Chapter 10, Problem 62P

To determine

The laplace equation of velocity potential function in spherical polar coordinate.

Answer to Problem 62P

The laplace equation of velocity potential function in spherical polar coordinate is $\left(\frac{1}{{\cos }^2\theta {\sin }^2\theta }\right)\left[\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)+\frac{1}{r^2}\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)\right]=0$.

Explanation of Solution

Given information:

The distance from origin is $r$ and angle of inclination between the radial vector and axis of rotation is $\theta$.

The radial vector about the coordinate axis is shown in Figure (1).

  

  Figure-(1)

Write the expression for component of vector $r$ along the x-axis.

  $x=r\cos \theta$   ...... (I)

Write the expression for component of vector $r$ along the y-axis.

  $y=r\sin \theta$   ...... (II)

Write the expression for Laplace equation for the velocity potential function Cartesian coordinate.

  $\begin{array}{l}{\nabla }^2\varphi =0\\ \frac{{\partial }^2\varphi }{\partial x^2}+\frac{{\partial }^2\varphi }{\partial y^2}=0\end{array}$   ...... (III)

Write expression for Laplace equation as function of $r$ and $\theta$.

  $\begin{array}{l}\left[\frac{\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)}{\left(\frac{\partial x^2}{\partial r^2}\right)}+\frac{\left(\frac{{\partial }^2\varphi }{\partial {\theta }^2}\right)}{\left(\frac{\partial x^2}{\partial {\theta }^2}\right)}\right]+\left[\frac{\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)}{\left(\frac{\partial y^2}{\partial r^2}\right)}+\frac{\left(\frac{{\partial }^2\varphi }{\partial {\theta }^2}\right)}{\left(\frac{\partial y^2}{\partial {\theta }^2}\right)}\right]=0\\ \left[\frac{\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)}{{\left(\frac{\partial x}{\partial r}\right)}^2}+\frac{\left(\frac{{\partial }^2\varphi }{\partial {\theta }^2}\right)}{{\left(\frac{\partial x}{\partial \theta }\right)}^2}\right]+\left[\frac{\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)}{{\left(\frac{\partial y}{\partial r}\right)}^2}+\frac{\left(\frac{{\partial }^2\varphi }{\partial {\theta }^2}\right)}{{\left(\frac{\partial y}{\partial \theta }\right)}^2}\right]=0\end{array}$   ...... (IV)

Differentiate Equation (I) with respect to $r$.

  $\begin{array}{c}\frac{\partial x}{\partial r}=\frac{\partial }{\partial r}\left(r\cos \theta \right)\\ =\cos \theta \end{array}$   ...... (V)

Differentiate Equation (I) with respect to $\theta$.

  $\begin{array}{c}\frac{\partial x}{\partial \theta }=\frac{\partial }{\partial \theta }\left(r\cos \theta \right)\\ =-r\sin \theta \end{array}$   ...... (VI)

Differentiate Equation (II) with respect to $r$.

  $\begin{array}{c}\frac{\partial y}{\partial r}=\frac{\partial }{\partial r}\left(r\sin \theta \right)\\ =\sin \theta \end{array}$   ...... (VII)

Differentiate Equation (II) with respect to $\theta$.

  $\begin{array}{c}\frac{\partial y}{\partial \theta }=\frac{\partial }{\partial \theta }\left(r\sin \theta \right)\\ =r\cos \theta \end{array}$   ...... (VIII)

Substitute $\cos \theta$ for $\frac{\partial x}{\partial r}$, $-r\sin \theta$ for $\frac{\partial x}{\partial \theta }$, $\sin \theta$ for $\frac{\partial y}{\partial r}$ and $r\cos \theta$ for $\frac{\partial y}{\partial \theta }$ in Equation (IV).

  $\begin{array}{c}\frac{\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)}{{\left(\cos \theta \right)}^2}+\frac{\left(\frac{{\partial }^2\varphi }{\partial {\theta }^2}\right)}{{\left(-r\sin \theta \right)}^2}+\frac{\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)}{{\left(\sin \theta \right)}^2}+\frac{\left(\frac{{\partial }^2\varphi }{\partial {\theta }^2}\right)}{{\left(r\cos \theta \right)}^2}=0\\ \left(\frac{{\partial }^2\varphi }{\partial r^2}\right)\left(\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }\right)+\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)\frac{1}{r^2}\left(\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }\right)=0\\ \left(\frac{{\partial }^2\varphi }{\partial r^2}\right)\left(\frac{1}{{\cos }^2\theta {\sin }^2\theta }\right)+\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)\frac{1}{r^2}\left(\frac{1}{{\cos }^2\theta {\sin }^2\theta }\right)=0\\ \left(\frac{1}{{\cos }^2\theta {\sin }^2\theta }\right)\left[\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)+\frac{1}{r^2}\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)\right]=0\end{array}$

Conclusion:

The laplace equation of velocity potential function in spherical polar coordinate is $\left(\frac{1}{{\cos }^2\theta {\sin }^2\theta }\right)\left[\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)+\frac{1}{r^2}\left(\frac{{\partial }^2\varphi }{\partial r^2}\right)\right]=0$.