2

5) Let D {0 < I < y 1} and be harmonic in D with the boundary conditions u(r, 0) 12 u(z.1) 12 _ 1, "(0,y) ~y} , u(1,y) = 1 y? a) What is the maximum ank...

Question

5) Let D {0 < I < y 1} and be harmonic in D with the boundary conditions u(r, 0) 12 u(z.1) 12 _ 1, "(0,y) ~y} , u(1,y) = 1 y? a) What is the maximum ankl minimum values of u in D? Assume that U 15 another harmonic function On the entire plane with v(0,0) = 1. Can v be zero for all values (I.y) outside of some disc?

5) Let D {0 < I < y 1} and be harmonic in D with the boundary conditions u(r, 0) 12 u(z.1) 12 _ 1, "(0,y) ~y} , u(1,y) = 1 y? a) What is the maximum ankl minimum values of u in D? Assume that U 15 another harmonic function On the entire plane with v(0,0) = 1. Can v be zero for all values (I.y) outside of some disc?



Answers

$5-15$ Use the Divergence Theorem to calculate the surface integral
$\iint_{S} \mathbf{F} \cdot d \mathbf{S} ;$ that is, calculate the flux of $\mathbf{F}$ across $S .$
$$\begin{array}{l}{\mathbf{F}=|\mathbf{r}| \mathbf{r}, \text { where } \mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k},} \\ {S \text { consists of the hemisphere } z=\sqrt{1-x^{2}-y^{2}} \text { and the disk }} \\ {x^{2}+y^{2} \leqslant 1 \text { in the } x y-\text { plane }}\end{array}$$

Were given a vector field F and surface s and were asked to use the divergence theorem to calculate the surface integral over s off f So f is the vector field magnitude of our times are where r is the vector x I plus yj plus zk and s consists of the hemisphere Z equals the square root of one minus X squared minus y squared and the disc X squared plus y squared is less than or equal to one in the X y plane. So first of all, it's right out f in its entirety as a function of X, y and Z. This is going to be well we have that Are is X i y Jay Z case. This is X times the norm of our which is the square root the X squared plus y squared plus C squared I plus Why times the square root of X squared plus y squared plus C squared Jenny plus z times the square root of X squared plus y squared plus Z squared. Okay. And therefore the divergence of f This is going to be taking a partial derivative. The first component with respect to X Ray used product rules. We have X times one half times X squared plus y squared plus Z squared to the negative one half times two X and then plus derivative of X is just one times the square root of X squared plus y squared plus C squared. And in fact, we repeat this pattern with Y and Z taking the role of X. So we add why times one half times x squared plus y squared plus z squared to the negative one half times to why plus X squared plus y squared plus z squared to the one half plus z times one half times X squared plus y squared plus C squared to the negative one half times to Z plus X squared plus y squared plus Z squared to the one half In combining terms and factoring out X squared plus y squared plus C script, the negative one half we get X squared plus y squared plus C squared to the negative one half times, and this is X squared plus X squared plus y squared plus z squared plus y squared plus X squared plus y squared plus C square plus Z squared plus X squared plus y squared plus Z squared and combining we get or times X squared plus y squared plus Z squared over the square root of X squared plus y squared plus C squared which of course, is the same as four times the square root of X squared plus y squared plus Z square Have you divergence? The're, um the surface integral over s of F is equal to be triple integral over the region. E bounded by S of the divergence of F which is four times the square root of X squared plus y squared plus C squared and the region E. This is a hemisphere which will be described using spherical coordinates easily. So making the switch This becomes thean aerated integral from zero to pi over to integral from 0 to 2 pi an integral from zero to the radius of our sphere, which is one of our function in terms of spherical coordinates. So we have four times rebel times the differential Rose Square to sign Phi dear Oh d theta be fine and using food Venus theorem. We can write this as a product into girls so we get integral from zero to pi over two. Sign Phi defy times and to grow from 0 to 2. Pi Do you see that times the integral from 01 of four row cubed hero and then taking anti derivatives. We get negative co sign five from zero to pi over two times Data from 0 to 2 pi times road to the fourth from 01 and evaluating kids negative one times it's actually opposite negative one. So it's positive. One times two pi times one mhm right, which is simply to pie yes.


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