp corners. The higher the order of continuity, the smoother the function is considered to be.

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