Question

\(A\) scalar-valued function \(\varphi\) is harmonic on a region \(D\) if \(\nabla^{2} \varphi=\nabla \cdot \nabla \varphi=0\) at all points of \(D\) Show that the potential function \(\varphi(x, y, z)=|\mathbf{r}|^{-p}\) is harmonic provided \(p=0\) or \(p=1,\) where \(\mathbf{r}=\langle x, y, z\rangle .\) To what vector fields do these potentials correspond?

Short answer

Question: Determine whether the potential function 𝜑(𝑥, 𝑦, 𝑧)=|𝐫|−𝑝 is harmonic for 𝑝=0 or 𝑝=1. If so, find the corresponding vector fields. Answer: The potential function 𝜑(𝑥, 𝑦, 𝑧)=|𝐫|−𝑝 is harmonic for both 𝑝=0 and 𝑝=1. The corresponding vector fields are a constant zero vector field for 𝑝=0, and an inverse cube law force field for 𝑝=1.

Step-by-step solution

Compute \(\nabla \varphi\)

First, we need to compute the gradient \(\nabla \varphi\) for \(\varphi(x, y, z)=|\mathbf{r}|^{-p}\). Recall that the gradient of a scalar-function is given by \(\nabla \varphi = \left( \frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial \varphi}{\partial z} \right)\). So, we need to compute these partial derivatives: - \(\frac{\partial \varphi}{\partial x} = \frac{\partial}{\partial x} \left( |\mathbf{r}|^{-p} \right) = -p |\mathbf{r}|^{-p-2} x\) - \(\frac{\partial \varphi}{\partial y} = \frac{\partial}{\partial y} \left( |\mathbf{r}|^{-p} \right) = -p |\mathbf{r}|^{-p-2} y\) - \(\frac{\partial \varphi}{\partial z} = \frac{\partial}{\partial z} \left( |\mathbf{r}|^{-p} \right) = -p |\mathbf{r}|^{-p-2} z\) Putting these together, we get: \(\nabla \varphi = \left( -p |\mathbf{r}|^{-p-2} x, -p |\mathbf{r}|^{-p-2} y, -p |\mathbf{r}|^{-p-2} z \right)\)

Compute \(\nabla^2 \varphi\)

Next, we need to compute the Laplacian of \(\varphi\), denoted by \(\nabla^2 \varphi\). Recall that the Laplacian is the divergence of the gradient, i.e., \(\nabla^2 \varphi = \nabla \cdot \nabla \varphi\). We computed \(\nabla \varphi\) in step 1, so now we just need to compute its divergence: \(\nabla^2 \varphi = \nabla \cdot \nabla \varphi = \frac{\partial}{\partial x} \left( -p |\mathbf{r}|^{-p-2} x \right) + \frac{\partial}{\partial y} \left( -p |\mathbf{r}|^{-p-2} y \right) + \frac{\partial}{\partial z} \left( -p |\mathbf{r}|^{-p-2} z \right)\) Computing these partial derivatives, we get: \(\nabla^2 \varphi = -p(p+2)|\mathbf{r}|^{-p-4} (x^2+y^2+z^2) + 3p|\mathbf{r}|^{-p-2}\)

Check conditions for harmonicity

To determine which values of \(p\) make \(\varphi\) harmonic, we need to find when \(\nabla^2 \varphi = 0\): \(0 = -p(p+2)|\mathbf{r}|^{-p-4} (x^2+y^2+z^2) + 3p|\mathbf{r}|^{-p-2}\) We can see that this equation holds for \(p=0\) and \(p=1\). So, we have shown that \(\varphi\) is harmonic for \(p = 0\) and \(p = 1\).

Corresponding vector fields

We can now determine the corresponding vector fields for these harmonic potential functions. Recall that the potential function \(\varphi\) is related to the vector field \(\mathbf{F}\) as \(\mathbf{F} = -\nabla \varphi\). For \(p=0\), \(\varphi(x, y, z)=|\mathbf{r}|^0 = 1\), and \(\nabla \varphi = (0, 0, 0)\). Therefore, the corresponding vector field is \(\mathbf{F} = -\nabla \varphi = (0, 0, 0)\), which is a constant zero vector field. For \(p=1\), \(\varphi(x, y, z)=|\mathbf{r}|^{-1}\), and \(\nabla \varphi = \left( -|\mathbf{r}|^{-3} x, -|\mathbf{r}|^{-3} y, -|\mathbf{r}|^{-3} z \right)\). Therefore, the corresponding vector field is \(\mathbf{F} = -\nabla \varphi = \left( |\mathbf{r}|^{-3} x, |\mathbf{r}|^{-3} y, |\mathbf{r}|^{-3} z \right)\), which corresponds to an inverse cube law force field like the electric field around a point charge. In conclusion, we have shown that the given potential function \(\varphi(x, y, z)=|\mathbf{r}|^{-p}\) is harmonic for \(p=0\) and \(p=1\). The corresponding vector fields are a constant zero vector field for \(p=0\), and an inverse cube law force field for \(p=1\).

Key concepts

Laplacian

The Laplacian is a key concept when dealing with harmonic functions. It operates on scalar-valued functions like \(\varphi(x, y, z)\) to determine properties such as how the function curves in space. For a harmonic function, the Laplacian must be zero at all points in the region \(D\). The Laplacian is given by \(abla^2 \varphi = abla \cdot abla \varphi\), which involves finding the divergence of the gradient of \(\varphi\). To compute this, you first find the gradient \(abla \varphi\) and then take its divergence. In the context of our exercise, calculating the Laplacian helps us verify the values of \(p\) that make the function harmonic. When we set \(abla^2 \varphi = 0\), we identified that \(p=0\) and \(p=1\) are solutions. This means that our given function \(\varphi(x,y,z) = |\mathbf{r}|^{-p}\) curves perfectly in space such that these values of \(p\) create no sources or sinks, aligning with the property of harmonic functions.

Gradient

Understanding the gradient of a function is crucial when exploring how scalar fields transform into vector fields. The gradient, denoted \(abla \varphi\), provides the direction and rate of the steepest ascent of the function \(\varphi\). This is expressed as a vector composed of all the partial derivatives of the function:

• \(\frac{\partial \varphi}{\partial x}\)
• \(\frac{\partial \varphi}{\partial y}\)
• \(\frac{\partial \varphi}{\partial z}\)

In the exercise, the gradient helps us transition from the scalar potential function to a vector field by taking these derivatives. Once we have \(abla \varphi\), we can determine corresponding vector fields, as seen with \(p=0\) and \(p=1\). Here, the gradient becomes a key player in defining the behavior and characteristics of the vector fields related to this potential.

Scalar-Valued Functions

Scalar-valued functions, like our potential function \(\varphi(x, y, z)\), associate a single value with every point in a space. These functions are pivotal in physics and mathematics, especially when dealing with phenomena that have magnitude but no specific direction, such as temperature or potential energy.In the exercise, \(\varphi(x, y, z) = |\mathbf{r}|^{-p}\) represents a scalar field. To be harmonic, the Laplacian of this scalar-valued function must equal zero. By studying these types of functions, you can understand how they affect vector fields. Through the derivatives in the gradient, scalar-valued functions serve as a foundation for defining influential fields, important in many areas of science and engineering.

Vector Fields

Vector fields extend scalar-valued functions by assigning a vector to every point in the space, depicting both magnitude and direction. These fields often model physical phenomena like gravitational or electric fields.The vector field \(\mathbf{F}\) related to a scalar potential function \(\varphi\) can be found using \(\mathbf{F} = -abla \varphi\). This operation finds the vector pointing in the direction of the greatest rate of decrease of \(\varphi\). In our exercise, for \(p=0\), the result is a constant zero vector field since the scalar function is constant. For \(p=1\), \(\mathbf{F}\) represents an inverse cube law force field. Understanding these vector fields gives insight into patterns like how forces act at a distance, a core idea in fields like electromagnetism and fluid dynamics.