Constant function

**daudaudaudau** · Jun14-09, 05:07 PM

My book (Saff&Snider) has the following theorem

Suppose u(x,y) is a real-valued function defined in a domain D. If the first partial derivatives of u satisfy

$LaTeX Code: \\frac{\\partial u}{\\partial x}=\\frac{\\partial u}{\\partial y}=0 $

at all points of D, then u=constant in D.

In the proof, the book says that since both partial derivatives are zero, u(x,y) is constant along any horizontal or vertical line segment. By definition two points in a domain can be connected by a polygonal path, and since such a path can be separated into horizontal and vertical line segments, any two points in the domain can be connected by a path consisting of horizontal and vertical lines. END PROOF.

But why not just do it like this

$LaTeX Code: u(x_1,y_1)-u(x_2,y_2) = \\int_{(x_2,y_2)}^{(x_1,y_1)}\\nabla u\\cdot d\\mathbf l $

i.e. the difference between two points is given by the line integral above, and because the gradient is zero, so is the difference. This way I don't have to worry about whether a polygonal path can be separated into horizontal and vertical line segments.

**HallsofIvy** · Jun14-09, 06:35 PM

You would first have to prove that the formula you give is true. The fact that the partial derivatives exist does NOT necessarily mean that the function itself is differentiable which you need to be able to integrate like that.. One can show that if the partial derivatives exist and are continuous then the function is differentiable but if that itself had not yet been proven, the books proof is simpler.

**daudaudaudau** · Jun14-09, 07:39 PM

I see. What about using the mean value theorem then?

$LaTeX Code: u(x_1,y_0)-u(x_0,y_0)=(x_1-x_0)\\frac{\\partial u(\\xi_x,y_0)}{\\partial x}=0 $

$LaTeX Code: u(x_1,y_1)-u(x_1,y_0)=(y_1-y_0)\\frac{\\partial u(x_1,\\xi_y)}{\\partial y}=0 $

And so it follows that

LaTeX Code: <BR>u(x_1,y_1)=u(x_0,y_0)<BR>

(This is the way Apostol does it when the domain is an open disk)

But I guess this won't work because the domain may have holes in it ?

**daudaudaudau** · Jun14-09, 07:58 PM

Or a better way would be to define the path between (x_0,y_0) and (x_1,y_1) to be r(t), with r(0)=(x_0,y_0) and r(1)=(x_1,y_1) and define f(t)=u(r(t)) and then the mean value theorem is
$LaTeX Code: f(1)-f(0)=fsingle-quote(\\xi)=\\nabla u(r(\\xi))\\cdot rsingle-quote(\\xi) $

But then you say that I still have to show that f(t) is differentiable? I guess this is true by the chain rule?

**Tibarn** · Jun14-09, 10:33 PM

Sounds more like they wanted a simple proof using an "obvious" statement rather than getting mired in integrals, gradients, and partial derivatives. If the book is aimed at beginners, that kind of proof is much easier to understand than something overly technical. Also, it depends on how much prior knowledge the book assumes. If they haven't defined line integrals yet or haven't proved a particular theorem, they couldn't use it in a proof.