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Chapter 4: Problem 4

Let us consider Laplace's equation \(\left(\partial^{2} \phi\right) /\left(\partial x^{2}\right)+\left(\partial^{2} \phi\right) /\left(\partial y^{2}\right)=0\), for \(-\infty0\), with the boundary conditions \((\partial \phi / \partial y)(x, 0)\) \(=h(x)\) and \(\phi(x, y) \rightarrow 0\) as \(x^{2}+y^{2} \rightarrow \infty\). Find the Fourier transform solution. Is there a constraint on the data \(h(x)\) for a solution to exist? If so, can this be explained another way?

Short Answer

Expert verified
Solution exists if \( \hat{h}(k) \) decays faster than \( k \) as \( k \rightarrow 0 \), implying \( h(x) \) must be integrable and decay fast.

Step by step solution

01

- Understand the Problem Statement

The given problem involves solving Laplace's equation: \[ \frac{\partial^{2} \phi}{\partial x^{2}} + \frac{\partial^{2} \phi}{\partial y^{2}} = 0 \].The domain is \[-\infty < x < \infty\] and \( y > 0\). The boundary conditions are \(\frac{\partial \phi}{\partial y}(x, 0) = h(x)\) and \( \phi(x, y) \rightarrow 0 \) as \( x^{2}+y^{2} \rightarrow \infty \). The task is to find the Fourier transform solution and determine any constraints on \( h(x) \).
02

- Apply the Fourier Transform

Take the Fourier transform of Laplace's equation with respect to \( x \). Let \( \hat{\phi}(k, y) \) be the Fourier transform of \( \phi(x, y) \). The transformed equation is:\[ -k^2 \hat{\phi}(k, y) + \frac{\partial^{2} \hat{\phi}(k, y)}{\partial y^{2}} = 0 \].
03

- Solve the Transformed Equation

Solve the ordinary differential equation obtained:\[ \frac{\partial^{2} \hat{\phi}(k, y)}{\partial y^{2}} = k^2 \hat{\phi}(k, y) \].The general solution is:\[ \hat{\phi}(k, y) = A(k) e^{ky} + B(k) e^{-ky} \].
04

- Apply Boundary Conditions in Fourier Domain

Use the boundary condition \( \phi(x, y) \rightarrow 0 \) as \( x^{2} + y^{2} \rightarrow \infty \). This implies that \(\hat{\phi}(k, y)\) must tend to 0 as \( \vert k \vert \rightarrow \infty \). Hence, \( A(k) = 0 \) and: \[ \hat{\phi}(k, y) = B(k) e^{-ky} \].
05

- Determine \(B(k)\) Using Initial Condition

The boundary condition \( \frac{\partial \phi}{\partial y}(x, 0) = h(x) \) in Fourier space gives:\[ -k B(k) = \hat{h}(k) \],where \( \hat{h}(k) \) is the Fourier transform of \( h(x) \).Therefore, \[ B(k) = -\frac{\hat{h}(k)}{k} \].
06

- Express the Solution in Fourier Space

The solution in Fourier space is:\[ \hat{\phi}(k, y) = -\frac{\hat{h}(k)}{k} e^{-ky} \].
07

- Inverse Fourier Transform to Obtain Solution

Use the inverse Fourier transform to retrieve \( \phi(x, y) \):\[ \phi(x, y) = \int_{-\infty}^{\infty} \hat{\phi}(k, y) e^{ikx} dk \].Substitute \( \hat{\phi}(k, y) \):\[ \phi(x, y) = -\int_{-\infty}^{\infty} \frac{\hat{h}(k)}{k} e^{-ky} e^{ikx} dk \].
08

- Constraint on \( h(x) \)

For the solution to exist, the term \( \frac{\hat{h}(k)}{k} \) must be finite. Therefore, \( \hat{h}(k) \) must decay faster than \( k \) as \( k \rightarrow 0 \). This implies that \( h(x) \) must be integrable and decay sufficiently fast to ensure a meaningful solution.
09

- Alternative Explanation of the Constraint

From a physical perspective, if \( h(x) \) does not decay sufficiently, the boundary condition will overpower the decay requirement at infinity, preventing a stable solution. This explains the necessity of constraints on \( h(x) \).

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Key Concepts

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

Laplace's Equation
Laplace's equation is a second-order partial differential equation (PDE) given by: \[ \frac{\partial^{2} \phi}{\partial x^{2}} + \frac{\partial^{2} \phi}{\partial y^{2}} = 0 \] It is often used to describe potential functions, like electric, gravitational, or fluid potentials in space. The main idea is that the sum of the second partial derivatives with respect to each variable is zero. This equation is significant in various fields of science and engineering because it helps predict the behavior of a potential function within a specified region. Laplace's equation is especially valuable because it can be applied in both two-dimensional and three-dimensional spaces.
Fourier Transform
The Fourier transform is a mathematical technique that transforms a function of time (or space) into a function of frequency. In simpler terms, it breaks down a complex function into its sine and cosine components. For a function \( f(x) \), its Fourier transform \( \hat{f}(k) \) is given by: \[ \hat{f}(k) = \int_{-\infty}^{\infty} f(x) e^{-ikx} \, dx \] In the context of solving Laplace's equation, we take the Fourier transform with respect to one variable to convert the partial differential equation into an ordinary differential equation (ODE). This makes it easier to solve, as we can leverage simpler mathematical techniques for ODEs.
Boundary Conditions
Boundary conditions specify the behavior of the function at the boundaries of the domain. For Laplace's equation, these conditions are crucial because they help determine a unique solution. In our problem, the boundary conditions are:
  • \( \frac{\partial \phi}{\partial y}(x, 0) = h(x) \)
  • \( \phi(x, y) \rightarrow 0 \) as \(x^2 + y^2 \rightarrow \infty \)
These conditions mean that the rate of change of the function \( \phi \) with respect to \( y \) at \( y = 0 \) is given by some known function \( h(x) \), and that \( \phi \) vanishes at infinity.
Ordinary Differential Equations
Ordinary differential equations (ODEs) involve functions of one variable and their derivatives. After applying the Fourier transform to Laplace鈥檚 equation, we convert it into an ODE: \[ \frac{\partial^{2} \hat{\phi}(k, y)}{\partial y^{2}} = k^2 \hat{\phi}(k, y) \] Solving this ODE, we find the general solution: \[ \hat{\phi}(k, y) = A(k) e^{ky} + B(k) e^{-ky} \] This form includes two terms representing solutions that grow and decay exponentially. By applying boundary conditions, we eliminate the non-physical solution, which leads us to a specific solution suitable for our problem.
Inverse Fourier Transform
To convert the solution back to the original space, we use the inverse Fourier transform, which is defined as: \[ f(x) = \int_{-\infty}^{\infty} \hat{f}(k) e^{ikx} \, dk \] Applying it to our transformed solution: \[ \phi(x, y) = -\int_{-\infty}^{\infty} \frac{\hat{h}(k)}{k} e^{-ky} e^{ikx} \, dk \] This step returns us from the frequency domain to the spatial domain, giving us the complete solution in terms of the original variables \( x \) and \( y \). It is important that the initial condition \( h(x) \) satisfies certain constraints to ensure the solution exists and is meaningful.

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Most popular questions from this chapter

Let \(f(z)\) be analytic outside a circle \(C_{R}\) enclosing the origin. (a) Show that $$ \frac{1}{2 \pi i} \oint_{C_{R}} f(z) d z=\frac{1}{2 \pi i} \oint_{C_{p}} f\left(\frac{1}{t}\right) \frac{d t}{t^{2}} $$ where \(C_{\rho}\) is a circle of radius \(1 / R\) enclosing the origin. For \(R \rightarrow \infty\) conclude that the integral can be computed to be Res \(\left(f(1 / t) / t^{2} ; 0\right)\). (b) Suppose \(f(z)\) has the convergent Laurent expansion $$ f(z)=\sum_{j=-\infty}^{-1} A_{j} z^{j} $$ Show that the integral above equals \(A_{-1 .}\) (See also Eq. (4.1.11).)

Show that the inverse Laplace transform of the function $$ \hat{F}(s)=\frac{1}{\sqrt{s^{2}+\omega^{2}}} $$ \(\omega>0\), is given by $$ f(x)=\frac{1}{\pi} \int_{-\omega}^{\omega} \frac{e^{i x r}}{\sqrt{\omega^{2}-r^{2}}} d r=\frac{2}{\pi} \int_{0}^{1} \frac{\cos (\omega x \rho)}{\sqrt{1-\rho^{2}}} d \rho $$ (The latter integral is a representation of \(J_{0}(\omega x)\), the Bessel function of order zero.) Hint: Deform the contour around the branch points \(s=\pm i \omega\), then show that the large contour at infinity and small contours encircling \(\pm i \omega\) are vanishingly small. It is convenient to use the polar representations \(s+i \omega_{1}=r_{1} e^{i \theta_{1}}\) and \(s-i \omega_{2}=r_{2} e^{i \theta_{2}}\), where \(-3 \pi / 2<\theta_{i} \leq \pi / 2, i=1,2\), and \(\left(s^{2}+\omega^{2}\right)^{1 / 2}=\sqrt{r_{1} r_{2}} e^{i\left(\theta_{1}+\theta_{2}\right) / 2} .\) The contributions on both sides of the cut add to give the result.

Show that $$ \int_{0}^{\infty} \frac{\cosh a x}{\cosh \pi x} d x=\frac{1}{2} \sec \left(\frac{a}{2}\right), \quad|a|<\pi $$ Use a rectangular contour with corners at \(\pm R\) and \(\pm R+i\).

Use Laplace transform methods to solve the ODE $$ \frac{d^{2} y}{d t^{2}}-k^{2} y=f(t), \quad k>0, \quad y(0)=y_{0}, \quad y^{\prime}(0)=y_{0}^{\prime} $$ (a) Let \(f(t)=e^{-k_{0 t}}, k_{0} \neq k, k_{0}>0\), so that the Laplace transform of \(f(t)\) is \(\hat{F}(s)=1 /\left(s+k_{0}\right)\), and find $$ \begin{aligned} y(t)=& y_{0} \cosh k t+\frac{y_{0}^{\prime}}{k} \sinh k t+\frac{e^{-k_{0} t}}{k_{0}^{2}-k^{2}} \\ &-\frac{\cosh k t}{k_{0}^{2}-k^{2}}+\frac{\left(k_{0} / k\right)}{k_{0}^{2}-k^{2}} \sinh k t \end{aligned} $$ (b) Suppose \(f(t)\) is an arbitrary continuous function that possesses a Laplace transform. Use the convolution product for Laplace transforms (Section \(4.5\) ) to find $$ y(t)=y_{0} \cosh k t+\frac{y_{0}^{\prime}}{k} \sinh k t+\int_{0}^{t} f\left(t^{\prime}\right) \frac{\sinh k\left(t-t^{\prime}\right)}{k} d t^{\prime} $$ (c) Let \(f(t)=e^{-k_{0} t}\) in \((\mathrm{b})\) to obtain the result in part (a). What happens when \(k_{0}=k ?\)

(a) Obtain the Fourier transform of \(f(x)=(\sin \omega x) /(x), \omega>0\). (b) Show that \(f(x)\) in part (a) is not \(L_{1}\), that is, \(\int_{-\infty}^{\infty}|f(x)| d x\) does not exist. Despite this fact, we can obtain the Fourier transform, so \(f(x) \in L_{1}\) is a sufficient condition, but is not necessary, for the Fourier transform to exist.
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