Lecture Notes in Mathematics. A First Course in Quasi-Linear Partial Differential Equations for Physical Sciences and Engineering Solution Manual

Lecture Notes in Mathematics A First Course in Quasi-Linear Partial Differential Equations for Physical Sciences and Engineering Solution Manual Marcel B. Finan Arkansas Tech University c All Rights Reserved May 1, 16

Preface This manuscript provides the complete and detailed solutions to A First Course in Partial Differential Equations for Physical Sciences and Engineering. Distribution of this book in any form is prohibited. Marcel B Finan August 9

Contents Preface.................................. Solutions to Section 1......................... 4 Solutions to Section......................... 11 Solutions to Section 3......................... 18 Solutions to Section 4......................... 4 Solutions to Section 5......................... 31 Solutions to Section 6.1...................... 36 Solution to Section 6....................... 39 Solutions to Section 7......................... 4 Solutions to Section 8......................... 46 Solutions to Section 9......................... 5 Solutions to Section 1......................... 55 Solutions to Section 11......................... 68 Solutions to Section 1......................... 76 Solutions to Section 13......................... 83 Solutions to Section 14......................... 9 Solutions to Section 15......................... 95 Solutions to Section 16......................... 16 Solutions to Section 17......................... 115 Solutions to Section 18......................... 15 Solutions to Section 19......................... 137 Solutions to Section......................... 158 Solutions to Section 1......................... 168 Solutions to Section......................... 18 Solutions to Section 3......................... 191 Solutions to Section 4......................... Solutions to Section 5......................... 6 3

4 CONTENTS Solutions to Section 1 Problem 1.1 Classify the following equations as either ODE or PDE. (a (y 4 + t =. (y +4 (b u x + y u y = y x y+x. (c y 4y =. (a ODE with dependent variable y and independent variabe x. (b PDE with dependent variable u and independent variabes x and y. (c ODE with dependent variable y and independent variabe x Problem 1. Write the equation in the coordinates s = x, t = x y. We have u xx + u xy + u yy = u x =u s s x + u t t x = u s + u t u xx =u ss s x + u st t x + u st s x + u tt t x = u ss + u st + u tt u xy =u ss s y + u st t y + u st s y + u tt t y = u st u tt u y =u s s y + u t t y = u t u yy = u st s y u tt t y = u tt. Substituting these expressions into the given equation we find Problem 1.3 Write the equation u ss = in the coordinates s = x + y, t = x. u xx u xy + 5u yy =

SOLUTIONS TO SECTION 1 5 We have u x =u s s x + u t t x = u s + u t u xx =u ss s x + u st t x + u st s x + u tt t x = u ss + 4u st + 4u tt u xy =u ss s y + u st t y + u st s y + u tt t y = u ss + u st u y =u s s y + u t t y = u s u yy =u ss s y + u st t y = u ss. Substituting these expressions into the given equation we find u ss + u tt = Problem 1.4 For each of the following PDEs, state its order and whether it is linear or non-linear. If it is linear, also state whether it is homogeneous or nonhomogeneous: (a uu x + x u yyy + sin x =. (b u x + e x u y =. (c u tt + (sin yu yy e t cos y =. (a Order 3, non-linear. (b Order 1, linear, homogeneous. (c Order, linear, non-homogeneous Problem 1.5 For each of the following PDEs, determine its order and whether it is linear or not. For linear PDEs, state also whether the equation is homogeneous or not; For nonlinear PDEs, circle all term(s that are not linear. (a x u xx + e x u = xu xyy. (b e y u xxx + e x u = sin y + 1xu y. (c y u xx + e x uu x = xu y + u. (d u x u xxy + e x uu y = 5x u x. (e u t = k (u xx + u yy + f(x, y, t. (a Linear, homogeneous, order 3.

6 CONTENTS (b Linear, non-homogeneous, order 3. The inhomogeneity is sin y. (c Non-linear, order. The non-linear term is e x uu x. (d Non-linear, order 3. The non-linear terms are u x u xxy and e x uu y. (e Linear, non-homogeneous, order. The inhomogeneity is f(x, y, t Problem 1.6 Which of the following PDEs are linear? (a Laplace s equation: u xx + u yy =. (b Convection (transport equation: u t + cu x =. (c Minimal surface equation: (1+Z yz xx Z x Z y Z xy +(1+Z xz yy =. (d Korteweg-Vries equation: u t + 6uu x = u xxx. (a Linear. (b Linear. (c Non-linear where all the terms are non-linear. (d Non-linear with non-linear term 6uu x Problem 1.7 Classify the following differential equations as ODEs or PDEs, linear or non-linear, and determine their order. For the linear equations, determine whether or not they are homogeneous. (a The diffusion equation for u(x, t : (b The wave equation for w(x, t : u t = ku xx. w tt = c w xx. (c The thin film equation for h(x, t : h t = (hh xxx x. (d The forced harmonic oscillator for y(t : y tt + ω y = F cos (ωt. (e The Poisson Equation for the electric potential Φ(x, y, z : Φ xx + Φ yy + Φ zz = 4πρ(x, y, z.

SOLUTIONS TO SECTION 1 7 where ρ(x, y, z is a known charge density. (f Burger s equation for h(x, t : h t + hh x = νh xx. (a PDE, linear, second order, homogeneous. (b PDE, linear, second order, homogeneous. (c PDE, quasi-linear (non-linear, fourth order. (d ODE, linear, second order, non-homogeneous. (e PDE, linear, second order, non-homogeneous. (f PDE, quasilinear (non-linear, second order Problem 1.8 Write down the general form of a linear second order differential equation of a function in three variables. A(x, y, zu xx + B(x, y, zu xy + C(x, y, zu yy + E(x, y, zu xz + F (x, y, zu yz + G(x, y, zu zz +H(x, y, zu x +I(x, y, zu y +J(x, y, zu z +K(x, y, zu = L(x, y, z Problem 1.9 Give the orders of the following PDEs, and classify them as linear or nonlinear. If the PDE is linear, specify whether it is homogeneous or nonhomogeneous. (a x u xxy + y u yy log (1 + y u = (b u x + u 3 = 1 (c u xxyy + e x u x = y (d uu xx + u yy u = (e u xx + u t = 3u. (a Order 3, linear, homogeneous. (b Order 1, non-linear. (c Order 4, linear, non-homogeneous (d Order, non-linear. (e Order, linear, homogeneous

8 CONTENTS Problem 1.1 Consider the second-order PDE u xx + 4u xy + 4u yy =. Use the change of variables v(x, y = y x and w(x, y = x to show that u ww =. Using the chain rule we find u x = u v + u w u xx =4u vv 4u vw + u ww u y =u v u yy =u vv u xy = u vv + u vw. Substituting these into the given PDE we find u ww = Problem 1.11 Write the one dimensional wave equation u tt = c u xx in the coordinates v = x + ct and w = x ct. We have Substituting we find u vw = Problem 1.1 Write the PDE u t =cu v cu w u tt =c u vv c u wv + c u ww u x =u v + u w u xx =u vv + u vw + u ww. u xx + u xy 3u yy = in the coordinates v(x, y = y 3x and w(x, y = x + y.

SOLUTIONS TO SECTION 1 9 We have u x = 3u v + u w u xx = 3( 3u v + u w v + ( 3u v + u w w = 9u vv 6u vw + u ww u xy = 3u vv + u vw 3u vw + u ww = 3u vv u vw + u ww u y =u v + u w u yy =(u v + u w v + (u v + u w w = u vv + u vw + u ww. Substituting into the PDE we find u vw = Problem 1.13 Write the PDE au x + bu y =, a in the coordinates s(x, y = bx ay and t(x, y = x. According to the chain rule for the derivative of a composite function, we have u x =u s s x + u t t x = bu s + u t u y =u s s y + u t t y = au s. Substituting these into the given equation to obtain abu s + au t abu s = or and since a we obtain au t = u t = Problem 1.14 Write the PDE u x + u y = 1 in the coordinates s = x y and t = x.

1 CONTENTS Using the chain rule we find u x = u s s x + u t t x = u s + u t u y = u s s y + u t t y = u s. Substituting these into the PDE to obtain u t = 1 Problem 1.15 Write the PDE au t + bu x = u, b in the coordinates v = ax bt and w = x. We have u t = bu v and u x = au v + u w. Substituting we find u w = 1 b u

SOLUTIONS TO SECTION 11 Solutions to Section Problem.1 Determine a and b so that u(x, y = e ax+by is a solution to the equation u xxxx + u yyyy + u xxyy =. We have u xxxx = a 4 e ax+by, u yyyy = b 4 e ax+by, and u xxyy = a b e ax+by. Thus, substituting these into the equation we find (a 4 + a b + b 4 e ax+by =. Since e ax+by, we must have a 4 + a b + b 4 = or (a + b =. This is true only when a = b =. Thus, u(x, y = 1 Problem. Consider the following differential equation tu xx u t =. Suppose u(t, x = X(xT (t. Show that there is a constant λ such that X = λx and T = λtt. Substituting into the differential equation we find tx T XT = or X X = T tt. The LHS is a function of x only whereas the RHS is a function of t only. This is true only when both sides are constant. That is, there is λ such that X X = T tt = λ and this leads to the two ODEs X = λx and T = λtt

1 CONTENTS Problem.3 Consider the initial value problem xu x + (x + 1yu y =, x, y > 1 u(1, 1 = e. Show that u(x, y = xex y is the solution to this problem. ( We have xu x +(x+1yu y = x y (ex +xe x +(x+1y xex = and u(1, 1 = e y Problem.4 Show that u(x, y = e y sin (x y is the solution to the initial value problem u x + u y + u =, x, y > 1 u(x, = sin x. We have u x +u y +u = e y cos (x y e y sin (x y e y cos (x y+ e y sin (x y = and u(x, = sin x Problem.5 Solve each of the following differential equations: (a du = where u = u(x. dx = where u = u(x, y. (b u x (a The general solution to this equation is u(x = C where C is an arbitrary constant. (b The general solution is u(x, y = f(y where f is an arbitrary function of y Problem.6 Solve each of the following differential equations: (a d u = where u = u(x. dx = where u = u(x, y. (b u x y

SOLUTIONS TO SECTION 13 (a The general solution to this equation is u(x = C 1 x + C where C 1 and C are arbitrary constants. (b We have u y = f(y where f is an arbitrary differentiable function of y. Hence, u(x, y = f(ydy + g(x Problem.7 Show that u(x, y = f(y + x + xg(y + x, where f and g are two arbitrary twice differentiable functions, satisfy the equation Let v(x, y = y + x. Then Hence, u xx 4u xy + 4u yy =. u x =f v (v + g(v + xg v (v u xx =4f vv (v + 4g v (v + 4xg vv (v u y =f v (v + xg v (v u yy =f vv (v + xg vv (v u xy =f vv (v + g v (v + xg vv (v. u xx 4u xy + 4u yy =4f vv (v + 4g v (v + 4xg vv (v 8f vv (v 4g v (v 8xg vv (v +4f vv (v + 4xg vv (v = Problem.8 Find the differential equation whose general solution is given by u(x, t = f(x ct+g(x+ct, where f and g are arbitrary twice differentiable functions in one variable. Let v = x ct and w = x + ct. We have u x =f v v x + g w w x = f v + g w u xx =f vv v x + g ww w x = f vv + g ww u t =f v v t + g w w t = cf v + cg w u tt = cf vv v t + cg ww w t = c f vv + c g ww Hence, u satisfies the wave equation u tt = c u xx

14 CONTENTS Problem.9 Let p : R R be a differentiable function in one variable. Prove that u t = p(uu x has a solution satisfying u(x, t = f(x + p(ut, where f is an arbitrary differentiable function. Then find the general solution to u t = (sin uu x. Let v = x + p(ut. Using the chain rule we find Thus u t = f v v t = f v (p(u + p u u t t. (1 tf v p u u t = f v p. If 1 tf v p u on any t interval I then f v p on I which implies that f v or p on I. But either condition will imply that tf v p u = and this will imply that 1 = 1 tf v p u =, a contradiction. Hence, we must have 1 tf v p u. In this case, Likewise, or u t = f v p 1 tf v p u. u x = f v (1 + p u u x t u x = f v 1 tf v p u. It follows that u t = p(uu x. If u t = (sin uu x then p(u = sin u so that the general solution is given by u(x, t = f(x + t sin u where f is an arbitrary differentiable function in one variable Problem.1 Find the general solution to the pde Hint: See Problem 1.. u xx + u xy + u yy =.

SOLUTIONS TO SECTION 15 Using Problem 1., we found u ss =. Hence, u(s, t = sf(t + g(t where f and g are arbitrary differentiable functions. In terms of x and y we find u(x, y = xf(x y + g(x y Problem.11 Let u(x, t be a function such that u xx exists and u(, t = u(l, t = for all t R. Prove that L u xx (x, tu(x, tdx. Using integration by parts, we compute L L u xx (x, tu(x, tdx = u x (x, tu(x, t L x= u x(x, tdx =u x (L, tu(l, t u x (, tu(, t = L u (x, tdx L u x(x, tdx Note that we have used the boundary conditions u(, t = u(l, t = and the fact that u x(x, t for all x [, L] Problem.1 Consider the initial value problem u t + u xx =, x R, t > u(x, = 1. (a Show that u(x, t 1 is a solution to this problem. (b Show that u n (x, t = 1 + en t sin nx is a solution to the initial value n problem u t + u xx =, x R, t > sin nx u(x, = 1 + n. (c Find sup{ u n (x, 1 : x R}. (d Find sup{ u n (x, t 1 : x R, t > }. (e Show that the problem is ill-posed.

16 CONTENTS (a This can be done by plugging in the equations. (b Plug in. (c We have sup{ u n (x, 1 : x R} = 1 n sup{ sin nx : x R} = 1 n. (d We have sup{ u n (x, t 1 : x R} = en t. n e (e We have lim t sup{ u n (x, t 1 : x R, t > } = lim n t t =. n Hence, the solution is unstable and thus the problem is ill-posed Problem.13 Find the general solution of each of the following PDEs by means of direct integration. (a u x = 3x + y, u = u(x, y. (b u xy = x y, u = u(x, y. (c u xtt = e x+3t, u = u(x, t. (a u(x, y = x 3 + xy + f(y, where f is an arbitrary differentiable function. (b u(x, y = x3 y + F (x + g(y, where F (x = f(xdx and g(y is an 6 arbitrary differentiable function. (c u(x, t = 1 18 ex+3t + t f 1 (xdx + f (xdx + g(t Problem.14 Consider the second-order PDE u xx + 4u xy + 4u yy =. (a Use the change of variables v(x, y = y x and w(x, y = x to show that u ww =. (b Find the general solution to the given PDE. (a Using the chain rule we find u x = u v + u w u xx =4u vv 4u vw + u ww u y =u v u yy =u vv u xy = u vv + u vw.

SOLUTIONS TO SECTION 17 Substituting these into the given PDE we find u ww =. (b Solving the equation u ww = we find u w = f(v and u(v, w = wf(v + g(v. In terms of x and y the general solution is u(x, y = xf(y x + g(y x Problem.15 Derive the general solution to the PDE u tt = c u xx by using the change of variables v = x + ct and w = x ct. We have u t =cu v cu w u tt =c u vv c u wv + c u ww u x =u v + u w u xx =u vv + u vw + u ww Substituting we find u vw = and solving this equation we find u v = f(v and u(v, w = F (v + G(w where F (v = f(vdv. Finally, using the fact that v = x + ct and w = x ct; we get d Alembert s solution to the one-dimensional wave equation: u(x, t = F (x + ct + G(x ct where F and G are arbitrary differentiable functions

18 CONTENTS Solutions to Section 3 Problem 3.1 Solve the IVP: y + ty = t, y( = Since p(t = t, we find µ(t = e tdt = e t. Multiplying the given equation by e t to obtain ( e t y = te t Integrating both sides with respect to t and using substitution on the righthand integral to obtain e t y = 1 et + C Dividing the last equation by e t to obtain y(t = Ce t + 1 Since y( =, we find C = 1. Thus, the unique solution to the IVP is given by y = 1 (1 e t Problem 3. Find the general solution: y + 3y = t + e t Since p(t = 3, the integrating factor is µ(t = e 3t. Thus, the general solution is y(t = e 3t e 3t (t + e t dt + Ce 3t = e 3t ( (te 3t + e t dt + Ce 3t = e 3t e 3t (3t 1 + 9 et + Ce 3t 3t 1 = + e t + Ce 3t 9 Problem 3.3 Find the general solution: y + 1 y = 3 cos t, t > t

SOLUTIONS TO SECTION 3 19 Since p(t = 1, the integrating factor is µ(t = t e dt t method of integrating factor we find 1 y(t = t 3t cos tdt + C t = 3 (t sin t + cos t + C t t = 3 sin t + 3 cos t + C t t Problem 3.4 Find the general solution: y + y = cos (3t. = e ln t = t. Using the We have p(t = so that µ(t = e t. Thus, y(t = e t e t cos (3tdt + Ce t But 13 9 Hence, e t cos (3tdt = et 3 sin (3t 3 e t sin (3tdt = et 3 sin (3t 3 ( et 3 cos (3t + 3 e t cos (3tdt = et (3 sin (3t + cos (3t 9 e t cos (3tdt = et (3 sin (3t + cos (3t 13 y(t = 1 (3 sin (3t + cos (3t + Ce t 13 e t cos (3tdt Problem 3.5 Find the general solution: y + (cos ty = 3 cos t. Since p(t = cos t we have µ(t = e sin t. Thus, y(t =e sin t e sin t ( 3 cos tdt + Ce sin t = 3e sin t e sin t + Ce sin t =Ce sin t 3

CONTENTS Problem 3.6 Given that the solution to the IVP ty + 4y = αt, y(1 = 1 3 interval < t <. What is the value of the constant α? exists on the Solving this equation with the integrating factor method with p(t = 4 t find µ(t = t 4. Thus, y = 1 t 4 (αtdt + C t 4 t 4 we = α 6 t + C t 4 Since the solution is assumed to be defined for all t, we must have C =. On the other hand, since y(1 = 1 we find α = 3 Problem 3.7 Suppose that y(t = Ce t + t + 1 is the general solution to the equation y + p(ty = g(t. Determine the functions p(t and g(t. The integrating factor is µ(t = e t. Thus, p(tdt = t and this implies that p(t =. On the other hand, the function t + 1 is the particular solution to the nonhomogeneous equation so that (t + 1 + (t + 1 = g(t. Hence, g(t = t + 3 Problem 3.8 Suppose that y(t = e t + e t + sin t is the unique solution to the IVP y + y = g(t, y( = y. Determine the constant y and the function g(t. First, we find y : y = y( = + 1 + = 1. Next, we find g(t : g(t = y + y = ( e t + e t + sin t + ( e t + e t + sin t = e t + e t + cos t e t + e t + sin t = e t + cos t + sin t Problem 3.9 Find the value (if any of the unique solution to the IVP y + (1 + cos ty = 1 + cos t, y( = 3 in the long run?

SOLUTIONS TO SECTION 3 1 The integrating factor is µ(t = e (1+cos tdt = e t+sin t. Thus, the general solution is y(t =e (t+sin t e t+sin t (t+sin t (1 + cos tdt + Ce (t+sin t =1 + Ce Since y( = 3, we find C = and therefore y(t = 1 + e (t+sin t. Finally, lim y(t = lim (1 + t t e sin t e t = 1 Problem 3.1 Solve the initial value problem ty = y + t, y(1 = 7 Rewriting the equation in the form y 1 t y = 1 we find p(t = 1 and µ(t = 1. Thus, the general solution is given by t t But y(1 = 7 so that C = 7. Hence, y(t = t ln t + Ct y(t = t ln t + 7t Problem 3.11 Show that if a and λ are positive constants, and b is any real number, then every solution of the equation y + ay = be λt has the property that y as t. Hint: Consider the cases a = λ and a λ separately. Since p(t = a we find µ(t = e at. Suppose first that a = λ. Then y + ay = be at

CONTENTS and the corresponding general solution is Thus, Now, suppose that a λ then y(t = bte at + Ce at lim t y(t = lim t ( bt + C e at b = lim t = ae at e at Thus, y(t = b a λ e λt + Ce at lim y(t = t Problem 3.1 Solve the initial-value problem y + y = e t y, y( = 1 using the substitution u(t = 1 y(t Substituting into the equation we find u u = e t, u( = 1 Solving this equation by the method of integrating factor with µ(t = e t we find u(t = te t + Ce t Since u( = 1, C = 1 and therefore u(t = te t + e t. Finally, we have y(t = ( te t + e t 1 Problem 3.13 Solve the initial-value problem ty + y = t t + 1, y(1 = 1 Rewriting the equation in the form y + t y = t 1 + 1 t

SOLUTIONS TO SECTION 3 3 Since p(t = t we find µ(t = t. The general solution is then given by y(t = t 4 t 3 + 1 + C t Since y(1 = 1 we find C = 1 1. Hence, Problem 3.14 Solve y 1 y = sin t, t integral, Si(t = t sin s s y(t = t 4 t 3 + 1 + 1 1t Since p(t = 1 t we find µ(t = 1 t. Thus, y(1 = 3. Express your answer in terms of the sine ds. = ( t y(t = Si(t + C ( 1 t y 1 t sin s s y(t = tsi(t + Ct Since y(1 = 3, C = 3 Si(1. Hence, y(t = tsi(t + (3 Si(1t

4 CONTENTS Solutions to Section 4 Problem 4.1 Solve the (separable differential equation y = te t ln y. At first, this equation may not appear separable, so we must simplify the right hand side until it is clear what to do. y =te t ln y =te t e ln ( 1 y =te t 1 y = t y et. Separating the variables and solving the equation we find 1 3 y y =te t (y 3 dt = te t dt 1 3 y3 = 1 et + C y 3 = 3 et + C Problem 4. Solve the (separable differential equation y = t y 4y t +.

SOLUTIONS TO SECTION 4 5 Separating the variables and solving we find y 4 y =t t + = t (ln y dt = (t dt ln y = t t + C y(t =Ce t t Problem 4.3 Solve the (separable differential equation ty = (y 4. Separating the variables and solving we find y y 4 = t (ln y 4 dt = t dt ln y 4 = ln t + C ln y 4 =C t y(t =Ct + 4 Problem 4.4 Solve the (separable differential equation y = y( y. Separating the variables and solving (using partial fractions in the process

6 CONTENTS we find 1 y y y( y = y + y ( y = (ln y dt 1 (ln y dt = dt ln y y =4t + C y y =Ce4t y(t = Ce4t 1 + Ce 4t Problem 4.5 Solve the IVP y = 4 sin (t, y( = 1. y Separating the variables and solving we find yy =4 sin (t (y =8 sin (t (y dt = 8 sin (tdt y = 4 cos (t + C y(t = ± C 4 cos (t. Since y( = 1, we find C = 5 and hence y(t = 5 4 cos (t Problem 4.6 Solve the IVP: yy = sin t, y( π =.

SOLUTIONS TO SECTION 4 7 Separating the variables and solving we find ( y dt = sin tdt y = cos t + C y = cos t + C. Since y( π =, we find C = 4. Thus, y(t = ± ( cos t + 4. From y( π =, we have y(t = ( cos t + 4 Problem 4.7 Solve the IVP: y + y + 1 =, y(1 =. Separating the variables and solving we find (ln (y + 1 = 1 ln (y + 1 = t + C y + 1 =Ce t y(t = Ce t 1. Since y(1 =, we find C = e. Thus, y(t = e 1 t 1 Problem 4.8 Solve the IVP: y ty 3 =, y( =.

8 CONTENTS Separating the variables and solving we find y y 3 dt = tdt ( y dt = t + C 1 y =t + C y 1 = t + C. Since y( =, we find C = 1 4. Thus, y(t = ± have y(t = 4t +1 4 4t +1. Since y( =, we Problem 4.9 Solve the IVP: y = 1 + y, y( π 4 = 1. Separating the variables and solving we find y 1 + y =1 arctan y =t + C y(t = tan (t + C. Since y( π 4 = 1, we find C = π. Hence, y(t = tan (t + π = cot t Problem 4.1 Solve the IVP: y = t ty, y( = 1.

SOLUTIONS TO SECTION 4 9 Separating the variables and solving we find y y y 1 = t y 1 y y + 1 = t ln y 1 y + 1 = t + C y 1 y + 1 =Ce t Since y( = 1, we find C = 1 3. Thus, y(t = 1 + Ce t 1 Ce t. y(t = 3 e t 3 + e t Problem 4.11 Solve the equation 3u y + u xy = by using the substitution v = u y. Letting v = u y we obtain 3v + v x =. Solving this ODE by the method of separation of variables we find v x v = 3 ln v(x, y = 3x + f(y v(x, y =f(ye 3x. Hence, u(x, y = f(ye 3x dy = F (ye 3x + G(x where F (y = f(ydy Problem 4.1 Solve the IVP (y sin yy = sin t t, y( =.

3 CONTENTS Separating the variables and solving we find f(ye 3x (y sin yy dt = (sin t tdt (4.1 f(ye 3x y + cos y = cos t t + C. (4. Since y( =, we find C =. Thus, y + cos y + cos t + t = Problem 4.13 State an initial value problem, with initial condition imposed at t =, having implicit solution y 3 + t + sin y = 4. Differentiating both sides of the given equation we find Problem 4.14 Can the differential equation 3y y + cos y + t =, y( = dy dx = x xy be solved by the method of separation of variables? Explain. If we try to factor the right side of the ODE, we get dy dx = x(x y. The second factor is a function of both x and y. The ODE is not separable

SOLUTIONS TO SECTION 5 31 Solutions to Section 5 Problem 5.1 Classify each of the following PDE as linear, quasi-linear, semi-linear, or nonlinear. (a xu x + yu y = sin (xy. (b u t + uu x = (c u x + u 3 u 4 y =. (d (x + 3u x + xy u y = u 3 (a Linear (b Quasi-linear, non-linear (c Non-linear (d Semi-linear, nonlinear Problem 5. Show that u(x, y = e x f(x y, where f is a differentiable function of one variable, is a solution to the equation u x + u y u =. Let w = x y. Then u x +u y u = e x f(w+e x f w (w e x f w (w e x f(w = Problem 5.3 Show that u(x, y = x xy satisfies the equation subject to the constraint We have xu x yu y = x y xu x yu y = u u(y, y = y, y. ( 3 x 1 y 1 y Problem 5.4 Show that u(x, y = cos (x + y satisfies the equation subject to the constraint ( 1 x 3 y 1 = x xy = u. Also, u(y, y = yu x + xu y = u(, y = cos y.

3 CONTENTS We have yu x + xu y = xy sin (x + y + xy sin (x + y =. Moreover, u(, y = cos y Problem 5.5 Show that u(x, y = y 1 (x y satisfies the equation subject to u(x, 1 = 1 (3 x. 1 x u x + 1 y u y = 1 y We have 1 x u x + 1 y u y = 1 x ( x + 1 y (1 + y = 1 y. Moreover, u(x, 1 = 1 (3 x Problem 5.6 Find a relationship between a and b if u(x, y = f(ax+by is a solution to the equation 3u x 7u y = for any differentiable function f such that f (x for all x. Let v = ax + by. We have d(ax + by u x =f v (v dx d(ax + by u y =f v (v dy Hence, by substitution we find 3a 7b = = af v (v = bf v (v. Problem 5.7 Reduce the partial differential equation au x + bu y + cu = to a first order ODE by introducing the change of variables s = bx ay and t = x.

SOLUTIONS TO SECTION 5 33 By the chain rule we find Thus, u x = u s s x + u t t x = bu s + u t u y = u s s y + u t t y = au s. = au x + bu y + cu = au t + cu. This is a first order linear ODE that can be solved using the method of separation of variables, assuming a Problem 5.8 Solve the partial differential equation u x + u y = 1 by introducing the change of variables s = x y and t = x. Using the chain rule we find u x = u s s x + u t t x = u s + u t u y = u s s y + u t t y = u s. Substituting these into the PDE to obtain u t = 1. Solving this ODE we find u(s, t = t + f(s where f is an arbitrary differentiable function in one variable. Now substituting for s and t we find u(x, y = x + f(x y Problem 5.9 Show that u(x, y = e 4x f(x 3y is a solution to the first-order PDE We have Thus, 3u x + u y + 1u =. u x = 4e 4x f(x 3y + e 4x f (x 3y u y = 3e 4x f (x 3y 3u x + u y + 1u = 1e 4x f(x 3y + 6e 4x f (x 3y 6e 4x f (x 3y + 1e 4x f(x 3y =

34 CONTENTS Problem 5.1 Derive the general solution of the PDE au t + bu x = u, b by using the change of variables v = ax bt and w = x. We have u t = bu v and u x = au v + u w. Substituting we find u w = 1 b u and solving this equation by the method of integrating factor, we find u(v, w = f(ve 1 b w where f is an arbitrary differentiable function in one variable. Thus, u(x, t = f(ax bte x b Problem 5.11 Derive the general solution of the PDE au x + bu y =, a by using the change of variables s = bx ay and t = x. According to the chain rule for the derivative of a composite function, we have u x =u s s x + u t t x = bu s + u t u y =u s s y + u t t y = au s. Substituting these into the given equation to obtain au t = and since a we obtain Solving this equation, we find u t =. u(s, t = f(s where f is an arbitrary differentiable function of one variable. Now, in terms of x and y we find u(x, y = f(bx ay

SOLUTIONS TO SECTION 5 35 Problem 5.1 Write the equation in the coordinates v = x ct, w = x. Using the chain rule, we find u t + cu x + λu = f(x, t, c u t =u v v t + u w w t = cu v u x =u v v x + u w w x = u v + u w Substituting these into the original equation we obtain the equation ( cu w + λu(v, w = f w, w v c Problem 5.13 Suppose that u(x, t = w(x ct is a solution to the PDE xu x + tu t = Au where A and c are constants. Let v = x ct. Write the differential equation with unknown function w(v. Using the chain rule we find u t = cw v and u x = w v. Substitution into the original PDE gives vw v (v = Aw(v

36 CONTENTS Solutions to Section 6.1 Problem 6.1.1 Find a b where a =< 4, 1, 1 4 > and b =< 6, 3, 8 >. We have: a b = 4(6 + 1( 3 + 1 ( 8 = 19 4 Problem 6.1. Find a b where a = 6, b = 5 and the angle between the two vectors is 1. We have: a b = a b cos ( ( π 3 = 6(5 1 = 15 Problem 6.1.3 If u is a unit vector, find u v and u w using the figure below. The figure shows that u = v = w = 1 so the triangle is an equilateral triangle. Hence, u v = u v cos 6 = 1. Likewise, u w = u v cos 1 = 1 Problem 6.1.4 Find the angle between the vectors a =< 4, 3 > and b =<, 1 >. We have: cos θ = a b a b = 5 5 5 = 1 ( 5. Hence, θ = cos 1 1 5 63 Problem 6.1.5 Find the angle between the vectors a =< 4, 3, 1 > and b =<,, 1 >. We have: cos θ = a b a b = 7 6 5 = 7 ( 13. Hence, θ = cos 1 7 13 5

SOLUTIONS TO SECTION 5 37 Problem 6.1.6 Determine whether the given vectors are orthogonal, parallel, or neither. (a a =< 5, 3, 7 > and b =< 6, 8, >. (b a =< 4, 6 > and b =< 3, >. (c a = i + j + k and b = 3 i + 4 j k. (d a = i + 6 j 4 k and b = 3 i 9 j + 6 k. (a a b = 4 so the vectors are not orthogonal. Also, or 1, the two vectors are not parallel. (b a b =, the vectors are orhtogonal. (c a b =, the vectors are orhtogonal. (d u = v, the vectors are parallel 3 a b a b = 4 83 14 Problem 6.1.7 Use vectors to decide whether the triangle with vertices P (1, 3,, Q(,, 4, and R(6,, 5 is right-angled. We have: QP =< 1, 3, > and QR =< 4,, 1 >. Since QP QR =, the vectors QP and QR are orthogonal so that the triangle P QR is a right triangle at Q Problem 6.1.8 Find a unit vector that is orthogonal to both i + j and i + k. Let u =< a, b, c > with a + b + c = 1. We are told that < a, b, c > < 1, 1, >= and < a, b, c > < 1,, 1 >=. These imply a + b = and a + c =. Hence, a + ( a + ( a = 1 or a = ± 1 3. It follows that either u =< 1 3, 1 3, 1 3 > or u =< 1 1 3, 3 1, 3 > Problem 6.1.9 Find the acute angle between the lines x y = 3 and 3x + y = 7. Since the points (1, 1 and (, 3 are on the line x y = 3, the vector v =< 1, > is a vector on the line. Likewise, using the points (, 7 and

38 CONTENTS (1, 4, the vector w =< 1, 3 > is on the line 3x + y = 7. The angle between the two lines is the angle between the two vectors given by Hence, θ = 45 cos θ = v w v w = 5 = 1. 5 1 Problem 6.1.1 Find the scalar and vector projections of the vector b =< 1,, 3 > onto a =< 3, 6, >. We have Comp a b = a b a = 9 7 Proj a a b 7 b = a a =< 49, 54 49, 18 49 > Problem 6.1.11 If a =< 3,, 1 >, find a vector b such that comp a b =. Suppose b =< x, y, z >. We are told that Comp a b = or a b a =. Hence, a b = a = 1. Thus, 3x z = 1. Hence, b =< x, y, 3x 1 > where y and z are arbitrary numbers Problem 6.1.1 Find the work done by a force F = 8 i 6 j + 9 k that moves an object from the point (, 1, 8 to the point (6, 1, along a straight line. The distance is measured in meters and the force in newtons. The displacement vector is D =< 6, 1 1, 8 >=< 6,, 1 >. The work done is W = F D = 8(6 + ( 6( + (9(1 = 144 Joules Problem 6.1.13 A sled is pulled along a level path through snow by a rope. A 3-lb force acting at an angle of 4 above the horizontal moves the sled 8 ft. Find the work done by the force. The work done by the force is W = 3(8 cos 4 1839 ft lb = 1839 slug

SOLUTIONS TO SECTION 5 39 Solution to Section 6. Problem 6..1 Find the gradient of the function We have F (x, y, z = e xyz + sin (xy. F x (x, y, z = yze xyz +y cos (xy, F y (x, y, z = xze xyz +x cos (xy, F z (x, y, z = xye xyz. Thus, F (x, y, z = (yze xyz + y cos (xy i + (xze xyz + x cos (xy j + xye xyz k Problem 6.. Find the gradient of the function We have F x (x, y, z = cos Thus, F (x, y, z = x cos ( y z. ( y, F y (x, y, z = x ( y z z sin, F z (x, y, z = xy ( y z z sin. z ( y F (x, y, z = cos i x ( y z z sin j + xy ( y z z sin k z Problem 6..3 Describe the level surfaces of the function f(x, y, z = (x + (y 3 + (z + 5. The level surfaces are spheres centered at (, 3, 5 and with radius C, C Problem 6..4 Find the directional derivative of u(x, y = 4x + y in the direction of a = i + j at the point (1, 1.

4 CONTENTS The unit vector in the direction of a is v = 1 5 i + 5 j. We have u x (x, y =8x u x (1, 1 =8 u y (x, y =y u y (1, 1 =. Hence, u v (1, 1 = ( 1, (8, = 1 5 5 5 Problem 6..5 Find the directional derivative of u(x, y, z = x z+y 3 z xyz in the direction of a = i + 3 k at the point (x, y, z. The unit vector in the direction of a is v = 1 1 i + 3 1 j. We have u x (x, y, z = xz yz, u y (x, y, z = 3y z xz, u z (x, y, z = x + y 3 z xy. Hence, u v (x, y, z = ( 1 3,, (xz yz, 3y z xz, x + y 3 z xy 1 1 = 1 1 (3x + 6y 3 z 3xy xz + yz Problem 6..6 Find the maximum rate of change of the function u(x, y = ye xy at the point (, and the direction in which this maximum occurs.

SOLUTIONS TO SECTION 5 41 The maximum rate of change is given by u(,. But Thus, u(x, y = y e xy i + (1 + xye xy j. u(, = 4 i + j so that the maximum rate of change at (, is given by u(, = 4 + 1 1 = 17. The maximum occurs in the direction of the unit vector u(, u(, = 4 i + 1 j 17 17 Problem 6..7 Find the gradient vector field for the function u(x, y, z = e z ln (x + y. We have Hence, u x (x, y, z = x x + y u y (x, y, z = y x + y u z (x, y, z =e z. u(x, y, z = x x + y i y x + y j + e z k

4 CONTENTS Solutions to Section 7 Problem 7.1 Solve u x + yu y = y with the initial condition u(, y = sin y. We have a = 1, b = y, and f = y. Solving dy = y we find y = k dx 1e x. Solving du = dx y = k1e x we find u = k 1 e x + f(k 1 = 1 y + f(k 1 = 1 y + f(ye x. Using the initial condition u(, y = sin y we find sin y 1 y = f(y. Hence, u(x, y = 1 y 1 y e x + sin (ye x Problem 7. Solve u x + yu y = u with the initial condition u(, y = sin y. We have a = 1, b = y, and f = u. Solving dy = y we find y = k dx 1e x. Solving du = dx u we find x + 1 = k 1 u. Thus, u(x, y =. Using the initial f(ye x x 1 condition u(, y = sin y we find f(y = csc y. Hence, u(x, y = csc (ye x x Problem 7.3 Find the general solution of yu x xu y = xyu. The system of ODEs is dy dx = x y, du dx = xu. Solving the first equation, we find x + y = k 1. Solving the second equation, we find u = k e x. Hence, u(x, y = e x f(x + y where f is an arbitrary differentiable function in one variable Problem 7.4 Find the integral surface of the IVP: xu x + yu y = u, u(x, 1 = + e x. The system of ODEs is dy dx = y x, du dx = u x.

SOLUTIONS TO SECTION 7 43 Solving the first equation, we find y = k 1 x. Solving the second equation, we find u = k x. Hence, u(x, y = xf ( y x where f is an arbitrary differentiable function in one variable. From the initial condition u(x, 1 = + e x we find f(x = x( + e 1 x. Hence, the integral surface is u(x, y = y( + e x y Problem 7.5 Find the unique solution to 4u x + u y = u, u(x, = 1 1+x. The system of ODEs can be written as dx 4 = dy 1 = du u. Solving the equation dx = dy we find x 4y = k 4 1 1. Solving the equation dy = du 1 1 we find u(x, y =. Using the initial condition u(x, = 1 u f(x 4y y 1+x we find f(x = 1 + x 1. Hence, u(x, y = (x 4y +1 y Problem 7.6 Find the unique solution to e y u x + xu y = xu, u(x, = e x. The system of ODEs can be written as dx e = dy y x = du xu. Thus, xdx = e y dy which implies x e y = k 1. Solving the equation du u we find y + 1 = k u = f(x e y. Hence, = dy u(x, y = 1 f(x e y y. Using the initial condition u(x, = e x we find f(x = e (x+1. Hence, u(x, y = 1 e x +e y 1 y

44 CONTENTS Problem 7.7 Find the unique solution to xu x + u y = 3x u, u(x, = tan 1 x. The system of ODEs can be written as dx x = dy 1 = du 3x u. Solving the equation dx x = dy 1 we find xe y = k 1. On the other hand, we have dx x = du 3x u = d(3x u u = (3x ud(3x u = udu. Thus, (3x u u = k = f(xe y which leads to u(x, y = 3 x 1 6x f(xe y. Using the initial condition, u(x, = tan 1 x we find f(x = 9x 6x tan 1 x. Hence, u(x, y = 3 x 9x e y 6xe y tan 1 (xe y 6x Problem 7.8 Solve: xu x yu y =, u(x, x = x 4. = 3 x 3 xe y +e y tan 1 (xe y Solving the equation dy = y we find xy = k dx x 1. Since the right-hand side is, u(x, y = k = f(k 1 = f(xy. But u(x, x = x 4 = f(x. Hence, f(x = x, x, Hence, u(x, y = x y, xy Problem 7.9 Find the general solution of yu x 3x yu y = 3x u. Solving the equation dy = dx 3x we find y + x 3 = k 1. Solving the equation du = dy we find uy = k u y = f(k 1 = f(y + x 3 where f is a differentiable function in one variable

SOLUTIONS TO SECTION 7 45 Problem 7.1 Find u(x, y that satisfies yu x + xu y = 4xy 3 subject to the boundary conditions u(x, = x 4 and u(, y =. Solving the equation dy = x we find dx y y x = k 1. On the other hand, du = 4y 3 dy so that u(x, y = y 4 + f(y x. Since u(x, = x 4, we have f( x = x 4 or f(x = x for x. Since u(, y = we find f(y = y 4 so that f(y = y for y. Hence, f(x = x for all x. Finally, u(x, y = y 4 (y x = x y x 4

46 CONTENTS Solutions to Section 8 Problem 8.1 Find the solution to u t + 3u x =, u(x, = sin x. Solving dt = 1 we find x 3t = k dx 3 1. Solving the equation du = we find dx u(x, t = k = f(x 3t where f is a differentiable function in one variable. Since u(x, = sin x, we find sin x = f(x. Hence, u(x, t = sin (x 3t Problem 8. Solve the equation au x + bu y + cu =. Solving the equation dy = b we find bx ay = k dx a 1. Solving the equation du = dx c u we find u(x, y = k a e c a x = f(bx aye c a x where f is a differentiable function in one variable Problem 8.3 Solve the equation u x +u y = cos (y x with the initial condition u(, y = f(y where f : R R is a given function. Solving the equation dy = we find x y = k dx 1. Solving the equation du = cos (y x = cos k dx 1 we find u(x, y = x cos k 1 + k = x cos (y x + g(x y where g is a differentiable function in one variable. Since u(, y = f(y, we obtain f(y = g( y or g(y = f( y. Thus, u(x, y = x cos (y x + f(y x Problem 8.4 Show that the initial value problem u t + u x = x, u(x, x = 1 has no solution. Solving the equation dy = 1 we find x y = k dx 1. Solving the equation du = x dx we find u(x, y = 1 x + f(x y where f is a differentiable function of one variable. Since u(x, x = 1 we find 1 = 1 x + f( or f( = 1 x which is impossible since f( is a constant. Hence, the given initial value problem has no solution

SOLUTIONS TO SECTION 8 47 Problem 8.5 Solve the transport equation u t + u x = 3u with initial condition u(x, = 1 1+x. Solving the equation dt dx = 1 we find x t = k 1. Solving the equation du dx te 3 x. Since u(x, = 1 1+x we find f(x = e 3 x. Hence, 1+x u(x, t = e 3t 1 + (x t Problem 8.6 Solve u t + u x 3u = t with initial condition u(x, = x. Solving the equation dt = 1 we find x t = k dx 1. Solving the equation du 3u + t by the method of integrating factor, we find u(x, t = 1 3 t 1 9 + k e 3t = 1 3 t 1 9 + f(x te3t. But u(x, = x which leads to f(x = x + 1. Hence, 9 [ u(x, t = e 3t (x t + 1 ] 1 9 3 t 1 9 Problem 8.7 Show that the decay term λu in the transport equation with decay u t + cu x + λu = can be eliminated by the substitution w = ue λt. dt = Using the chain rule we find w t = u t e λt + λue λt and w x = u x e λt. Substituting these equations into the original equation we find w t e λt λu + cw x e λt + λu = or w t + cw x =

48 CONTENTS Problem 8.8 (Well-Posed Let u be the unique solution to the IVP u t + cu x = u(x, = f(x and v be the unique solution to the IVP u t + cu x = u(x, = g(x where f and g are continuously differentiable functions. (a Show that w(x, t = u(x, t v(x, t is the unique solution to the IVP u t + cu x = u(x, = f(x g(x (b Write an explicit formula for w in terms of f and g. (c Use (b to conclude that the transport problem is well-posed. That is, a small change in the initial data leads to a small change in the solution. (a w(x, t is a solution to the equation follows from the principle of superposition. Moreover, w(x, = u(x, v(x, = f(x g(x. (b w(x, t = f(x ct g(x ct. (c From (b we see that sup x,t { u(x, t v(x, t } = sup{ f(x g(x }. Thus, small changes in the initial data produces small changes in the solution. Hence, the problem is a well-posed problem Problem 8.9 Solve the initial boundary value problem u t + cu x = λu, x >, t > u(x, =, u(, t = g(t, t >. x

SOLUTIONS TO SECTION 8 49 Solving dt = 1 we find x ct = k dx c 1. Solving the equation du = λ u we find dx c u(x, t = f(x cte λ c x. From the condition u(, t = g(t we find f( ct = g(t or f(t = g ( c t. Thus, ( u(x, t = g t x e λ c x. c This is valid only for x < ct since g is defined on (,. Also, this expression will not satisfy u(x, =. So we define u(x, t = for x ct. That is, the solution to the initial boundary value problem is { ( g t x u(x, t = c e λ c x if x < ct if x ct Problem 8.1 Solve the first-order equation u t +3u x = with the initial condition u(x, = sin x. Solving the equation dt = we find x 3t = k dx 3 1. Solving the equation du = we find u(x, t = k dx = f(x 3t where f is an arbitrary differentiable function. Using the initial condition we find f(x = sin x or f(x = sin ( x. The final answer is u(x, t = sin ( x 3t Problem 8.11 Solve the PDE u x + u y = 1. Solving the equation dy = 1 we find x y = k dx 1. Solving the equation du = 1 dx we find u(x, y = x + f(x y where f is an arbirary differentiable function in one variable

5 CONTENTS Solutions to Section 9 Problem 9.1 Find the general solution of the PDE ln (y + uu x + u y = 1. dx The characteristic equations are = dy = du. Using the second and ln (y+u 1 1 third fractions we find that y + u = k 1. Now, from the first and second fractions we have dx ln k 1 = dy so that x + k 1 = y ln k 1. Hence, y ln (y + u x = k. Hence, the general solution is given by u = y +f(y ln (y + u x where f is an arbitrary differentiable function Problem 9. Find the general solution of the PDE x(y uu x + y(u xu y = u(x y. The characteristic equations are given by or dx = x(y u dx + dy + du x(y u + y(u x + u(x y = d(x + y + u = du u(x y. Hence. x + y + u = k 1. On the other hand we have dx x + dy y (x y = du u x y. dy y(u x = du u(x y du. We have u(x y This implies that or dx x + dy y = du u ln xyu = k that is xyu = k. Hence, the general solution is given by u = f(x+y+u xy f is an arbitrary differentiable function where Problem 9.3 Find the general solution of the PDE u(u + xy(xu x yu y = x 4.

SOLUTIONS TO SECTION 9 51 The characteristic equations are dx = dy = du ux(u +xy yu(u +xy. From the first x 4 and second fractions we get dx = dy. Upon integration we find xy = k x y 1. From first and third fractions we get x 3 dx = (u 3 + uxydu or x 3 dx = (u 3 + k 1 udu. Integration leads to x4 = u4 + k 1 4 4 u + k or x 4 u 4 k 1 u = k. Substituting for k 1 we find x 4 u 4 xyu = k. Hence, the general solution is given by x 4 u 4 xyu = f(xy where f is an arbitrary differentiable function Problem 9.4 Find the general solution of the PDE (y + xuu x (x + yuu y = x y. The characteristic equations are dx = dy = y+xu x+yu xdx + ydy udu xy + x u xy y u ux + uy = du. We have x y du x y. Thus, xdx + ydy udu = or x + y u = k 1. On the other hand, ydx + xdy + du y + xyu x xyu + x y = du x y. That is, ydx + xdy + udu =. Hence, xy + u = k. The general solution is xy + u = f(x + y u where f is an arbitrary differentiable function Problem 9.5 Find the general solution of the PDE (y + u u x xyu y + xu =. The characteristic equations are dx = dy = du. Using the last two y +u xy xu fractions we find dy y = du u which leads to y u = k 1. On the other hand, we have xdx + ydy + udu xy + xu xy xu = du xu. Thus, xdx + ydy + udu = or x + y + u = k. The general solution is x + y + u = f ( y u where f is an arbitrary differentiable function Problem 9.6 Find the general solution of the PDE u t + uu x = x.

5 CONTENTS The characteristic equations are dx u = dt 1 = du x. Solving dx = du we find u x u x = k 1. Solving dt we find x + u = x+u k e t = e t f(u x where f is an arbitrary differentiable function 1 = d(x+u Problem 9.7 Find the general solution of the PDE (y uu x + (u xu y = x y. The characteristic equations are We have dx y u = dy u x = dx + dy + du y u + u x + x y du x y. = dx + dy + du so that dx + dy + du =. Hence, x + y + u = k 1. Likewise, xdx + ydy + udu x(y u + y(u x + u(x y = xdx + ydy + udu so that xdx + ydy + udu =. Hence, x + y + u = k. The general solution is given by x + y + u = f(x + y + u where f is an arbitrary differentiable function Problem 9.8 Solve x(y + uu x y(x + uu y = (x y u. The characteristic equations are dx x(y + u = dy y(x + u = du (x y u.

SOLUTIONS TO SECTION 9 53 We first note that Thus, dx x y + u = dy y (x + u = du dx u x y = x dx x + dy y + du u = which gives xyu = k 1. Likewise, we have + dy y + du u. xdx + ydy du x (y + u y (x + u (x y u xdx + ydy du =. Thus, xdx+ydy du = and this implies that x +y u = k. The general solution is given by x + y u = f(xyu where f is an arbitrary differentiable function Problem 9.9 Solve 1 x u x + u y =. The characteristic equations are dx 1 x = dy 1 = du. From the last fraction, we have u(x, y = k 1. From the first two fractions, we have y = sin 1 x + k = sin 1 x + f(u where f is a differentiable function Problem 9.1 Solve u(x + yu x + u(x yu y = x + y. The characteristic equations are dx u(x + y = dy u(x y = du x + y.

54 CONTENTS Each of these ratio is equivalent to ydx + xdy udu = xdx ydy udu or d(xy u 1 = d(x y u. Hence, 1 (x y u = f(xy u where f is a differentiable function

SOLUTIONS TO SECTION 1 55 Solutions to Section 1 Problem 1.1 Solve (y uu x + (u xu y = x y with the condition u ( x, 1 x =, x, x 1. Let Γ : x (t = t, y (t = 1 t, u (t =. Then a(x, y, u dy dt b(x, y, u dx dt = t 1 t. 3 Hence, the problem has a unique solution. The characteristic equations are We have dx y u = dy u x = dx + dy + du y u + u x + x y du x y. = dx + dy + du so that dx + dy + du =. Hence, x + y + u = c 1. Likewise, xdx + ydy + udu x(y u + y(u x + u(x y = xdx + ydy + udu so that xdx + ydy + udu =. Hence, x + y + u = c. The general solution is given by x + y + u = f(x + y + u where f is an arbitrary differentiable function. Now, using the Cauchy data u ( ( x, x 1 = we find f x + 1 x = x + 1 = (. x + 1 x x Hence, f(x = x. Hence, the integral surface is described by and the unique solution is given by (x + y + u = x + y + u + u(x, y = 1 xy x + y, x + y

56 CONTENTS Problem 1. Solve the linear equation yu x + xu y = u, with the Cauchy data u(x, = x 3, x > Let Γ : x (t = t, y (t =, u (t = t 3. Then a(x, y, u dy dt b(x, y, u dx = t. dt Hence, the problem has a unique solution. The characteristic equations are dx y = dy x = du u. Using the first two fractions we find x y = c 1 Now, since du u = dx + dy x + y we can write u = c (x + y. Hence, the general solution is given by u = (x + yf(x y where f is an arbitrary differentiable function. Using the Cauchy data u(x, = x 3, we find f(x = x, that is f(x = x. Consequently, the unique solution is given by u(x, y = (x + y(x y Problem 1.3 Solve x(y + uu x y(x + uu y = (x y u with the Cauchy data u(x, x = 1, x >. Let Γ : x (t = t, y (t = t, u (t = 1.

SOLUTIONS TO SECTION 1 57 Then a(x, y, u dy dt b(x, y, u dx dt = t(t + 1. Hence, the problem has a unique solution. The characteristic equations are We first note that dx x(y + u = dy y(x + u = du (x y u. Thus, dx x y + u = dy y (x + u = du dx u x y = x dx x + dy y + du u = which gives xyu = c 1. Likewise, we have + dy y + du u. xdx + ydy du x (y + u y (x + u (x y u xdx + ydy du =. Thus, xdx+ydy du = and this implies that x +y u = c. The general solution is given by f(xyu = x + y u where f is an arbitrary differentiable function. Using the Cauchy data u(x, x = 1 we find f(x = x. Hence, the unique solution is given by Problem 1.4 Solve xyu + x + y u + = xu x + yu y = xe u with the Cauchy data u(x, x =, x >. Let Γ : x (t = t, y (t = t, u (t =.

58 CONTENTS Then a(x, y, u dy dt b(x, y, u dx dt = t. Hence, the problem has a unique solution. The characteristic equations are dx x = dy y = du xe u. Using the first two fractions we find y = c x 1. Using the first and the last fractions we find dx = e u du or x e u = c. Hence, the general solution is given by ( y x e u = f x where f is an arbitrary differentiable function. Using the Cauchy data u(x, x =, we find f(x = x 1. Hence, the unique integral surface is described by y x + x eu + 1 = or Problem 1.5 Solve the initial value problem ( u(x, y = ln x + 1 y x xu x + u y =, The initial curve in parametric form is Since u(x, = f(x. Γ : x (t = t, y (t =, u (t = f(t. a(x (t, y (t, u (t dy dt (t b(x (t, y (t, u (t dx (t = 1 dt the initial value problem has a unique solution. The characteristic equations are dx x = dy 1 = du. Hence xe y = c 1 and u(x, y = c = g(xe y where g is a differentiable function in one variable. Using the Cauchy data u(x, = f(x we find g(x = f(x. Hence, the unique solution is u(x, y = f(xe y

SOLUTIONS TO SECTION 1 59 Problem 1.6 Solve the initial value problem u t + au x =, u(x, = f(x. The initial curve parametrization is given by Γ : x (w = w, t (w =, u (w = f(w. Since a(x (w, t (w, u (w dt dw (w b(x (w, t (w, u (w dx (w = 1 dw the initial value problem has a unique solution. The characteristic equations are dt 1 = dx a = du. Thus, x at = c 1 and u(t, x = c = g(x at where g is a differentiable function in one variable. Hence, the unique solution is given by u(t, x = f(x at Problem 1.7 Solve the initial value problem au x + u y = u, u(x, = cos x The initial curve parametrization is given by Since Γ : x (t = t, y (t =, u (t = cos t. a(x (t, y (t, u (t dy dt (t b(x (t, y (t, u (t dx (t = 1 dt the initial value problem has a unique solution. The characteristic equations are dx a = dy 1 = du u.

6 CONTENTS Hence, x ay = c 1 and y + 1 = c u = f(x ay where f is a differentiable function in one variable. Using the Cauchy data u(x, = cos x we find f(x = sec x. The unique solution is u(x, y = Problem 1.8 Solve the initial value problem 1 sec (x ay y u x + xu y = u, u(1, y = h(y. The initial curve parametrization is given by Γ : x (t = 1, y (t = t, u (t = h(t. Since a(x (t, y (t, u (t dy dt (t b(x (t, y (t, u (t dx dt (t = 1 the initial value problem has a unique solution. The characteristic equations are dx 1 = dy x = du u. Thus, y x = c 1 and u = f(y x e x. Using the Cauchy data u(1, y = h(y, we find f(y = e 1 h ( y+1. Hence, u(x, y = h (y x + 1 e x 1 Problem 1.9 Solve the initial value problem uu x + u y =, u(x, = f(x where f is an invertible function.

SOLUTIONS TO SECTION 1 61 The initial curve parametrization is given by Γ : x (t = t, y (t =, u (t = f(t. Since a(x (t, y (t, u (t dy dt (t b(x (t, y (t, u (t dx (t = 1 dt the initial value problem has a unique solution. The characteristic equations are dx u = dy 1 = du. Thus, u(x, y = c 1 and x uy = c = g(u. Using the Cauchy data u(x, = f(x we find g(f(x = x and so g(x = f 1 (x. Hence, the unique solution is given by u(x, y = f(x uy Problem 1.1 Solve the initial value problem 1 x u x + u y =, u(, y = y. The initial curve parametrization is given by Since Γ : x (t =, y (t = t, u (t = t. a(x (t, y (t, u (t dy dt (t b(x (t, y (t, u (t dx dt (t = 1 the initial value problem has a unique solution. The characteristic equations are dx 1 x = dy 1 = du. Thus, y arcsin x = c 1 and u(x, y = c = f(y arcsin x. Using the Cauchy data u(, y = y we find f(y = y. Hence, the unique solution is given by u(x, y = y arcsin x

6 CONTENTS Problem 1.11 Consider xu x + yu y =. (i Find and sketch the characteristics. (ii Find the solution with u(1, y = e y. (iii What happens if you try to find the solution satisfying either u(, y = g(y or u(x, = h(x for given functions g and h? (i The characterisitcs are solutions to the ODE dy = y. Solving we find dx x y = Cx. Thus, the characteristics are parobolas in the plane centered at the origin. See figure below. (ii The general solution is u(x, y = f(yx, and so the solution satisfying the condition at u(1, y = e y is u(x, y = e yx. (iii In the first case, we cannot substitute x = into yx (the argument of the function f, above because x is not defined at. Similarly, in the second case, we d need to find a function f so that f( = h(x. If h is not constant, it is not possible to satisfy this condition for all x R Problem 1.1 Solve the equation u x + u y = u subject to the condition u(x, = cos x.