Paper 3, Section II, B

Vector Calculus | Part IA, 2009

Let SS be a bounded region of R2\mathbb{R}^{2} and S\partial S be its boundary. Let uu be the unique solution to Laplace's equation in SS, subject to the boundary condition u=fu=f on S\partial S, where ff is a specified function. Let ww be any smooth function with w=fw=f on S\partial S. By writing w=u+δw=u+\delta, or otherwise, show that

Sw2 dASu2 dA\int_{S}|\nabla w|^{2} \mathrm{~d} A \geqslant \int_{S}|\nabla u|^{2} \mathrm{~d} A

Let SS be the unit disc in R2\mathbb{R}^{2}. By considering functions of the form g(r)cosθg(r) \cos \theta on both sides of ()(*), where rr and θ\theta are polar coordinates, deduce that

01(r(dg dr)2+g2r)dr1\int_{0}^{1}\left(r\left(\frac{\mathrm{d} g}{\mathrm{~d} r}\right)^{2}+\frac{g^{2}}{r}\right) \mathrm{d} r \geqslant 1

for any differentiable function g(r)g(r) satisfying g(1)=1g(1)=1 and for which the integral converges at r=0r=0.

[f(r,θ)=(fr,1rfθ),2f(r,θ)=1rr(rfr)+1r22fθ2]\left[\nabla f(r, \theta)=\left(\frac{\partial f}{\partial r}, \frac{1}{r} \frac{\partial f}{\partial \theta}\right), \quad \nabla^{2} f(r, \theta)=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial f}{\partial r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} f}{\partial \theta^{2}} \cdot\right]

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