Strauss PDEs 2e: Section Exercise 3 Page 1 of 13. u tt c 2 u xx = cos x. ( 2 t c 2 2 x)u = cos x. v = ( t c x )u

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1 Strauss PDEs e: Setion Exerise 3 Page 1 of 13 Exerise 3 Solve u tt = u xx + os x, u(x, ) = sin x, u t (x, ) = 1 + x. Solution Solution by Operator Fatorization Bring u xx to the other side. Write the left side as an operator ating on u. u tt u xx = os x ( t x)u = os x The operator is a differene of squares, so it an be fatored. Let so that the PDE beomes ( t + x )( t x )u = os x v = ( t x )u ( t + x )v = os x. The seond-order PDE we started with has thus been redued to the following system of first-order PDEs that an be solved with the method of harateristis. u t u x = v (1) v t + v x = os x () We will solve the seond one for v first, and one that is known, the first equation for u will be solved. For a funtion of two variables φ = φ(x, t), its differential is defined as dφ = φ φ dt + t x dx. If we divide both sides by dt, then we get the relationship between the ordinary derivative of φ and its partial derivatives. dφ dt = φ t + φ dx (3) x dt Comparing this with equation (), we see that along the urves in the xt-plane that satisfy the PDE for v(x, t) redues to an ODE. dx dt =, (4) dv dt = os x (5) Beause is a onstant, equation (4) an be solved by integrating both sides with respet to t. x = t + ξ, (6)

2 Strauss PDEs e: Setion Exerise 3 Page of 13 where ξ is a harateristi oordinate. Substitute this expression for x into equation (5) to obtain an ODE that only involves t (ξ is regarded as a onstant). Integrate both sides with respet to t. dv dt = os(t + ξ) v(ξ, t) = 1 sin(t + ξ) + f(ξ), where f is an arbitrary funtion of the harateristi oordinate ξ. In order to write v in terms of x and t, solve equation (6) for ξ. x = t + ξ ξ = x t Hene, v(x, t) = 1 sin x + f(x t). As a result, equation (1) beomes u t u x = 1 sin x + f(x t). Comparing this equation with equation (3), we see that along the urves in the xt-plane that satisfy dx =, (7) dt the PDE for u(x, t) redues to an ODE. du dt = 1 sin x + f(x t) (8) Beause is a onstant, equation (7) an be solved by integrating both sides with respet to t. x = t + η, (9) where η is another harateristi oordinate. Substitute this expression for x into equation (8) to obtain an ODE that only involves t (η is regarded as a onstant). du dt = 1 sin( t + η) + f( t + η t) du dt = 1 sin( t + η) + f(η t) Integrate both sides with respet to t. u(η, t) = 1 os( t + η) + ˆ t f(η s) ds + g(η), where g is an arbitrary funtion of the harateristi oordinate η. The integral of an arbitrary funtion is another arbitrary funtion. u(η, t) = 1 os( t + η) + F (η t) + g(η),

3 Strauss PDEs e: Setion Exerise 3 Page 3 of 13 In order to write u in terms of x and t, solve equation (9) for η. Hene, x = t + η η = x + t u(x, t) = 1 os( t + η) + F (x + t t) + g(x + t) = 1 os x + F (x t) + g(x + t). This is the general solution to u tt = u xx + os x. If we apply the two initial onditions, we an determine F and g. Before doing so, take a derivative of the solution with respet to t. u t (x, t) = F (x t) + g (x + t) From the initial onditions we obtain the following system of equations. u(x, ) = 1 os x + F (x) + g(x) = sin x u t (x, ) = F (x) + g (x) = 1 + x Even though this system is in terms of x, it s really in terms of w, where w is any expression we hoose. 1 os w + F (w) + g(w) = sin w F (w) + g (w) = 1 + w Differentiating both sides of the first equation with respet to w, we get 1 sin w + F (w) + g (w) = os w g (w) = os w + 1 sin w F (w). Plug this expression for g (w) into the seond equation. [ F (w) + os w + 1 ] sin w F (w) = 1 + w F (w) = 1 ( os w + 1 ) sin w w 1 Solve for F (w) and obtain an expression for F (x t). F (w) = 1 ) 1 w ( sin w os w w + C 1 F (x t) = 1 [ sin(x t) 1 os(x t) (x t) ] (x t) + C 1 Use the first equation to solve for g(w) and obtain an expression for g(x + t). g(w) = sin w 1 os w F (w) = sin w 1 os w 1 ( sin w ) 1 w os w w C 1

4 Strauss PDEs e: Setion Exerise 3 Page 4 of 13 g(w) = 1 ) 1 w ( sin w os w + + w C 1 g(x + t) = 1 [ sin(x + t) 1 os(x + t) + (x + t) ] + (x + t) C 1 The general solution for u(x, t) beomes u(x, t) = 1 os x + F (x t) + g(x + t) = 1 os x + 1 ] 1 (x t) [ sin(x t) os(x t) (x t) + C ] 1 (x + t) [ sin(x + t) os(x + t) + + (x + t) C 1 = 1 os x [sin(x t) + sin(x + t)] [os(x t) + os(x + t)] + 1 [ ] (x + t) (x t) + (x + t) (x t) = 1 os x + 1 (sin x os t os x sin t + sin x os t + os x sin t) 1 (os x os t + sin x sin t + os x os t sin x sin t) + 1 [t(x + 1)] = 1 os x + sin x os t 1 os x os t + t(x + 1). Therefore, u(x, t) = 1 os x(1 os t) + sin x os t + t(x + 1).

5 Strauss PDEs e: Setion Exerise 3 Page 5 of 13 Solution by the Method of Charateristis Bring u xx to the left side of the PDE. u tt u xx = os x Comparing this 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 =, C =, D =, E =, F =, and G = os x. The harateristi equations for a seond-order PDE are given by dx dt = 1 A (B ± B 4AC), the solutions of whih are known as the harateristis. Sine B 4AC = 4 >, the PDE is hyperboli, so the solutions to these equations are two real and distint families of harateristi urves in the xt-plane. dx dt = 1 (± 4 ) dx dt = 1 (±) dx dt = or dx dt = Integrate both sides of eah equation with respet to t. Now make the substitutions, x = t + C or x = t + C 3 ξ = x t = C η = x + t = C 3, so that the PDE takes the simplest form. The aim is to write u tt, u xx, and os x in terms of the new variables, ξ and η. Solving these two equations for x and t with elimination gives x = 1 (η + ξ) t = 1 (η ξ). Use the hain rule to write the old derivatives in terms of the new variables. t = ξ ξ t + η η t = u ξ( ) + u η () = (u η u ξ ) x = ξ ξ x + η η x = u ξ(1) + u η (1) = u ξ + u η Find the seond derivatives by using the hain rule again. u t = ( ξ t (u η u ξ ) = t ξ + η ) [ ξ (u η u ξ ) = t η t ξ (u η u ξ ) + η ] t η (u η u ξ ) u x = ( ξ x (u ξ + u η ) = x ξ + η ) (u ξ + u η ) = ξ x η x ξ (u ξ + u η ) + η x η (u ξ + u η )

6 Strauss PDEs e: Setion Exerise 3 Page 6 of 13 Hene, u t = [( )(u ξη u ξξ ) + ()(u ηη u ξη )] = (u ξξ u ξη + u ηη ) u x = (1)(u ξξ + u ξη ) + (1)(u ξη + u ηη ) = u ξξ + u ξη + u ηη. Substituting these expressions into the PDE, u tt u xx = os x, we obtain [ ] 1 ( u ξξ u ξη + u ηη ) ( u ξξ + u ξη + u ηη ) = os (η + ξ). Simplify the left side. Divide both sides by 4. [ ] 1 4 u ξη = os (η + ξ) u ξ η = 1 [ ] 1 4 os (η + ξ) This is known as the first anonial form of the PDE. Integrate both sides of it partially with respet to η. ˆ η ˆ u η ξ η ds = 1 [ ] 1 η=s 4 os (s + ξ) ds + f(ξ), where f is an arbitrary funtion of ξ. ξ = 1 [ 1 4 sin ξ = 1 sin [ 1 (η + ξ) Now integrate both sides partially with respet to ξ. ˆ ξ ˆ ξ { ξ ds = 1 [ ] 1 ξ=s sin (η + s) where g is an arbitrary funtion of η. u(ξ, η) = ] η (s + ξ) ] + f(ξ) + f(ξ) } + f(s) ds + g(η), { [ ] } 1 1 ξ os (η + s) + F (s) + g(η) u(ξ, η) = 1 os [ 1 (η + ξ) ] + F (ξ) + g(η) Sine u has been solved for, hange bak to the original variables, x and t, by substituting the expressions for ξ and η. u(x, t) = 1 os x + F (x t) + g(x + t) This is the general solution to u tt = u xx + os x. If we apply the two initial onditions, we an determine F and g. Before doing so, take a derivative of the solution with respet to t. u t (x, t) = F (x t) + g (x + t)

7 Strauss PDEs e: Setion Exerise 3 Page 7 of 13 From the initial onditions we obtain the following system of equations. u(x, ) = 1 os x + F (x) + g(x) = sin x u t (x, ) = F (x) + g (x) = 1 + x Even though this system is in terms of x, it s really in terms of w, where w is any expression we hoose. 1 os w + F (w) + g(w) = sin w F (w) + g (w) = 1 + w Differentiating both sides of the first equation with respet to w, we get 1 sin w + F (w) + g (w) = os w g (w) = os w + 1 sin w F (w). Plug this expression for g (w) into the seond equation. [ F (w) + os w + 1 ] sin w F (w) = 1 + w F (w) = 1 ( os w + 1 ) sin w w 1 Solve for F (w) and obtain an expression for F (x t). F (w) = 1 ) 1 w ( sin w os w w + C 1 F (x t) = 1 [ sin(x t) 1 os(x t) (x t) ] (x t) + C 1 Use the first equation to solve for g(w) and obtain an expression for g(x + t). g(w) = sin w 1 os w F (w) = sin w 1 os w 1 ( sin w ) 1 w os w w C 1 g(w) = 1 ) 1 w ( sin w os w + + w C 1 g(x + t) = 1 [ sin(x + t) 1 os(x + t) + (x + t) ] + (x + t) C 1

8 Strauss PDEs e: Setion Exerise 3 Page 8 of 13 The general solution for u(x, t) beomes u(x, t) = 1 os x + F (x t) + g(x + t) = 1 os x + 1 ] 1 (x t) [ sin(x t) os(x t) (x t) + C ] 1 (x + t) [ sin(x + t) os(x + t) + + (x + t) C 1 = 1 os x [sin(x t) + sin(x + t)] [os(x t) + os(x + t)] + 1 [ ] (x + t) (x t) + (x + t) (x t) = 1 os x + 1 (sin x os t os x sin t + sin x os t + os x sin t) 1 (os x os t + sin x sin t + os x os t sin x sin t) + 1 [t(x + 1)] = 1 os x + sin x os t 1 os x os t + t(x + 1). Therefore, u(x, t) = 1 os x(1 os t) + sin x os t + t(x + 1).

9 Strauss PDEs e: Setion Exerise 3 Page 9 of 13 Solution by Green s Theorem u tt u xx = os x, < x <, t > u(x, ) = sin x u t (x, ) = 1 + x The harateristis were found to be straight lines, ξ = x t and η = x + t, with slopes ±. Suppose (x, t ) is the point in the xt-plane we want to evaluate u at. The equations of the lines going through this point are x x = (t t ) x x = (t t ). Integrate both sides of the inhomogeneous wave equation over the triangular domain D enlosed by these lines (from left to right as indiated below). Write the double integral expliitly on the right side. ˆ t (u tt u xx ) da = Rewrite the left side. D D D [ x ( u x ) ] ˆ t t (u t) da = ˆ x (t t ) Multiply both sides by 1. [ x ( u x ) ] ˆ t t (u t) da = x +(t t ) ˆ x (t t ) x +(t t ) ˆ x (t t ) x +(t t ) Apply Green s theorem (essentially the divergene theorem in two dimensions) to the double integral on the left to turn it into a ounterlokwise line integral around the triangle s boundary bdy D. ffi ˆ t ˆ x (u t dx + (t t ) u x dt) = bdy D x +(t t )

10 Strauss PDEs e: Setion Exerise 3 Page 1 of 13 Let L 1, L, and L 3 represent the legs of the triangle. The line integral is the sum of three integrals, one over eah leg. ˆ (u t dx + u x dt) + L 1 ˆ (u t dx + u x dt) + L ˆ (u t dx + u x dt) = L 3 ˆ t ˆ x (t t ) x +(t t ) On L 1 On L On L 3 t = x x = (t t ) x x = (t t ) dt = dx = dt dx = dt Replae the differentials in the integrals over L and L 3. ˆ x +t ˆ ( u t (x, ) dx + [u t ( dt) + u x dx )] ˆ + x t L L 3 [ u t ( dt) + u x ( dx In this exerise u t (x, ) = 1 + x. ˆ x +t ˆ ( ) ˆ ( ) ˆ (1+x) dx dt + x t L t x dx t + dt + L 3 t x dx = )] ˆ t = ˆ x (t t ) x +(t t ) ˆ x (t t ) x +(t t ) The seond and third integrands on the left side are how the differential of u = u(x, t) is defined. ˆ x +t ˆ ˆ ˆ t ˆ x (t t ) (1 + x) dx du + du = L L 3 x +(t t ) x t Evaluate the seond and third integrals on the left side. ˆ x +t x t (1+x) dx [u(x, t ) u(x +t, )] + [u(x t, ) u(x, t )] = ˆ t ˆ x (t t ) x +(t t ) In this exerise u(x, ) = sin x, so u(x + t, ) = sin(x + t ) and u(x t, ) = sin(x t ). ˆ x +t x t (1 + x) dx u(x, t ) + [sin(x + t ) + sin(x t )] = Solve this equation for u(x, t ). u(x, t ) = [sin(x + t ) + sin(x t )] + ˆ x +t x t (1 + x) dx + ˆ t ˆ x (t t ) x +(t t ) ˆ t ˆ x (t t ) x +(t t )

11 Strauss PDEs e: Setion Exerise 3 Page 11 of 13 Divide both sides by. u(x, t ) = 1 [sin(x + t ) + sin(x t )] + 1 ˆ x +t x t (1 + x) dx + 1 Finally, swith the roles of x and t with those of x and t, respetively. u(x, t) = 1 [sin(x + t) + sin(x t)] + 1 Proeed to evaluate the last integrals. ˆ x+t x t ˆ x+t (1 + x ) dx + 1 ˆ t ˆ x (t t ) x +(t t ) ˆ t ˆ x (t t) u(x, t) = 1 [sin(x + t) + sin(x t)] + 1 (1 + x ) dx + 1 x t = 1 ( os t sin x) + 1 ( ) x + x x+t + 1 ˆ t x+(t t ) sin x dt x t x (t t ) ˆ t x+(t t) ˆ t ˆ x+(t t ) x (t t ) os x dx dt os x dx dt = os t sin x + 1 (t + tx) + 1 {sin[x + (t t )] sin[x (t t )]} dt = os t sin x + t + tx + 1 ( ) os x os(x + t) os(x t) os x = os t sin x + t(x + 1) 1 [os(x + t) + os(x t)] + 1 os x = os t sin x + t(x + 1) 1 ( os t os x) + 1 os x = os t sin x + t(x + 1) 1 os t os x + 1 os x Therefore, u(x, t) = 1 os x(1 os t) + sin x os t + t(x + 1).

12 Strauss PDEs e: Setion Exerise 3 Page 1 of 13 Solution by Duhamel s Priniple u tt u xx = os x, < x <, t > u(x, ) = sin x u t (x, ) = 1 + x Use the fat that the PDE is linear to split up the problem. Let u(x, t) = v(x, t) + w(x, t), where v and w satisfy the following initial value problems. v tt v xx = w tt w xx = os x v(x, ) = sin x v t (x, ) = 1 + x w(x, ) = w t (x, ) = The solution for v is given by d Alembert s formula in setion.1 on page 36. ˆ x+t v(x, t) = 1 [sin(x + t) + sin(x t)] + 1 = os t sin x + t(x + 1) x t (1 + x ) dx Aording to Duhamel s priniple, the solution to the inhomogeneous wave equation is w(x, t) = ˆ t W (x, t s; s) ds, where W = W (x, t; s) is the solution to the assoiated homogeneous equation with a partiular hoie for the initial onditions. W tt W xx =, < x <, t > W (x, ; s) = The solution for W is given by d Alembert s formula. W (x, t; s) = 1 ˆ x+t x t = 1 sin r x+t W t (x, ; s) = os x os r dr x t = 1 [sin(x + t) sin(x t)] = 1 [ os x sin t] = 1 os x sin t The solution to the inhomogeneous wave equation is then ˆ t 1 w(x, t) = os x sin[(t s)] ds = 1 ˆ t os x { sin[(s t)]} ds = 1 os x ( 1 os t ) = 1 os x(1 os t).

13 Strauss PDEs e: Setion Exerise 3 Page 13 of 13 Therefore, u(x, t) = 1 os x(1 os t) + sin x os t + t(x + 1). We an hek that the Duhamel solution satisfies the wave equation. Use the Leibnitz rule to differentiate the integrals. w tt w xx = [ ˆ t ] ˆ t W (x, t s; s) ds t t x W (x, t s; s) ds = t = = ˆ t ˆ t [ˆ t t ] ˆ t W (x, t s; s) ds + W (x, ; t) 1 W (x, t; ) W xx (x, t s; s) ds }{{} = t W (x, t s; s) ds + W t(x, ; t) 1 W t (x, t; ) [W tt (x, t s; s) W xx (x, t s; s) ] ds + W t (x, ; t) = os x }{{} = ˆ t W xx (x, t s; s) ds