Chapter 16, Problem 16.16P

(a)

To determine

Prove that ${\nabla }^{2}f=\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(rf\right)$.

Answer to Problem 16.16P

Proved that ${\nabla }^{2}f=\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(rf\right)$.

Explanation of Solution

As $f\left(r\right)$ is spherically symmetric, so $r$ should satisfy $r=\sqrt{x^2+y^2+z^2}$, where $x,y,z$ are the three Cartesian coordinates.

Write the expression for ${\nabla }^{2}f$.

    ${\nabla }^{2}f=\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}+\frac{{\partial }^{2}f}{\partial z^2}$        (I)

Take double derivative of $f\left(r\right)$.

    $\begin{array}{c}\frac{{\partial }^2}{\partial x^2}\left[f\left(r\right)\right]=\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}f\left(r\right)\right)\\ =\frac{\partial }{\partial x}\left[f^{\prime }\left(r\right)\cdot \frac{\partial r}{\partial x}\right]\end{array}$        (II)

Use $\sqrt{x^2+y^2+z^2}$ for $r$ in the equation (II).

    $\begin{array}{c}\frac{{\partial }^2}{\partial x^2}\left[f\left(r\right)\right]=\frac{\partial }{\partial x}\left[f^{\prime }\left(r\right)\cdot \frac{\partial }{\partial x}\sqrt{x^2+y^2+z^2}\right]\\ =\frac{\partial }{\partial x}\left[f^{\prime }\left(r\right)\cdot \left(\frac{1}{2}\right)\sqrt{x^2+y^2+z^2}\left(2x\right)\right]\\ =\left[f^{\prime }\left(r\right)\frac{1}{\sqrt{x^2+y^2+z^2}}\frac{\partial x}{\partial x}\right]+\left[f^{\prime }\left(r\right)\frac{\partial }{\partial x}{\left(x^2+y^2+z^2\right)}^{\frac{1}{2}}x\right]\\ +\left[\frac{\partial }{\partial x}{f}^{\prime }\left(r\right){\left(x^2+y^2+z^2\right)}^{-\frac{1}{2}}x\right]\\ =\left[f^{\prime }\left(r\right)\frac{1}{\sqrt{x^2+y^2+z^2}}\right]+\left[f^{\prime }\left(r\right)\left(\frac{-1}{2}\right)\frac{\partial }{\partial x}{\left(x^2+y^2+z^2\right)}^{-\frac{3}{2}}2x\cdot x\right]\\ +\left[\frac{\partial }{\partial x}{f}^{″}\left(r\right){\left(x^2+y^2+z^2\right)}^{-\frac{1}{2}}x\right]\end{array}$        (III)

Solve equation (III).

    $\begin{array}{c}\frac{{\partial }^2}{\partial x^2}\left[f\left(r\right)\right]=\frac{f^{\prime }\left(r\right)}{r}-\frac{f^{\prime }\left(r\right)x^2}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}+\left[f^{″}\left(r\right)\left(\frac{1}{2}\right){\left(x^2+y^2+z^2\right)}^{-\frac{1}{2}}2x\cdot {\left(x^2+y^2+z^2\right)}^{-\frac{1}{2}}\cdot x\right]\\ =\frac{f^{\prime }\left(r\right)}{r}-\frac{f^{\prime }\left(r\right)x^2}{r^3}+\frac{f^{″}\left(r\right)x^2}{r^2}\end{array}$        (IV)

Similarly, second derivative of function with respect to $y$ can be written as,

    $\frac{{\partial }^2}{\partial y^2}\left[f\left(r\right)\right]=\frac{f^{\prime }\left(r\right)}{r}-\frac{f^{\prime }\left(r\right)y^2}{r^3}+\frac{f^{″}\left(r\right)y^2}{r^2}$        (V)

Similarly, second derivative of function with respect to $z$ can be written as,

    $\frac{{\partial }^2}{\partial z^2}\left[f\left(r\right)\right]=\frac{f^{\prime }\left(r\right)}{r}-\frac{f^{\prime }\left(r\right)z^2}{r^3}+\frac{f^{″}\left(r\right)z^2}{r^2}$        (VI)

Use equation (IV), (V), and (VI) in (I).

    $\begin{array}{c}{\nabla }^{2}f=\frac{f^{\prime }\left(r\right)}{r}-\frac{f^{\prime }\left(r\right)x^2}{r^3}+\frac{f^{″}\left(r\right)x^2}{r^2}+\frac{f^{\prime }\left(r\right)}{r}-\frac{f^{\prime }\left(r\right)y^2}{r^3}+\frac{f^{″}\left(r\right)y^2}{r^2}+\frac{f^{\prime }\left(r\right)}{r}-\frac{f^{\prime }\left(r\right)z^2}{r^3}+\frac{f^{″}\left(r\right)z^2}{r^2}\\ =3\frac{f^{\prime }\left(r\right)}{r}-\frac{f^{\prime }\left(r\right)\left[x^2+y^2+z^2\right]}{r^3}+\frac{f^{″}\left(r\right)\left[x^2+y^2+z^2\right]}{r^2}\\ =3\frac{f^{\prime }\left(r\right)}{r}-\frac{f^{\prime }\left(r\right)r^2}{r^3}+\frac{f^{″}\left(r\right)r^2}{r^2}\\ =2\frac{f^{\prime }\left(r\right)}{r}+f^{″}\left(r\right)\end{array}$         (VII)

Solve $\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(rf\right)$

    $\begin{array}{c}\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(rf\right)=\frac{1}{r}\frac{\partial }{\partial r}\left[\frac{\partial }{\partial x}\left(rf\right)\right]\\ =\frac{1}{r}\frac{\partial }{\partial r}\left[r{f}^{\prime }\left(r\right)+1\cdot f\left(r\right)\right]\\ =\frac{1}{r}\left[r\cdot f^{″}\left(r\right)+1\cdot f^{\prime }\left(r\right)+f^{\prime }\left(r\right)\right]\\ =\frac{1}{r}\left[r\cdot f^{″}\left(r\right)+2f^{\prime }\left(r\right)\right]\\ =\frac{2f^{\prime }\left(r\right)}{r}+f^{″}\left(r\right)\end{array}$        (VIII)

Equate equation (VII) and (VIII).

    ${\nabla }^{2}f=\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(rf\right)$        (IX)

Conclusion:

Therefore, ${\nabla }^{2}f=\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(rf\right)$.

(b)

To determine

Prove that ${\nabla }^{2}f=\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(rf\right)$ using the formula inside the back cover for ${\nabla }^2$ in spherical polar co-ordinates.

Answer to Problem 16.16P

Proved that ${\nabla }^{2}f=\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(rf\right)$ using the formula inside the back cover for ${\nabla }^2$ in spherical polar co-ordinates..

Explanation of Solution

Write the expression for ${\nabla }^{2}f$ in spherical polar co-ordinates.

    ${\nabla }^{2}f=\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(rf\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial f}{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^{2}f}{\partial {\varphi }^2}$        (X)

$f$ is a function of $r$ only, so the second term and third term will vanishes. Hence the equation (XI) becomes,

    $\begin{array}{c}{\nabla }^{2}f=\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(rf\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \times 0\right)+\frac{1}{r^2{\sin }^2\theta }\times 0\\ =\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(rf\right)\end{array}$        (XI)

Conclusion:

Therefore, it is proved that ${\nabla }^{2}f=\frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(rf\right)$ using the formula inside the back cover for ${\nabla }^2$ in spherical polar co-ordinates.