ENGI 940 Lecture Notes 8 - PDEs Page 8.01 8. Partial Differential Equations Partial differential equations (PDEs) are equations involving functions of more than one variable and their partial derivatives with respect to those variables. Most (but not all) physical models in engineering that result in partial differential equations are of at most second order and are often linear. (Some problems such as elastic stresses and bending moments of a beam can be of fourth order). In this course we shall have time to look at only a small subset of second order linear partial differential equations. Sections in this Chapter: 8.01 Major Classifications of Common PDEs 8.0 The Wave Equation Vibrating Finite String 8.03 The Wave Equation Vibrating Infinite String 8.04 d Alembert Solution 8.05 The Maximum-Minimum Principle 8.06 The Heat Equation
ENGI 940 8.01 - Classification of PDEs Page 8.0 8.01 Major Classifications of Common PDEs A general second order linear partial differential equation in two Cartesian variables can be written as u u u u u Axy (, ) + Bxy (, ) + C( xy, ) = f xyu,,,, x x y y x y Three main types arise, based on the value of D = B 4AC (a discriminant): Hyperbolic, wherever (x, y) is such that D > 0; Parabolic, wherever (x, y) is such that D = 0; Elliptic, wherever (x, y) is such that D < 0. Among the most important partial differential equations in engineering are: u The wave equation: = c u u u or its one-dimensional special case = c x (where u is the displacement and c is the speed of the wave); u The heat (or diffusion) equation: µρ = K u + K u a one-dimensional special case of which is u K u = µρ x (where u is the temperature, μ is the specific heat of the medium, ρ is the density and K is the thermal conductivity); The potential (or Laplace s) equation: a special case of which is u x u y + = 0 u = 0 The complete solution of a PDE requires additional information, in the form of initial conditions (values of the dependent variable and its first partial derivatives at t = 0), boundary conditions (values of the dependent variable on the boundary of the domain) or some combination of these conditions.
ENGI 940 8.0 - Finite String Page 8.03 8.0 The Wave Equation Vibrating Finite String The wave equation is u = c u If u(x, t) is the vertical displacement of a point at location x on a vibrating string at time t, then the governing PDE is u u = c x If u(x, y, t) is the vertical displacement of a point at location (x, y) on a vibrating membrane at time t, then the governing PDE is u u u = c u = c + x y or, in plane polar coordinates (r, θ), (appropriate for a circular drum), u u 1 u 1 u = c + + r r r r θ 3 For spherically symmetric waves in, the wave equation is u c u u u = r c = + r r r r r r The latter two cases lead to solutions involving Bessel functions. Example 8.0.1 An elastic string of length L is fixed at both ends (x = 0 and x = L). The string is displaced into the form y = f (x) and is released from rest. Find the displacement y(x, t) at all locations on the string (0 < x < L) and at all subsequent times (t > 0). The boundary value problem for the displacement function y(x, t) is: y y = c for 0 < x < L and t > 0 x Both ends fixed for all time: y(0, t) = y(l, t) = 0 for t > 0 Initial configuration of string: y(x, 0) = f (x) for 0 < x < L String released from rest: y ( x,0) = 0 for 0 x L
ENGI 940 8.0 - Finite String Page 8.04 Example 8.0.1 (continued) Separation of Variables (or Fourier Method) Attempt a solution of the form y(x, t) = X(x) T(t) Substitute y(x, t) = X(x) T(t) into the PDE: ( X( xt ) ( t) ) c ( X( xt ) ( t) ) = x
ENGI 940 8.0 - Finite String Page 8.05 Example 8.0.1 (continued) Therefore our complete solution is L nπu nπ x nπct y ( x, t) = f ( u) sin du sin cos L 0 L L L n = 1
ENGI 940 8.0 - Finite String Page 8.06 Example 8.0.1 (continued) This solution is valid for any initial displacement function f (x) that is continuous with a piece-wise continuous derivative on [0, L] with f (0) = f (L) = 0. nπ x If the initial displacement is itself sinusoidal f ( x) = asin for some n L, then the complete solution is a single term from the infinite series, nπ x nπct y( xt, ) = asin cos L L. Example 8.0. Suppose that the initial configuration is triangular: ( 0 1 ) 1 ( ) x x L y( x,0) = f ( x) = L x L< x L Then the Fourier sine coefficients are L nπ u Cn = f ( u) sin du L 0 L
ENGI 940 8.0 - Finite String Page 8.07 Example 8.0. (continued) See the web page "www.engr.mun.ca/~ggeorge/940/demos/ex80.html" for an animation of this solution.
ENGI 940 8.0 - Finite String Page 8.08 Example 8.0. (continued) Some snapshots of the solution are shown here: These graphs were generated from the Fourier series, truncated after the fifth non-zero term.
ENGI 940 8.0 - Finite String Page 8.09 Example 8.0.3 An elastic string of length L is fixed at both ends (x = 0 and x = L). The string is initially in its equilibrium state [y(x, 0) = 0 for all x] and is released with the initial velocity y = g( x). Find the displacement y(x, t) at all locations on the string (0 < x < L) ( x,0) and at all subsequent times (t > 0). The boundary value problem for the displacement function y(x, t) is: y y = c for 0 < x < L and t > 0 x Both ends fixed for all time: y(0, t) = y(l, t) = 0 for t > 0 Initial configuration of string: y(x, 0) = 0 for 0 < x < L y String released with initial velocity: ( x,0) ( ) = g x for 0 x L As before, attempt a solution by the method of the separation of variables. Substitute y(x, t) = X(x) T(t) into the PDE: dt d X ( X( xt ) ( t )) = c ( X( xt ) ( t) ) X = c T x dt dx Again, each side must be a negative constant. 1 dt 1 d X = = λ c T dt X dx We now have the pair of ODEs d X dt + λ X = 0 and + λ ct = 0 dx dt The general solutions are X x = A cos λx + B sin λx and T t = C cos λct + Dsin λct ( ) ( ) ( ) ( ) ( ) ( ) respectively, where A, B, C and D are arbitrary constants.
ENGI 940 8.0 - Finite String Page 8.10 Example 8.0.3 (continued) Consider the boundary conditions: y 0, t = X 0 T t = 0 t 0 ( ) ( ) ( ) For a non-trivial solution, this requires X ( ) ( ) ( ) ( ) ( ) 0 = 0 A= 0. y L, t = X L T t = 0 t 0 X L = 0 nπ Bsin L = 0 n =, n L ( λ ) λ ( ) We now have a solution only for a discrete set of eigenvalues λ n, with corresponding eigenfunctions nπ x Xn ( x) = sin, ( n = 1,, 3, ) L and nπ x yn( x, t) = Xn( x) Tn( t) = sin Tn( t), ( n = 1,, 3, ) L So far, the solution has been identical to Example 8.0.1. Consider the initial condition y(x, 0) = 0 : y x,0 = 0 X xt 0 = 0 x T 0 = 0 ( ) ( ) ( ) ( ) The initial value problem for T(t) is now nπ T + λ ct = 0, T( 0) = 0, where λ = L the solution to which is nπ ct Tn( t) = Cnsin, ( n ) L Our eigenfunctions for y are now nπ x nπct yn xt, = Xn xtn t = Cnsin sin, n L L ( ) ( ) ( ) ( )
ENGI 940 8.0 - Finite String Page 8.11 Example 8.0.3 (continued) Differentiate term by term and impose the initial velocity condition: y n π c n sin n π x = C = g( x) ( x,0) n = 1 L L which is just the Fourier sine series expansion for the function g(x). The coefficients of the expansion are L nπc nπu Cn = g ( u) sin du L L 0 L which leads to the complete solution n = 1 L 1 nπu nπ x nπct y ( x, t) = g ( u) sin du sin sin πc n 0 L L L This solution is valid for any initial velocity function g(x) that is continuous with a piece-wise continuous derivative on [0, L] with g(0) = g(l) = 0. The solutions for Examples 8.0.1 and 8.0.3 may be superposed. Let y1 (, ) Let y (, ) Then y( xt, ) y( xt, ) y ( xt, ) xt be the solution for initial displacement f (x) and zero initial velocity. xt be the solution for zero initial displacement and initial velocity g(x). = + satisfies the wave equation 1 (the sum of any two solutions of a linear homogeneous PDE is also a solution), and satisfies the boundary conditions y(0, t) = y(l, t) = 0 : ( ) ( ) ( ) ( ) y x,0 = y1 x,0 + y x,0 = f x + 0, which satisfies the condition for initial displacement f (x). yt ( x,0 ) = y1t( x,0 ) + yt( x,0) = 0+ g( x), which satisfies the condition for initial velocity g(x). Therefore the sum of the two solutions is the complete solution for initial displacement f (x) and initial velocity g(x): n = 1 nπu nπ x nπct y ( x, t) = f ( u) sin du sin cos L 0 L L L 1 nπu nπ x nπct + g ( u) sin du sin sin πc n 0 L L L n = 1 L L
ENGI 940 8.0 - Finite String Page 8.1 Example 8.0.4 An elastic string of length 1 m is fixed at both ends (x = 0 and x = 1). The string is initially in the shape of an arc of a parabola [y(x, 0) = x x for 0 < x < 1] and is released with the initial velocity y ( x,0) ( 0 1) x x x =. It is known that the wave speed is c = 5 m s 1. Find the displacement y(x, t) at all locations on the string (0 < x < 1) and at all subsequent times (t > 0). In the formula for the complete solution of the wave equation, n = 1 nπu nπ x nπct y ( x, t) = f ( u) sin du sin cos L 0 L L L 1 nπu nπ x nπct + g ( u) sin du sin sin πc n 0 L L L n = 1 we know that L = 1, c = 5 and L L A Maple file for this solution is available at "www.engr.mun.ca/~ggeorge/940/demos/ex804.mws".
ENGI 940 8.03 - Infinite String Page 8.13 8.03 The Wave Equation Vibrating Infinite String Example 8.03.1 An elastic string of infinite length is displaced into the form y = f (x) and is released from rest. Find the displacement y(x, t) at all locations on the string x and at all subsequent times (t > 0). The boundary value problem for the displacement function y(x, t) is: y y = c for < x < and t > 0 x Initial configuration of string: y(x, 0) = f (x) for x String released from rest: y ( x,0) = 0 for x We no longer have the additional boundary conditions of fixed endpoints. However, it is reasonable to insist upon a bounded solution. Separation of Variables (or Fourier Method) Attempt a solution of the form y(x, t) = X(x) T(t) Again we find the linked pair of ordinary differential equations X + ω X = 0 and T + ω ct = 0 If ω = 0 then X(x) = ax + b. However, for a bounded solution, we require a = 0. For other values of ω, X(x) = a cos ωx + b sin ωx, which is bounded for all x and all ω. The ω = 0 case is a special case of this solution. We have a continuum of eigenvalues ω with corresponding eigenfunctions Xω ( x) = aω cos( ωx) + bωsin ( ωx) It then follows that T ( t) = c cosωct + d sinωct ω ω ω Imposing the initial condition of zero velocity, y = X ( x) T ( 0) = X ( x) dωωc = 0 x dω = 0 ( x,0)
ENGI 940 8.03 - Infinite String Page 8.14 Example 8.03.1 (continued) Therefore we have, for any real ω, a solution of the wave equation and the initial velocity condition, yω ( x, t) = Xω ( x) Tω ( t) = ( aω cos( ωx) + bωsin ( ωx) ) cos( ωct ) [where c ω has been absorbed into the other arbitrary constants a ω and b ω.] The superposition of solutions now leads to an integral, not a discrete sum. 0 ω 0 ω ω (, ) = (, ) = ( cos( ) + sin ( )) cos( ) y xt y xt dω a ωx b ωx ωct dω Imposing the remaining condition, (,0) = ( cos( ω ) + sin ( ω )) ω = ( ) y x aω x bω x d f x 0 But this is just the Fourier integral representation of f (x) on (, ). Therefore a ω and b ω are just the Fourier integral coefficients 1 1 aω = f ( u) cos( ωu) du and bω f ( u) sin ( ωu) du π = π The complete solution is 1 y ( x, t) = f ( u) cos( ωu) du cos( ωx) 0 π 1 + f ( u) sin ( ωu) du sin ( ωx) cos( ωct ) dω π which, after re-iteration (interchanging the order of integration) is 1 y ( x, t) = ( cos( ωu) cos( ωx) sin ( ωu) sin ( ωx) ) f ( u) cos ( ωct ) dωdu π + 0 1 y ( x, t) = f ( u) cos( ω( u x) ) cos( ωct ) dωdu π 0 1 = f ( u) ( cos( ω( u x ct )) cos( ω( u x ct ))) dω du π + + 0 ω = 1 sin ( ω( u x + ct )) sin ( ω( u x ct )) = f ( u) + du π u x + ct u x ct ω = 0
ENGI 940 8.04 - d Alembert Solution Page 8.15 8.04 d Alembert Solution One form of the solution to Example 8.03.1, ω = 1 sin ( ω( u x + ct )) sin ( ω( u x ct )) y ( x, t) = f ( u) + du π u x + ct u x ct ω = 0 suggests that, in general, one might seek solutions to the wave equation of the form f ( x + ct ) + f ( x ct ) y( xt, ) = f ( r) + f ( s) Let r = x + ct and s = x ct, then yrs (, ) = and y x = y x = y = y = y 1 y = x c ( + ) + ( ) f x ct f x ct Therefore y( xt, ) = is a solution to the wave equation for all twice differentiable functions f (u). This is part of the d Alembert solution. This d Alembert solution satisfies the initial displacement condition: f ( x+ 0) + f ( x 0) y( x,0) = = f ( x) Also y( xt, ) t t = 0 = The d Alembert solution therefore satisfies both initial conditions.
ENGI 940 8.04 - d Alembert Solution Page 8.16 A more general d Alembert solution to the wave equation for an infinitely long string is (, ) y x t ( + ) + ( ) f x ct f x ct 1 = + c x+ ct x ct ( ) g u du This satisfies the wave equation y y = c x for < x < and t > 0 and Initial configuration of string: y(x, 0) = f (x) for x and Initial speed of string: y ( x,0) for any twice differentiable functions f (x) and g(x). ( ) = g x for x Physically, this represents two identical waves, moving with speed c in opposite directions along the string. x+ ct 1 = g u du satisfies both initial conditions: c x ct 1 x+ ct 1 x y( x, t) = g( u) du y( x,0) = g( u) du = 0 c x ct c x Proof that y( x, t) ( ) Using a Leibnitz differentiation of the integral: y =
ENGI 940 8.04 - d Alembert Solution Page 8.17 Example 8.04.1 An elastic string of infinite length is displaced into the form y = cos π x/ on [ 1, 1] only (and y = 0 elsewhere) and is released from rest. Find the displacement y(x, t) at all locations on the string x and at all subsequent times (t > 0). For this solution to the wave equation we have initial conditions π x cos ( 1 x 1) y( x,0) = f ( x) = 0 ( otherwise) and y ( x,0) = g ( x ) = 0 The d Alembert solution is f ( x + ct ) + f ( x ct ) 1 x+ ct f ( x + ct ) + f ( x ct ) y( x, t) = + g( u) du = + 0 c x ct π ( x + ct ) cos ( 1 ct x 1 ct ) where f ( x + ct ) = 0 ( otherwise) ( x ct ) π cos ( 1+ ct x 1+ ct ) and f ( x ct ) = 0 ( otherwise) We therefore obtain two waves, each of the form of a single half-period of a cosine function, moving apart from a superposed state at x = 0 at speed c in opposite directions. See the web page "www.engr.mun.ca/~ggeorge/940/demos/ex8041.html" for an animation of this solution.
ENGI 940 8.04 - d Alembert Solution Page 8.18 Example 8.04.1 (continued) Some snapshots of the solution are shown here:
ENGI 940 8.04 - d Alembert Solution Page 8.19 A more general case of a d Alembert solution arises for the homogeneous PDE with constant coefficients u u u A + B + C = 0 x x y y The characteristic (or auxiliary) equation for this PDE is Aλ + Bλ + C = 0 This leads to the complementary function (which is also the general solution for this homogeneous PDE) u x, y = f y+ λ x + f y+ λ x, ( ) 1( 1 ) ( ) where B D 1 and B + λ = λ D = A A and D = B 4AC and f1, f are arbitrary twice-differentiable functions of their arguments. λ 1 and λ are the roots (or eigenvalues) of the characteristic equation. In the event of equal roots, the solution changes to uxy (, ) = f1( y+ λx) + hxy (, ) f( y+ λx) where h(x, y) is any non-trivial linear function of x and/or y (except y + λx). The wave equation is a special case with y = t, A = 1, B = 0, C = 1/c and λ = ± 1/c.
ENGI 940 8.04 - d Alembert Solution Page 8.0 Example 8.04. u u u 3 + x x y y = 0 u(x, 0) = x u y (x, 0) = 0 (a) Classify the partial differential equation. (b) Find the value of u at (x, y) = (0, 1). (a) Compare this PDE to the standard form u u u A B C x x y y + + = 0 A = 1, B = 3, C = D = 9 4 = 1 > 0 Therefore the PDE is (b)
ENGI 940 8.04 - d Alembert Solution Page 8.1 Example 8.04. (continued)
ENGI 940 8.04 - d Alembert Solution Page 8. Example 8.04.3 Find the complete solution to u u u 6 5 + = 14, x x y y u(x, 0) = x + 1, u y (x, 0) = 4 6x. This PDE is non-homogeneous. For the particular solution, we require a function such that the combination of second partial derivatives resolves to the constant 14. It is reasonable to try a quadratic function of x and y as our particular solution. Try u = ax + bxy + cy P
ENGI 940 8.04 - d Alembert Solution Page 8.3 Example 8.04.3 (continued)
ENGI 940 8.04 - d Alembert Solution Page 8.4 Example 8.04.3 - Alternative Treatment of the Particular Solution Find the complete solution to u u u 6 5 + = 14, x x y y u(x, 0) = x + 1, u y (x, 0) = 4 6x. This PDE is non-homogeneous. For the particular solution, we require a function such that the combination of second partial derivatives resolves to the constant 14. It is reasonable to try a quadratic function of x and y as our particular solution. Try u = ax + bxy+ cy P u x u ax by and bx cy y P P = + = + up up up = a, = b and = c x x y y up up up 6 5 + = 1a 5b+ c = 14 x x y y We have one condition on three constants, two of which are therefore a free choice. Let us leave the free choice unresolved for now. Complementary function: + 5 ± 1 1 1 λ = = or 1 3 The complementary function is C (, ) = ( + 1 1 ) + ( + ) u xy f y x g y x and the general solution is 3 (, ) = ( + 1 1 ) + ( + ) + + + u x y f y x g y x ax bxy cy 3
ENGI 940 8.04 - d Alembert Solution Page 8.5 Example 8.04.3 (continued) (, ) = ( + 1 1 ) + ( + ) + + + u x y f y x g y x ax bxy cy 3 u = f 1 1 ( y+ 3 x) + g ( y+ x) + bx + cy y Imposing the two boundary conditions: 1 1 ( ) ( ) ( ) u x,0 = f x + g x + ax + 0+ 0 = x+ 1 (A) and 3 ( ) 1 ( ) 1 ( ) u x,0 = f x + g x + bx + 0 = 4 6x (B) y d dx 3 A = 1 f 1 3 x + g x + ax = (C) 3 ( ) 1 ( ) 1 ( )
ENGI 940 8.04 - d Alembert Solution Page 8.6 Example 8.04.3 (continued)
ENGI 940 8.04 - d Alembert Solution Page 8.7 Example 8.04.4 Find the complete solution to u u u + + x x y y = 0, u = 0 on x = 0, u = x on y = 1. A = 1, B =, C = 1 D = 4 4 1 = 0 Therefore the PDE is
ENGI 940 8.04 - d Alembert Solution Page 8.8 Example 8.04.4 - Alternative Treatment of the Complementary Function Find the complete solution to u u u + + x x y y = 0, u = 0 on x = 0, u = x on y = 1. A = 1, B =, C = 1 D = 4 4 1 = 0 ± 0 λ = = 1 or 1 The complementary function (and general solution) is uxy (, ) = f( y x) + hxyg (, ) ( y x) where h(x, y) is any convenient non-trivial linear function of (x, y) except a multiple of (y x). The most general choice possible is h( x, y) = ax+ by, with the restriction a λb. (, ) = ( ) + ( + ) ( ) ( 0, ) ( ) ( ) 0 ( ) ( ) u x y f y x ax by g y x u y = f y + by g y = f y = by g y (A) and u x,1 = f 1 x + ax + b g 1 x = x ( ) ( ) ( ) ( )
ENGI 940 8.04 - d Alembert Solution Page 8.9 Example 8.04.4 Extension (continued) It is easy to confirm that this solution is correct: (, ) ( 1) uxy = xx y+
ENGI 940 8.04 - d Alembert Solution Page 8.30 Two-dimensional Laplace Equation u x u y + = 0 A = C = 1, B = 0 D = 0 4 < 0 This PDE is elliptic everywhere. 0 ± 4 λ = = ± j The general solution is (, ) = ( ) + ( + ) u x y f y jx g y jx where f and g are any twice-differentiable functions. A function f (x, y) is harmonic if and only if f = 0 everywhere inside a domain Ω. Example 8.04.5 Is u = e x sin y harmonic on?
ENGI 940 8.04 - d Alembert Solution Page 8.31 Example 8.04.6 Find the complete solution uxy (, ) to the partial differential equation additional information 3 ( 0, ) u y = y and u x x = 0 = 0 u = 0, given the The PDE is u x u y u = + = 0 (which means that the solution uxy (, ) is an harmonic function). A = C = B = D = B AC = < The PDE is 1, 0 4 4 0
ENGI 940 8.05 - Max-Min Principle Page 8.3 8.05 The Maximum-Minimum Principle Let Ω be some finite domain on which a function u(x, y) and its second derivatives are defined. Let Ω be the union of the domain with its boundary. Let m and M be the minimum and maximum values respectively of u on the boundary of the domain. If u 0 in Ω, then u is subharmonic and u r < M u r M r in Ω ( ) or ( ) If u 0 in Ω, then u is superharmonic and u r > m u r m r in Ω ( ) or ( ) If u = 0 in Ω, then u is harmonic (both subharmonic and superharmonic) and u is either constant on Ω or m < u < M everywhere on Ω. Example 8.05.1 u = 0 in Ω : x + y < 1 and u(x, y) = 1 on C : x + y = 1. Find u(x, y) on Ω.
ENGI 940 8.05 - Max-Min Principle Page 8.33 Example 8.05. u = 0 in the square domain Ω : < x < +, < y < +. On the boundary C, on the left and right edges (x = ±), u(x, y) = 4 y, while on the top and bottom edges (y = ±), u(x, y) = x 4. Find bounds on the value of u(x, y) inside the domain Ω. For < y < +, 0 < 4 y < 4. For < x < +, 4 < x 4 < 0. Therefore, on the boundary C of the domain Ω, 4 < u(x, y) < +4 so that m = 4 and M = +4. u(x, y) is harmonic (because u = 0). Therefore, everywhere in Ω, 4 < u(x, y) < +4 Note: u(x, y) = x y is consistent with the boundary condition and u = x 0, u = 0 y u =, u = = u x y x y x u u u = + = 0 x y Contours of constant values of u are hyperbolas. A contour map illustrates that 4 < u(x, y) < +4 within the domain is indeed true.
ENGI 940 8.06 - Heat Equation Page 8.34 8.06 The Heat Equation For a material of constant density ρ, constant specific heat μ and constant thermal conductivity K, the partial differential equation governing the temperature u at any location (x, y, z) and any time t is u K = k u, where k = µρ Example 8.06.1 Heat is conducted along a thin homogeneous bar extending from x = 0 to x = L. There is no heat loss from the sides of the bar. The two ends of the bar are maintained at temperatures T 1 (at x = 0) and T (at x = L). The initial temperature throughout the bar at the cross-section x is f (x). Find the temperature at any point in the bar at any subsequent time. The partial differential equation governing the temperature u(x, t) in the bar is u = k u x together with the boundary conditions u( 0, t) T1 u Lt, = T and the initial condition u(x, 0) = f (x) = and ( ) [Note that if an end of the bar is insulated, instead of being maintained at a constant u u temperature, then the boundary condition changes to ( 0, t) = 0 or ( Lt, ) = 0.] x x Attempt a solution by the method of separation of variables. u(x, t) = X(x) T(t) T X XT = k X T = k = c T X Again, when a function of t only equals a function of x only, both functions must equal the same absolute constant. Unfortunately, the two boundary conditions cannot both be satisfied unless T 1 = T = 0. Therefore we need to treat this more general case as a perturbation of the simpler ( T 1 = T = 0) case.
ENGI 940 8.06 - Heat Equation Page 8.35 Example 8.06.1 (continued) Let u(x, t) = v(x, t) + g(x) Substitute this into the PDE: v v ( v( xt, ) + g( x) ) = k v ( xt, ) + g( x) = k x x + g x This is the standard heat PDE for v if we choose g such that g"(x) = 0. g(x) must therefore be a linear function of x. ( ) ( ) We want the perturbation function g(x) to be such that u( 0, t) = T1 and u( Lt, ) = T and v(0, t) = v(l, t) = 0 g 0 It follows that T T1 g( x) = x + T1 L and we now have the simpler problem Therefore g(x) must be the linear function for which ( ) = T1 and g( L) T =. together with the boundary conditions v(0, t) = v(l, t) = 0 and the initial condition v(x, 0) = f (x) g(x) v = k v x Now try separation of variables on v(x, t) : v(x, t) = X(x) T(t) 1 T X XT = k X T = = c kt X But v(0, t) = v(l, t) = 0 X(0) = X(L) = 0 This requires c to be a negative constant, say λ. The solution is very similar to that for the wave equation on a finite string with fixed ends nπ (section 8.0). The eigenvalues are λ = and the corresponding eigenfunctions are L any non-zero constant multiples of nπ x Xn ( x) = sin L
ENGI 940 8.06 - Heat Equation Page 8.36 Example 8.06.1 (continued) The ODE for T(t) becomes nπ T + kt = 0 L whose general solution is nπ kt/ L Tn( t) = cne Therefore nπx n π kt vn( xt, ) = Xn( xt ) n( t) = cnsin exp L L If the initial temperature distribution f (x) g(x) is a simple multiple of sin n π x L for nπx n π kt some integer n, then the solution for v is just v( xt, ) = cn sin exp. L L Otherwise, we must attempt a superposition of solutions. nπx n π kt v( xt, ) = cn sin exp n = 1 L L nπ x such that v( x,0) = cn sin = f ( x) g( x) L. n = 1 nπ z The Fourier sine series coefficients are cn = ( f ( z) g ( z) ) sin dz L 0 L so that the complete solution for v(x, t) is L T T1 nπz nπx n π kt v( x, t) = f ( z) z T1 sin dz sin exp L n = 1 0 L L L L and the complete solution for u(x, t) is T T1 u( xt, ) = v( xt, ) + x + T L Note how this solution can be partitioned into a transient part v(x, t) (which decays to zero as t increases) and a steady-state part g(x) which is the limiting value that the temperature distribution approaches. L 1
ENGI 940 8.06 - Heat Equation Page 8.37 Example 8.06.1 (continued) As a specific example, let k = 9, T 1 = 100, T = 00, L = and f (x) = 145x 40x + 100, (for which f (0) = 100, f () = 00 and f (x) > 0 x). 00 100 Then g( x) = x + 100 = 50 x + 100 The Fourier sine series coefficients are (( nπ z cn = 145z 40z + 100) ( 50z + 100) ) sin dz 0 nπ z cn = 145 ( z z) sin dz 0 After an integration by parts (details omitted here), ( z ) ( ) 16 nπz 8 1 nπz cn = 145 ( z + z) + 3 cos + sin n π ( nπ) nπ 30 c 3 (( 1) n n = 1) nπ ( ) The complete solution is z= z= 0 ( ) π π 30 1 1 n x 9n t u( xt, ) = 50x + 100 sin exp 3 3 π n = 1 n 4 Some snapshots of the temperature distribution (from the tenth partial sum) from the Maple file at "www.engr.mun.ca/~ggeorge/940/demos/ex8061.mws" are shown on the next page. n
ENGI 940 8.06 - Heat Equation Page 8.38 Example 8.06.1 (continued) The steady state distribution is nearly attained in much less than a second! END OF CHAPTER 8 END OF ENGI. 940!