# Laplace transform in spherical coordinates

• A
• lvotapka
In summary, Michael Wertheim's 1963 paper uses a one-sided Laplace transform in spherical coordinates to obtain a final equation (Eq. 3) that involves a convolution property and a special math trick. The derivation process also involves converting the Laplace transform to spherical coordinates, integrating over the angles, and using a Taylor expansion or integration by parts.
lvotapka
TL;DR Summary
A 1963 paper by Michael Wertheim uses a Laplace transform in spherical coordinates. How is the resulting equation obtained?
Summary: A 1963 paper by Michael Wertheim uses a Laplace transform in spherical coordinates. How is the resulting equation obtained?

In 1963, Michael Wertheim published a paper (relevant page attached here), where he presented the following equation (Eq. 1):

$$y(\bar{r}) = 1 + n \int_{|\bar{r}'|<R} y(\bar{r}') d\bar{r}' - n \int_{|\bar{r}'|<R, |\bar{r} - \bar{r}'|<R} y(\bar{r}') y(\bar{r} - \bar{r}') d\bar{r}'$$

Where ##\bar{r}## and ##\bar{r}'## are 3-vectors (x, y, z) and ##y(\bar{r})## is a function of one of those vectors. R and n are constants. Note that ##\bar{r}## and ##\bar{r}'## are two different vector variables, the prime does NOT indicate a derivative. He then says he uses a one-sided Laplace transform to obtain the following equation (Eq. 3):

$$t[F(t) + G(t)] = t^{-1}[1 + 24 \eta K ] - 12 \eta [F(-t) - F(t)]G(t)$$

where

$$F(t) = R^{-2} \int^R_0 ry(r) exp(-sr) dr$$
$$G(t) = R^{-2} \int^\infty_R ry(r) exp(-sr) dr$$
$$K = R^{-3} \int^R_0 r^2 y(r) dr$$
$$\eta = \dfrac{1}{6} \pi R^3 n$$
$$t=sR$$

A 3D Laplace transform in spherical coordinates seems rare enough that it is difficult to find resources to reach the latter equations from the first one. I've been able to convert the Laplace transform to spherical coordinates, integrate over the angles, and obtain some equations similar to the latter equations, but not exactly. Clearly, the convolution property of Laplace transforms are used to obtain the RHS of Eq. 3. Any further suggestions or guidance that anyone could provide would be much appreciated.

#### Attachments

• wertheim1963_pg1.pdf
95.3 KB · Views: 369
Last edited:
Have you tried to use a Taylor expansion to prove it, or simply integration by parts?

Thanks for the response. I haven't tried either of those approaches. I'll try to think about how they might help. If you have any suggestions for moving forward using those, I'd be interested to hear them.

I've thought some more about it, and I think I'm pretty close, but still uncertain about the last step...

First of all, I can get the Laplace transform of ##y(\bar{r})## and get something similar to F(t) + G(t) on the LHS of the equation:

$$\int_0^{\infty} dx \int_0^{\infty} dy \int_0^{\infty} dz y(\bar{r})exp(-\bar{r}\cdot\bar{s})$$

When converted to spherical coordinates:

$$\int_0^{\infty} dr \int_0^{\frac{\pi}{2}} d\theta sin(\theta) r^2 y(r)exp(-rs\cdot cos(\theta)) \int_0^{\frac{\pi}{2}} d\phi$$

Then using u-substitution and integrating over phi and theta:

$$-\frac{\pi}{2}\int_0^{\infty} dr \frac{r}{s} y(r) + \frac{\pi}{2}\int_0^{\infty} dr \frac{r}{s} exp(-sr) y(r)$$

The first and second term on the RHS are equivalent to ##y(\bar{r})## equalling 1 and a constant (since the integration is constant with respect to ##y(\bar{r})##).

That just leaves the final term on the RHS:

$$-n\int_{} d\bar{r} \int_{\substack{|\bar{r}'| < R}{|\bar{r}-\bar{r}'| > R}} y(\bar{r}') y(\bar{r}-\bar{r}') d\bar{r} exp(-\bar{r}\cdot\bar{s})$$

This is where I seem to be stuck. I'm sure that there is some way to use the convolution property of Laplace transforms to move forward but everything I try seems to be a dead end.

I finally found the answer to my own question. Since it is fairly complicated, I'm not going to give the full resolution here, though if I ever publish the results, I'll try to remember to add a link from this thread to the publication. Alternatively, feel free to message me if you need more specifics.

Indeed, one has to use the convolution property of Laplace transforms in order to get all of the zero boundaries of the Laplace transform to cancel out. Then, one must use a nice math trick for doing a convolution in 3D of a radially symmetric function, which reduces the triple integral to a double integral. After a change in variables, and by assuming that y(r) = y(-r), one obtains the desired result.

## 1. What is the Laplace transform in spherical coordinates?

The Laplace transform in spherical coordinates is a mathematical tool used to convert a function of time into a function of frequency in three-dimensional space. It is commonly used in physics and engineering to solve differential equations and analyze systems in spherical coordinates.

## 2. How is the Laplace transform in spherical coordinates different from the Laplace transform in Cartesian coordinates?

The main difference between the Laplace transform in spherical coordinates and Cartesian coordinates is the coordinate system used. In spherical coordinates, the function is expressed in terms of radius, polar angle, and azimuthal angle, while in Cartesian coordinates, it is expressed in terms of x, y, and z coordinates. Additionally, the Laplace transform in spherical coordinates involves an extra term, the spherical Bessel function, which is not present in the Cartesian transform.

## 3. What is the significance of using the Laplace transform in spherical coordinates?

The Laplace transform in spherical coordinates allows for the analysis of systems that are not easily described in Cartesian coordinates, such as those with spherical symmetry. It also simplifies the solution of certain differential equations, making it a valuable tool in many fields of science and engineering.

## 4. How is the Laplace transform in spherical coordinates applied in real-world problems?

The Laplace transform in spherical coordinates has many practical applications, including in electromagnetics, fluid dynamics, and quantum mechanics. It is used to solve partial differential equations and to analyze the behavior of physical systems, such as antennas, waves, and particles, in three-dimensional space.

## 5. Are there any limitations to using the Laplace transform in spherical coordinates?

While the Laplace transform in spherical coordinates is a powerful tool, it does have some limitations. It is not applicable to all types of functions, and it may not always provide a unique solution. Additionally, the spherical Bessel function can be difficult to solve for certain values, making the transform more complex to apply in some cases.

• Calculus
Replies
2
Views
1K
• Calculus
Replies
3
Views
1K
Replies
2
Views
300
• Calculus
Replies
4
Views
1K
• Calculus
Replies
2
Views
811
• Calculus
Replies
3
Views
1K
• Calculus
Replies
3
Views
2K
• Calculus
Replies
3
Views
1K
• Calculus
Replies
1
Views
1K
• Calculus
Replies
4
Views
11K