Module 7: The Laplace Equation

Transcription

1 Module 7: The Laplace Equation In this module, we shall study one of the most important partial differential equations in physics known as the Laplace equation 2 u = 0 in Ω R n, (1) where 2 u := n i=1 2 u is the Laplacian of the function u. The theory of the solutions of x 2 i Laplace s equation is called potential theory. The equation (1) is often referred to as the potential equation as the function u is frequently a potential function. Solutions of (1) that have continuous second-order partial derivatives are called harmonic functions. For easy of exposition, we shall study Laplace s equation in two dimensions. This module consists of five lectures. The first lecture introduces some basic concepts and the maximum and minimum principle for boundary value problems (BVP). In the second lecture, we discuss the Green s identities, fundamental solution of the Laplace equation and the Poisson integral formula. The solution of the Laplace equation for rectangular region is discussed in the third lecture. The mixed BVP for a rectangle is discussed in the fourth lecture. In the fifth lecture, we solve the Laplace equations for the annular region between concentric circles. Finally, the sixth lecture is devoted to the interior and exterior Dirichlet problems for the Laplace equations. 1

2 MODULE 7: THE LAPLACE EQUATION 2 Lecture 1 Basic Concepts and The Maximum/Minimum Principle Let Ω be an open region in R 2. The Laplace equation in two dimension is of the form 2 u(x, y) = 0, (x, y) Ω, (1) where 2 := is the Laplace operator or the Laplacian. The equation of the type x 2 x 2 (1) plays an important role in a variety of physical contexts such as in Gravitation theory, electrostatics, steady-state heat conduction problems and fluid flow problems. Some examples of physical problems(cf. [10]): EXAMPLE 1. (Gravitation theory) The force of attraction F, both inside and outside the attracting matter, can be expressed in terms of a gravitational potential u by the equation F = u. In empty space u satisfies Laplace s equation 2 u = 0. EXAMPLE 2. (Steady-state heat flow problem) In the theory of heat conduction if the temperature u does not vary with the time, then u satisfies the equation (κ u) = 0, where κ is the thermal conductivity. If κ is a constant throughout the medium then 2 u = 0. EXAMPLE 3. (Fluid flow problem) The velocity q of a perfect fluid in irrotational motion can be expressed in terms of a velocity potential u by the equation q = u. If there are no sources or sinks at all points of the fluid the function u satisfies Laplace s equation 2 u = 0. The inhomogeneous Laplace equation 2 u(x, y) = f(x, y) in Ω, where f is a given function is known as the Poisson equation.

3 MODULE 7: THE LAPLACE EQUATION 3 1 Types of BVP Because these solutions do not depend on time, initial conditions are irreverent and only boundary conditions are specified. There are three basic types of boundary conditions that are usually associated Laplace s equation. They are Dirichlet BVP: If the BC are of Dirichlet type i.e., if the solution u(x, y) to Laplace equation in a domain Ω is specified on the boundary i.e., u(x, y) = f(x, y) on, where f(x, y) is a given function. The Laplace equation together with Dirichlet BC are called the Dirichlet problem / Dirichlet BVP. The Dirichlet problem for Laplace equation is of the form 2 u(x, y) = 0 in Ω; u(x, y) = f(x, y) on. Neumann BVP: We know the BC are of Neumann type if the directional derivative along the outward normal to the boundary is specified on i.e., (x, y) = g(x, y) for (x, y). In physical terms, the normal component of the solution gradient is known on the boundary. In steady-state heat flow problem, Neumann BC means the rate of heat loss or gain through the boundary points is prescribed. The Laplace equation together with Neumann BC are called the Neumann BVP/ Neumann problem which is written as 2 u = 0 in Ω; (x, y) = g(x, y) for (x, y). The Neumann problem will have no solution unless we assume that the average value of the function g on is zero. This assumption is known as the compatibility condition = g = 0, which will be discussed in the next lecture. Robin s BVP. The boundary conditions are called Robin s type or mixed type if Dirichlet BC are specified on part of the boundary and Neumann type BC are specified on the remaining part of the boundary. For example, + c(u g) = 0,

4 MODULE 7: THE LAPLACE EQUATION 4 where c is a constant and g is a given function that can vary over the boundary. The Laplace equation together with the Rabin s/mixed BC known as Rabin s BVP / Mixed BVP. 2 The maximum/minimum principle The maximum/minimum principle for Laplace s equation is stated in the following theorem. THEOREM 4. (The maximum/minimum principle for Laplace s equation) Let u(x, y) C 2 (Ω) C() be a solution of Laplace s equation 2 u(x, y) := u xx + u yy = 0 (2) in a bounded region Ω with boundary. Then the maximum and minimum values of u attain on. That is, Proof. max u(x, y) = max u(x, y); and min u(x, y) = min u(x, y). Since u is continuous in it attains its maximum either in Ω or on. Suppose u achieves its maximum at some point (x 0, y 0 ) Ω. Let u(x 0, y 0 ) = max Ω u(x, y) = M 0 > M b, where M b = max u(x, y). Consider the function v(x, y) = u(x, y) + ϵ[(x x 0 ) 2 + (y y 0 ) 2 ], (3) for some ϵ > 0. Note that v(x 0, y 0 ) = u(x 0, y 0 ) = M 0 and max v(x, y) M b + ϵd 2, where d is the diameter of Ω. For such ϵ (0 < ϵ < (M 0 M b )/d 2 ), the maximum of v can not occur on because M 0 = v(x 0, y 0 ) > max v(x, y). This implies there may be points in Ω where v > M 0. Let At (x 1, y 1 ), we must have v(x 1, y 1 ) = max v(x, y). Ω v xx 0 and v yy 0 = v xx + v yy 0. (4)

5 MODULE 7: THE LAPLACE EQUATION 5 From (3), we observe that v xx + v yy = u xx + u yy + 2ϵ + 2ϵ = 4ϵ > 0, where we have used the fact that u xx + u yy = 0. This led to a contradiction to (4). Thus, max So, the maximum of u attains on. v(x, y) max v(x, y). Ω To prove that the minimum of u is also achieved on the boundary, replace u by u in the above argument to obtain This completes the proof. min u = max ( u) = max( u) = min (u). We now discuss the maximum and minimum principle for Poisson s equation 2 u(x, y) = f(x, y) in Ω. (5) THEOREM 5. (The maximum/minimum principle for Poisson s equation) Let Ω be a bounded domain in R 2 with boundary. Then the maximum values of a solution u of (5) attain on if f(x, y) > 0 in Ω and the minimum values of u occur on if f(x, y) < 0 in Ω. Proof. Since u is continuous in a closed and bounded domain, it must assume its maximum in Ω or in. Suppose that the maximum is assumed at a point (x 0, y 0 ) in Ω, i.e., u(x 0, y 0 ) = max u(x, y). Suppose that f(x, y) > 0 in Ω. Then at (x 0, y 0 ) Ω, we must have As f > 0, it follows from (5) that u xx (x 0, y 0 ) 0, u yy (x 0, y 0 ) 0. u xx + u yy > 0, which is a contradiction. Hence, the maximum of u(x, y) must occur on. To show that the minimum of u(x, y) attains on if f(x, y) < 0 in Ω, replace u by u in the preceding argument. This is equivalent to replacing f by f in (4). Since f < 0, we obtain f > 0 and conclude that u assumes its maximum on. Therefore, u assumes its minimum on and this completes the proof. The maximum/minimum principle can be used to prove uniqueness and continuous dependence of the solution for the Dirichlet s problems.

6 MODULE 7: THE LAPLACE EQUATION 6 THEOREM 6. Let Ω be a bounded domain in R 2 with boundary. The solution of the Dirichlet s problem 2 u(x, y) = f(x, y) in Ω, u(x, y) = g(x, y) on (6) if it exists, is unique. Proof. Let u 1 (x, y) and u 2 (x, y) be two solutions of (6). Set v(x, y) = u 1 (x, y) u 2 (x, y). Then v satisfies 2 v = 0 in Ω, v = 0 on. The maximum/minimum principle yields (cf. Theorem 4) v = 0 in Ω = u 1 u 2 = 0 in Ω. Thus, we have u 1 = u 2, which proves the uniqueness. Next, we shall prove the continuous dependence of the solution on the boundary data. THEOREM 7. The solution of the Dirichlet problem depends continuously on the boundary data. Proof. Let u i, i = 1, 2 be the solutions of Then the function v = u 1 u 2 solves 2 u i = F in Ω R 2, u i = f i on. 2 v = 0 in Ω with v = f 1 f 2 on. By the maximum/minimum principle v attains its maximum/minimum on. Thus, for all (x, y), we have max ( f 1 f 2 ) min (f 1 f 2 ) v(x, y) max (f 1 f 2 ) max ( f 1 f 2 ). If f 1 f 2 < ϵ then ϵ < min v(x, y) v(x, y) max v(x, y) < ϵ. Therefore, f 1 f 1 < ϵ = v(x, y) < ϵ for all (x, y). This completes the proof.

7 MODULE 7: THE LAPLACE EQUATION 7 Practice Problems 1. Let u satisfy the Laplace equation in a disk Ω = {(x, y) x 2 +y 2 < 1} and continuous on. If u(cos θ, sin θ) sin θ + cos(2θ), then show that u(x, y) y + x 2 y 2, (x, y). 2. Consider the elliptic equation (α u) = F, α > 0, in a bounded region Ω R 2 with the boundary. Show that if F < 0 in Ω, the solution u assumes its maximum on and if F > 0 in Ω, the solution u assumes its minimum on. 3. Let Ω be a bounded region R 2. Use the maximum principle to prove continuous dependence on the data for the Dirichlet problem for the elliptic equation (α u) = F in Ω with α > 0.