Lecture6. Partial Differential Equations

Transcription

1 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring ecture6. Partial Differential Equations 6.1 Review of Differential Equation We have studied the theoretical aspects of the solution of linear homogeneous differential equations with constant coefficient equation ay + by + cy = 0 in practice. We have the sum of a function and its derivatives equal to zero, so the derivatives must have the same form as the function. Therefore we expect the function to be e x. If we insert this in the equation, we obtain: am 2 + bm + c = 0 This is called the auxiliary equation of the homogeneous differential equation ay + by + cy = 0. If we solve the characteristic equation, we will see three different possibilities: Two real roots, double real root and complex conjugate roots. 6.2 Partial Differential Equations A differential equation in which partial derivatives occur is called a partial differential equation. Mathematical models of physical phenomena involving more than one independent variable often include partial differential equations. They also arise in such diverse areas as epidemiology (for example, multivariable predator/prey models of AIDS), traffic flow studies and the analysis of economies. 1

2 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring We will be primarily concerned in this part with three broadly defined kinds of phenomena: wave motion, radiation or conduction of energy, and potential theory. Models of these phenomena involve partial differential equations called, respectively, the wave equation, the heat equation, and the potential equation, or aplace's equation. Partial differential equations are much more complicated compared with the ordinary differential equations. There is no universal solution technique for nonlinear equations, even the numerical simulations are usually not straightforward. Thus, we will mainly focus on the linear partial differential equations and the equations of special interests in engineering and computational sciences. A partial differential equation (PDE) is a relationship containing one or more partial derivatives. Similar to the ordinary differential equation, the highest nth partial derivative is referred to as the order n of the partial differential equation. The general form of a partial differential equation can be written as φ(x, y, u x, u y, 2 u x 2, 2 u y 2, 2 u x y, ) = 0 where u is the dependent variable and x, y,... are the independent variables. A simple example of partial differential equations is the linear first order partial differential equation, which can be written as a x, y u u + b x, y = f(x, y) x y for two independent variables and one dependent variable u. If the right hand side is zero or simply f(x, y) = 0, then the equation is said to be homogeneous. The equation is said to be linear if a, b and f are unctions of x, y only, not u itself. For simplicity in notations in the studies of partial differential equations, compact subscript forms are often used in the literature. They are u x u x, u y u y, u xx 2 u x 2, u yy 2 u y 2, u xy 2 u x y and thus we can write the general of one dimensional PDE as a u x + b u y = f. In this chapter generally we deal with second order differential equation. 6.3 Classification of PDE A linear second-order partial differential equation can be written in the generic form in terms of two independent variables x andy, a u xx + b u yx + c u yy + g u x + h u y +k u = f. where a, b, c, g, h, k and f are functions of x and y only. If f ( x, y, u) is also a function of u, then we say that this equation is quasi-linear. If, = b 2 4ac < 0 the equation is elliptic. One famous example is the aplace equation u xx + u yy = 0 If Δ > 0, it is hyperbolic. One example is the wave equation u tt =c 2 u xx. If Δ = 0, it is parabolic. Diffusion and heat conduction equations are of the parabolic type u t =k u xx. Examples of PDEs Many physical processes in engineering are governed by three classic partial differential equations so they are widely used in a vast range of applications. Among the most frequently encountered PDEs are the following: 1. aplace's equation 2 φ = 0 This very common and very important equation occurs in studies of a. electromagnetic phenomena including electrostatics, dielectrics, steady 2

3 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring currents, and magnetostatics, b. hydrodynamics (irrotational flow of perfect fluid and surface waves), с. heat flow, and d. gravitation. 2. Poisson's equation 2 φ = ρ/ε 0. In contrast to the homogeneous aplace equation, Poisson's equation is non-homogeneous with a source term -ρ/ε The wave (Helmholtz) and time-independent diffusion equations 2 φ k 2 φ = 0 These equations appear in such diverse phenomena as a. elastic waves in solids including vibrating strings, bars, membranes, b. sound or acoustics, с electromagnetic waves, and d. nuclear reactors. 4. The time-dependent diffusion equation 2 φ = 1 a 2 φ t and the corresponding four-dimensional forms involving the d'alembertian, a four-dimensional analog of the aplacian in Minkowski space 5. The time-dependent wave equation, 2 φ = 0 6. The scalar potential equation 2 φ = ρ ε 0. ike Poisson's equation this equation is nonhomogeneous with a source term ρ ε The Klein-Gordon equation 2 φ = μ 2 φ and the corresponding vector equations in which the scalar function φ is replaced by a vector function. Other more complicated forms are common. 8. The Schrodinger wave equation and for the time-independent case. 9. The equations for elastic waves and viscous fluids and the telegraphy equation. 10. Maxwell's coupled partial differential equations for the electric and magnetic fields and those of Dirac for relativistic electron wavefunctions. 6.4 Techniques for Solving PDEs Different types of equations usually require different solution techniques. However, there are some methods that work for most of the linearly partial differential equations with appropriate boundary conditions on a regular domain. These methods include separation of variables, series expansions, similarity solutions, hybrid methods, and integral transform methods. 3

4 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring i) Solution by direct integration The simplest form of partial differential equation is such that a solution can be determined by direct partial integration. Example 1: Solve the differential equation u xx = 12 x 2 (t+1) given that x = 0, u = cos(2t) and u x = sint. Notice that the boundary conditions are functions of t and not just constants. Integrating the equation u xx = 12 x 2 (t+1) partially with respect to x, we have u x = 4 x 3 (t+1) + f(t) where the arbitrary function f(t) takes the place of the normal arbitrary constant in ordinary integration. Integrating partially again with respect to x gives u = x 4 (t+1) + x f(t)+ g(t) where g(t) is a second arbitrary function. To find the two arbitrary functions, we apply the given initial conditions that x = 0, u = cos(2t) and u x = sint. Subtituting in the relevant equations gives then the solution becomes f(t) = sint and g(t) = cos(2t) u = x 4 (t+1) + x sint + cos(2t) Example 2: Solve the differential equation u xy = sin(x+y) given that at y = 0, u x =1 and at x = 0, u = (y-1) 2. Initial conditions and boundary conditions: As with any differential equation, the arbitrary constants or arbitrary functions in any particular case are determines from the additional information given concerning the variables of the equation. These extra facts are called the initial conditions or, more generally the boundary conditions since they do not lways refer to zero values of the independent variables. Example 3: Solve the differential equation u xy = sin(x)cos(y), subject to the boundary conditions that at y = π/2, u x =2x and at x = π, u = 2sin(y). 4

5 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring ii) Separation of Variables The separation of variables attempts a solution of the form u = X(x)Y(y)Z(z)T(t) where X(x), Y(y), Z(z) and T(t) are functions of x, y, z, t, respectively. In order to determine these functions, they have to satisfy the partial differential equation and the required boundary conditions. As a result, the partial differential equation is usually transformed into two or three ordinary differential equations (ODEs), and these ordinary differential equations often appear as eigenvalue problems. The final solution is then obtained by solving these ODEs. A solution that has this form is said to be separable in x, y, z and t, and seeking solutions of this form is called the method of separation of variables. As simple examples we may observe that, of the functions (i) xyz 2 sin bt, (ii) xy + zt, (iii) (x 2 + y 2 )z cos ωt, (i) is completely separable, (ii) is inseparable in that no single variable can be separated out from it and written as a multiplicative factor, whilst (iii) is separable in z and t but not in x and y. The following examples are to illustrate the method of solution by studying the wave equation, the heat equation and the aplace equation. a) The wave equation The wave equation is c 2 2 u = 2 φ 2 t = u tt governs a wide variety a wave phenomena such as electromagnetic waves, water waves, supersonic flow, pulsatile blood flow, acoustics, elastic waves in solids, and vibrating strings and membranes. In this introductory section we derive the wave equations governing the vibrating string and vibrating membrane and outline, in the exercises, several other such cases leading to wave equations. Vibrating String with Zero Initial Velocity Consider an elastic string of length, fastened at its ends on the x axis at x = 0 and x =. The string is displaced, then released from rest to vibrate in the x, y plane. We want to find the displacement function y(x, t), whose graph is a curve in the x, y plane showing the shape of the string at time t. If we took a snapshot of the string at time t, we would see this curve. The boundary value problem for the displacement function is 5

6 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring The Fourier method, or separation of variables, consists of attempting a solution of the form y(x, t) = X(x)T(t). Substitute this into the wave equation to obtain XT" = c 2 X" T, where T' = dt/dt and X' = dx/dx. Then The left side of this equation depends only on x, and the right only on t. Because x and t are independent, we can choose any to we like and fix the right side of this equation at the constant value T"(t o )/c 2 T(t o ), while varying x on the left side. Therefore X"/X must be constant for all x in (0, ). But then T"/c 2 T must equal the same constant for all t > 0. Denote this constant -. (The negative sign is customary and convenient, but we would arrive at the same final solution if we used just ). A is called the separation constant, and we now have Then X" + X = 0 and T" + c 2 T = 0. The wave equation has separated into two ordinary differential equations. Now consider the boundary conditions. First, y(0, t) = X(0)T(t) = 0 for t 0. If T(t) = 0 for all t 0, then y(x, t) = 0 for 0 x and t 0. This is indeed the solution if f(x) = 0, since in the absence of initial velocity or a driving force, and with zero displacement, the string remains stationary for all time. However, if T(t) 0 for any time, then this boundary condition can be satisfied only if X(0)=0 Similarly, y(, t) = X()T(t) = 0 for t 0 requires that X()=0. We now have a boundary-value problem for X: X" + X = 0 ; X (0) = X () = 0. The values of for which this problem has nontrivial solutions are the eigenvalues of this problem, and the corresponding nontrivial solutions for X are the eigenfunctions. By solving we get The eigenfunctions are nonzero constant multiples of X n x = sin nπx for n = 1, 2,.... At this point we therefore have infinitely many possibilities for the separatio n 6

7 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring constant and for X(x). Now turn to T(t). Since the string is released from rest, This requires that T' (0) = 0. The problem to be solved for T is therefore T" + c 2 T = 0; T' (0) = 0. However, we now know that A can take on only values of the form n 2 π 2 / 2, so this problem is really The differential equation for T has general solution Now T" +( n 2 π 2 c 2 / 2 )T = 0; T (0) = 0. T t = a cos nπct T 0 = nπc b = 0 so b = 0. We therefore have solutions for T(t) of the form + b sin nπct T n t = c n cos nπct for each positive integer n, with the constants c, as yet undetermined. We now have, f o r n = 1, 2,...., functions y n x, t = c n sin nπx cos nπct Each of these functions satisfies both boundary conditions and the initial condition y t (x, 0) = 0. We need to satisfy the condition y(x, 0) = f(x). It may be possible to choose some n so that y n (x, t) is the solution for some choice of c n. For example, suppose the initial displacement is f x = 14sin 3πx Now choose n = 3 and c 3 = 14 to obtain the solution y x, t = 14 sin 3πx cos 3πct This function satisfies the wave equation, the conditions y(0) = y() = 0, the initial condition y(x, 0) = 14sin(3 x/), and the zero initial velocity condition y t (x, 0) = 0 However, depending on the initial displacement function, we may not be able to get by simply by picking a particular n and c n. For example, if we initially pick the string up in the middle and have initial displacement function (as in Figure ), then we can never satisfy y(x, 0) = f(x) with one of the y n s. Even if we try a finite linear combination 7

8 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring we cannot choose c 1,..., c N to satisfy y(x, 0) = f(x) for this function, since f(x) cannot be written as a finite sum of sine functions. We are therefore led to attempt an infinite superposition We must choose the c N to satisfy y x, t = c n sin nπx cos nπct y x, 0 = c n sin nπx We can do this! The series on the right is the Fourier sine expansion of f(x) on [0, ]. Thus choose the Fourier sine coefficients c n = 2 f(x)sin nπx dx 0 With this choice, we obtain the solution y x, t = 2 0 f(x)sin nπx dx sin nπx cos nπct This strategy will work for any initial displacement function f which is continuous with a piecewise continuous derivative on [0, ], and satisfies f(0) = f() = 0. These conditions ensure that the Fourier sine series of f(x) on [0, ] converges to f(x) for 0 x. In specific instances, where f(x) is given, we can of course explicitly compute the coefficients in this solution. For example, if = π and the initial position function is f(x) = x cos(5x/π) on [0, π], then the nth coefficient in the solution is c n = 2 π cos 5x nπx 8 n( 1) n+1 sin dx = π 0 2 π π(5 + 2n) 2 (5 2n) 2 The solution for this initial displacement function, and zero initial velocity, is y x, t = 8 n( 1) n+1 π π(5 + 2n) 2 (5 2n) 2 sin (nx)cos (nct) Vibrating String with Given Initial Velocity and Zero Initial Displacement Now consider the case that the string is released from its horizontal position (zero initial displacement), but with an initial velocity given at x by g(x). The boundary value problem for the displacement function is 8

9 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring We begin as before with separation of variables. Put y(x, t) = X(x)T(t). Since the partial differential equation and boundary conditions are the same as before, we again obtain with eigenvalues X" + X = 0; X (0) = X() = 0, and eigenfunctions constant multiples of X n x = sin nπx Now, however, the problem for T is different and we have so T(0) = 0. The problem for T is y(x, 0) = 0 = X(x)T(0), T" +( n 2 π 2 c 2 / 2 )T = 0; T(0) = 0. (In the case of zero initial velocity we had T'(0) = 0). The general solution of the differential equation for T is T t = a cos nπct + b sin nπct Since T(0) = a = 0, solutions for T(t) are constant multiples of sin(nπct/). Thus, for n = 1, 2,..., we have functions y n x, t = c n sin nπx cos nπct Each of these functions satisfies the wave equation, the boundary conditions and the zero initial displacement condition. To satisfy the initial velocity condition y t (x, 0) = g(x), we generally must attempt a superposition y x, t = c n sin nπx Assuming that we can differentiate this series term by term, then y x x, 0 = c n sin nπx sin nπct nπc = g(x) This is the Fourier sine expansion of g(x) on [0, ]. Choose the entire coefficient of sin(nπx/) to be the Fourier sine coefficient of g(x) on [0, ] : c n nπc = 2 0 g x sin nπx 9 dx

10 Or EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring The solution is y x, t = c n = 2 nπc 2 nπc 0 0 g x sin nπx g x sin nπx dx dx sin nπx sin nπct For example, suppose the string is released from its horizontal position with an initial velocity given by g(x) = x (1 + cos(a π x/)). Compute 0 g x sin nπx dx = 0 x(1 + cos (πx/)) sin nπx dx The solution corresponding to this initial velocity function is If we let c = 1 and = π, the solution becomes Example 4: 10

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13 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring b) The Heat equation The conduction of heat in a uniform bar depends on the initial distribution of temperature and on the physical properties of bar, i.e. the thermal conductivity and specific heat of material, and the mass per unit length of the bar. Ends of the Bar Kept at Temperature Zero Suppose we want the temperature distribution u(x, t) in a thin, homogeneous (constant density) bar of length, given that the initial temperature in the bar at time zero in the cross section at x perpendicular to the x axis is f(x). The ends of the bar are maintained at temperature zero for all time. The one dimensional heat equation is written as where k = K/ is a positive constant depending on the material of the bar. The number k is called the diffusivity of the bar. The boundary value problem modeling this temperature distribution is We will use separation of variables. Substitute u(x, t) = X(x)T(t) into the heat equation to get X T' = k X" T or The left side depends only on time, and the right side only on position, and these variables are independent. Therefore for some constant, Now u(0, t) = X(0)T(t) = 0. If T(t) = 0 for all t, then the temperature function has the constant value zero, which occurs if the initial temperature f(x) = 0 for 0 x. Otherwise T(t) cannot be identically zero, so we must have X(0) = 0. Similarly, u(, t) = X()T(t) = 0 implies that X() = 0. The problem for X is therefore X" + X = 0; X(0) = X() = 0. We seek values of A (the eigenvalues) for which this problem for X has nontrivial solutions (the eigenfunctions). This problem for X is exactly the same one encountered for the space-dependent function in separating variables in the wave equation. There we found that the eigenvalues are for n = 1, 2,..., and corresponding eigenfunctions are nonzero constant multiples of 13

14 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring The problem for T becomes X n x = sin nπx T +( n 2 π 2 k / 2 )T = 0 which has general solution T n t = c n e n2 π 2 kt / 2 For n = 1, 2,, we now have functions u n x, t = c n sin nπx e n2 π 2 kt / 2 which satisfy the heat equation on [0, ] and the boundary conditions u(0, t) = u(, t) = 0. There remains to find a solution satisfying the initial condition. We can choose n and c n, so that u n x, 0 = c n sin nπx = f(x) only if the given initial temperature function is a multiple of this sine function. This need not be the case. In general, we must attempt to construct a solution using the superposition u n x, t = e n2 π 2 kt / 2. c n sin nπx Now we need u x, 0 = c n sin nπx = f(x) which we recognize as the Fourier sine expansion of f(x) on [0, ]. Thus choose c n = 2 f(x)sin nπx dx 0 With this choice of the coefficients, we have the solution for the temperature distribution function: u x, t = 2 0 f(x)sin nπx dx sin nπx e n2 π 2 kt / 2 Example 5: Suppose the initial temperature function is constant A for 0 < x <, while the temperature at the ends is maintained at zero. To write the solution for the temperature distribution function, we need to compute Then, the solution is c n = 2 0 Asin nπx dx = 2A nπ 1 cos nπ = 2A nπ 1 ( 1)n u x, t = 2A nπ 1 ( 1)n sin nπx e n 2 π 2 kt / 2 Since 1 - (-1) n is zero if n is even, and equals 2 if n is odd, we need only sum over the odd integers and can write 14

15 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring u x, t = 4A π 1 2n 1 sin (2n 1)πx e (2n 1)2 π 2 kt / 2 Temperature in a Bar with Insulated Ends Consider heat conduction in a bar with insulated ends, hence no energy loss across the ends. If the initial temperature is f(x), the temperature function is modeled by the boundary value problem Attempt a separation of variables by putting u(x, t) = X(x)T(t). We obtain, as in the preceding subsection, Now X" + X = 0,T' + k T=0. implies (except in the trivial case of zero temperature) that X'(0) = 0. Similarly, implies that X'() = 0. The problem for X(x) is therefore The eigenvalues are X" + X = 0 ; X' (0) = X' () = 0. for n = 0, 1, 2,..., with eigenfunctions nonzero constant multiples of X n x = cos nπx The equation for T is now T +( n 2 π 2 k / 2 )T = 0 When n = 0 we get T o (t) = constant. For n = 1, 2,..., We now have functions T n t = c n e n2 π 2 kt / 2 u n x, t = c n cos nπx e n2 π 2 kt / 2 15

16 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring for n = 0, 1, 2,..., each of which satisfies the heat equation and the insulation boundary conditions. To satisfy the initial condition we must generally use a superposition u x, t = 1 2 c 0 + c n cos nπx e n2 π 2 kt / 2 Here we wrote the constant term (n=0) as c o /2 in anticipation of a Fourier cosine expansion. Indeed, we need u x, 0 = f(x) = 1 2 c 0 + c n cos nπx the Fourier cosine expansion of f(x) on [0, ]. (This is also the expansion of the initial temperature function in the eigenfunctions of this problem.) We therefore choose c n = 2 f(x)cos nπx dx 0 With this choice of coefficients, equation of differential equation gives the solution of this boundary value problem as u x, t = 1 2 c f(x)cos nπx dx cos nπx e n2 π 2 kt / 2 Example 6: Suppose the left half of the bar is initially at temperature A, and the right half is kept at temperature zero. Thus Then and, for n = 1, 2,..., c 0 = 2 /2 Adx = A 0 c n = 2 0 /2 Acos nπx dx = 2A nπ sin (nπ/2) The solution for this temperature function is u x, t = 1 2 A + 2A nπx sin (nπ/2) cos nπ e n2 π 2 kt / 2 Now sin(nπ/2) is zero if n is even. Further, if n = 2k - 1 is odd, then sin(nπ/2) = (-1) k+ 1. The solution may therefore be written u x, t = 1 2A A + 2 nπ ( 1) n+1 2n 1 (2n 1)πx cos e (2n 1)2 π 2 kt / 2 16

17 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring a) The Potential (aplace) equation Harmonic Functions and the Dirichlet Problem The partial differential equation 2 u x u y 2 = 0 is called aplace's equation in two dimensions. In 3-dimensions this equation is 2 u x u y u z 2 = 0 The aplacian 2 (read "del squared") is defined in 2-dimensions by and in three dimensions by 2 = 2 u x u y 2 2 = 2 u x u y u z 2 In this notation, aplace's equation 2 u = 0 A function satisfying aplace's equation in a certain region is said to be harmonic on that region. For example, x 2 - y 2 and 2xy are both harmonic over the entire plane. aplace's equation is encountered in problems involving potentials, such as potentials for force fields in mechanics, or electromagnetic or gravitational fields. aplace's equation is also known as the steady-state heat equation. The heat equation in 2- or 3-space dimensions is k 2 u = u t In the steady-state case (the limit as t ) the solution becomes independent of t, so u equation becomes aplace's equation. t = 0 and the heat In problems involving aplace's equation there are no initial conditions. However, we often encounter the problem of solving 2 u(x, y) = 0 for (x, y) in some region D of the plane, subject to the condition that u(x, y) = f(x, y), for (x, y) on the boundary of D. This boundary is denoted D. Here f is a function having given values on D, which is often a curve, or made up of several curves (Figure). The problem of determining a harmonic function having given boundary values is called a Dirichlet problem, and f is called the boundary data of the problem. There are versions of this problem in higher dimensions, but we will be concerned primarily with dimension 2. 17

18 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring Typical boundary D of a region D. The difficulty of a Dirichlet problem is usually dependent on how complicated the region D is. In general, we have a better chance of solving a Dirichlet problem for a region that possesses some type of symmetry, such as a disk or rectangle. We will begin by solving the Dirichlet problem for some familiar regions in the plane. Dirichlet Problem for a Rectangle et R be a solid rectangle, consisting of points (x, y) with 0 x, 0 y K. We want to find a function that is harmonic at points interior to R, and takes on prescribed values on the four sides of R, which form the boundary R of R. This kind of problem can be solved by separation of variables if the boundary data is nonzero on only one side of the rectangle. We will illustrate this kind of problem, and then outline a strategy to follow if the boundary data is nonzero on more than one side. Example 7: Consider the Dirichlet problem 2 u(x,y)=0 for 0 < x <, 0 < y < K, u(x,0)=0 for 0 x, u(0, y) = u(, y) = 0 for 0 y K, u(x, K) = ( - x) sin(x) for 0 x. where the boundary data are shown in figure. et u(x, y) = X(x)Y(y) and substitute into aplace's equation to obtain Then From the boundary conditions, X" + X = 0 and Y"- Y = 0. u(x, 0) = X(x)Y(0) = 0 Boundary data given on boundary sides of the rectangle. 18

19 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring so Y(0) = 0. Similarly, X (0) = X () = 0. The problem for X(x) is a familiar one, with eigenvalues n = n 2 π 2 / 2 and eigenfunctions that are nonzero constant multiples of sin(nπx/). The problem for Y is now Y -( n 2 π 2 / 2 )Y = 0; Y(0)=0 Solutions of this problem are constant multiples of sinh(nπy/). For each positive integer n = 1, 2,..., we now have functions u n x, y = b n sin nπx sinh nπy which are harmonic on the rectangle, and satisfy the zero boundary conditions on the top, bottom and left sides of the rectangle. To satisfy the boundary condition on the side y = K, we must use a superposition Choose the coefficients so that u x, y = b n sin nπx sinh nπy u x, K = b n sin nπx sinh nπk = ( x)sin (x) This is a Fourier sine expansion of ( - x) sin(x) on [0, ], so we must choose the entire coefficient to be the sine coefficient: b n sinh ( nπx ) = 2 ( x)sin (x)sin ( nπx )dx 0 b n sinh ( nπk ) = nπ 1 ( 1)n cos () n 2 π 2 + n 4 π 4 Then, The solution is u x, y = b n = 4 2 nπ 1 ( 1) n cos () sinh ( nπx ) ( 2 n 2 π 2 ) nπ 1 ( 1) n cos () sinh ( nπx ) ( 2 n 2 π 2 ) 2 sin nπx sin nπy Dirichlet Problem for a Disk We will solve the Dirichlet problem for a disk of radius R centered about the origin. In polar coordinates, the problem is aplace's equation in polar coordinates is It is easy to check that the functions 19

20 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring 1, r n cos(nθ), and r n sin(nθ) are all harmonic on the entire plane. Thus attempt a solution To satisfy the boundary condition, we need to choose the coefficients so that But this is just the Fourier expansion of f(θ) on [-π, π], leading us to choose and Then and The solution is This can also be written or Example 8: Solve the Dirichlet problem for The solution is 20

21 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring Example 8: Solve the Dirichlet problem for Convert the problem to polar coordinates, using x = rcos(θ) and y = r sin(θ). et u(x, y) = u(rcos(θ), rsin(θ)) = U(r, θ). The condition on the boundary, where r = 3, becomes The solution is Now and Therefore To convert this solution back to rectangular coordinates, use the fact that Then 21

22 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring Example 9: 22

23 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring 6.5 Exercises 1. Answer: 2. Answer: 23

24 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring 3. Answer: 4. Answer: 5. Answer: 6. Answer: 24

25 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring 7. Answer: 8. Answer: 9. Answer: 10. Answer: 11. Answer: 12. Answer: 25

26 EP219 ecture notes - prepared by- Assoc. Prof. Dr. Eser OĞAR 2012-Spring 13. Answer: 14. Answer: 15. Answer: 16. Answer: 17. Answer: 18. Answer: 19. Answer: 20. Answer: 21. Answer: 22. Answer: 23. Answer: References: 1. Advanced Engineering Mathematics, International Student Edition, Peter V. O'Neil. 2. Advanced Engineering Mathematics, K.A. Stroud, 4th edition. 3. Mathematical methods for physics and engineering, K.F. Riley, M.P. Hobson and S. J. Bence, 3th edition. 26