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

by yungman
Tags: continuous, derivatives, function, meaning
 P: 3,830 The definitions of a harmonic function u are: It has continuous 1st and 2nd derivatives and it satisfies $\nabla^2 u = 0$. Is the second derivative equal zero consider continuous? Example: $$u=x^2+y^2 ,\; \hbox{ 1st derivative }=u_x + u_y = 2x+2y,\; \hbox{ 2nd derivative }=u_{xx} + u_{yy} + u_{xy} + u_{yx} = 2+2=4$$ is continuous. How about $$u=x+y,\; \hbox{ 1st derivative }= 1+1=2,\; \hbox{ 2nd derivative }= 0$$, Is the 2nd derivative continuous if equal to zero? For $\; u=x+y,\; \nabla^2u=0$!!!
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P: 20,433
 Quote by yungman The definitions of a harmonic function u are: It has continuous 1st and 2nd derivatives and it satisfies $\nabla^2 u = 0$. Is the second derivative equal zero consider continuous?
Yes, and why would it not be considered continuous?
 Quote by yungman Example: $u=x^2+y^2 ,\;u' = 2x+2y,\; u''= 2+2=4$ 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 $$u_x = \frac{\partial u}{\partial x}$$

 Quote by yungman How about $u=x+y,\; u' = 1+1=2,\; u''= 0$, is the 2nd derivative continuous? For $\; u=x+y,\; \nabla^2u=0$!!!
P: 3,830
 Quote by Mark44 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 $$u_x = \frac{\partial u}{\partial x}$$
I change the symbol on the original post already using the full PDE form. But my question remain the same.

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P: 20,433

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

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
$$\frac{\partial u}{\partial x}$$
and
$$\frac{\partial u}{\partial x}$$.

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
$$\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt}$$.

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.
 P: 3,830 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 $r^2=x^2+y^2+z^2$. So what is the meaning of Harmonic function having continuous 1st and 2nd derivatives. This is being talked in the PDE text book 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.
 P: 3,830 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.
 Mentor P: 20,433 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.
P: 3,830
 Quote by Mark44 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.

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 $\nabla^2 u=0$, 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.
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P: 20,433
 Quote by yungman 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.
 Quote by yungman Also $\nabla^2 u=0$, 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?

 Quote by yungman 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.
P: 3,830
 Quote by Mark44 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 highschool.

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 refering 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!!!
 Mentor P: 20,433 Well, keep "plugging" away!

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