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Find the Green's function for the three-dimensional Neumann problem $$ \nabla^{2} \phi=0 \quad \text { for } z>0 \quad \text { and } \quad \frac{\partial \phi}{\partial z}=f(x, y) \quad \text { on } z=0 $$ Determine \(\phi(x, y, z)\) if $$ f(x, y)= \begin{cases}\delta(y) & \text { for }|x|

Short Answer

Expert verified
The solution is \ \phi(x, y, z) = \begin{cases} \delta(y) z &\text{ for } |x| < a, \ 0 &\text{ for } |x| \geq a. \end{cases}.

Step by step solution

01

General Solution of Laplace's Equation

The general form for the solution to Laplace's equation \( abla^{2} \phi = 0 \) in three dimensions can be written as \[ \phi(x, y, z) = A(x, y) z + B(x, y) \]. Since \( \abla^{2} \phi = 0 \), \( A \) and \( B \) must independently satisfy \( \abla_{\text{2D}} \) when applied to them.
02

Applying the Neumann Boundary Condition

We have the boundary condition \( \frac{\text{\partial} \phi}{\text{\partial} z} = f(x, y) \) on \( z = 0 \). Differentiating the general solution \phi(x, y, z) \ with respect to \( z \) gives us \( A(x, y) \). Therefore, we have \( A(x, y) = f(x, y) \. \)
03

Substituting the Specific Boundary Condition

According to the specific boundary condition, \( f(x, y) \) is \[ f(x,y) = \begin{cases} \delta(y) &\text{ for } |x| < a. \ 0 &\text{ for } |x| \geq a. \end{cases} \]. Therefore, \( A(x, y) \) is exactly this piecewise function.
04

Constructing the Solution

Substitute \( A(x, y) \) back into the general form. Thus, the specific solution is \( \phi(x, y, z) = f(x, y) z + B(x, y). \) To satisfy the boundary conditions fully, \( B(x, y) \) must vanish. This gives \[ \phi(x, y, z) = f(x, y)z. \]
05

Piecewise Solution for \( \phi(x, y, z)\)

Thus, using the specific form of \( f(x, y) \), the solution \phi(x, y, z)\ is \ \phi(x, y, z) = \begin{cases} \delta(y) z &\text{ for } |x| < a, \ 0 &\text{ for } |x| \geq a. \end{cases} \.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Green's function
Green's function is a fundamental solution used to solve differential equations, particularly in physics and engineering. Understanding Green's function is crucial for solving problems involving boundary conditions. It essentially allows you to find a particular solution to inhomogeneous differential equations based on the given boundary conditions. For the Neumann problem in three dimensions, Green's function helps in finding the potential \( \phi \) that satisfies the given Laplace's equation and boundary conditions. This method essentially converts a complex boundary problem into an easier-to-handle problem.
Laplace's equation
Laplace's equation, denoted as \( abla^{2} \phi = 0 \), is a second-order partial differential equation. It's widely used in physics and engineering to describe the behavior of electric, gravitational, and fluid potentials. In this problem, you solve Laplace's equation to find the potential function \( \phi(x, y, z) \). In a three-dimensional space, this equation implies that the sum of the second partial derivatives in all three coordinate directions is zero. The general solution involves finding functions \( A(x, y) \) and \( B(x, y) \) that satisfy this condition.
Boundary conditions
Boundary conditions are essential in solving differential equations because they specify the solution's behavior on the boundary of the domain. In the Neumann problem, the boundary condition is given as \( \frac{\partial \phi}{\partial z} = f(x, y) \) on \( z = 0 \). This condition informs us about the rate of change of the potential \( \phi \) with respect to \( z \) at the boundary. To solve the problem, you need to apply this condition to the general solution of Laplace's equation. By doing so, you ensure that the derived solution conforms to the physical or geometrical constraints of the problem.
Three-dimensional solutions
Three-dimensional solutions to differential equations provide a more comprehensive understanding of physical phenomena. In this context, you are dealing with a three-dimensional Neumann problem where the potential function \( \phi(x, y, z) \) depends on all three spatial coordinates. The general solution is a linear combination of functions \( A(x, y) \) and \( B(x, y) \), which need to satisfy Laplace's equation. By including the specific boundary conditions, you can determine the exact form of these functions, leading to a complete solution that describes the system's behavior in three dimensions.

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Most popular questions from this chapter

(a) Show that the gravitational potential due to a uniform disc of radius \(a\) and mass \(M\), centred at the origin, is given for \(ra\) by $$ \frac{G M}{r}\left[1-\frac{1}{4}\left(\frac{a}{r}\right)^{2} P_{2}(\cos \theta)+\frac{1}{8}\left(\frac{a}{r}\right)^{4} P_{4}(\cos \theta)-\cdots\right] $$ where the polar axis is normal to the plane of the disc. (b) Reconcile the presence of a term \(P_{1}(\cos \theta)\), which is odd under \(\theta \rightarrow \pi-\theta\), with the symmetry with respect to the plane of the disc of the physical system. (c) Deduce that the gravitational field near an infinite sheet of matter of constant density \(\rho\) per unit area is \(2 \pi G \rho\).

A sphere of radius \(a\) and thermal conductivity \(k_{1}\) is surrounded by an infinite medium of conductivity \(k_{2}\) in which, far away, the temperature tends to \(T_{\infty}\). A distribution of heat sources \(q(\theta)\) embedded in the sphere's surface establish steady temperature fields \(T_{1}(r, \theta)\) inside the sphere and \(T_{2}(r, \theta)\) outside it. It can be shown, by considering the heat flow through a small volume that includes part of the sphere's surface, that $$ k_{1} \frac{\partial T_{1}}{\partial r}-k_{2} \frac{\partial T_{2}}{\partial r}=q(\theta) \quad \text { on } \quad r=a $$ Given that $$ q(\theta)=\frac{1}{a} \sum_{n=0}^{\infty} q_{n} P_{n}(\cos \theta) $$ find complete expressions for \(T_{1}(r, \theta)\) and \(T_{2}(r, \theta)\). What is the temperature at the centre of the sphere?

Consider the PDE \(\mathcal{L} u(\mathbf{r})=\rho(\mathbf{r})\), for which the differential operator \(\mathcal{L}\) is given by $$ \mathcal{L}=\nabla \cdot[p(\mathbf{r}) \nabla]+q(\mathbf{r}) $$ where \(p(\mathbf{r})\) and \(q(\mathbf{r})\) are functions of position. By proving the generalised form of Green's theorem, $$ \int_{V}(\phi \mathcal{L} \psi-\psi \mathcal{L} \phi) d V=\oint_{S} p(\phi \nabla \psi-\psi \nabla \phi) \cdot \hat{\mathbf{n}} d S $$ show that the solution of the PDE is given by $$ u\left(\mathbf{r}_{0}\right)=\int_{V} G\left(\mathbf{r}, \mathbf{r}_{0}\right) \rho(\mathbf{r}) d V(\mathbf{r})+\oint_{S} p(\mathbf{r})\left[u(\mathbf{r}) \frac{\partial G\left(\mathbf{r}, \mathbf{r}_{0}\right)}{\partial n}-G\left(\mathbf{r}, \mathbf{r}_{0}\right) \frac{\partial u(\mathbf{r})}{\partial n}\right] d S(\mathbf{r}) $$ where \(G\left(\mathbf{r}, \mathbf{r}_{0}\right)\) is the Green's function satisfying \(\mathcal{L} G\left(\mathbf{r}, \mathbf{r}_{0}\right)=\delta\left(\mathbf{r}-\mathbf{r}_{0}\right)\).

Solve the following first-order partial differential equations by separating the variables: (a) \(\frac{\partial u}{\partial x}-x \frac{\partial u}{\partial y}=0\); (b) \(x \frac{\partial u}{\partial x}-2 y \frac{\partial u}{\partial y}=0\).

A membrane is stretched between two concentric rings of radii \(a\) and \(b(b>a)\). If the smaller ring is transversely distorted from the planar configuration by an amount \(c|\phi|,-\pi \leq \phi \leq \pi\), show that the membrane then has a shape given by $$ u(\rho, \phi)=\frac{c \pi}{2} \frac{\ln (b / \rho)}{\ln (b / a)}-\frac{4 c}{\pi} \sum_{m \text { odd }} \frac{a^{m}}{m^{2}\left(b^{2 m}-a^{2 m}\right)}\left(\frac{b^{2 m}}{\rho^{m}}-\rho^{m}\right) \cos m \phi $$.

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