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HOMEWORK 1 SHUANGLIN SHAO 1. P5. #1. Proof. Equations 1, 2, 4, 7, 8 are linear equations. Equations 3, 5, 6 are nonlinear equations. 2. P5. #2. Proof. (a). linear. L(au + bv) = (au + bv)x + x(au + bv)y = aux + bvx + xauy + xbvy = a(ux + xuy ) + b(vx + xvy ) = aL(u) + bL(v). This proves that L is linear. (b). nonlinear. L(au + bv) = (au + bv)x + (au + bv)(au + bv)y = aux + bvx + (au + bv)(auy + bvy ) = aux + bvx + a2 uuy + b2 vvy + abuvy + bavuy . On the other hand, aL(u) + bL(v) = a(ux + uuy ) + b(vx + vvy ) = aux + bvx + auuy + bvvy . If a = 2, b = 0, these two identities above not equal; otherwise uuy = 0, implies that (u2 )y = 0, i.e., u2 (x, y) = u2 (x, 0). The last condition put a constraint on u, which is not known. So the linearity fails. 1 (c). nonlinear. L(au + bv) = (au + bv)x + ((au + bv)y )2 = aux + bvx + (auy + bvy )2 = aux + bvx + a2 u2y + b2 vy2 + 2abuy vy . On the other hand, aL(u) + bL(v) = a(ux + u2y ) + b(vx + vy2 ) = aux + bvx + au2y + bvy2 . Similarly as in (b), if a = 2, b = 0, if L is linear, u2y = 0. So uy = 0. This is not known. So linearity fails. (d). nonlinear. L(au + bv) = (au + bv)x + (au + bv)y + 1 = a(ux + uy ) + b(vx + vy ) + 1. On the other hand aL(u) + bL(v) = a(ux + uy + 1) + b(vx + vy + 1) = a(ux + uy ) + b(vx + vy ) + a + b. So if a = b = 1, if L is linear, we have 1 = 2. So linearity fails. Another way to see this is that L does not map zero functions to zero. (e). linear. The proof is similar to that in (a). So we omit it. 3. P5. # 3. Proof. (a). It is a linear inhomogeneous equation. Define L(u) = ut − uxx , g = −1. (b). It is a linear homogeneous equation. Define L(u) = ut − uxx + xu. (c). nonlinear. The proof is similar to that in (b) in #2; so we omit it. (d). It is a linear inhomogeneous equation. Define L(u) = ut − uxx , g = −x2 . 2 (e). It is a linear homogeneous equation. Define L(u) = iut − uxx + 1 u. x (f ). nonlinear. If the operator is defined, uy ux L(u) = p +q , 1 + u2x 1 + u2y then L(2u) 6= 2L(u). Otherwise 2ux 1 1 ! p −p 1 + u2x 1 + 4u2x 1 1 = 0. + 2uy q −q 2 2 1 + uy 1 + 4uy This equation puts a constraint on u, which is not known. (g). It is a linear homogeneous equation. Define L(u) = ux + ey uy . (h). It is a linear inhomogeneous equation. If we define √ L(u) = ut + uxxxx + 1 + u, then L(0) = 1 6= 0. So linearity fails. 4. P5. # 4. Proof. Let u and v solve the linear homogeneous equation L(u) = g. Then we have L(u) = g, L(v) = g. So we have L(u − v) = 0. So the difference u − v solves the equation L(u) = 0. 3 5. P5. #10. 000 00 000 00 Proof. Define L(u) = u − 3u + 4u. Then the equation u − 3u + 4u = 0 reduces to the linear homogeneous equation L(u) = 0. Let V be the solution space of L(u) = 0. Let u, v ∈ V. For any a, b ∈ R, then au + bv ∈ V because L(au + bv) = aL(u) + bL(v) = 0. So V is a vector space over R. From the theory of ODE, V is spanned by the basis {e−x , e2x , xe2x }. 6. P5. #11. Proof. To the equation, uuxy = ux uy , LHS = f (x)g(y) (f (x)g(y))xy = f (x)fx (x)g(y)gy (y). On the other hand, RHS = ux uy = (f (x)g(y))x (f (x)g(y))y = f (x)fx (x)g(y)gy (y). Thus u(x, y) = f (x)g(y) is a solution. 7. P9. #1. Proof. By formula (2) in Section 1.2, u(x, t) = f (2x − 3t), where f is a function of one variable. Given the condition, we have u(x, 0) = f (2x) = sin x. So f (x) = sin x3 . Thus u(x, t) = sin 2x − 3t 3 = sin(x − t). 2 2 4 8. P9. #2. Proof. Let v = uy . Then 3v + vx = 0. We multiply both sides by e3x , vx e3x + 3e3x v = 0. By using the chain rule of differentiation, (ve3x )x = 0. Thus by the fundamental theorem of calculus, v(x, y)e3x − v(0, y) = 0. Set g(y) = v(0, y). Then v(x, y) = e−3x g(y). To solve for u, uy = e−3x g(y). Again by the fundamental theorem of calculus, Z y −3x g(t)dt. u(x, y) − u(x, 0) = e 0 Let f (x) = u(x, 0) and h(y) = Ry 0 g(t)dt. We have u(x, y) = f (x) + e−3x h(y), where f and h are both functions of one variable. 9. P10. # 3. Proof. We follow the method in solving Equation (4) in section 1.2. The equation is 1 ux + uy = 0. 1 + x2 1 So the directional derivative of u along the direction (1, 1+x 2 ) is zero. We look for the equations of these characteristic curves y = y(x) in the plane 1 such that their directions are (1, 1+x 2 ). Their equations are dy(x) 1 = . dx 1 + x2 Thus integrating from 0 to x, we obtain Z x 1 y(x) − y(0) = dt = tan−1 x. 2 1 + t 0 Hence y(x) = tan−1 x + y(0); 5 that is to say, y − tan−1 x = C, where C = y(0). On each of these curves u(x, y) is a constant because uy d u(x, tan−1 x + C) = ux + = 0. dx 1 + x2 Thus u(x, tan−1 x + C) = u(0, tan−1 0 + C) = u(0, C) is independent of x. Therefore there exists a function of one variable f such that u(x, y) = f (y − tan−1 x). 10. P10. #5. Proof. We still use the method in solving Equation (4) in section 1.2. The equation is y ux + uy = 0. x So the directional derivative of u along the direction (1, xy ) is zero. We look for the equations of these characteristic curves y = y(x) in the plane such that their directions are (1, xy ). Their equations are dy(x) y = . dx x To solve this equation, y 0 (x) y(x) = x1 . Thus d(ln y(x)) 1 = . dx x Integrating this equation from 1 to x, we see that ln y(x) = ln x + ln y(1); that is to say, ln y(x) = ln y(1)x, where y(x) = Cx, where C = y(1). On each of these curves u(x, y) is a constant because d y u(x, Cx) = ux + Cuy = ux + uy = 0. dx x Thus u(x, Cx) = u(1, C) is independent of x. Therefore there exists a function of one variable f such that u(x, y) = f (y/x). 6 11. P10. # 8. Proof. The equation is (1) aux + buy + cu = 0. Thus uy ux +b = −c. u u Let v = ln u. The equation above is a avx + bvy + c = 0. Thus without loss of generality, we assume that a 6= 0. Thus c c a(v + x)x + b(v + x)y = 0. a a Therefore we have c v + x = f (bx − ay), a where f is a function of one variable. Since u = ev , c u(x, y) = e− a x+f (bx−ay) . (2) Another way is to use the coordinate method. From the equation a uux +b −c. Let v = ln u. Then avx + bvy = −c. 0 We change variables: x = bx − ay and y 0 = ax + by. Then vx = bvx0 + avy0 , vy = −avx0 + bvy0 . Thus −c = avx + bvy = (a2 + b2 )vy0 . Hence we have vy 0 = − c . a2 + b2 Integrating from 0 to y 0 , we have v(x0 , y 0 ) − v(x0 , 0) = − cy 0 . a2 + b2 Therefore − u(x0 , y 0 ) = u(x0 , 0)e cy 0 a2 +b2 , i.e., (3) u(x, y) = f (bx − ay)e − c (ax+by) a2 +b2 . Remark. Equation (2) and (3) is in the same form. Writing c − e− a x = e c bc (ax+by)− (bx−ay) a2 +b2 a(a2 +b2 ) 7 . uy u = Then we write − ac x+f (bx−ay) e − =e c (ax+by) a2 +b2 bc − 2 +b2 ) (bx−ay) f (bx−ay) a(a e , e where the second factor is a function of bx − ay. 12. P10. # 9. Proof. Let v = u − x. Then the original equation is vx + vy = 0. Thus we see that v(x, y) = f (x − y), where f is a function of one variable. Therefore v(x, y) = x + f (x − y). 13. P10. # 10. Proof. Let x0 = −x + y and y 0 = x + y. Then ux = −ux0 + uy0 , uy = ux0 + uy0 , y 0 − x0 , 2 x0 + y 0 y= . 2 Thus the equation that ux + uy + u = ex+2y becomes x= 1 x0 3 0 1 uy0 + u = e 2 + 2 y . 2 2 y0 Multiplying both sides by e 2 , 0 y d 1 x0 +2y0 u 2 e = e2 . dy 0 2 This implies that y0 1 x0 e 2 u(x0 , y 0 ) − u(x0 , 0) = e 2 2 Hence Z 0 y0 1 x0 0 e2t dy 0 = e 2 (e2y − 1). 4 y0 1 x0 −y0 0 u(x0 , y 0 ) = e 2 (e2y − 1) − e− 2 u(x0 , 0). 4 In terms of (x, y)-variables, x+y 1 x+2y u(x, y) = e − e−x − e− 2 u(y − x, 0). 4 8 One can verify that u(x, y) satisfies that ux +uy +u = ex+2y . But if imposing the condition that u(x, 0) = 0, we obtain a contradiction. Indeed, since u(x, 0) = 0 for any x, u(y − x, 0) = 0. Thus 1 (4) u(x, y) = (ex+2y − e−x ). 4 This is also a solution to the equation. Let y = 0 in (4). Thus 1 u(x, 0) = (ex − e−x ), 4 which is generally not zero. Therefore there is no solution for this equation with u(x, 0) = 0. Department of Mathematics, KU, Lawrence, KS 66045 E-mail address: [email protected] 9