em2sp13e2s

# Solution as mentioned this is a calculus exercise

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Solution. As mentioned, this is a Calculus exercise. Suppose u ( r,θ,t ) = J 2 ( ξr ) cos(2 θ ) sin( ξt ). By a celebrated calculus result that goes by the name of chain rule , ∂u ∂t ( r,θ,t ) = ∂t ( J 2 ( ξr ) cos(2 θ ) sin( ξt )) = J 2 ( ξr ) cos(2 θ ) ∂t (sin( ξt )) = J 2 ( ξr ) cos(2 θ ) ξ cos( ξt ) = ξJ 2 ( ξr ) cos(2 θ ) cos( ξt ) , 2 u ∂t 2 ( r,θ,t ) = ξJ 2 ( ξr ) cos(2 θ ) ∂t (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 ) ² 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 ) ² 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 ) = 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
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Solution As mentioned this is a Calculus exercise Suppose u...

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