Strauss PDEs 2e: Section Exercise 1 Page 1 of 6

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1 Strauss PDEs e: Setion.1 - Exerise 1 Page 1 of 6 Exerise 1 Solve u tt = u xx, u(x, 0) = e x, u t (x, 0) = sin x. Solution Solution by Operator Fatorization By fatoring the wave equation and making a substitution, we an write it as an equivalent system of unoupled first-order PDEs and solve it using the methods of the previous hapter. Start off by bringing all terms of the PDE to one side. u tt u xx = 0 Write the left side as an operator ating on u, L u. Now fator the operator. ( t x)u = 0 ( t x )( t + x )u = 0 If we let v = ( t + x )u, then what remains is ( t x )v = 0. That is, { ut + u x = v v t v x = 0. On the paths defined by =, x(ξ, 0) = ξ (1) the PDE on the bottom redues to an ODE, dv = 0. () That is, v = v(x, t) is onstant on the harateristis defined by (1). Integrating (), we find that v(ξ, t) = f(ξ), where f is an arbitrary funtion of the harateristi oordinate, ξ. Solving (1) by integration gives Solve now for ξ. Therefore, x = t + ξ ξ = x + t v(x, t) = f(x + t). We an hek that this is the solution of the bottom PDE. v t = f v x = f

2 Strauss PDEs e: Setion.1 - Exerise 1 Page of 6 v t v x = 0, so this is the orret solution. Now we use this solution to solve the PDE on the top. Our task is to solve u t + u x = f(x + t). On the paths defined by the PDE on top redues to an ODE, =, x(η, 0) = η (3) du = f(x + t). (4) That is, u = u(x, t) is onstant on the harateristis defined by (3). Solving (3) by integration, we find that Solve now for η. x = t + η. η = x t Now that we know x in terms of η, we an solve (4) for u. du = f(t + η) Integrate both sides now with respet to t. ˆ t u(η, t) = f(s + η) ds + g(η), where g is an arbitrary funtion of η. Now that the integration is done, hange bak to the original variables. ˆ t u(x, t) = f(s + x t) ds + g(x t) To solve the integral, use a substitution. Plug in s = t to get the upper limit of integration. w = s + x t dw = ds 1 dw = ds It beomes u(x, t) = ˆ x+t f(w) 1 dw + g(x t) u(x, t) = 1 F (x + t) + g(x t). Now that we have the general solution for u(x, t), we use the initial onditions to determine the unknown funtions, F and g. u(x, 0) = e x 1 F (x) + g(x) = ex u t (x, 0) = sin x 1 F (x) g (x) = sin x

3 Strauss PDEs e: Setion.1 - Exerise 1 Page 3 of 6 Multiply both sides of the top equation by and differentiate both sides with respet to x. Adding these two equations gives us 1 F (x) + g (x) = e x 1 F (x) g (x) = sin x F (x) = e x + sin x. Subtrating the two equations gives us Divide both sides by to isolate g. Solve for F and g by integrating both sides. g (x) = e x sin x. g (x) = 1 ex 1 sin x F (x) = e x os x g(x) = 1 ex + 1 os x What we solved for are atually F (w) and g(w), where w is any expression. Thus, F (x + t) = e x+t os(x + t) g(x t) = 1 ex t + 1 os(x t). Plugging these into u(x, t), we obtain the answer to the initial value problem. u(x, t) = 1 [ 1 [ex+t os(x + t)] + ex t + 1 ] os(x t) u(x, t) = 1 ( e x+t + e x t) + 1 [os(x t) os(x + t)] u(x, t) = 1 (ex osh t) + 1 ( sin x sin t) Therefore, u(x, t) = e x osh t + 1 sin x sin t.

4 Strauss PDEs e: Setion.1 - Exerise 1 Page 4 of 6 Solution by the Method of Charateristis Comparing this equation with the general form of a seond-order PDE, Au tt + Bu xt + Cu xx + Du t + Eu x + F u = G, we see that A = 1, B = 0, C =, D = 0, E = 0, F = 0, and G = 0. The harateristi equations of this PDE are given by = 1 ( B ± ) B A 4AC = 1 (± ) = or =. Note that the disriminant, B 4AC = 4, is greater than 0, whih means that the PDE is hyperboli. Therefore, the solutions to the ordinary differential equations are two real and distint families of harateristi urves in the xt-plane. Solving for the onstants of integration, x = t + C 1 or x = t + C. C 1 = x t = φ(x, t) C = x + t = ψ(x, t). Now we make the hange of variables, ξ = φ(x, t) = x t and η = ψ(x, t) = x + t, so that the PDE takes the simplest form. With these new variables the PDE beomes where, using the hain rule, A u ξξ + B u ξη + C u ηη + D u ξ + E u η + F u = G, A = Aξ t + Bξ t ξ x + Cξ x B = Aξ t η t + B(ξ t η x + ξ x η t ) + Cξ x η x C = Aη t + Bη t η x + Cη x D = Aξ tt + Bξ xt + Cξ xx + Dξ t + Eξ x E = Aη tt + Bη xt + Cη xx + Dη t + Eη x F = F G = G. Plugging in the numbers and derivatives to these formulas, we find that A = 0, B = 4, C = 0, D = 0, E = 0, F = 0, and G = 0. Thus, the PDE simplifies to Solving for u ξη gives 4 u ξη = 0. u ξη = 0. This is known as the first anonial form of the wave equation. We an solve it by integrating both sides with respet to η and then integrating both sides again with respet to ξ. u ξ = f(ξ),

5 Strauss PDEs e: Setion.1 - Exerise 1 Page 5 of 6 where f is an arbitrary funtion of ξ. u(ξ, η) = F (ξ) + G(η), where F and G are arbitrary funtions of ξ and η, respetively. Now hange bak to the original variables with the substitutions, ξ = x t and η = x + t, to get the general solution of the wave equation. u(x, t) = F (x t) + G(x + t) We determine these unknown funtions by using the initial onditions. u(x, 0) = e x F (x) + G(x) = e x u t (x, 0) = sin x F (x) + G (x) = sin x Multiply both sides of the top equation by and differentiate both sides with respet to x. Adding these two equations gives us Subtrating the two equations gives us Divide both sides by to isolate F and G. Solve for F and G by integrating both sides. F (x) + G (x) = e x F (x) + G (x) = sin x G (x) = e x + sin x. F (x) = e x sin x. F (x) = 1 ex 1 sin x G (x) = 1 ex + 1 sin x F (x) = 1 ex + 1 os x G(x) = 1 ex 1 os x What we solved for are atually F (w) and G(w), where w is any expression. Thus, F (x + t) = 1 ex+t + 1 os(x + t) G(x t) = 1 ex t 1 os(x t). Plugging these into u(x, t), we obtain the answer to the initial value problem. u(x, t) = 1 ex+t + 1 os(x + t) + 1 ex t 1 os(x t) u(x, t) = 1 ( e x+t + e x t) + 1 [os(x t) os(x + t)]

6 Strauss PDEs e: Setion.1 - Exerise 1 Page 6 of 6 u(x, t) = 1 (ex osh t) + 1 ( sin x sin t) Therefore, u(x, t) = e x osh t + 1 sin x sin t. This is the same answer that we obtained with operator fatorization. We an hek that this is the solution to the wave equation. u t = e x sinh t + sin x os t u tt = e x osh t sin x sin t u x = e x osh t + 1 os x sin t u xx = e x osh t 1 sin x sin t u tt = u xx, so this is the solution to the wave equation. Now hek the initial onditions. u(x, 0) = e x = e x u t (x, 0) = e x 0 + sin x 1 = sin x To learn about anonial forms and the method of harateristis for seond-order PDEs, read hapter 4 of Debnath s Linear Partial Differential Equations for Sientists and Engineers, 4th Edition.