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Homework Help: Gauss's Mean Value Property problem.

  1. Aug 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Identify integral as the mean value of a harmonic function at a point and evaluate the integral:

    [tex]\frac{1}{2\pi} \int_0^{2\pi} \; cos(1+cos(t)) cosh(2+sin(t)) \; dt[/tex]


    [tex] u(x_0,y_0) = \frac{1}{2\pi} \int_0^{2\pi} \; u[x_0+rcos(t) , \; y_0+rsin(t)] \; dt[/tex]

    2. Relevant equations

    [tex] u(x_0,y_0) = \frac{1}{2\pi} \int_0^{2\pi} \; u[x_0+rcos(t) , \; y_0+rsin(t)] \; dt[/tex]
    This equation show if [itex]u[/tex] is harmonic function in a region, then the value of u at any point [itex](x_0, y_0)[/itex] equal to the average of u on the line integral of a circle inside the region centered at [itex] (x_0, y_0) [/itex].

    3. The attempt at a solution

    [tex]\frac{1}{2\pi} \int_0^{2\pi} \; cos( 1+cos(t) ) cosh( 2+sin(t)) \; dt[/tex]

    [tex] x(t) = 1+cos(t) \hbox { and } y(t) = 2+sin(t) [/tex]

    [itex]\nabla^2 u=0 [/tex] and has continuous 1st and 2nd partial derivatives implies [itex]u[/tex] is harmonic function which implies:

    [tex] u(x_0,y_0) = \frac{1}{2\pi} \int_0^{2\pi} \; u[x_0+rcos(t) , \; y_0+rsin(t)] \; dt[/tex]

    [tex] x(t) = 1+cos(t) \hbox { and } y(t) = 2+sin(t) \;\; \Rightarrow\;\; x_0=1 \;\hbox{ and }\; y_0=2\hbox{ Therefore } u(x_0, y_0)=u(1,2) \; \hbox{ and } r=1[/tex]

    Therefore the line integral is a circle center at (1,2).

    [tex]u(1,2)= cos(1)cosh(2) = \frac{1}{2\pi} \int_0^{2\pi} \; cos(1+cos(t)) cosh(2+sin(t)) \; dt[/tex]

    Which the answer is [itex]cos(1)cosh(2)[/itex].

    But the answer of the book is [itex] 2\pi u(0,0) = 2\pi cos(1) cosh(2)[/itex] which implies the center of the circle is (0,0). I don't see what I done wrong. Please help.
  2. jcsd
  3. Aug 6, 2010 #2
    Anyone please?
  4. Aug 6, 2010 #3
    I really think the answer of the book is wrong. Compare the given equation:

    [tex] u(x_0,y_0) = \frac{1}{2\pi} \int_0^{2\pi} \; u[x_0+rcos(t) , \; y_0+rsin(t)] \; dt[/tex]


    [tex]\frac{1}{2\pi} \int_0^{2\pi} \; cos( 1+cos(t) ) cosh( 2+sin(t)) \; dt[/tex]

    [tex] u(x,y) = u[x_0+rcos(t) , \; y_0+rsin(t)] = cos( 1+cos(t) ) cosh( 2+sin(t)) \Rightarrow\; x(t) = 1+cos(t) \hbox { and } y(t) = 2+sin(t) [/tex]

    Where [itex](x_0,y_0)=(1,2) \hbox { and } r=1[/itex]

    If I just decrease r to zero, then the point is the center of the circle. which is (1,2). The book seems to use (0,0) as the center. Also from the answer of the book, the value of the integral has [itex]cos(1)cosh(2)[/itex] which pretty much agree with me that the center is (1,2).
  5. Aug 6, 2010 #4


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    Yes, I think the book has a typo. They meant to say the integral is 2*pi*u(1,2)=2*cos(1)*cosh(2). You did show u(x,y)=cos(x)*cosh(y) is harmonic, right?
  6. Aug 6, 2010 #5
    Yes I have proofed that u(x,y)=cos(x)*cosh(y) is harmonic.

    I want to make sure you mean [itex]2\pi u(1,2) = 2\pi cos(1)cosh(2)[/itex]


    This is a brand new section added to the book on the last revision, that's what I think it's typo. But you never can be certain, these math text books seldom make error. Always have to be sure by posting here first.

  7. Aug 6, 2010 #6


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    Yes, 2*pi*u(1,2)=2*pi*cos(1)*cosh(2), ack I did make a mistake. I skipped the pi in the second part. Sorry. As I did, it's not that rare for math textbooks to contain errors.
  8. Aug 7, 2010 #7
    Thanks for your time.

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