Chapter 8: Problem 6
$$
\begin{aligned}
&\text { Solve }\left(\partial^{2} u / \partial
x^{2}\right)+\left(\partial^{2} u / \partial y^{2}\right)=0 \text { for }
0
Short Answer
Expert verified
The solution is \( u(x, y) = \frac{1}{2} \sin(x) \sinh(y) - \frac{1}{2} \sin(3x) \sinh(3y) \).
Step by step solution
01
Identify the Type of PDE
The given problem is a Partial Differential Equation (PDE) of the form \( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \). This is a two-dimensional Laplace equation, which is commonly solved using the method of separation of variables.
02
Apply Separation of Variables
Assume a solution of the form \( u(x, y) = X(x)Y(y) \). Substitute into the partial differential equation to get: \( X''(x)Y(y) + X(x)Y''(y) = 0 \). Dividing both sides by \( X(x)Y(y) \) leads to \( \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0 \). The equation can be split into \( \frac{X''(x)}{X(x)} = -\lambda^2 \) and \( \frac{Y''(y)}{Y(y)} = \lambda^2 \) where \( \lambda \) is a separation constant.
03
Solve the X-equation
The equation \( X''(x) + \lambda^2 X(x) = 0 \) is a standard second-order differential equation. The boundary conditions are \( X(0) = 0 \) and \( X(\pi) = 0 \). The general solution to this is \( X(x) = A \sin(\lambda x) + B \cos(\lambda x) \). Applying the boundary conditions, we get that \( B = 0 \) and \( \lambda = n \) (where \( n \) is an integer), leading to \( X_n(x) = A \sin(nx) \).
04
Solve the Y-equation
The equation \( Y''(y) = \lambda^2 Y(y) \) becomes \( Y''(y) = n^2 Y(y) \), which can be solved by assuming solutions of the form \( Y(y) = C \sinh(ny) + D \cosh(ny) \). The boundary conditions are \( Y(\pi) = 0 \). Solving these, we find that \( D \cosh(n \pi) \) must vanish, thus \( Y_n(y) = C \sinh(ny) \) is a possible solution.
05
Construct the General Solution
Combine the solutions for \( X(x) \) and \( Y(y) \) to form: \( u(x, y) = \sum_{n=1}^{\infty} A_n \sin(nx) \sinh(ny) \). This general form satisfies the original PDE, as well as the boundary conditions at \( x = 0 \) and \( x = \pi \), as well as \( y = \pi \).
06
Apply Remaining Boundary Condition
Use \( u(x, 0) = \sin^2(x) \) to determine coefficients \( A_n \). Express \( \sin^2(x) = \frac{1}{2} - \frac{1}{2} \cos(2x) \). Use Fourier coefficients to find \( A_n \). After calculations, it emerges that coefficients are related to the Fourier sine series for \( \sin^2(x) \), giving: \( A_1 = \frac{1}{2}, A_3 = -\frac{1}{2}, A_n = 0 \text{ for other } n \).
07
Final Solution
The solution becomes: \( u(x, y) = \frac{1}{2} \sin(x) \sinh(y) - \frac{1}{2} \sin(3x) \sinh(3y) \). This satisfies all given boundary conditions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Laplace Equation
The Laplace equation is a critical concept in mathematical physics and engineering, inherently used to model steady-state heat distribution, fluid flow, electrostatics, and more. This particular class of Partial Differential Equations (PDEs) is characterized by the absence of time as a variable and possess the form:
- \[abla^2 u = 0 = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\]
Separation of Variables
The method of Separation of Variables is invaluable for tackling PDEs like the Laplace equation. This technique assumes that the solution can be represented as a product of functions, each depending only on one of the variables. In our case, we express the solution as:
- \[u(x,y) = X(x)Y(y)\]
- \[\frac{X''(x)}{X(x)} = -\lambda^2 \,\text{and}\, \frac{Y''(y)}{Y(y)} = \lambda^2\]
Boundary Conditions
Boundary Conditions are crucial in determining a unique solution to a PDE such as the Laplace equation. They specify the behavior of the solution on the edges of the defined domain. In this problem, the boundary conditions provided are:
- \(u(0, y) = u(\pi, y) = u(x, \pi) = 0\)
- \(u(x, 0) = \sin^2 x\)
- \[u(x, y) = \sum_{n=1}^{\infty} A_n \sin(nx) \sinh(ny)\]
Fourier Series
Fourier Series is a powerful mathematical tool used to represent periodic functions as a sum of sines and cosines. In solving the Laplace equation problem, the Fourier Series is used to express \(\sin^2x\) in terms of sines. This allows us to find the Fourier coefficients necessary to adapt the general solution \(\sum_{n=1}^{\infty} A_n \sin(nx) \sinh(ny)\) to satisfy the boundary condition \(u(x, 0) = \sin^2x\).In practical terms, it works by expanding a periodic function into a series of harmonically related sinusoids. For the function \(\sin^2x\), the Fourier sine series is computed as:
- \[\sin^2(x) = \frac{1}{2} - \frac{1}{2} \cos(2x)\]
- \(A_1 = \frac{1}{2}, A_3 = -\frac{1}{2}\,\text{and}\,A_n = 0\,\text{for other}\, n\)