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[tex]\int_{\Omega }^{}f^{5}(f_{x}+2f_{y})=0[/tex]

where f is some function

Can somebody explain how?

- Thread starter Weilin Meng
- Start date

- #1

- 25

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[tex]\int_{\Omega }^{}f^{5}(f_{x}+2f_{y})=0[/tex]

where f is some function

Can somebody explain how?

- #2

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What does the integration area omega mean here? Perhaps I can try to solve your question if I know that.

greetings Janm

- #3

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In this case, omega is just some closed domain lying in R^2

- #4

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So the equation has no poles = singularities?In this case, omega is just some closed domain lying in R^2

- #5

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- #6

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Hello Weilin

my function theory was a lot of years ago and my book complex variables is somewhere lost for a long time. At the time I bought it together with a collegue student... The same thing as with comic books. Just the nice ones get lost... I still want to find a intuitive answer to your question. May I do that with a counter question.

Would the integral of f^4*(2f_x+f_y) also be zero?

So has it to do with odd or even powers of f? If so we could reduce it to the integral f*(f_x+2f_y)...

greetings Janm

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- #8

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So what if you put f=x, is it then zero? Not as far as I can tell, if Omega is any domain in R^2.

- #9

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In this case, omega is just some closed domain lying in R^2

Is it a standard double integral over the region omega?

[tex]\int_{\Omega}\int f^{5}\left(f_{x}+2f_{y}\right)dA[/tex]

If so, then it is not necessarily 0. Please give more notation, path integral, line integral, double integral, Mdx + Ndy, F dot ds, F dot n ds, etc...

- #10

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Hello daudaudaudauSo what if you put f=x, is it then zero? Not as far as I can tell, if Omega is any domain in R^2.

if you put f(z)=f(x+iy)=Re(z)=x, then f_x=1 and f_y=0 so the integrand would be x^5(1+2*0)=x^5.

You get the integral of x^5 dA

f(z)= x^5 is a surface bended in one direction and can be made of flat paper.

greetings Janm