Laplace equation in polar coordinates.

In summary, the Laplace operator in polar coordinates can be expressed as ##\nabla^2 u=\frac {\partial ^2 u}{\partial r^2}+\frac{1}{r}\frac {\partial u}{\partial r}+\frac{1}{r^2}\frac {\partial ^2 u}{\partial \theta^2}##, and can be derived through several methods such as covariant integrals, brute force, or the action principle. The choice of using cartesian or polar form for the function u depends on which form is given. Otherwise, the chain rule must be used to convert between the two forms.
  • #1
5,707
240
[tex]\nabla^2 u=\frac {\partial ^2 u}{\partial x^2}+\frac {\partial ^2 u}{\partial y^2}=\frac {\partial ^2 u}{\partial r^2}+\frac{1}{r}\frac {\partial u}{\partial r}+\frac{1}{r^2}\frac {\partial ^2 u}{\partial \theta^2}[/tex]

I want to verify ##u=u(r,\theta)##, not ##u(x,y)##

Because for ##u(x,y)##, it will just be ##\frac {\partial ^2 u}{\partial x^2}+\frac {\partial ^2 u}{\partial y^2}##

Thanks
 
Last edited:
Physics news on Phys.org
  • #3
I know the derivation, I just want to confirm as none of the books or notes specify this.

It should be ##u(r,\theta)## not ##u(x,y)##.
 
  • #4
Of course for [itex]\vec{r} \neq 0[/itex] you have
[tex]u(\vec{r})=u[\vec{r}(r,\theta)]=u[\vec{r}(x,y)].[/tex]
The vector operators are better defined through covariant integrals than by brute force as in UltrafastPED's source, although it's of course a valid way to find the expression for the covariant differential operators.

A third, more physicist's way, is to use the action principle. Define the action
[tex]A=\int_{\mathbb{R}^2} \mathrm{d}^2 \vec{r} \left [\frac{1}{2}(\vec{\nabla} u)^2+u \Phi \right],[/tex]
where [itex]\Phi[/itex] is an arbitrary external field. Taking the variation you see that the stationary point is
[tex]\frac{\delta A}{\delta u}=0 \; \Rightarrow \; \Delta u=\Phi.[/tex]
To evaluate thus the Laplace operator in polar coordinates, you simply write out the action in these coordinates. You only need the gradient in polar coordinates, which is very easy to derive by using differential forms:
[tex]\mathrm{d} u=\mathrm{d} \vec{x} \cdot \vec{\nabla} u=\mathrm{d} r \vec{e}_r \cdot \vec{\nabla} u + r \mathrm{d} \theta \vec{e}_\vartheta \cdot \vec{\nabla} u=\mathrm{d} r \partial_r u + \mathrm{d} \theta \partial_{\theta} u.[/tex]
Comparison of both sides of the last equation of this line shows that
[tex]\vec{\nabla} u = \vec{e}_r \partial_r u + \vec{e}_{\theta} \frac{1}{r} \partial_{\theta} u.[/tex]
Further you have
[tex]\mathrm{d}^2 \vec{r}=r \mathrm{d} r \mathrm{d} \theta,[/tex]
and thus
[tex]A=\int_0^{\infty} \mathrm{d} r \int_0^{2 \pi} \mathrm{d} \theta r \left [\frac{1}{2}(\partial_r u)^2+\frac{1}{2r^2} (\partial_{\theta}) u + u \Phi \right].[/tex]
Then via the Euler-Lagrange equations you get the stationary point of the action as given by
[tex]\frac{\partial}{\partial r} \left (r \frac{\partial u}{\partial r} \right )+\frac{1}{r} \frac{\partial^2 u}{\partial \theta^2}=r \Phi,[/tex]
and finally dividing by [itex]r[/itex]
[tex]\Phi=\Delta u=\frac{1}{r} \frac{\partial}{\partial r} \left (r \frac{\partial u}{\partial r} \right )+\frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}.[/tex]
 
  • Like
Likes 1 person
  • #5
See steps (4) and (5).

If your function is expressed as u(x,y) you should use the cartesian form.
If it is in polar form then you should use the polar form.

Otherwise you have to go through all of the steps (4) and (5), making use of the chain rule.
 

1. What is the Laplace equation in polar coordinates?

The Laplace equation in polar coordinates is a mathematical equation that describes the behavior of a scalar field in two-dimensional polar coordinates. It is often used in physics, engineering, and mathematics to model phenomena such as heat conduction, fluid flow, and electrostatics.

2. How is the Laplace equation derived in polar coordinates?

The Laplace equation in polar coordinates is derived by applying the polar coordinate system to the standard Laplace equation in Cartesian coordinates. This involves converting the Laplace operator (the second partial derivative of the scalar field with respect to the Cartesian coordinates) into polar coordinates and simplifying the resulting equation.

3. What are the applications of the Laplace equation in polar coordinates?

The Laplace equation in polar coordinates has many applications in physics and engineering. It is commonly used to solve boundary value problems in electrostatics, heat conduction, and fluid dynamics. It is also used in mathematical modeling of physical systems, such as predicting the behavior of heat flow in a circular object or the potential of an electric charge in a circular region.

4. What are the boundary conditions for solving the Laplace equation in polar coordinates?

The boundary conditions for solving the Laplace equation in polar coordinates depend on the specific problem being solved. In general, they describe the behavior of the scalar field at the edges of the domain, and can include conditions such as fixed values, gradients, or fluxes.

5. How is the Laplace equation solved in polar coordinates?

The Laplace equation in polar coordinates can be solved using various techniques, including separation of variables, conformal mapping, and numerical methods. The specific method used depends on the complexity of the problem and the desired level of accuracy. In some cases, analytical solutions may be possible, while in others, numerical approximations are necessary.

Suggested for: Laplace equation in polar coordinates.

Replies
5
Views
1K
Replies
4
Views
558
Replies
1
Views
943
Replies
3
Views
2K
Replies
10
Views
3K
Replies
5
Views
2K
Replies
3
Views
1K
Back
Top