Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 9 pot

42.3 Solutions Solution 42.1 We write the derivatives of u(x, t) in terms of derivatives of F(η). u t = αxt α−1 F  = α η t F  u x = t α F  u xx = t 2α F  = η 2 x 2 F  We substitite these expressions into the h eat equation. α η t F  = ν η 2 x 2 F  F  = α ν x 2 t 1 η F  We can write this equation in terms of F and η only if α = −1/2. We make this substitution and solve the ordinary differential equation for F (η). F  F  = − η 2ν log(F  ) = − η 2 4ν + c F  = c exp  − η 2 4ν  F = c 1  exp  − η 2 4ν  dη + c 2 We can write F in terms of the error function. F = c 1 erf  η 2 √ ν  + c 2 1894 We write this solution in terms of x and t. u(x, t) = c 1 erf  x 2 √ νt  + c 2 This is not the general solution of the heat equation. There are many other solutions. Note that since x and t do not explicitly appear in the heat equation, u(x, t) = c 1 erf  x − x 0 2  ν(t − t 0 )  + c 2 is a solution. Solution 42.2 We write the derivatives of φ in terms of f. φ t = ∂ξ ∂t ∂ ∂ξ f = x α f  = t −1 ξf  φ x = ∂ξ ∂x ∂ ∂ξ f = αx α−1 tf  φ xx = f  ∂ ∂x  αx α−1 t  + αx α−1 tαx α−1 t ∂ ∂ξ f  φ xx = α 2 x 2α−2 t 2 f  + α(α − 1)x α−2 tf  φ xx = x −2  α 2 ξ 2 f  + α(α − 1)ξf   We substitute these expressions into the di ffus ion equation. ξf  = x −2 t  α 2 ξ 2 f  + α(α − 1)ξf   In order for this equation to depend only on the variable ξ, we must have α = −2. For this choice we obtain an ordinary 1895 differential equation for f(ξ). f  = 4ξ 2 f  + 6ξf  f  f  = 1 4ξ 2 − 3 2ξ log(f  ) = − 1 4ξ − 3 2 log ξ + c f  = c 1 ξ −3/2 e −1/(4ξ) f(ξ) = c 1  ξ t −3/2 e −1/(4t) dt + c 2 f(ξ) = c 1  1/(2 √ ξ) e −t 2 dt + c 2 f(ξ) = c 1 erf  1 2 √ ξ  + c 2 1896 Chapter 43 Method of Characteristics 43.1 First Order Linear Equations Consider the following first order wave equation. u t + cu x = 0 (43.1) Let x(t) be some path in the phase plane. Perhaps x(t) describes the position of an observer who is noting the value of the solution u(x(t), t) at their current location. We differentiate with respect to t to see how the solution varies for the observer. d dt u(x(t), t) = u t + x  (t)u x (43.2) We note that if the observer is moving with velocity c, x  (t) = c, then the solution at their current location does not change because u t + cu x = 0. We will examine this more carefully. By comparing Equations 43.1 and 43.2 we obtain ordinary differential equations representing the position of an observer and the value of the solution at that position. dx dt = c, du dt = 0 1897 Let the observer start at the position x 0 . Then we have an initial value problem for x(t). dx dt = c, x(0) = x 0 x(t) = x 0 + ct These lines x(t) are called characteristics of Equation 43.1. Let the initial condition be u(x, 0) = f(x). We have an initial value problem for u(x(t), t). du dt = 0, u(0) = f(x 0 ) u(x(t), t) = f(x 0 ) Again we see that the solution is constant along the characteristics. We substitute the equation for the characteristics into this expression. u(x 0 + ct, t) = f(x 0 ) u(x, t) = f(x − ct) Now we see that the solution of Equation 43.1 is a wave moving with velocity c. The solution at time t is the initial condition translated a distance of ct. 43.2 First Order Quasi-Linear E quations Consider the following quasi-linear equation. u t + a(x, t, u)u x = 0 (43.3) We will solve this equation with the method of characteristics. We differentiate the solution along a path x(t). d dt u(x(t), t) = u t + x  (t)u x (43.4) 1898 By comparing Equations 43.3 and 43.4 we obtain ordinary differential equations for the characteristics x(t) and the solution along the characteristics u(x(t), t). dx dt = a(x, t, u), du dt = 0 Suppose an initial condition is specified, u(x, 0) = f(x). Then we have ordinary differential equation, initial value problems. dx dt = a(x, t, u), x(0) = x 0 du dt = 0, u(0) = f(x 0 ) We see that the solution is constant along the characteristics. The solution of Equation 43.3 is a wave moving with velocity a(x, t, u). Example 43.2.1 Consider the inviscid Burger equation, u t + uu x = 0, u(x, 0) = f(x). We write down the differential equations for the solution along a characteristic. dx dt = u, x(0) = x 0 du dt = 0, u(0) = f(x 0 ) First we solve the equation for u. u = f(x 0 ). Then we solve for x. x = x 0 + f(x 0 )t. This gives us an implicit solution of the Burger equation. u(x 0 + f(x 0 )t, t) = f(x 0 ) 1899 43.3 The Method of Characteristics and the Wave Equation Consider the one dimensional wave equation, u tt = c 2 u xx . We make the change of variables, a = u x , b = u t , to obtain a c oupled system of first order equations. a t − b x = 0 b t − c 2 a x = 0 We write this as a matrix equation.  a b  t +  0 −1 −c 2 0  a b  x = 0 The eigenvalues and eigenvectors of the matrix are λ 1 = −c, λ 2 = c, ξ 1 =  1 c  , ξ 2 =  1 −c  . The matrix is diagonalized by a similarity transformation.  −c 0 0 c  =  1 1 c −c  −1  0 −1 −c 2 0  1 1 c −c  We make a change of variables to diagonalize the system.  a b  =  1 1 c −c  α β   1 1 c −c  α β  t +  0 −1 −c 2 0  1 1 c −c  α β  x = 0 1900 Now we left multiply by the inverse of the matrix of eigenvectors to obtain an uncoupled system that we can solve directly.  α β  t +  −c 0 0 c  α β  x = 0. α(x, t) = p(x + ct), β(x, t) = q(x −ct), Here p, q ∈ C 2 are arbitrary functions. We change variables back to a and b. a(x, t) = p(x + ct) + q(x − ct), b(x, t) = cp(x + ct) − cq(x − ct) We could integrate either a = u x or b = u t to obtain the solution of the wave equation. u = F(x − ct) + G(x + ct) Here F, G ∈ C 2 are arbitrary functions. We see that u(x, t) is the sum of a waves moving to the right and left with speed c. This is the general solution of the one-dimensional wave equation. Note that for any given problem, F and G are only determined to whithin an additive constant. For any constant k, adding k to F and subtracting it from G does not change the solution. u = (F(x − ct) + k) + (G(x − ct) − k) 43.4 The Wave Equation for an Infinite Domain Consider the Cauchy problem for the wave equation on −∞ < x < ∞. u tt = c 2 u xx , −∞ < x < ∞, t > 0 u(x, 0) = f(x), u t (x, 0) = g(x) We know that the solution is the sum of right-moving and left-moving waves. u(x, t) = F (x − ct) + G(x + ct) (43.5) 1901 The initial conditions give us two constraints on F and G. F (x) + G(x) = f(x), −cF  (x) + cG  (x) = g(x). We integrate the second equation. −F (x) + G(x) = 1 c  g(x) dx + const Here Q(x) =  q(x) dx. We solve the system of equations for F and G. F (x) = 1 2 f(x) − 1 2c  g(x) dx −k, G(x) = 1 2 f(x) + 1 2c  g(x) dx + k Note that the value of the constant k does not affect the solution, u(x, t). For simplicity we take k = 0. We substitute F and G into Equation 43.5 to determine the solution. u(x, t) = 1 2 (f(x − ct) + f(x + ct)) + 1 2c   x+ct g(x) dx −  x−ct g(x) dx  u(x, t) = 1 2 (f(x − ct) + f(x + ct)) + 1 2c  x+ct x−ct g(ξ) dξ u(x, t) = 1 2 (u(x − ct, 0) + u(x + ct, 0)) + 1 2c  x+ct x−ct u t (ξ, 0) dξ 43.5 The Wave Equation for a Semi-Infinite Domain Consider the wave equation for a semi-infinite domain. u tt = c 2 u xx , 0 < x < ∞, t > 0 u(x, 0) = f(x), u t (x, 0) = g(x), u(0, t) = h(t) 1902 [...]... satisfy y(s) ≥ 0 for all real s Figure 43.4 shows some characteristics in the (x, y) plane with starting points from ( 5, 1) to (5, 1) and a plot of the solution 2 1. 75 1 .5 1. 25 1 0. 75 0 .5 0. 25 -10-7 .5 -5 -2 .5 -2 -1 x 0 1 2 15 10 u 5 0 0 .5 2 .5 5 7 .5 10 1 y Figure 43.4: Some characteristics and the solution u(x, y) 191 7 1 .5 2 Chapter 44 Transform Methods 44.1 Fourier Transform for Partial Differential... zero for x > c2 t We apply these restrictions to the solution u(x, t) = F (x − c1 t) + g(x + c1 t), x < 0 f (x − c2 t), x>0 We use the continuity of u and ux at x = 0 to solve for f and g F (−c1 t) + g(c1 t) = f (−c2 t) F (−c1 t) + g (c1 t) = f (−c2 t) We integrate the second equation F (−t) + g(t) = f (−c2 t/c1 ) c1 −F (−t) + g(t) = − f (−c2 t/c1 ) + a c2 191 1 We solve for f for x < c2 t and for g for. .. 0, where f (t) is a given function 1 Obtain the formula for the Laplace transform of φ(x, t), Φ(x, s) and use the convolution theorem for Laplace transforms to show that t x 1 x2 f (t − τ ) 3/2 exp − 2 dτ φ(x, t) = √ τ 4a τ 2a π 0 192 3 2 Discuss the special case obtained by setting f (t) = 1 and also that in which f (t) = 1 for 0 < t < T , with f (t) = 0 for t > T Here T is some positive constant Exercise... δ(x − ξ) as y → 0 19 25 44 .5 Hints Hint 44.1 ∞ The desired solution form is: u(x, y) = −∞ K(x − ξ, y)f (ξ) dξ You must find the correct K Take the Fourier transform with respect to x and solve for u(ω, y) recalling that uxx = −ω 2 u By uxx we denote the Fourier transform ˆ ˆ ˆ ˆ with respect to x of uxx (x, y) Hint 44.2 Use the Fourier convolution theorem and the table of Fourier transforms in the appendix... Equation 43 .9 to obtain differential equations for x, y and u dx = y + u, ds dy = y, ds 19 15 du =x−y ds (43 .9) x -2 0 2 10 u 0 -10 -2 0 y 2 Figure 43.3: The solution u(x, y) We parametrize the initial data in terms of s x(s = 0) = α, y(s = 0) = 1, u(s = 0) = 1 + α We solve the equation for y subject to the inital condition y(s) = es This gives us a coupled set of differential equations for x and u dx =... 43.4 191 0 43.10 Solutions Solution 43.1 1 u(x, 0) = ut (x, 0) = F (x), x < 0 0, x>0 −c1 F (x), x < 0 0, x>0 2 Regardless of the initial condition, the solution has the following form u(x, t) = f1 (x − c1 t) + g1 (x + c1 t), x < 0 f2 (x − c2 t) + g1 (x + c2 t), x > 0 For x < 0, the right-moving wave is F (x − c1 t) and the left-moving wave is zero for x < −c1 t For x > 0, there is no left-moving wave and. .. Transform Consider the problem ∂u ∂u − + u = 0, ∂t ∂x −∞ < x < ∞, 192 0 t > 0, dξ u(x, 0) = f (x) Taking the Fourier Transform of the partial differential equation and the initial condition yields ∂U − ıωU + U = 0, ∂t 1 U (ω, 0) = F (ω) = 2π ∞ f (x) e−ıωx dx −∞ Now we have a first order differential equation for U (ω, t) with the solution U (ω, t) = F (ω) e(−1+ıω)t Now we apply the inverse Fourier transform... u and ux at x = 0 Suppose for t < 0 a wave approaches the junction x = 0 from the left, i.e as t approaches 0 from negative values: u(x, t) = F (x − c1 t) x < 0, t ≤ 0 0 x > 0, t ≤ 0 As t increases further, the wave reaches x = 0 and gives rise to reflected and transmitted waves 190 8 1 Formulate the appropriate initial values for u at t = 0 2 Solve the initial-value problem for −∞ < x < ∞ , t > 0 3 Identify... reflected and transmitted waves in your solution and determine the reflection and transmission coefficients for the junction in terms of c1 and c2 Comment also on their values in the limit c1 → c2 Exercise 43.2 Consider a semi-infinite string, x > 0 For all time the end of the string is displaced according to u(0, t) = f (t) Find the motion of the string, u(x, t) with the method of characteristics and then... Hint 44.4 Hint 44 .5 Use the Fourier convolution theorem The transform pairs, F[π(δ(x + τ ) + δ(x − τ ))] = cos(ωτ ), sin(ωτ ) F[π(H(x + τ ) − H(x − τ ))] = , ω will be useful Hint 44.6 Hint 44.7 Hint 44.8 v(x, t) satisfies the same partial differential equation You can solve the problem for v(x, t) with the Fourier sine 192 6 transform Use the convolution theorem to invert the transform To show that ∞ . the wave reaches x = 0 and gives rise to reflected and transmitted waves. 190 8 1. Formulate the appropriate initial values for u at t = 0. 2. Solve the initial-value problem for −∞ < x < ∞. (−t) + g(t) = f(−c 2 t/c 1 ) −F (−t) + g(t) = − c 1 c 2 f(−c 2 t/c 1 ) + a 191 1 We solve for f for x < c 2 t and for g for x > −c 1 t. f(−c 2 t/c 1 ) = 2c 2 c 1 + c 2 F (−t) + b, g(t) = c 2 −. x −2 t  α 2 ξ 2 f  + α(α − 1)ξf   In order for this equation to depend only on the variable ξ, we must have α = −2. For this choice we obtain an ordinary 18 95 differential equation for f(ξ). f  = 4ξ 2 f  +