Also notice that y y and y y are both less than zero Then G x 1 2 \u03c0 x x x x 2 y

Also notice that y y and y y are both less than zero

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Also, notice that ( y < - y > ) and - ( y < + y > ) are both less than zero. Then ∂G ∂x = - 1 2 π x - x 0 ( x - x 0 ) 2 + ( y < - y > ) 2 - x - x 0 ( x - x 0 ) 2 + ( y < + y > ) 2 = G = 1 2 π Z ( x - x 0 ) dx ( x - x 0 ) 2 + ( y < + y > ) 2 - Z ( x - x 0 ) dx ( x - x 0 ) 2 + ( y < - y > ) 2 = 1 4 π Z d (( x - x 0 ) 2 + ( y < + y > ) 2 ) ( x - x 0 ) 2 + ( y < + y > ) 2 - Z d (( x - x 0 ) 2 + ( y < - y > ) 2 ) ( x - x 0 ) 2 + ( y < - y > ) 2 = 1 4 π log ( x - x 0 ) 2 + ( y < + y > ) 2 ( x - x 0 ) 2 + ( y < - y > ) 2 Therefore, G = 1 4 π log ( x - x 0 ) 2 + ( y < + y > ) 2 ( x - x 0 ) 2 + ( y < - y > ) 2 (33) = G = 1 4 π log ( x - x 0 ) 2 + ( y + y 0 ) 2 ( x - x 0 ) 2 + ( y - y 0 ) 2 (34) = - 1 2 π log p ( x - x 0 ) 2 + ( y - y 0 ) 2 - log p ( x - x 0 ) 2 + ( y + y 0 ) 2 (35) = G = - 1 2 π log | r - r 0 | - 1 2 π log | r - r 00 | (36)
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, which is obtained from the method of images with r = ( x, y ) , r 0 = ( x 0 , y 0 ), and r 00 = ( x 0 , - y 0 ). Therefore, the Green’s function from the equation (33) is equivalent to the one obtained from the method of images. Note that: the Green’s function from the Equation (34) works for either case y < y 0 or y > y 0 . (Problem 6) By taking a Fourier transform in x , solve 2 u = 0 - ∞ < x < , 0 < y < a with u ( x, 0) = f ( x ) and u ( x, a ) = g ( x ). Answer: From the given Laplace equation, we have u xx + u yy = 0 = F u xx ; x ξ ] + F u yy ; x ξ ] = 0 = ⇒ - ξ 2 ¯ u + ¯ u yy = 0 = ¯ u = Ae ξy + Be - ξy Notice that, we cannot take the Fourier inverse of ¯ u now since A and B can be functions of ξ . In order to find A ( ξ ) and B ( ξ ), we first take the Fourier transform of the boundary conditions such that F u ( x, 0); x ξ = 1 2 π Z -∞ f ( x ) e iξx dx = ¯ f ( ξ ) = ¯ u | y =0 = A + B , and F u ( x, a ); x ξ = 1 2 π Z -∞ g ( x ) e iξx dx = ¯ g ( ξ ) = ¯ u | y = a = Ae ξa + Be - ξa = ( A = ¯ f ( ξ ) - B ( ¯ f ( ξ ) - B ) e ξa + Be - ξa = ¯ g ( ξ ) = ( A = ¯ f ( ξ ) - B ¯ f ( ξ ) e ξa - 2 B sinh( ξa ) = ¯ g ( ξ ) = ( A = - ¯ f ( ξ ) e - ξa g ( ξ ) 2 sinh( ξa ) B = ¯ f ( ξ ) e ξa - ¯ g ( ξ ) 2 sinh( ξa ) = ¯ u = - ¯ f ( ξ ) e - ξa + ¯ g ( ξ ) 2 sinh( ξa ) e ξy + ¯ f ( ξ ) e ξa - ¯ g ( ξ ) 2 sinh( ξa ) e - ξy = ¯ u = 1 2 sinh( ξa ) ¯ f ( ξ )( e ( a - y ) ξ - e - ( a - y ) ξ ) + 2¯ g ( ξ ) sinh( ξy ) = ¯ u = sinh[( a - y ) ξ ] sinh( ξa ) ¯ f ( ξ ) + sinh( ) sinh( ξa ) ¯ g ( ξ ) = u = F - 1 ¯ u ; x ξ = F - 1 sinh[( a - y ) ξ ] sinh( ) ¯ f ( ξ ) + sinh( ) sinh( ξa ) ¯ g ( ξ ); ξ x (37) = u = F - 1 ¯ h 1 ( ξ ) ¯ f ( ξ ) + ¯ h 2 ( ξ g ( ξ ); ξ x (38)
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, where ¯ h 1 ( ξ ) = sinh[( a - y ) ξ ] sinh( ) , and ¯ h 2 ( ξ ) = sinh( ) sinh( ) , whereby h 1 ( x ) = F - 1 ¯ h 1 ( ξ ); ξ x = r 1 2 π Z -∞ sinh[( a - y ) ξ ] sinh( ) e - ixξ = r 1 2 π Z -∞ sinh[( a - y ) ξ ] sinh( ) cos( ) + i r 1 2 π Z -∞ sinh[( a - y ) ξ ] sinh( ) sin( ) = 2 2 π Z 0 sinh[( a - y ) ξ ] sinh( ) cos( ) = 2 2 π π 2 a sin π ( a - y ) a cosh πx a + cos π ( a - y ) a = h 1 ( x ) = π a 2 sin π ( a - y ) a cosh πx a + cos π ( a - y ) a , and h 2 ( x ) = F - 1 ¯ h 1 ( ξ ); ξ x = r 1 2 π Z -∞ sinh( ) sinh( ) e - ixξ = 2 2 π Z 0 sinh( ) sinh( ) cos( ) = 2 2 π π 2 a sin( π a y ) cosh πx a + cos( π a y ) = h 2 ( x ) = π a 2 sin( π a y ) cosh πx a + cos( π a y ) Now applying the convolution theorem to the Equation (38), then we have u ( x, y ) = 1 2 π Z -∞ dx 0 f ( x 0 ) h 1 ( x - x 0 ) + g ( x 0 ) h 2 ( x - x 0 ) = 1 2 π Z -∞ dx 0 ( f ( x 0 ) π a 2 sin π ( a - y ) a cosh h π a ( x - x 0 ) i + cos π ( a - y ) a + g ( x 0 ) π a 2 sin( π a y ) cosh h π a ( x - x 0 ) i + cos( π a y ) ) Therefore, u ( x, y ) = 1 2 a Z -∞ dx 0 ( f ( x 0 ) sin π ( a - y ) a cosh h π a ( x - x 0 ) i + cos π ( a - y ) a + g ( x 0 ) sin( π a y ) cosh h π a ( x - x 0 ) i + cos( π a y ) )
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  • Prof. B.A. Shadwick
  • Theoretical Physics, Cos, Trigraph, Emoticon, R0, rn rn

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