Chem Differential Eq HW Solutions Fall 2011 54

Chem Differential Eq HW Solutions Fall 2011 54 - 54 Chapter...

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Unformatted text preview: 54 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates Solutions to Exercises 4.1 1. We could use Cartesian coordinates and compute ux , uy , uxx , and uyy directly from the definition of u. Instead, we will use polar coordinates, because the expression x2 + y 2 = r2, simplifies the denominator, and thus it is easier to take derivatives. In polar coordinates, u(x, y ) = x r cos θ cos θ = = = r−1 cos θ . x2 + y 2 r2 r So ur = −r−2 cos θ, urr = 2r−3 cos θ, uθ = −r−1 sin θ, uθθ = −r−1 cos θ. Plugging into (1), we find 2 1 1 2 cos θ cos θ cos θ u = urr + ur + 2 uθθ = − 3 − 3 = 0 (if r = 0). r r r3 r r If you used Cartesian coordinates, you should get uxx = 2x(x2 − 3y 2 ) (x2 + y 2 )2 and uyy = − 2x(x2 − 3y 2 ) . (x2 + y 2)2 5. In spherical coordinates: u(r, θ, φ) = r3 ⇒ urr = 6r, uθ = 0, uθθ = 0, uφφ = 0. Plugging into (3), we find 2 u= ∂ 2u 2 ∂u ∂u ∂ 2u 1 ∂ 2u + + cot θ +2 + csc2 θ 2 ∂r2 r ∂r r ∂θ2 ∂θ ∂φ = 6r + 6r = 12r. 9. (a) If u(r, θ, φ) depends only on r, then all partial derivatives of u with respect to θ and φ are 0. So (3) becomes 2 u= ∂ 2u 2 ∂u ∂ 2u 2 ∂u ∂u ∂ 2u 1 ∂ 2u = + + cot θ + +2 + csc2 θ . ∂r2 r ∂r r ∂θ2 ∂θ ∂φ2 ∂r2 r ∂r (b) If u(r, θ, φ) depends only on r and θ, then all partial derivatives of u with respect to φ are 0. So (3) becomes 2 u= ∂ 2 u 2 ∂u ∂u 1 ∂ 2u + + cot θ +2 . 2 2 ∂r r ∂r r ∂θ ∂θ ...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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