em2sp13e2s

# Cos ξt ξ 2 j 2 ξr cos2 θ sin ξt u r r θ t ξj 2

• Notes
• 5

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(cos( ξt )) = - ξ 2 J 2 ( ξr ) cos(2 θ ) sin( ξt ) , ∂u ∂r ( r, θ, t ) = ξJ 2 ( ξr ) cos(2 θ ) sin( ξt ) , 2 u ∂r 2 ( r, θ, t ) = ξ 2 J ′′ 2 ( ξr ) cos(2 θ ) sin( ξt ) , 2 u ∂θ 2 ( r, θ, t ) = - 4 J 2 ( ξr ) cos(2 θ ) sin( ξt ) thus (from the last three equalities) u rr ( r, θ, t ) + 1 r u r ( r, θ, t ) + 1 r 2 u θθ ( r, θ, t ) = ( ξ 2 J ′′ 2 ( ξr ) + ξ r J 2 ( ) - 4 r 2 J 2 ( ξr ) ) cos(2 θ ) sin( ξt ) and proving that u tt = u rr + r 1 u r + r 2 u θθ reduces to showing that - ξ 2 J 2 ( ξr ) cos(2 θ ) sin( ξt ) = ( ξ 2 J ′′ 2 ( ξr ) + ξ r J 2 ( ) - 4 r 2 J 2 ( ξr ) ) cos(2 θ ) sin( ξt ) . Cancelling cos(2 θ ) sin( ξt ), this is equivalent to showing that - ξ 2 J 2 ( ξr ) = ξ 2 J ′′ 2 ( ξr ) + ξ r J 2 ( ) - 4 r 2 J 2 ( ξr ); equivalently that ( ) 2 J ′′ 2 ( ξr ) + rξJ 2 ( ) + (( ) 2 - 4) J 2 ( ξr ) = 0 . This last equation is just the defining ODE for J 2 , evaluated at z = ; thus it holds. Finally, u (1 , θ, t ) = J 2 ( ξ ) cos(2 θ ) sin( ξt ) = 0 since J 2 ( ξ ) = 0. 3. (20 points) Solve by separation of polar variables u ( x, y ) = 0 , x 2 + y 2 < 1 , u ( x, y ) = x + y, x 2 + y 2 = 1 . This problem is not formulated in the way the text trained you. It is the same problem as u rr + 1 r u r + 1 r 2 u θθ = 0 , 0 r < 1 , - π θ π, u (1 , θ ) = cos θ + sin θ, - π θ π Solve the second problem (which is the same as the first). It is OK to leave the solution in the form u ( r, θ ) = function of r and θ . But 5 bonus points will be given for returning it to cartesian coordinates. Will this be an aha! moment?

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4 Solution. We look for all nonzero solutions of the form u ( r, θ ) = R ( r )Θ( θ ). They will have to satisfy R ′′ Θ + 1 r R Θ + 1 r 2 R Θ ′′ = 0; Multiplying by r 2 / ( R Θ) and separating variables we get r 2 R ′′ R + rR R = - Θ ′′ Θ = λ, for constant λ . We have the equations Θ ′′ + λ Θ = 0 , r 2 R ′′ + rR - λR = 0 . For geometry reasons, the solutions of the first equation must be periodic of period 2 π in θ ; this forces λ = n 2 , n = 0 , 1 , 2 , 3 , . . . and the corresponding solutions are Θ 0 ( θ ) = 1 2 a 0 , a 0 a constant , Θ n ( θ ) = a n cos + b n sin nθ, constants a n , b n , n = 1 , 2 , 3 , . . . . Looking at the equation for R , with λ = n 2 it has the form r 2 R ′′ + rR - n 2 R = 0 .
• Spring '13
• Schonbek
• Sin, Boundary value problem, J2

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