Let u=f(x,y) where x=rcosθ and y=rsinθ. How to show that ∂²u/∂x²+∂²u/∂y² = ∂²u/∂r²+(1/r)∂u/∂r+(1/r²)∂²u/∂θ²
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HOME > > Let u=f(x,y) where x=rcosθ and y=rsinθ. How to show that ∂²u/∂x²+∂²u/∂y² = ∂²u/∂r²+(1/r)∂u/∂r+(1/r²)∂²u/∂θ²

Let u=f(x,y) where x=rcosθ and y=rsinθ. How to show that ∂²u/∂x²+∂²u/∂y² = ∂²u/∂r²+(1/r)∂u/∂r+(1/r²)∂²u/∂θ²

[From: ] [author: ] [Date: 11-12-08] [Hit: ]
............
∂²u/∂x² + ∂²u/∂y² = ∂²u/∂r² + (1/r)∂u/∂r + (1/r²)∂²u/∂θ²

Thanks for the help

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By the Chain Rule,
∂u/∂r = ∂u/∂x ∂x/∂r + ∂u/∂y ∂y/∂r = (cos θ) ∂u/∂x + (sin θ) ∂u/∂y

∂²u/∂r² = (∂/∂r) [(cos θ) ∂u/∂x + (sin θ) ∂u/∂y]
..........= (cos θ) (∂/∂r) ∂u/∂x + (sin θ) (∂/∂r) ∂u/∂y
..........= cos θ [(cos θ) ∂²u/∂x² + (sin θ) ∂²u/∂x∂y] + sin θ [(cos θ) ∂²u/∂y∂x + (sin θ) ∂²u/∂y²]
..........= (cos² θ) ∂²u/∂x² + (2 sin θ cos θ) ∂²u/∂x∂y + (sin² θ) ∂²u/∂y².

∂u/∂θ = ∂u/∂x ∂x/∂θ + ∂u/∂y ∂y/∂θ = (-r sin θ) ∂u/∂x + (r cos θ) ∂u/∂y

∂²u/∂θ² = (∂/∂θ) [(-r sin θ) ∂u/∂x + (r cos θ) ∂u/∂y]
...........= [(-r cos θ) ∂u/∂x + (-r sin θ) (∂/∂θ)∂u/∂x] + [(-r sin θ) ∂u/∂y + (r cos θ) (∂/∂θ)∂u/∂y]
...........= (-r cos θ) ∂u/∂x + (-r sin θ) [(-r sin θ) ∂²u/∂x² + (r cos θ) ∂²u/∂x∂y] + (-r sin θ) ∂u/∂y
+ (r cos θ) [(-r sin θ) ∂²u/∂x∂y + (r cos θ) ∂²u/∂y²]
...........= (-r cos θ) ∂u/∂x + (r² sin² θ) ∂²u/∂x² - (2r² sin θ cos θ) ∂²u/∂x∂y + (-r sin θ) ∂u/∂y
+ (r² cos² θ) ∂²u/∂y²

Therefore, ∂²u/∂r² + (1/r) ∂u/∂r + (1/r²) ∂²u/∂θ²
= [(cos² θ) ∂²u/∂x² + (2 sin θ cos θ) ∂²u/∂x∂y + (sin² θ) ∂²u/∂y²]
+ (1/r) [(cos θ) ∂u/∂x + (sin θ) ∂u/∂y]
+ (1/r²) [(-r cos θ) ∂u/∂x + (r² sin² θ) ∂²u/∂x² - (2r² sin θ cos θ) ∂²u/∂x∂y + (-r sin θ) ∂u/∂y
+ (r² cos² θ) ∂²u/∂y²]

= [(cos² θ) ∂²u/∂x² + (2 sin θ cos θ) ∂²u/∂x∂y + (sin² θ) ∂²u/∂y²]
+ (1/r)(cos θ) ∂u/∂x + (1/r)(sin θ) ∂u/∂y]
+ [((-1/r) cos θ) ∂u/∂x + (sin² θ) ∂²u/∂x² - (2 sin θ cos θ) ∂²u/∂x∂y - ((1/r) sin θ) ∂u/∂y + (cos² θ) ∂²u/∂y²]

= (cos² θ + sin² θ) ∂²u/∂x² + (sin² θ + cos² θ) ∂²u/∂y²
= ∂²u/∂x² + ∂²u/∂y².

I hope this helps!
1
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