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Chem Differential Eq HW Solutions Fall 2011 54

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

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54 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates Solutions to Exercises 4.1 1. We could use Cartesian coordinates and compute u x , u y , u xx , and u yy directly from the definition of u . Instead, we will use polar coordinates, because the expression x 2 + y 2 = r 2 , simplifies the denominator, and thus it is easier to take derivatives. In polar coordinates, u ( x, y ) = x x 2 + y 2 = r cos θ r 2 = cos θ r = r - 1 cos θ . So u r = - r - 2 cos θ, u rr = 2 r - 3 cos θ, u θ = - r - 1 sin θ, u θθ = - r - 1 cos θ. Plugging into (1), we find 2 u = u rr + 1 r u r + 1 r 2 u θθ = 2 cos θ r 3 - cos θ r 3 - cos θ r 3 = 0 (if r = 0) . If you used Cartesian coordinates, you should get u xx = 2 x ( x 2 - 3 y 2 ) ( x 2 + y 2 ) 2 and u yy = - 2 x ( x 2 - 3 y 2 ) ( x 2 + y 2 ) 2 . 5. In spherical coordinates: u ( r, θ, φ ) = r 3 u rr = 6 r, u θ = 0 , u θθ = 0 , u φφ = 0 . Plugging into (3), we find 2 u = 2 u ∂r 2 + 2
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