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Chapter 4: Problem 9

Use multiple Fourier transforms to solve $$ \frac{\partial \phi}{\partial t}-\left(\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}\right)=0 $$ on the infinite domain \(-\infty0\), with \(\phi(x, y) \rightarrow 0\) as \(x^{2}+y^{2} \rightarrow \infty\), and \(\phi(x, y, 0)=f(x, y) .\) How does the solution simplify if \(f(x, y)\) is a function of \(x^{2}+y^{2} ?\) What is the Green's function in this case?

Short Answer

Expert verified
The solution is \( \phi(x, y, t) = \int_{0}^{\infty} \hat{f}(k) e^{-k^2 t} J_0(kr) k \, \mathrm{d}k \). The Green's function is \( G(x, y, t) = \frac{1}{4\pi t} e^{-(x^2 + y^2)/(4t)} \).

Step by step solution

01

Write down the given PDE and initial conditions

The given partial differential equation (PDE) is: \[ \frac{\partial \phi}{\partial t} - \left(\frac{\partial^{2} \phi}{\partial x^{2}} + \frac{\partial^{2} \phi}{\partial y^{2}} \right) = 0 \]with the initial condition:\[ \phi (x, y, 0) = f(x, y) \]where \( \phi(x, y) \rightarrow 0 \) as \( x^{2} + y^{2} \rightarrow \infty \).
02

Apply the Fourier transform in the x and y directions

The 2D Fourier transform of \( \phi(x, y, t) \) is defined by: \[ \hat{\phi}(k_x, k_y, t) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-i(k_x x + k_y y)} \phi(x, y, t) \mathrm{d}x \mathrm{d}y \]Taking the Fourier transform of both sides of the PDE, we obtain:\[ \frac{\partial \hat{\phi}}{\partial t} - (k_x^2 + k_y^2) \hat{\phi} = 0 \]
03

Solve the ordinary differential equation (ODE) for \( \hat{\phi} \)

This transforms the PDE into a simpler ODE:\[ \frac{\partial \hat{\phi}}{\partial t} = (k_x^2 + k_y^2) \hat{\phi} \]which is a first-order linear ODE. Solving this ODE, we get:\[ \hat{\phi}(k_x, k_y, t) = \hat{\phi}(k_x, k_y, 0) e^{-(k_x^2 + k_y^2) t} \]Using the initial condition, we have:\[ \hat{\phi}(k_x, k_y, 0) = \hat{f}(k_x, k_y) \]
04

Apply the inverse Fourier transform

To find \( \phi(x, y, t) \), apply the inverse Fourier transform:\[ \phi(x, y, t) = \frac{1}{(2\pi)^2} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{i(k_x x + k_y y)} \hat{f}(k_x, k_y) e^{-(k_x^2 + k_y^2) t} \mathrm{d}k_x \mathrm{d}k_y \]
05

Address simplification when \( f(x, y) = f(x^2 + y^2) \)

When \( f(x, y) \) is a function of \( r^2 = x^2 + y^2 \), \( \hat{f}(k_x, k_y) \) depends only on \( k = \sqrt{k_x^2 + k_y^2} \). The solution now simplifies to:\[ \phi(x, y, t) = \frac{1}{(2\pi)} \int_{0}^{\infty} \int_{0}^{2\pi} e^{ikr\cos(\theta-\varphi)} \hat{f}(k) e^{-k^2 t} k \mathrm{d}\theta \mathrm{d}k \]Simplifying the angular part using the orthogonality of exponentials, we get:\[ \phi(x, y, t) = \int_{0}^{\infty} \hat{f}(k) e^{-k^2 t} J_0(kr) k \mathrm{d}k \] where \( J_0 \) is the Bessel function of the first kind.
06

Determine the Green's function

The Green's function solution for the problem when \( f(x, y) \) is a delta function \( \delta(x) \delta(y) \) is given by:\[ G(x, y, t) = \frac{1}{4\pi t} e^{-(x^2 + y^2)/(4t)} \]Therefore, the solution for a general initial condition \( f(x, y) \) is:\[ \phi(x, y, t) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} G(x - \xi, y - \eta, t) f(\xi, \eta) \,\mathrm{d}\xi \,\mathrm{d}\eta \]

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Key Concepts

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

Fourier Transform
A Fourier transform is a powerful mathematical technique that transforms a function of time (or space) into a function of frequency. This method is vital in solving partial differential equations (PDEs) as it converts complex differential operations into simpler algebraic forms.
In solving your given PDE, we applied a 2D Fourier transform \({\phi}(k_x, k_y, t)\). This helps reduce the PDE into an ordinary differential equation (ODE), which is easier to solve.
Green's Function
A Green's function is a type of solution used to solve inhomogeneous differential equations. It essentially represents the response of the system to a point source, such as a delta function.
In your exercise, the Green's function for the problem when the initial condition is a delta function \(\delta(x) \delta(y)\) is \[ G(x, y, t) = \frac{1}{4\pi t} e^{-(x^2 + y^2)/(4t)} \]. This allows us to construct the solution for any general initial condition \(f(x, y)\) by integrating this Green's function with the initial condition.
Partial Differential Equations
Partial Differential Equations (PDEs) are equations that involve rates of change with respect to continuous variables. They are fundamental in describing various physical phenomena such as heat conduction or wave propagation.
In your case, the given PDE is \[ \frac{\partial \phi}{\partial t} - \left(\frac{\partial^{2} \phi}{\partial x^{2}} + \frac{\partial^{2} \phi}{\partial y^{2}} \right) = 0 \], which depicts a diffusion or heat equation in two dimensions.
Bessel Functions
Bessel functions, particularly \(J_0\), appear frequently when solving PDEs with radial symmetry. They are special types of functions that provide solutions to Bessel's differential equation.
When the initial condition \(f(x, y)\) is radially symmetric, meaning it depends only on \(x^2 + y^2\), the solution involves Bessel functions: \[ \phi(x, y, t) = \int_{0}^{\infty} \hat{f}(k) e^{-k^2 t} J_0(kr) k \, \mathrm{d}k \]. This simplifies the representation of the solution significantly.
Initial Conditions
Initial conditions are the values of the solution and sometimes its derivatives at the initial time. They are crucial to solving PDEs since they define the starting scenario of the problem.
In your problem, the initial condition is: \( \phi(x, y, 0) = f(x, y) \). This initial condition is used to determine \( \hat{\phi}(k_x, k_y, 0) \) in the Fourier domain and directly influences the form of the solution at later times.

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Most popular questions from this chapter

Evaluate the following real integrals by residue integration: (a) \(\int_{-\infty}^{\infty} \frac{x \sin x}{\left(x^{2}+a^{2}\right)} d x ; \quad a^{2}>0\) (b) \(\int_{-\infty}^{\infty} \frac{\cos k x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x ; \quad a^{2}, b^{2}, k>0\) (c) \(\int_{-\infty}^{\infty} \frac{x \cos k x}{x^{2}+4 x+5} d x ; \quad k>0\) (d) \(\int_{0}^{\infty} \frac{\cos k x}{x^{4}+1} d x, \quad k\) real (e) \(\int_{0}^{\infty} \frac{x^{3} \sin k x}{x^{4}+a^{4}} d x ; \quad k\) real, \(a^{4}>0\) (f) \(\int_{0}^{2 \pi} \frac{d \theta}{1+\cos ^{2} \theta}\) (g) \(\int_{0}^{\pi / 2} \sin ^{4} \theta d \theta\) (h) \(\int_{0}^{2 \pi} \frac{d \theta}{(5-3 \sin \theta)^{2}}\) (i) \(\int_{-\infty}^{\infty} \frac{\cos k x \cos m x}{\left(x^{2}+a^{2}\right)} d x, \quad a^{2}>0, k, m\) real.

Show that the Fourier transform of the "Gaussian" \(f(x)=\exp \left(-\left(x-x_{0}\right)^{2} / a^{2}\right) x_{0}, a\) real, is also a Gaussian: $$ \hat{F}(k)=a \sqrt{\pi} e^{-(k a / 2)^{2}} e^{-i k x_{0}} $$

Evaluate the following real integrals. (a) \(\int_{0}^{\infty} \frac{d x}{x^{2}+a^{2}}, \quad a^{2}>0\) (Verify your answer by using usual antiderivatives.) (b) \(\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a^{2}\right)^{2}}, \quad a^{2}>0\) (c) \(\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}, \quad a^{2}, b^{2}>0\) (d) \(\int_{0}^{\infty} \frac{d x}{x^{6}+1}\)

Show explicitly that the Laplace transform of the second derivative of a function of \(x\) satisfies $$ \int_{0}^{\infty} f^{\prime \prime}(x) e^{-s x} d x=s^{2} \hat{F}(s)-s f(0)-f^{\prime}(0) $$

Use the transformation \(t=(x-1) /(x+1)\) on the principal branch of the following functions to show that (a) \(\quad \int_{-1}^{1}\left(\frac{1+t}{1-t}\right)^{k-1} d t=\frac{2(1-k) \pi}{\sin k \pi}, \quad 0
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