Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/208

 Hence

Now, since $$V_i$$ is a homogeneous function of negative degree $$i+1$$,

The first two terms therefore of the right hand member of equation (44) destroy each other, and, since $$V_i$$ satisfies Laplace’s equation, the third term is zero, so that $$H_i$$ also satisfies Laplace’s equation, and is therefore a solid harmonic of degree $$i$$.

We shall next shew that the value of $$H_i$$ thus derived from $$V_i$$ is of the most general form.

A homogeneous function of $$x, y, z$$ of degree $$i$$ contains

terms. But

is a homogeneous function of degree $$i-2$$, and therefore contains $$\tfrac{1}{2}i(i-1)$$ terms, and the condition $$\nabla^{2}H_{i}=0$$ requires that each of these must vanish. There are therefore $$\tfrac{1}{2}i(i-1)$$ equations between the coefficients of the $$\tfrac{1}{2}(i+1)(i+2)$$ terms of the homogeneous function, leaving $$2i+1$$ independent constants in the most general form of $$H_i$$.

But we have seen that $$V_i$$ has $$2i+1$$ independent constants, therefore the value of $$H_i$$ is of the most general form.

134.] The function $$V_i$$ satisfies the condition of vanishing at infinity, but does not satisfy the condition of being everywhere finite, for it becomes infinite at the origin.

The function $$H_i$$ satisfies the condition of being finite and continuous at finite distances from the origin, but does not satisfy the condition of vanishing at an infinite distance.

But if we determine a closed surface from the equation

and make $$H_i$$ the potential function within the closed surface and