We calculate the derivatives of σ and τ. σ x = 0 σ y = 2y τ x = 2x τ y = 0 Then we calculate the derivatives of u. u x = 2xu τ u y = 2yu σ u xx = 4x 2 u ττ + 2u τ u yy = 4y 2 u σσ + 2u σ Finally we transform the equation to canonical form. y 2 u xx + x 2 u yy = 0 σ(4τu ττ + 2u τ ) + τ(4σu σσ + 2u σ ) = 0 u σσ + u ττ = − 1 2σ u σ − 1 2τ u τ 36.2 Equilibrium Solutions Example 36.2.1 Consider the equilibrium solution for the following problem. u t = u xx , u(x, 0) = x, u x (0, t) = u x (1, t) = 0 Setting u t = 0 we have an ordinary differential equation. d 2 u dx 2 = 0 This equation has the solution, u = ax + b. 1694 Applying the boundary conditions we see that u = b. To determine the constant, we note that the heat energy in the rod is constant in time.  1 0 u(x, t) dx =  1 0 u(x, 0) dx  1 0 b dx =  1 0 x dx Thus the equilibrium solution is u(x) = 1 2 . 1695 36.3 Exercises Exercise 36.1 Classify and transform the following equation into canonical form. u xx + (1 + y) 2 u yy = 0 Hint, Solution Exercise 36.2 Classify as hyperb olic, parabolic, or elliptic in a region R each of the equations: 1. u t = (pu x ) x 2. u tt = c 2 u xx − γu 3. (qu x ) x + (qu t ) t = 0 where p(x), c(x, t), q(x, t), and γ(x) are given functions that take on only positive values in a region R of the (x, t) plane. Hint, Solution Exercise 36.3 Transform each of the following equations for φ(x, y) into canonical form in appropriate regions 1. φ xx − y 2 φ yy + φ x − φ + x 2 = 0 2. φ xx + xφ yy = 0 The equation in part (b) is known as Tricomi’s equation and is a mo del for transonic fluid flow in which the flow speed changes from supersonic to subsonic. Hint, Solution 1696 36.4 Hints Hint 36.1 Hint 36.2 Hint 36.3 1697 36.5 Solutions Solution 36.1 For y = −1, the equation is parabolic. For this case it is already in the canonical form, u xx = 0. For y = −1, the equation is elliptic. We find new variables that will put the equation in the form u ξψ = G(ξ, ψ, u, u ξ , u ψ ). dy dx = ı  (1 + y) 2 = ı(1 + y) dy 1 + y = ıdx log(1 + y) = ıx + c 1 + y = c e ıx (1 + y) e −ıx = c ξ = (1 + y) e −ıx ψ = ξ = (1 + y) e ıx The variables that will put the equation in canonical form are σ = ξ + ψ 2 = (1 + y) cos x, τ = ξ −ψ ı2 = (1 + y) sin x. We calculate the derivatives of σ and τ. σ x = −(1 + y) sin x σ y = cos x τ x = (1 + y) cos x τ y = sin x Then we calculate the derivatives of u. u x = −(1 + y) sin(x)u σ + (1 + y) cos(x)u τ u y = cos(x)u σ + sin(x)u τ u xx = (1 + y) 2 sin 2 (x)u σσ + (1 + y) 2 cos 2 (x)u ττ − (1 + y) cos(x)u σ − (1 + y) sin(x)u τ u yy = cos 2 (x)u σσ + sin 2 (x)u ττ 1698 We substitute these results into the differential equation to obtain the canonical form. u xx + (1 + y) 2 u yy = 0 (1 + y) 2 (u σσ + u ττ ) − (1 + y) cos(x)u σ − (1 + y) sin(x)u τ = 0  σ 2 + τ 2  (u σσ + u ττ ) − σu σ − τu τ = 0 u σσ + u ττ = σu σ + τu τ σ 2 + τ 2 Solution 36.2 1. u t = (pu x ) x pu xx + 0u xt + 0u tt + p x u x − u t = 0 Since 0 2 − p0 = 0, the equation is parabolic. 2. u tt = c 2 u xx − γu u tt + 0u tx − c 2 u xx + γu = 0 Since 0 2 − (1)(−c 2 ) > 0, the equation is hyperbolic. 3. (qu x ) x + (qu t ) t = 0 qu xx + 0u xt + qu tt + q x u x + q t u t = 0 Since 0 2 − qq < 0, the equation is elliptic. 1699 Solution 36.3 1. For y = 0, the equation is hyperbolic. We find the new independent variables. dy dx =  y 2 1 = y, y = c e x , e −x y = c, ξ = e −x y dy dx = −  y 2 1 = −y, y = c e −x , e x y = c, ψ = e x y Next we determine x and y in terms of ξ and ψ. ξψ = y 2 , y =  ξψ ψ = e x  ξψ, e x =  ψ/ξ, x = 1 2 log  ψ ξ  We calculate the derivatives of ξ and ψ. ξ x = − e −x y = −ξ ξ y = e −x =  ξ/ψ ψ x = e x y = ψ ψ y = e x =  ψ/ξ Then we calculate the derivatives of φ. ∂ ∂x = −ξ ∂ ∂ξ + ψ ∂ ∂ψ , ∂ ∂y =  ξ ψ ∂ ∂ξ +  ψ ξ ∂ ∂ψ φ x = −ξφ ξ + ψφ ψ , φ y =  ξ ψ φ ξ +  ψ ξ φ ψ φ xx = ξ 2 φ ξξ − 2ξψφ ξψ + ψ 2 φ ψψ + ξφ ξ + ψφ ψ , φ yy = ξ ψ φ ξξ + 2φ ξψ + ψ ξ φ ψψ 1700 Finally we transform the equation to canonical form. φ xx − y 2 φ yy + φ x − φ + x 2 = 0 −4ξψφ ξψ + ξφ ξ + ψφ ψ − ξφ ξ + ψφ ψ − φ + log  ψ ξ  = 0 φ ξψ = 1 2ξ φ ψ + φ − log  ψ ξ  For y = 0 we have the ordinary differential equation φ xx + φ x − φ + x 2 = 0. 2. For x < 0, the equation is hyperbolic. We find the new independent variables. dy dx = √ −x, y = 2 3 x √ −x + c, ξ = 2 3 x √ −x − y dy dx = − √ −x, y = − 2 3 x √ −x + c, ψ = 2 3 x √ −x + y Next we determine x and y in terms of ξ and ψ. x = −  3 4 (ξ + ψ)  1/3 , y = ψ −ξ 2 We calculate the derivatives of ξ and ψ. ξ x = √ −x =  3 4 (ξ + ψ)  1/6 , ξ y = −1 ψ x =  3 4 (ξ + ψ)  1/6 , ψ y = 1 1701 Then we calculate the derivatives of φ. φ x =  3 4 (ξ + ψ)  1/6 (φ ξ + φ ψ ) φ y = −φ ξ + φ ψ φ xx =  3 4 (ξ + ψ)  1/3 (φ ξξ + φ ψψ ) + (6(ξ + ψ)) 1/3 φ ξψ + (6(ξ + ψ)) −2/3 (φ ξ + φ ψ ) φ yy = φ ξξ − 2φ ξψ + φ ψψ Finally we transform the equation to canonical form. φ xx + xφ yy = 0 (6(ξ + ψ)) 1/3 φ ξψ + (6(ξ + ψ)) 1/3 φ ξψ + (6(ξ + ψ)) −2/3 (φ ξ + φ ψ ) = 0 φ ξψ = − φ ξ + φ ψ 12(ξ + ψ) For x > 0, the equation is elliptic. The variables we defined before are complex-valued. ξ = ı 2 3 x 3/2 − y, ψ = ı 2 3 x 3/2 + y We choose the new real-valued variables. α = ξ − ψ, β = −ı(ξ + ψ) We write the derivatives in terms of α and β. φ ξ = φ α − ıφ β φ ψ = −φ α − ıφ β φ ξψ = −φ αα − φ ββ 1702 [...]... z, t) has the form u(x, y, z, t) = v(x, y, z) eıωt , which satisfy the conditions: u(x, y, z, t) = 0 for x = 0, L, y = 0, L, lim |u| = ∞ and = 0 z > 0, z→∞ Indicate in terms of inequalities involving k = ω/c and appropriate eigenvalues, λn,m say, for which n and m the solutions un,m satisfy the conditions Hint, Solution Exercise 37.30 Find the modes of oscillation and their frequencies for a rectangular... λ) By looking at Figure 37.1, (the plot shows the functions f (x) = x, f (x) = cot x and has lines at x = nπ), we see that there are an infinite number of positive eigenvalues and that λn → (nπ)2 as n → ∞ The eigenfunctions are ψn = cos( 17 14 λn x) 10 8 6 4 2 2 4 6 8 10 -2 Figure 37.1: Plot of x and cot x The solution for φ is φn = an cos( λn t) + bn sin( λn t) Thus the solution to the differential equation... with boundary and initial conditions given by φ(x, 0) = 0, φ(0, t) = t, 1722 φx (l, t) = −cφ(l, t) where c > 0 is a constant Hint, Solution Exercise 37.16 Let φ(x, t) satisfy the equation φt = a2 φxx for 0 < x < l, t > 0 with initial conditions φ(x, 0) = 0 for 0 < x < l, with boundary conditions φ(0, t) = 0 for t > 0, and φ(l, t) + φx (l, t) = 1 for t > 0 Obtain two series solutions for this problem,... the right side in only dependent on x, and the relation is valid for all t and x, both sides of the equation must be constant T X = = −λ κT X Here −λ is an arbitrary constant (You’ll see later that this form is convenient.) u(x, t) = X(x)T (t) will satisfy the partial differential equation if X(x) and T (t) satisfy the ordinary differential equations, T = −κλT and X = −λX Now we see how lucky we are... the right end is insulated and the initial temperature distribution is known at time t = 0 To find the temperature we solve the problem: ∂u ∂2u = κ 2, 0 < x < h, t > 0 ∂t ∂x u(0, t) = ux (h, t) = 0 u(x, 0) = f (x) 1 Why h? Because l looks like 1 and we use L to denote linear operators 17 04 We look for special solutions of the form, u(x, t) = X(x)T (t) Substituting this into the partial differential equation... this problem, one which is useful for large t and the other useful for small t Hint, Solution Exercise 37.17 A rod occupies the portion 1 < x < 2 of the x-axis The thermal conductivity depends on x in such a manner that the temperature φ(x, t) satisfies the equation φt = A2 (x2 φx )x (37.6) where A is a constant For φ(1, t) = φ(2, t) = 0 for t > 0, with φ(x, 0) = f (x) for 1 < x < 2, show that the appropriate... included as a time dependent forcing term of the form: s(x, t) = v cos π(x−ξ) 2d sin πt δ , for |x − ξ| < d, 0 0 < t < δ, otherwise Find the motion of the string for t > δ Discuss the effects of the width of the hammer and duration of the blow with regard to the energy in overtones Hint, Solution Exercise 37.29 Find the propagating modes in a square waveguide of side L for harmonic signals of frequency... Hint, Solution Exercise 37. 14 Solve the heat equation of Exercise 37.13 with the same initial conditions but with the boundary conditions φ(0, t) = 0, cφ(l, t) + φx (l, t) = 0 Here c > 0 is a constant Although it is not possible to solve for the eigenvalues λ in closed form, show that the eigenvalues assume a simple form for large values of λ Hint, Solution Exercise 37. 15 Use a series expansion technique... (h) = 0 We have a regular Sturm-Liouville problem for X(x) X + λX = 0, X(0) = X (h) = 0 The eigenvalues and orthonormal eigenfunctions are λn = 2 (2n − 1)π 2h 2 , Xn = 2 sin h Actually luck has nothing to do with it I planned it that way 17 05 (2n − 1)π x , 2h n ∈ Z+ Now we solve the equation for T (t) T = −κλn T T = c e−κλn t The eigen-solutions of the partial differential equation that satisfy the homogeneous... variables for a linear partial differential equation for u(x, y, z, ) 1 Substitute u(x, y, z, ) = X(x)Y (y)Z(z) · · · into the partial differential equation Separate the equation into ordinary differential equations 2 Translate the boundary conditions for u into boundary conditions for X, Y , Z, The continuity of u may give additional boundary conditions and boundedness conditions 3 Solve the differential . Solution 1696 36 .4 Hints Hint 36.1 Hint 36.2 Hint 36.3 1697 36 .5 Solutions Solution 36.1 For y = −1, the equation is parabolic. For this case it is already in the canonical form, u xx = 0. For y = −1,. we determine x and y in terms of ξ and ψ. x = −  3 4 (ξ + ψ)  1/3 , y = ψ −ξ 2 We calculate the derivatives of ξ and ψ. ξ x = √ −x =  3 4 (ξ + ψ)  1/6 , ξ y = −1 ψ x =  3 4 (ξ + ψ)  1/6 ,. f(x) 1 Why h? Because l looks like 1 and we use L to denote linear operators 17 04 We look for special solutions of the form, u(x, t) = X(x)T (t). Substituting this into the partial differential equation yields X(x)T  (t)