Department of mathematics MA201 Mathematics III

Department of mathematics MA201 Mathematics III Academic Year 2015-2016 Model Solutions: Quiz-II (Set - B) 1. Obtain the bilinear transformation which maps the points z 0, 1, onto the points w i, 1, i respectively. Find also the fixed points of the transformation. (4 marks) We know that the bilinear (Möbius) transformation which maps the points (z 1, z 2, z 3 ) onto (w 1, w 2, w 3 ) is given by (w w 1 )(w 2 w 3 ) (w w 3 )(w 2 w 1 ) (z z 1)(z 2 z 3 ) (z z 3 )(z 2 z 1 ). To avoid the substitution of z 3 directly, we put z 3 1/z 3, where z 3 0. Therefore, the above relation becomes (w w 1 )(w 2 w 3 ) (w w 3 )(w 2 w 1 ) (z z 1)(z 2 z 3 1) (zz 3 1)(z 2 z 1 ). Now, using the given values z 1 0, z 2 1, z 3 0 w 1 i, w 2 1, w 3 i, we have (w i)(1 + i) (w + i)(1 i) z( 1) ( 1)(1) w i (1 i)z w + i (1 + i). That is, 2w 2i (1 i)z + (1 + i) (1 + i) (1 i)z which is the required transformation. The fixed points are given by w (1 + i)z + (i 1) (1 + i) (1 i)z z + i iz + 1. z z + i iz + 1 iz 2 + z z + i that is, z 2 1, which gives the fixed points are z ±1. Alternate Method : Let a,b,c denote four complex constants with the restriction that ad bc 0 Then the function w(z) az + b cz + d Given points z 0, 1, substitute in (0.1) then we get (0.1) w(0) i i b d w(1) 1 1 a + b d + c w( ) i i a c. [2 marks]

From the above three equations (after solving them), we get The fixed points are given by w(z) az + b d( iz + 1)(i 1) cz + d dz(i 1) + d(i + 1) z(i + 1) + (i 1) z(i 1) + (i + 1). z z(i + 1) + (i 1) z(i 1) + (i + 1) z2 (i 1) + z(i + 1) z(i + 1) + (i 1), that is, z 2 1, which gives the fixed points are z ±1. 2. Show that the transformation w (z i)/(1 iz) maps the upper half of the z-plane onto the interior of the circle w 1 in the w-plane. (3 marks) The given transformation is which becomes w z i 1 iz w iwz z i z(1 + iw) w + i, (0.2) that is, z w + i 1 + iw. (0.3) Equation (0.3) is the inverse transformation of (0.2), which can be written as x + iy u + i(v + 1) (1 v) + iu [u + i(v + 1)][(1 v) iu] (1 v) 2 + u 2 u[1 v + v + 1] + i[1 v2 u 2 ] u 2 + (1 v) 2 2u + i(1 u2 v 2 ) u 2 + (1 v) 2. Therefore, we have x 2u u 2 + (1 v) 2, y 1 u2 v 2 u 2 + (1 v) 2. The upper half of the z-plane is given by y > 0. Its image is given by y 1 u2 v 2 u 2 + (1 v) 2 > 0 u2 + v 2 < 1, that is, w 2 < 1, which gives w < 1, the interior of the circle w 1. Alternate Method : Observe the following z 1 1 w 1 1 z 2 0 w 2 i z 3 1 w 3 1. Page 2

The points z 1 1, z 2 0, z 3 1 lies on the line Im(z) 0. So, the upper half of the z plane i.e., Im(z) > 0 lies on the left region of the line Im(z) 0 uniquely determined by the points z 1 1, z 2 0, z 3 1. Note that, the corresponding image points w 1 1, w 2 i, w 3 1 lie on the circle w < 1. So, the interior of the circle w 1 in the w plane lies on the left region of the circle w 1 uniquely determined by the points w 1 1, w 2 i, w 3 1. By orientation principle of Möbius transformation, it follows that the Möbius transformation maps left region onto left region right region onto right region respectively. Therefore we conclude that the given Möbius transformation maps the upper half of the z-plane onto the interior of the circle w 1 in the w-plane. 3. Derive a first-order partial differential equation by eliminating the arbitrary function f from the relation f(x 2 + y 2, x 2 z 2 ) 0. (3 marks) Let u x 2 + y 2, v x 2 z 2. (0.4) Then the given relation becomes f(u, v) 0. (0.5) Differentiating (0.5) with respect to x y, we get [ ] f u u x + p u + f [ ] v z v x + p v 0 z [ f u u y + q u ] + f [ v z v y + q v ] 0. z (0.6) Here, u x 2x, u y 2y, u z 0, v x 2x, After substituting these values, the equations in (0.6) become 2x f + (2x 2zp) f u v 0 2y f f 2zq u v 0. Eliminating f/ u f/ v from equations (0.7), we obtain 2x 2x 2pz 2y 2qz 0, v y 0, v z 2z. (0.7) which gives yp xq xy z, Page 3

this is the required first-order PDE. Alternate Method 1: We can express the given relation as u ϕ(v) as x 2 + y 2 ϕ(x 2 z 2 ) (0.8) where ϕ is an arbitrary function. Differentiating (0.8) w.r.to x y respectively,we get Eliminating the function ϕ from (0.9) (0.10) we obtain 2x ϕ (v)(2x 2xp) (0.9) 2y ϕ (v)( 2zq) (0.10) 2x 2y 2x 2zp 2zq py qx xy z. x y x zp zq xy z(py qx) Alternate Method 2: The given relation is f(x 2 + y 2, x 2 z 2 ) 0, which can be written as f(u, v) 0, where u x 2 + y 2 v x 2 z 2. The required PDE is given by Now calculate the Jacobian (u, v) (y, z) (u, v) (z, x) (u, v) (x, y) (u, v) (u, v) (u, v) p + q (y, z) (z, x) (x, y) u x v y u z v z u x v x u z v z u x v x u y v y 2y 0 0 2z 4yz, 0 2x 2z 0 4xz, 2x 2y 2x 0 4xy. Therefore, the required partial differrential equation is p( 4yz) + q(4xz) 4xy pyz qxz xy py qx xy z. 4. Find the integral surface of the partial differential equation x(y 2 + z)p y(x 2 + z)q (x 2 y 2 )z which passes through the straight lines x + y 0, z 1. (5 marks) Page 4

The given first-order quasilinear PDE is in the Lagrange form, its auxiliary equations are x(y 2 + z) y(x 2 + z) Using the multipliers 1/x, 1/y 1/z in equations (0.11), we obtain that integrating this, we obtain that x + y + dz z 0, dz (x 2 y 2 )z. (0.11) xyz c 1. (0.12) Using the multipliers x, y 1 in equations (0.11), we obtain that which gives x + y dz 0, The general integral of the first-order PDE is given by where u xyz v x 2 + y 2 2z. x 2 + y 2 2z c 2. (0.13) F (u, v) 0, The particular surface, which we are looking for passes through the line x + y 0 (0.14) z 1. (0.15) Now, using (0.14) (0.15) in (0.12) (0.13), we obtain the following relations Eliminating x from the previous equations, we get x 2 c 1, (0.16) 2x 2 2 c 2. (0.17) 2c 1 + c 2 + 2 0. Replacing the values of c 1 c 2 in the above equation, from (0.12) (0.13), we get x 2 + y 2 + 2xyz 2z + 2 0, which is the equation of the required surface passes through the given initial curves. 5. Solve the nonlinear first order partial differential equation xpq yq 2 1 0 by Charpit s method. Page 5

(5 marks) The given first-order PDE is nonlinear. Let therefore, the Charpit s auxiliary equations become f p f q xq f xpq yq 2 1 0 (0.18) dz pf p + qf q xp 2yq dp f x + pf z dz 2(xpq yq 2 ) From the last two relations from the above equations, we get dq f y + qf z dp pq dq q 2. (0.19) dp pq dq q 2 dp dq p q pq a p a q. Using p a q in the given PDE (0.18), we get ax 1 ax yq 2 1 yq 2 ax 1 q ±. y ax 1 y Let us take the positive value, that is, q, which gives p a y ax 1. Substituting the values of p q in the following ODE y ax 1 dz p + q a ax 1 +. y Integrating the above ODE, we obtain that z 2 y(ax 1) + b (z b) 2 4y(ax 1). which is the required complete integral. Alternate Method 1: From the first last relation from equations (0.19), we get x dq q q ax Using q ax in the given PDE (0.18), we get p a2 x 2 y + 1 ax 2 Substituting the values of p q in the following ODE dz p + q a2 x 2 y + 1 ax 2 + ax Page 6

Integrating the above ODE, we obtain that which is the required complete integral. Alternate Method 2: z 1 + axy + b. ax From the first fourth relation from equations (0.19), we get x dp p p a x Using p a x in the given PDE (0.18), we get q a ± a 2 4y 2y Let us take the positive value of q. Substituting the values of p q in the following ODE dz p + q a x + a + a 2 4y 2y Integrating the above ODE, we obtain that z a log x + a 2 log y + ( a 2 4y a 2 log a + ) a 2 4y a + b. a 2 4y which is the required complete integral. 6. Show that the first-order PDEs xp yq x, x 2 p + q xz are compatible. Also, find the common solution of the PDEs. (5 marks) Let f xp yq x 0 (0.20) g x 2 p + q xz 0. (0.21) We know that in order to show the equations are compatible, we have to prove that [f, g] (f, g) g) (f, g) (f, g) + p (f, + + q (x, p) (z, p) (y, q) (z, q) 0. The LHS becomes p 1 x 2xp z x 2 + p 0 x x x 2 + q y 0 1 + q 0 y x 1 (p 1)x 2 x(2xp z) + px 2 q xyq x 2 + xz q xyq x 2 + [x 2 p + q] q xyq, using (0.21) x 2 + x[x + yq] xyq using (0.20) 0 Page 7

Therefore, the given two first-order PDEs are compatible. (or directly [2 marks]) Observe that (f, g) (p, q) (1+xy) 0, on the domain D, which does not contain the points such that xy + 1 0. Solving the equations (0.20) (0.21) for p q, we get p 1 + yz 1 + xy q Substituting the values of p q in the ODE (0.22) x(z x) 1 + xy. (0.23) dz p + q 1 + yz x(z x) + 1 + xy 1 + xy dz dz z x y(z x) x(z x) + 1 + xy 1 + xy y + x 1 + xy z x + c(1 + xy). This is the required one-parameter family of common solutions of the first-order PDEs. 7. Classify the second-order partial differential equation y 2 u xx x 2 u yy 0 in R 2 excluding the x-axis y-axis. Further, obtain its canonical form. (5 marks) Given the second-order PDE is y 2 u xx x 2 u yy 0, (0.24) comparing this with the stard second-order linear PDE, we obtain the coefficients as A y 2, B 0, C x 2, D 0, E 0, F 0, G 0. The discriminant B 2 4AC 0 + 4x 2 y 2 > 0, for all x, y. Therefore, the given second-order PDE is of hyperbolic type. It has two real characteristic, the characteristic equations are given by ξ x B B 2 4AC x ξ y 2A y y x 0 y 2 x 2 c 1 η x B + B 2 4AC x η y 2A y y + x 0 y 2 + x 2 c 2. Page 8

Hence, the new coordinates are ξ y 2 x 2 η y 2 + x 2. [1 + 1 mark] To find the canonical equation, we have to calculate the first second derivatives of ξ η: ξ y 2 x 2, ξ x 2x, ξ xx 2, ξ y 2y, ξ yy 2, ξ xy ξ yx 0. η y 2 + x 2, η x 2x, η xx 2, η y 2y, η yy 2, η xy η yx 0. To determine the new coefficients, we substituting these values in the following: A Aξx 2 + Bξ x ξ y + Cξy 2 0. Similarly, we have B 2Aξ x η x + B(ξ x η y + ξ y η x ) + 2Cξ y η y 16x 2 y 2 C Aη 2 x + Bη x η y + Cη 2 y y 2 4x 2 x 2 4y 2 0, D Aξ xx + Bξ xy + Cξ yy + Dξ x + Eξ y 2(x 2 + y 2 ) E Aη xx + Bη xy + Cη yy + Dη x + Eη y 2(y 2 x 2 ), F G 0. Substituting these values in the stard PDE, we obtain the following canonical form: u ξη (x2 + y 2 )u ξ + (y 2 x 2 )u η 8x 2 y 2, the denominator is non-zero, because the domain excludes x 0 y 0. Replacing x, y by ξ, η, the canonical form becomes which is the required canonical form. u ξη ηu ξ ξu η 2(ξ 2 η 2 ), Page 9