1 MATH443 PARTIAL DIFFERENTIAL EQUATIONS Second Midterm Exam-Solutions December 6 2017 Wednesday 10:40-12:30 SA-Z02 QUESTIONS: Solve any four of the following five problems [25]1. Solve the initial and boundary value problem u t ku xx = 0 k > 0 t > 0 x (0 1) u(t 0) = 0 u(t 1) = 0 t > 0 u(0 x) = A x(1 x) x (0 1) and discuss the validity of your solution. Here A is a given constant Solution: We have solved this problem in class and also in lecture notes. We proved a theorem saying that if the initial temperature vanishes at the end poins and has continuous second derivatives in (0 1) then there exist unique solution. This problem is a verification of this theorem. Since f(x) = Ax(1 x) satisfy the conditions f(0) = f(1) = 0 and has constant second derivative in (0 1) we expect that this initial and boundary value problem has unique solution. In class we have found that (L = 1) u(t x) = B n sin(nπx) e kn2 π 2t B n = 2A n=1 ˆ 1 0 (x x 2 ) sin(nπx)dx n = 1 2 By using the integration by part we find that B n = 0 for n = even and B n = n = odd. Hence the solution of the initial and boundary value problem is 4A n 3 π 3 for u(t x) = 4A π 3 n=odd sin(nπx) n 3 e kn2 π2t for all x [0 1]. The sums in the RHS above and below x(1 x) = 4 π 3 n=odd sin(nπx) n 3
2 are both uniformly convergent in (0 1). Hence the solution given above is a valid solution of the problem. [25]2. Let xy 3 z xx + x 2 y 2 z xy 2x 3 y z yy y 3 z x + 2yx 3 z y = 0. (a) Determine the type of this partial differential equation (b) find its canonical form and (c) find the solution of this equation. Solution: (a) P = xy 3 S = x 2 y 2 and T = 2yx 3. Hence 4RT S 2 = 9x 4 y 4 0. This means that differential equation is hyperbolic in region xy 0. For xy = 0 differential equation reduces to a condition on z on the x and y -axis. Hence we shall only consider the hyperbolic case. (b) To find the canonical form we need the coordinate transformations. The quadratic equation P λ 2 + Sλ + T = 0 has two real roots λ 1 = ξ x ξ y = x y λ 2 = η x η y = 2x y Hence we find ξ = x 2 + y 2 and η = x 2 1 2 y2. With this change of variables we have z(x y) = ζ(ξ η) and z x = 2xζ ξ + 2xζ η z y = 2yζ ξ yζ η z xx = 4x 2 (ζ ξξ + 2ζ ξη + ζ ηη ) + 2ζ ξ + 2ζ η z xy = 2xy(2ζ ξξ + ζ ξη ζ ηη ) z yy = y 2 (4ζ ξξ 4ζ ξη + ζ ηη ) + 2ζ ξ ζ η. Inserting these into the differential equation above we find 18x 3 y 3 ζ ξη = 0 since xy 0 we have the resulting form of the equation ζ ξη = 0 (c) The general solution is then given by z(x y) = f(x 2 + y 2 ) + g(x 2 1 2 y2 ) where f and g are twice differentiable functions.
3 [25]3. Find the solution of the equation z xxx + 2λ z xxy + λ 2 z xyy = e 2x where λ is any real constant. Solution: Differential operator L correspond to this differential equation is reducible this means that z xxx + 2λ z xxy + λ 2 z xyy = D (D + λd ) 2 z = e 2x Hence the solution is z = z 1 + z 2 + z p. Here z p is the particular solution which can be easily found by letting z p = Ae 2x. Inserting it into the differential equation we find that A = 1. 8 z 1 and z 2 are the solutions of Dz 1 = 0 (D + λd ) 2 z 2 = 0 We find that z 1 = f 1 (y) z 2 = x f 2 (λx y) + f 3 (λx y) Hence the solution of the problem is z(x y) = 1 8 e2x + f 1 (y) + x f 2 (λx y) + f 3 (λx y) where f 1 f 2 and f 3 are twice differentiable functions. [25]4. Solve the initial value problems (a) z xx z yy = 0 with z(0 y) = h(y) and z x (0 y) = H(y) where h and H are some differentiable functions (b) z xx z yy = 0 with z(y y) = w(y) and z x (y y) = W (y) where w and W are some differentiable functions. Solution: There are two characteristic curves of this differential equation x y = const and x + y = const. (a) The initial curve in this problem is x = 0 y = t which is not one of the characteristic curve. Hence there should exist a unique solution of this problem. Differential equation has the general solution z(x y) = f(x + y) + g(x y) where f and g are twice differentiable functions. Using the initial conditions we get f(y) + g( y) = h(y) f y (y) g y ( y) = H(y) Then we find f and g in terms of the given functions h and H as
4 f(y) = 1 2 h(y) + 1 2 g y (y) = 1 2 h( y) 1 2 ˆ y H(ξ) dξ a ˆ y where a is an arbitrary constant. Then the solution is a H(ξ) dξ z(x y) = 1 2 (h(x + y) + h(y x)) + 1 2 ˆ y+x y x H(ξ) dξ (b) The initial curve in this problem is x = t y = t. Hence it is one of the characteristic curve then either there exists no solution or there exists infinitely many solutions. The general solution is same as part (a). Using the initial conditions we get f(2y) + g(0) = w(y) f ξ (2y) + g η 0 = W (y) 1 2 f y(2y) + g η 0 = W (y) We find that f(y) = g(0) + w(y/2) and we have the constraint 2W y = w yy. Hence we have the conclusion: If 2W y w yy there exists no solution of this problem. If 2W y = w yy there exists infinitely many solutions The function g is left arbitrary. z(x y) = w( x + y ) g(0) + g(x y) 2 [25]5. Find the second order linear partial differential equation whose general integral is z(x y) = f(2x + y 2 ) + g(x 2y 2 ) where f and g are at least twice differentiable functions. Solution: Taking the partial derivatives of z we obtain z x (x y) = 2f + g z y (x y) = 2yf 4yg z xx = 4f + g z xy = 4yf 4yg z yy = 2f 4g + 4y 2 f + 16y 2 g Solving f g f and g from the first four equations above we get (for y 0)
5 f = z y + 4yz x 10y g = yz x z y 5y f = z xy + 4yz xx 20y g = 4yz xx 4z xy 20y and inserting them in to the last equation we obtain 4y 3 z xx 3y 2 z xy y z yy + z y = 0