Ordinary & Partial Differential Equations - Reynolds (2000) - Chapter 13 - The Laplace and Poiss

Ordinary & Partial Differential Equations - Reynolds (2000) - Chapter 13 - The Laplace and Poiss

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 13 The Laplace and Poisson Equations U p to now, we have dealt almost exclusively with pde s for which one independent variable corresponds to time . Now, we will analyze a pde for which this is not the case: Laplace’s equation. To motivate where Laplace’s equation comes from and what it models, consider the heat equation with periodic boundary conditions: u t = κ u xx ( - L x L ) u ( - L , t )= u ( L , t ) u x ( - L , t u x ( L , t ) . Physically, this system models diffusion of heat within a thin “one-dimensional” ring of wire. You can imagine “bending” the interval - L x L into a ring. With the two ends x = - L and x = L in contact, they correspond to the same physical location (hence the two boundary conditions). Using the separation of variables technique, you can show that the solution of the above system is given by u ( x , t A 0 2 + n = 1 $ A n cos ± n π x L ² + B n sin ± n π x L ²% e - n 2 π 2 κ t / L 2 . Given an initial condition u ( x ,0 φ ( x ) for - L x L , the constants appear- ing in the formula for u ( x , t ) are given by the Fourier coef±cients A n = 1 L # L - L φ ( x ) cos ± n π x L ² d x and B n = 1 L # L - L φ ( x ) sin ± n π x L ² d x . 367
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
368 Question: How do solutions behave as t ? Intuitively, we would expect the exponential factor in the above series to decay rapidly to 0 as t . This would suggest that only the leading term survives as t : lim t u ( x , t )= A 0 2 = 1 2 L # L - L φ ( x ) d x for all x [ - L , L ] . Notice that this integral represents the average value of the initial temperature distribution φ ( x ) . Consequently, we expect the solution of the heat equation on a ring-shaped domain (periodic boundary conditions) to approach a steady-state in which heat is uniformly distributed around the ring. Moving beyond the above example, let us ask a more general question: What are the steady-state solutions of the heat equation (or the wave equation)? In other words, are there solutions the heat equation that are time-independent? Physical intuition suggests that if we apply a Fxed temperature distribution along the boundary of the spatial domain, then the heat proFle should equilibrate to some steady conFguration throughout the domain as t . Example 13 . 0 . 22 . In one spatial dimension, Fnding steady-state solutions of the heat equation is rather easy. Suppose that we wish to Fnd steady-state solutions of u t = κ u xx on the domain 0 < x < L , subject to homogeneous Dirichlet boundary conditions u ( 0, t 0 = u ( L , t ) . Since such solutions are time-independent, we should set u t = 0, reducing the pde to an ode κ u ± ( x 0. Integrating twice, we Fnd that u ( x C 1 x + C 2 , where C 1 and C 2 are constants. The boundary conditions imply that both of the constants are zero, which means that the only steady-state solution of this Dirichlet problem is the constant function u = 0. As an exercise, Fnd the steady-state solution of the heat equation with the more general Dirichlet conditions u ( 0, t τ 1 and u ( L , t τ 2 . You should Fnd that temperature varies linearly between τ 1 and τ 2 over the domain 0 x L .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/27/2012 for the course MATH 532 taught by Professor Reynolds during the Fall '11 term at VCU.

Page1 / 37

Ordinary & Partial Differential Equations - Reynolds (2000) - Chapter 13 - The Laplace and Poiss

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online