# 29 a if we replace r by y this becomes x2 n2 y 0 145 y

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Unformatted text preview: that ∂2u ∂ ∂u ∂u ∂ ∂ ∂u ( )= cos θ + sin θ = cos θ = 2 ∂r ∂r ∂r ∂r ∂x ∂y ∂r = cos2 θ ∂u ∂x + sin θ ∂ ∂r ∂u ∂y ∂2u ∂2u ∂2u + 2 cos θ sin θ + sin2 θ 2 . ∂x2 ∂x∂y ∂y Similarly, ∂2u ∂ ∂u ∂u ∂ ∂u ( )= −r sin θ + r cos θ = ∂θ2 ∂θ ∂θ ∂θ ∂x ∂y = −r sin θ ∂ ∂θ = r2 sin2 θ ∂u ∂x + r cos θ ∂ ∂θ ∂u ∂y − r cos θ ∂u ∂u − r sin θ ∂x ∂y ∂2u ∂2u ∂2u ∂u − 2r2 cos θ sin θ + r2 cos2 θ 2 − r , 2 ∂x ∂x∂y ∂y ∂r which yields 1 ∂2u ∂2u ∂2u ∂ 2 u 1 ∂u + cos2 θ 2 − . = sin2 θ 2 − 2 cos θ sin θ r2 ∂θ2 ∂x ∂x∂y ∂y r ∂r Adding these results together, we obtain ∂2u 1 ∂2u ∂ 2 u ∂ 2 u 1 ∂u + 2 2= + 2− , 2 ∂r r ∂θ ∂x2 ∂y r ∂r or equivalently, ∆u = ∂2u 1 ∂ 2 u 1 ∂u . + 2 2+ 2 ∂r r ∂θ r ∂r 137 Finally, we can write this result in the form ∆u = 1∂ r ∂r r ∂u ∂r + 1 ∂2u . r2 ∂θ2 (5.24) This formula for the Laplace operator, together wit...
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