MAT334-2018F > Thanksgiving bonus

Thanksgiving bonus 1

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Victor Ivrii:
Read section 2.1.1* and solve the problem (10 karma points). As a sample see solutions to odd numbered problems

Problem 2. Decide whether or not the given function represents a locally sourceless and/or irrotational flow. For those that do, decide whether the flow is globally sourceless and/or irrotational. Sketch some of the streamlines.
$$\frac{x}{x^2+y^2}+i\frac{y}{x^2+y^2},\qquad \text{as } x^2+y^2>0.$$

Meng Wu:
Let$$u(x,y)=\frac{x}{x^2+y^2}, v(x,y)=\frac{y}{x^2+y^2}$$
$$\frac{\partial{u}}{\partial{x}}=\frac{-(x^2-y^2)}{(x^2+y^2)^2}, \frac{\partial{u}}{\partial{y}}=\frac{2xy}{-(x^2+y^2)^2}$$
$$\frac{\partial{v}}{\partial{x}}=\frac{-2xy}{(x^2+y^2)^2}, \frac{\partial{v}}{\partial{y}}=\frac{x^2-y^2}{(x^2+y^2)^2}$$
$$\frac{\partial{u}}{\partial{x}}+\frac{\partial{v}}{\partial{y}}=0, \frac{\partial{v}}{\partial{x}}+\frac{\partial{u}}{\partial{y}}=0, x^2+y^2>0$$
Thus,  given function represents a locally sourceless and irrotational flow.

Victor Ivrii:
you checked local conditions. You need to check integral conditions as well since there is a singular point $(0,0)$,

Hyun Woo Lee:
Let $$f = u + iv$$
Then, $$u(x, y) = \frac{x}{x^2+y^2}, v(x, y) = \frac{y}{x^2+y^2}$$
Hence, $$\frac{\partial u}{\partial x} = \frac{-(x^2-y^2)}{(x^2+y^2)^2}, \frac{\partial u}{\partial y} = \frac{-2xy}{(x^2+y^2)2}$$
And
$$\frac{\partial v}{\partial x} = \frac{-2xy}{(x^2+y^2)^2}, \frac{\partial v}{\partial y} = \frac{(x^2 - y^2)}{(x^2+y^2)^2}$$
And,
$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0, \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} = 0$$ This means that our function is locally sourceless and irrotational flux on domain D that does not include the origin (where our function is not defined).

Now note that our function can be written as $$f(z) = \frac{1}{\overline{z}}$$
Since $$\frac{1}{\overline{z}} = \frac{x}{x^2+y^2} + i\frac{y}{x^2+y^2}$$
Now, lets take a look on the circle $$|z| = 1$$
The normal component of f is $$f\cdot n = \cos{\theta}\cos{\theta} + \sin{\theta}sin{\theta}$$
Since $$f(z) = \frac {\cos{\theta} +i\sin{\theta}}{r}, r = 1, \theta = arg(\frac{1}{\overline{z}})$$
And $$n = \cos{\theta} - i\sin{\theta}$$
Then,
$$\int_{|z| =1} f\cdot n ds = 1\int_{|z| = 1} ds = 2\pi$$
Hence, our function is locally sourceless and irrotational flow but is not globally sourceless.