ENGI 4430 PDEs - d Alembert Solutions Page 11.01 11. 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 very small subset of second order linear partial differential equations. 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 Ax, y B x, y C x, y f x, y, u,, 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: The wave equation: u t c u u u or its one-dimensional special case c [which is hyperbolic everywhere] t x (where u is the displacement and c is the speed of the wave); u The heat (or diffusion) equation: K u K u t a one-dimensional special case of which is u K u [which is parabolic everywhere] t x (where u is the temperature, μ is the specific heat of the medium, ρ is the density and K is the thermal conductivity);
ENGI 4430 PDEs - d Alembert Solutions Page 11.0 The potential (or Laplace s) equation: a special case of which is u x u y 0 u 0 [which is elliptic everywhere] 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. d Alembert Solution Example 11.01 In Term 3 (example 4.8.5 in ENGI 344) we saw that f x ct f x ct yx, t is a solution to the wave equation y 1 y 0 x c t This solution also satisfies the initial conditions yx,0 f x and y x t for any twice differentiable function f x. t, 0 t 0 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 xct xct g u du This satisfies the wave equation y y c t x and for x and t 0 y x,0 f x for x Initial configuration of string: and y Initial speed of string: t x,0 g x for any twice differentiable functions f (x) and g(x). for x
ENGI 4430 PDEs - d Alembert Solutions Page 11.03 Physically, this represents two identical waves, moving with speed c in opposite directions along the string. xct 1 g u du satisfies both initial conditions: c xct xct x y x 1 1, t,0 0 g u du y x c c g u du Proof that yx, t xct x For the other condition, we require Leibnitz differentiation of a definite integral: d bx db da bx f f x, t dt f x, bx f x, ax dt dxax dx dx ax x
ENGI 4430 PDEs - d Alembert Solutions Page 11.04 Example 11.0 x An elastic string of infinite length is displaced into the form y cos on 1,1 only (and y = 0 elsewhere) and is released from rest. Find the displacement yx, t at all locations on the string x and at all subsequent times (t > 0). See the web page "www.engr.mun.ca/~ggeorge/4430/demos/ex110.html" for an animation of this solution.
ENGI 4430 PDEs - d Alembert Solutions Page 11.05 Example 11.0 (continued) Some snapshots of the solution are shown here:
ENGI 4430 PDEs - d Alembert Solutions Page 11.06 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 A B C 0 This leads to the complementary function (which is also the general solution for this homogeneous PDE) where 1 1 u x, y f y x f y x, B D 1 and B D A A and D B 4AC and f1, f are arbitrary twice-differentiable functions of their arguments. 1 and are the roots (or eigenvalues) of the characteristic equation. In the event of equal roots, the solution changes to u x, y f y x h x, y f y x 1 where hx, y is any non-trivial linear function of x and/or y (except any multiple of y x). 1 The wave equation is a special case with y t, A 1, B 0, C and c 1. c Note how this method has some similarities with a method for solving second order linear ordinary differential equations with constant coefficients (in ENGI 344 or equivalent).
ENGI 4430 PDEs - d Alembert Solutions Page 11.07 Example 11.03 u u u 3 0 x x y y u x,0 u y x 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
ENGI 4430 PDEs - d Alembert Solutions Page 11.08 Example 11.03 (continued)
ENGI 4430 PDEs - d Alembert Solutions Page 11.09 Example 11.04 Find the complete solution to u u u 6 5 14, x x y y u x,0 x 1, u x,0 4 6 x. y 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 4430 PDEs - d Alembert Solutions Page 11.10 Example 11.04 (continued)
ENGI 4430 PDEs - d Alembert Solutions Page 11.11 Example 11.04 - 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 x,0 4 6 x. y 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 or 1 3 The complementary function is u x, y f y 1 1 3 x g y x and the general solution is C
ENGI 4430 PDEs - d Alembert Solutions Page 11.1 Example 11.04 (continued)
ENGI 4430 PDEs - d Alembert Solutions Page 11.13 Example 11.04 (continued)
ENGI 4430 PDEs - d Alembert Solutions Page 11.14 Example 11.05 Find the complete solution to u u u 0, x x y y u = 0 on x = 0, u = x on y = 1.
ENGI 4430 PDEs - d Alembert Solutions Page 11.15 Example 11.05 - Alternative Treatment of the Complementary Function Find the complete solution to u u u 0, x x y y u = 0 on x = 0, u = x on y = 1. A 1, B, C 1 D 4 41 0 0 1 or 1 The complementary function (and general solution) is ux, y f y x hx, y g y x where hx, y is any convenient non-trivial linear function of xy, except a multiple of y x. The most general choice possible is
ENGI 4430 PDEs - d Alembert Solutions Page 11.16 Example 11.05 Extension (continued)
ENGI 4430 PDEs - d Alembert Solutions Page 11.17 Two-dimensional Laplace Equation u x u y 0 A function f, x y is harmonic if and only if f 0 everywhere inside a domain. Example 11.06 Is x u e sin y harmonic on?
ENGI 4430 PDEs - d Alembert Solutions Page 11.18 Example 11.07 Find the complete solution u x, y 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 u x, y is an harmonic function). A C B D B AC The PDE is elliptic everywhere. A.E.: 1, 0 4 4 0 1 0 j C.F.: u x y f y jx g y jx C, The PDE is homogeneous u x y P.S.: P, 0 G.S.: ux, y f y jx g y jx, u x y j f y jx j g y jx x Using the additional information, 3 3 0, 3 u y f y g y y g y y f y g y y f y and u 0, y 0 j f y j g y j f y 3y f y x 3 3 f y 1 3y f y y f y y 3 3 3 g 1 1 y y y y Therefore the complete solution is u x, y 1 1 y jx y jx 3 3 1 3 3 3 y 3 jx y 3 jx y jx y 3 jx y 3 jx y jx 3 y 3 j x y 3, 3 u x y y x y 3 Note that the solution is completely real, even though the eigenvalues are not real.
ENGI 4430 PDEs - d Alembert Solutions Page 11.19 Summary: u u u For the partial differential equation A B C r x, y x x y y the general solution is the sum of a complementary function and a particular solution u x, y u x, y u x, y Complementary function: B D Find D B 4AC then A u x, y f y x g y x C 1 unless 0 D, in which case uc x, y f y x hx, y g y x C P where h x, y is any convenient linear function that is not a constant multiple of y x. Particular solution: If r x, y 0 (homogeneous PDE) then trivially up x, y 0. If, up x, y ax bxy cy If r x, y is a linear function r x, y lx my, then r x y is a non-zero constant, then try 3 3 try up x, y ax bx y cxy dy In this course no other forms of r x, y will be examinable. Apply initial / boundary conditions only after finding the general solution, in order to f y x, g y x and therefore the complete solution. find the arbitrary functions 1
ENGI 4430 PDEs - d Alembert Solutions Page 11.0 [Space for Additional Notes] [End of Chapter 11]