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# Our product solutions are of the form u nm x y t a nm

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Our product solutions are of the form u nm ( x, y, t ) = ( a nm cos n 2 + m 2 πt + b nm sin n 2 + m 2 πt ) sin nπx sin nπy. The general solution to the PDE and boundary conditions coming from the superposition of these solutions is u ( x, y, t ) = n,m =1 ( a nm cos n 2 + m 2 πt + b nm sin n 2 + m 2 πt ) sin nπx sin nπy. We have to satisfy the initial conditions. Setting t = 0 and equating to u ( x, y, 0), we get sin( πx ) sin(2 πy ) + 3 sin(5 πx ) sin( πy ) = n,m =1 a nm sin nπx sin nπ, from which we get a 12 = 1, a 51 = 3, all other a nm = 0. Differentiating with respect to t and equating to u t ( x, y, 0), - sin(3 πx ) sin(3 πy ) = n,m =1 b nm n 2 + m 2 π sin nπx sin from which b 33 = - 1 / ( π 18), all other b nm = 0. The solution is u ( x, t ) = cos( 5 πt ) sin( πx ) sin(2 πy ) + 3 cos( 26 πt ) sin(3 πx ) sin( πy ) - 1 π 18 sin( 18 πt ) sin(3 πx ) sin(3 πy ). Grading Note: There being NO excuse for not doing this exercise correctly and completely, I graded: 50 points for doing it correctly (or almost correctly). That included showing how the separation of variables was done, and showing all work. 30 points for doing something reasonable. 0 points for doing nothing or almost nothing. A few grades fell between these parameters. 2. (10 points) Show that u ( r, θ, t ) = J 2 ( ξr ) cos(2 θ ) sin( ξt ), where ξ is one of the positive zeroes of the Bessel function J 2 , satisfies u tt ( r, θ, t ) = u rr ( r, θ, t ) + 1 r u r ( r, θ, t ) + 1 r 2 u θθ ( r, θ, t ) , 0 r < 1 , - π θ π, t > 0 , u (1 , θ, t ) = 0 , - π θ π, t > 0 .

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3 This is really a calculus exercise. As a hint, recall that the equation satisfied by J 2 ( z ) is z 2 J ′′ 2 ( z ) + zJ 2 ( z ) + ( z 2 - 4) J 2 ( z ) = 0 so if by some miracle you happen to have somewhere a factor of the form ξ 2 J ′′ 2 ( ξr ) + ξ r J ( ξr ) - 4 r 2 J 2 ( ξr ); that is, ( ( ) 2 J ′′ 2 ( ξr ) + ( ) J ( ξr ) - 4 J 2 ( ξr ) ) 1 r 2 , it will equal (can be replaced by) - ξ 2 J 2 ( ). Or vice-versa. 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
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