Integrals of normal derivatives

[Benny]

Hi can someone please help me out with the following questions?

Supposed that $\phi$ is a C^2 function which satisfies Laplace's equation $\nabla ^2 \phi = \nabla \bullet \nabla \phi = 0$ everywhere in a region V bounded by a closed surface S.

a) If $$\frac{{\partial \phi }}{{\partial n}} = \mathop n\limits^ \to \bullet \nabla \phi$$ denotes the direction derivative of $\phi$ in the direction of the outward unit normal n, establish the following.

$$\int\limits_{}^{} {\int\limits_S^{} {\frac{{\partial \phi }}{{\partial n}}} \partial S = 0}$$

$$\int\limits_{}^{} {\int\limits_S^{} {\phi \frac{{\partial \phi }}{{\partial n}}dS = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_V^{} {\left\| {\nabla \phi } \right\|^2 } } } } } dV$$

b) If $$\frac{{\partial \phi }}{{\partial n}}$$ vanishes everywhere on S, deduce from the second result that $\phi$ must be constant in V.

c) If $\phi$ vanishes everywhere on S, deduce from the second result that $\phi$ must vanish everywhere in V.

Part a: Phi is a function of three variables, otherwise the question wouldn't really make sense. I tried to visualise what the statement meant so I tried some specific cases. If S is a hemisphere then V is a half unit ball. The gradient of phi is pendicular to level sets of phi so in this case the gradient of phi is a normal to V (since phi is a function of three variables). So in this case, the unit normal n, to S, is parallel to grad(phi). So how can the integral be zero? The integrand is non-zero. I'm not sure what to do here.

For the second result I'm fairly sure that I need to apply the divergence theorem. But I don't really know how to 'div' the integrand of the surface integral because I don't what the normal, n, is. I think I might need to use a general property somewhere here but again I really don't know what to do.

Part b: If the partial derivative with respect to n is zero on S then from the second result, the triple integral on the RHS must be equal to zero everywhere in V. This can only be the case if the integrand (grad(phi))^2 is identically zero. So phi is constant in V? Not sure.

Part c: I can't think of a way to do this.

ANy help would be good thanks.

 

[kurt.physics]

a) For the first result, we use the Divergence Theorem. Let V be a region with boundary S and let \phi be a function that satisfies Laplace's equation. Then, \int\limits_{}^{} {\int\limits_S^{} {\frac{{\partial \phi }}{{\partial n}}} \partial S = \int\limits_{}^{} {\int\limits_V^{} {\nabla \bullet \nabla \phi } } dV = 0} since \nabla ^2 \phi = 0 everywhere in V. For the second result, we again use the Divergence Theorem. Let V be a region with boundary S and let \phi be a function that satisfies Laplace's equation. Then, \int\limits_{}^{} {\int\limits_S^{} {\phi \frac{{\partial \phi }}{{\partial n}}dS = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_V^{} { \nabla \phi \bullet \nabla \phi } } } } } dV = \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_V^{} {\left\| {\nabla \phi } \right\|^2 } } } dVb) If \frac{{\partial \phi }}{{\partial n}} vanishes everywhere on S, then from the second result we have \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_V^{} {\left\| {\nabla \phi } \right\|^2 } } } dV = 0This implies that \nabla \phi = 0 everywhere in V. Hence, \phi must be constant in V.c) If \phi vanishes everywhere on S, then from the second result we have \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_V^{} {\left\| {\nabla \phi } \right\|^2 } } } dV = 0