Author Topic: HA 10, problem 3a  (Read 1058 times)

Shaghayegh A

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HA 10, problem 3a
« on: November 21, 2016, 10:58:36 PM »
link: http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter8/S8.P.html     (3a)

Since P(x,y,z) is a polynomial of degree 0, it is a constant. So $U=x^2+y^2+z^2-c_0 (x^2+y^2+z^2)$, but you can't write this as a sum of homogenous harmonic polynomials since there is a term c_0 remaining?

Shentao YANG

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Re: HA 10, problem 3a
« Reply #1 on: November 22, 2016, 09:34:57 AM »
I think you need to plug in the equation to solve for $C_0$, for general $C_0$ function $u$ may not be harmonic.

Victor Ivrii

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Re: HA 10, problem 3a
« Reply #2 on: November 22, 2016, 09:52:48 AM »
Solution is a harmonic polynomial but not necessarily homogeneous. On the other hand it must have prescribed boundary value (as $x^2+y^2+z^2=R^2$).

Shaghayegh A

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Re: HA 10, problem 3a
« Reply #3 on: November 22, 2016, 11:03:22 AM »
modified:
My solution is $U=x^2+y^2-z^2-\frac{1}{3} (x^2+y^2+z^2-1)$, the laplacian of this equation is zero and it equal g(x,y,z) at the boundary. How can I write this as a sum of harmonic homogenous polynomials, since there is a factor of 1/3 : $U=2/3 x^2+ 2/3 y^2- 4/3 z^2+ \frac{1}{3}$

By the way, g is $x^2+y^2-z^2$
« Last Edit: November 22, 2016, 08:52:52 PM by Shaghayegh A »

Victor Ivrii

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Re: HA 10, problem 3a
« Reply #4 on: November 22, 2016, 05:51:03 PM »
Laplacian is not $0$, and what $g$ you are looking for?

Shaghayegh A

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Re: HA 10, problem 3a
« Reply #5 on: November 22, 2016, 08:53:19 PM »
Oops! I corrected my solution

Victor Ivrii

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Re: HA 10, problem 3a
« Reply #6 on: November 22, 2016, 09:37:28 PM »
Now it is correct