Question

In the circular cylindrical coordinates $r,\theta ,z$ defined by $$x=r\cos \theta ,y=r\sin \theta ,z=z$$ Laplace's equation is $$u_{rr}+(1/r)u_r+\left(1/r^2\right)u_{\theta \theta }+u_{zz}=0$$ (a) Show that if $u(r,\theta ,z)=R(r)\mathrm{\Theta }(\theta )Z(z),$ then $R,\mathrm{\Theta },$ and $Z$ satisfy the ordinary differential equations $$\begin{array}{rl}{r}^2{R}^{\prime \prime }+r{R}^{\prime }+({\lambda }^2{r}^2-n^2)R& =0\\ {\mathrm{\Theta }}^{\prime \prime }+n^2\mathrm{\Theta }& =0\\ Z^{\prime \prime }-{\lambda }^{2}Z& =0\end{array}$$ (b) Show that if $u(r,\theta ,z)$ is independent of $\theta ,$ then the first equation in part (a) becomes $$r^2{R}^{\prime \prime }+r{R}^{\prime }+{\lambda }^2{r}^{2}R=0$$ the second is omitted altogether, and the third is unchanged.

Short answer

Answer: When u(r, θ, z) is independent of θ, the equation in r simplifies to: $$r^2{R}^{″}(r)+r{R}^{\prime }(r)+{\lambda }^2{r}^{2}R=0$$

Step-by-step solution

Express u as a product of functions

Take u(r, θ, z) as a product of functions of r, θ, and z: $$u(r,\theta ,z)=R(r)\mathrm{\Theta }(\theta )Z(z)$$

Compute the second derivatives

Compute the second derivatives of u(r, θ, z) with respect to r, θ, and z: $$\begin{gathered} u_{rr}=(R^{″}(r)\mathrm{\Theta }(\theta )Z(z)) \\ {u}_{\theta \theta }=(R(r){\mathrm{\Theta }}^{″}(\theta )Z(z)) \\ {u}_{zz}=(R(r)\mathrm{\Theta }(\theta )Z^{″}(z)) \end{gathered}$$

Plug the second derivatives into Laplace's equation

Substitute these second derivatives into Laplace's equation and divide by RΘZ: $$\frac{R^{″}(r)}{R(r)}+\frac{1}{r}\frac{R^{\prime }(r)}{R(r)}+\frac{{\mathrm{\Theta }}^{″}(\theta )}{r^2\mathrm{\Theta }(\theta )}+\frac{Z^{″}(z)}{Z(z)}=0$$

Separate the variables in the equation

Separate the variables r, θ, and z by setting the expression in r, θ, and z equal to a constant, respectively: $$\frac{Z^{″}(z)}{Z(z)}-{\lambda }^2=0$$ $$\frac{{\mathrm{\Theta }}^{″}(\theta )}{\mathrm{\Theta }(\theta )}+n^2=0$$ $$\frac{R^{″}(r)}{R(r)}+\frac{1}{r}\frac{R^{\prime }(r)}{R(r)}+\frac{{\lambda }^2{r}^2-n^2}{r^2}=0$$

Obtain ordinary differential equations

Rearrange the equation in r to get the ordinary differential equations: $$r^2{R}^{″}(r)+r{R}^{\prime }(r)+({\lambda }^2{r}^2-n^2)R=0$$ $${\mathrm{\Theta }}^{″}(\theta )+n^2\mathrm{\Theta }=0$$ $$Z^{″}(z)-{\lambda }^{2}Z=0$$

Address the case where u(r, θ, z) is independent of θ

When u(r, θ, z) is independent of θ, Θ becomes a constant, say Θ = 1. This way, the equation in θ disappears completely, and the equation in r simplifies as follows: $$r^2{R}^{″}(r)+r{R}^{\prime }(r)+{\lambda }^2{r}^{2}R=0$$ In this case, the equation in z remains unchanged.

Key concepts

Cylindrical Coordinates

Cylindrical coordinates represent a point in space by specifying its distance from a chosen reference axis, the angle with respect to a chosen reference direction on a chosen plane perpendicular to the axis, and the distance from the chosen plane to the point along the axis. In simpler terms, just like spherical coordinates are handy to solve problems involving spheres, cylindrical coordinates are ideal for dealing with cylindrical symmetry.

To understand how these coordinates come into play in Laplace's equation, visualize a scenario where you have electrical potential or temperature distribution around a wire or a tube. In defining cylindrical coordinates $r,\theta ,z$, each point in space is given by the radial distance $r$, the angular position $\theta$, and the height $z$, such that $x=r\cos \theta$, $y=r\sin \theta$, and $z=z$. These provide a more natural way to analyze our problem because they align with the symmetry of the physical scenario we might be modeling.

Separation of Variables

The technique of separation of variables is a powerful method for solving partial differential equations, such as Laplace's equation, which involve multiple variables. It hinges on the assumption that a multi-variable problem can be broken down into simpler, single-variable problems.

To give an analogy, imagine trying to solve a difficult puzzle—you might want to divide it into smaller sections, solve each one, and then put them all together. In mathematical terms, we look for solutions that can be expressed as the product of functions of each individual variable: $u(r,\theta ,z)=R(r)\mathrm{\Theta }(\theta )Z(z)$.

By using this method, we can transform a complex equation into a set of ordinary differential equations. Each equation then depends only on one variable and the original problem becomes more manageable. This is precisely what was done in the textbook solution, allowing the problem to segregate neatly into solvable bits.

Ordinary Differential Equations

An ordinary differential equation (ODE) is an equation involving functions of one independent variable and its derivatives. These equations describe numerous physical phenomena and are fundamental to understanding certain aspects of mathematics and physics.

Imagine an ODE as a recipe for how a certain quantity changes over time or space. Just as different recipes give rise to a variety of dishes, different ODEs describe a variety of dynamic systems. The ODEs shown in the exercise solutions are recipes instructing us how the functions $R(r)$, $\mathrm{\Theta }(\theta )$, and $Z(z)$ should behave.

Crucially, the simplicity of ODEs in comparison to partial differential equations like Laplace’s equation comes from the fact that they only involve derivatives with respect to one variable. The three ODEs derived using the separation of variables provide a straightforward path to find the solutions for the function $u(r,\theta ,z)$, given certain boundary conditions.