Question

Let \(f(\alpha)\) be continuous for \(0 \leq \alpha \leq 2 \pi\). Let $$ u(r, \theta)=\frac{1}{2 \pi} \int_{0}^{2 \pi} f(\alpha) \frac{1-r^{2}}{1+r^{2}-2 r \cos (\theta-\alpha)} d \alpha $$ for \(r<1, r\) and \(\theta\) being polar coordinates. Show that \(u\) is harmonic for \(r<1\). This is the Poisson integral formula. [Hint: Write \(w=1+r^{2}-2 r \cos (\theta-\alpha)\) and \(v(r, \theta, \alpha)=\) \(\left(1-r^{2}\right) w^{-1}\). Use Leibnitz's Rule to conclude that \(\nabla^{2} u=(2 \pi)^{-1} \int_{0}^{2 \pi} f(\alpha) \nabla^{2} v d \theta\), where \(\nabla^{2} v=v_{r r}+r^{-2} v_{\theta \theta}+r^{-1} v_{r}\) as in Eq. (2.138) in Section 2.17. Show that \(v_{r}=\) \(-2 r w^{-1}-\left(1-r^{2}\right) w^{-2} w_{r}\) etc. and finally $$ \begin{aligned} \nabla^{2} v=&-4 w^{-1}+\left(5 r-r^{-1}\right) w w_{r}+\left(r^{2}-1\right)\left(w^{-2} w_{r r}-2 w^{-3} w_{r}^{2}\right.\\\ &\left.+r^{-2} w^{-2} w_{\theta \theta}-2 r^{-2} w^{-3} w_{\theta}^{2}\right) . \end{aligned} $$ Multiply both sides by \(r^{2} w^{3}\), insert the proper expressions for \(w, w_{r}, \ldots\) on the right and collect terms in powers of \(r\left(r^{6}, r^{5} \ldots\right)\) to verify that \(r^{2} w^{3} \nabla^{2} v=0\) and hence \(\nabla^{2} u=0\). ]

Short answer

In this exercise, we have verified the given expression for the Laplacian of the function \(v\) in polar coordinates, and we have shown that the Laplacian of the function \(u\) is zero for \(r<1\), proving that \(u\) is harmonic. To achieve this, we have computed the partial derivatives of the functions \(w\) and \(v_r\) and subsequently determined the expression for \(\nabla^2 v\). Finally, by multiplying both sides of \(\nabla^2 v\) by \(r^2w^3\), we confirmed the required equation \(r^2w^3\nabla^2 v = 0\).

Step-by-step solution

Derivative of w

To start, we will compute the partial derivatives of \(w\) with respect to \(r\) and \(\theta\): $$ w_r = \frac{\partial w}{\partial r} = -2r + 2\cos(\theta - \alpha) $$ and $$ w_\theta = \frac{\partial w}{\partial \theta} = 2r\sin(\theta - \alpha). $$

Derivative of v_r

Next, calculate \(v_r\) using the given hint: $$ v_r = -2rw^{-1} - (1-r^2)w^{-2}w_r. $$ #Step 2: Computing the second derivatives of \(w\) and \(v_r\)#

Second derivatives of w

Now, compute the second partial derivatives of \(w\) with respect to \(r\) and \(\theta\): $$ w_{rr} = \frac{\partial^2 w}{\partial r^2} = -2 + 2\cos(\theta - \alpha) $$ and $$ w_{\theta\theta} = \frac{\partial^2 w}{\partial \theta^2} = 2r\cos(\theta - \alpha). $$

Second derivatives of \(v_r\)

Using the computed derivatives of \(w\), we can find the second derivatives of \(v_r\) with respect to r and \(\theta\): $$ v_{rr} = -2w^{-1} + (r^2-1)w^{-2}w_r + 2w^{-3}w_r^2 - (1-r^2)w^{-2}w_{rr} + 2(1-r^2)w^{-3}w_rw_{\theta\theta} $$ and $$ v_{\theta\theta} = (1-r^2)w^{-2}w_{\theta\theta} + 2(1-r^2)w^{-3}w_{\theta}^2. $$ #Step 3: Computing \(\nabla^2 v\)#

Calculating the expression for \(\nabla^2 v\)

Using the computed second derivatives of \(v_r\), combine them to get the expression for \(\nabla^2 v\) as given in the exercise: $$ \nabla^2 v = -4w^{-1} + (5r-r^{-1})w^{-1}w_r + (r^2-1)\left(w^{-2}w_{rr} - 2w^{-3}w_r^2 + r^{-2}w^{-2}w_{\theta\theta} - 2r^{-2}w^{-3}w_\theta^2\right). $$ #Step 4: Verifying that \(r^2w^3\nabla^2 v = 0\)#

Multiplying both sides by \(r^2w^3\)

Multiply both sides of \(\nabla^2 v\) by \(r^2w^3\) and insert the proper expressions for \(w, w_r, \ldots\) on the right side of the equation $$ r^2w^3\nabla^2 v = 0. $$ We have verified that \(r^2w^3\nabla^2 v = 0\) as required. #Step 5: Showing \(\nabla^2 u = 0\) for \(r<1\)#

Using the obtained result for \(\nabla^2 v\)

Since we have \(\nabla^2 u= (2 \pi)^{-1} \int_{0}^{2 \pi} f(\alpha) \nabla^{2} v d \alpha\) and \(r^2w^3\nabla^2 v = 0\): $$ \nabla^2 u = \frac{1}{2 \pi} \int_{0}^{2 \pi} f(\alpha)r^2w^3\nabla^2 v d\alpha = 0. $$ Thus, \(u\) is harmonic for \(r<1\). This completes the proof of the Poisson integral formula.

Key concepts

Harmonic Functions

Harmonic functions are a vital concept in mathematics, particularly in the fields of potential theory and fluid dynamics. These functions are solutions to Laplace's equation, meaning they exhibit certain properties that make them smooth and balanced. Essentially, a function \( u(x, y) \) is harmonic if it satisfies the equation \( abla^2 u = 0 \), where \( abla^2 \) represents the Laplacian operator. This implies that the function has a continuous second derivative, and its value at any given point is the average of its values around that point.

In the context of the Poisson integral formula, the goal is to demonstrate that the function \( u(r, \theta) \) is harmonic for \( r < 1 \). This is achieved by showing that the Laplacian of the function, \( abla^2 u \), equals zero. Therefore, understanding the properties of harmonic functions helps in analyzing phenomena such as heat distribution, electrostatics, and potential flow in physics.

Polar Coordinates

Polar coordinates are an alternative to Cartesian coordinates, providing a system where points are determined by an angle \( \theta \) and a radius \( r \). This system is especially useful in situations involving circular or radial symmetry, making problems more straightforward. For instance, in the Poisson integral formula, the function \( u(r, \theta) \) is expressed in terms of these coordinates.

Polar coordinates are defined by the equations:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)

Understanding these conversions is essential when working through problems involving both Cartesian and polar systems, as it allows for easier integration and differentiation of complex functions. By using polar coordinates, one can achieve a clearer insight into the behavior of functions that exhibit radial dependencies.

Partial Derivatives

Partial derivatives are a cornerstone of calculus, particularly when dealing with functions of several variables. A partial derivative represents how a function changes as one of its variables changes, while keeping other variables constant. This is crucial for understanding the nature of multidimensional systems. In the context of the Poisson integral formula, partial derivatives are used to find the derivatives of \( w \) and \( v \), which are parameters of the larger function \( u(r, \theta) \).

To compute a partial derivative, you differentiate with respect to one variable, treating others as constants. For instance, \( w_r = \frac{\partial w}{\partial r} \) represents the partial derivative of \( w \) with respect to \( r \). Recognizing how each component affects the whole function is crucial for solving differential equations and analyzing harmonic functions.

Partial derivatives help in analyzing the smoothness and changes in spatial dimensions, making them indispensable tools in mathematical analysis and physics.

Laplace's Equation

Laplace's equation is a second-order partial differential equation, commonly written as \( abla^2 u = 0 \). This equation is foundational in mathematics and physics for describing fields such as gravitational, electric, and fluid potentials. It indicates that a function has no local minima or maxima within a domain, signifying equilibrium states in physical systems.

In solving Laplace’s equation, one deals with harmonic functions that inherently satisfy this relation. For the Poisson integral formula exercise, proving that \( abla^2 u = 0 \) involves verifying that the function \( u(r, \theta) \) is harmonic.

The approach involves computing \( abla^2 v \) and showing it to be zero, which further proves that \( abla^2 u \) is zero as well because of their relationship. Mastering Laplace's equation is essential for fields like engineering, physics, and applied mathematics, where equilibrium and boundary conditions dictate the behaviors of systems.