A METHOD OF SOLVING LAGRANGE S FIRST-ORDER PARTIAL DIFFERENTIAL EQUATION WHOSE COEFFICIENTS ARE LINEAR FUNCTIONS

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1 International Journal of Differential Equations and Applications Volume 14 No , ISSN: url: doi: PA acadpubl.eu A METHOD OF SOLVING LAGRANGE S FIRST-ORDER PARTIAL DIFFERENTIAL EQUATION WHOSE COEFFICIENTS ARE LINEAR FUNCTIONS Syed Md Himayetul Islam 1, J. Das Née Chaudhuri 2 1 Fatullapur Adarsha High School Vill.- Fatullapur, P.O.- Nimta Dist.- North 24-Parganas, Kolkata, , Department of Pure Mathematics Calcutta University 35, Ballygunge Circular Road, Kolkata, West Bengal, INDIA Abstract: A method of solving Lagrange s first-order partial differential equation of the form Pp+Qq = R, where P, Q, R are linear functions of x, y, z, has been presented below. AMS Subject Classification: 35F05, 35A99 Key Words: Lagrange s first-order partial differential equation, linear functions, simultaneous ordinary differential equations, linear homogeneous algebraic equations Received: May 20, 2014 Correspondence author c 2015 Academic Publications, Ltd. url:

2 66 S.M.H. Islam, J. Das N. Chaudhuri 1. Introduction In the book [1] page 46-47, the partial differential equation of the form a 1 x+b 1 y +c 1 z +d 1 p+a 2 x+b 2 y +c 2 z +d 2 q = a 3 x+b 3 y +c 3 z +d 3 1 where p = z x, q = z y and a i, b i, c i, d i i = 1,2,3 are all real numbers, has been discussed in brief. The present paper comprises a detailed discussion of the same including the cases of failure of the method adopted there. 2. The Method For determining the solutions of1.1, it is required to consider the simultaneous ordinary differential equations dx a 1 x+b 1 y +c 1 z +d 1 = dy a 2 x+b 2 y +c 2 z +d 2 = dz a 3 x+b 3 y +c 3 z +d 3. 2 Suppose it is possible to find numbers, µ, ν, ρ such that each ratio of 2 dx+µdy +νdz a 1 +a 2 µ+a 3 νx+b 1 +b 2 µ+b 3 νy +c 1 +c 2 µ+c 3 νz +d 1 +d 2 µ+d 3 ν = dx+µdy +νdz ρx+µy+νz. 3 Clearly, 3 holds if: a 1 ρ+a 2 µ+a 3 ν =0, b 1 +b 2 ρµ+b 3 ν =0, c 1 +c 2 µ+c 3 ρν =0, d 1 +d 2 µ+d 3 ν =0. 4 Considering the first three equations of 4 comprising a system of linear homogeneous algebraic equations in the three variables, µ, ν, for a non-trivial solution,µ,ν 0,0,0 of 4, one notes that the rank of the coefficient matrix should not exceed two. So we should have a 1 ρ a 2 a 3 b 1 b 2 ρ b 3 = 0. 5 c 1 c 2 c 3 ρ

3 A METHOD OF SOLVING LAGRANGE S FIRST-ORDER This leads to the following cubic in ρ: ρ 3 a 1 +b 2 +c 3 ρ 2 +b 2 c 3 b 3 c 2 +a 2 b 1 a 1 b 2 +a 3 c 1 a 1 c 3 ρ +b 2 c 3 b 3 c 2 a 1 +b 3 c 1 b 1 c 3 a 2 +b 1 c 2 b 2 c 1 a 3 = 0. Let the roots of the equation 2.5 be ρ 1, ρ 2, ρ 3. The following cases arise: Case I: ρ 1, ρ 2, ρ 3 are real and distinct, Case II: One of ρ 1, ρ 2, ρ 3 is real, the other two are complex, Case III: ρ 1, ρ 2, ρ 3 are real, but ρ 1 = ρ 2 ρ 3, Case IV: ρ 1, ρ 2, ρ 3 are real, and ρ 1 = ρ 2 = ρ 3. In the following sections the four cases I-IV have been discussed citing an example in each case Case I: Roots of 6 are Real and Distinct Let i,µ i,ν i be the solutions of the first three equations of the system 4 corresponding to ρ = ρ i i = 1,2,3. Then, from 3, we get 1 dx+µ 1 dy +ν 1 dz ρ 1 1 x+µ 1 y +ν 1 z = 2dx+µ 2 dy +ν 2 dz ρ 2 2 x+µ 2 y +ν 2 z = 3dx+µ 3 dy +ν 3 dz ρ 3 3 x+µ 3 y +ν 3 z. 7 If each of i,µ i,ν i i = 1,2,3 satisfies the last equation of 4, the above relations lead to the required solution of the PDE 1.1 as F 1 x+µ 1 y+ν 1 z ρ 2 2 x+µ 2 y+ν 2 z ρ 1, 2 x+µ 2 y+ν 2 z ρ 3 3 x+µ 3 y+ν 3 z ρ 2 where F is an arbitrary real-valued function of two real variables. = 0, Example 1. y +2zp+z + 1 xq = x y. 8 2 For this PDE, the first three equations of 4 become ρ 1 µ ν =0, 2 +ρµ+ν =0, 2 µ+ρν =0. 9

4 68 S.M.H. Islam, J. Das N. Chaudhuri In this case the fourth equation is automatically satisfied and the equation 5 becomes ρ ρ 1 = 0, ρ leading to the equation ρ ρ = 0. Roots of the above equation are 0, ± 6 2. Using ρ = 0 in 9 we get µ+2ν = 0, +ν = 0, whence we have Using ρ = 6 2 in 9 we get 1 = µ 2 = ν 1. 6 µ 2ν = 0, 2+ 6µ+2ν = 0. It then follows that hence we have = µ = ν 6 2, 5 = µ = 6 4 ν Similarly, using ρ = 6 2 So the relations 7 become dx 2dy +dz 0 in 9 we get 5 = µ 6+4 = ν = 5dx+ 6 4dy dz 6 2 5x+ 6 4y z = 5dx+ 6+4dy dz 6 2 5x+ 6+4y z. 11

5 A METHOD OF SOLVING LAGRANGE S FIRST-ORDER From the above relations the required solution of the PDE 8 can be written as F x 2y +z, 6y +z 2 5x 4y +2z 2 = 0, where F is an arbitrary real-valued function of two real variables. 4. Case II: ρ 1 is real, ρ 2 = ρ 2 + iρ 2 and ρ 3 = ρ 2 iρ 2,ρ 2,ρ 2 R Let 1,µ 1,ν 1 be the solution of the first three equations of the system 4 corresponding to the root ρ 1 and 2 +i 2,µ 2 +iµ 2,ν 2 +iν 2, 2, 2, µ 2, µ 2, ν 2, ν 2 R, be the solution of the above mentioned first three equations of the system 4 corresponding to the root ρ 2 + iρ 2. Then the solution of the system 4 corresponding to the root ρ 2 iρ 2 will be 2 i 2,µ 2 iµ 2,ν 2 ν 2. Now, from 3, we get 1 dx+µ 1 dy +ν 1 dz ρ 1 1 x+µ 1 y +ν 1 z 2 +i 2 dx+µ 2 +iµ 2 dy +ν 2 +iν 2 dz = ρ 2 +iρ 2 2 +i 2 x+µ 2 +iµ 2 y +ν 2 +iν 2 z 2 i 2 dx+µ 2 iµ 2 dy +ν 2 iν 2 dz =. ρ 2 iρ 2 2 i 2 x+µ 2 iµ 2 y +ν 2 iν 2 z 12 It is assumed that each of 1,µ 1,ν 1, 2 ±i 2,µ 2 ±iµ 2,ν 2 ±iν 2 satisfies the last equation of 4. From the first equation of 12 we have a solution of the simultaneous equations 2, say u 1, as u 1 = 1 x+µ 1 y+ν 1 z ρ 2 +iρ 2 2 +i 2 x+µ 2 +iµ 2 y+ν 2 +iν ρ1 2 z This implies that = constant = C 1,say. lnu 1 = ρ 2 +iρ 2 ln 1 x+µ 1 y +ν 1 z ρ 1 ln 2 +i 2 x+µ 2 +iµ 2 y +ν 2 +iν 2 z

6 70 S.M.H. Islam, J. Das N. Chaudhuri = ρ 2 +iρ 2 ln 1 x+µ 1 y +ν 1 z ρ 1 ln 2 x+µ 2 y +ν 2 z+i 2 x+µ 2 y +ν 2 z = lnc 1. Writing ln 2 x+µ 2 y +ν 2 z+i 2 x+µ 2 y +ν 2 z = z 1 +iz 2 it is noted that and we get z 1 = 1 2 ln 2 x+µ 2 y +ν 2 z x+µ 2 y +ν 2 z 2, 13 z 2 = arctan 2 x+µ 2 y +ν 2 z 2 x+µ 2 y +ν 2 z, 14 lnu 1 = ρ 2 +iρ 2 ln 1 x+µ 1 y +ν 1 z ρ 1 z 1 +iz 2 = ρ 2 ln 1 x+µ 1 y +ν 1 z ρ 1 z 1 +i ρ 2 ln 1 x+µ 1 y +ν 1 z ρ 1 z 2 = lnc 1. Hence u 1 =exp ρ 2 ln 1 x+µ 1 y +ν 1 z ρ 1 z 1 cos ρ 2 ln 1 x+µ 1 y +ν 1 z ρ 1 z 2 +isin ρ 2 ln 1 x+µ 1 y +ν 1 z ρ 1 z 2 = 1 x+µ 1 y +ν 1 z ρ 2 exp ρ 1 z 1 cos ρ 2 ln 1 x+µ 1 y +ν 1 z ρ 1 z 2 +isin ρ 2 ln 1 x+µ 1 y +ν 1 z ρ 1 z 2 =C 1. So one solution of the simultaneous equations 12, say u, can be taken as u = 1 x+µ 1 y +ν 1 z ρ 2 exp ρ 1 z 1 cos ρ 2 ln 1 x+µ 1 y +ν 1 z ρ 1 z 2

7 A METHOD OF SOLVING LAGRANGE S FIRST-ORDER = constant. 15 From the last equation of 12 we have another solution of the simultaneous equations 12, say v 1, as v 1 = 2 x+µ 2 y +ν 2 z+i 2 x+µ 2 y +ν 2 z 2 x+µ 2 y +ν 2 z i 2 x+µ 2 y +ν 2 z =C 2 say, ρ2 iρ 2 ρ2 iρ 2 = constant or, lnv 1 = ρ 2 iρ 2 z 1 +iz 2 ρ 2 +iρ 2 z 1 iz 2 = lnc 2, where z 1, z 2 are given by 13 and 14 respectively. This shows that lnv 1 = 2iρ 2 z 2 ρ 2 z 1 = lnc 2. Hence another solution of the simultaneous equations 12, say v, can be taken as v = ρ 2 z 2 ρ 2 z 1 = constant. 16 Therefore, in this case, the required solution of the PDE 1.1 is given by F 1 x+µ 1 y +ν 1 z ρ 2 exp ρ 1 z 1 cos ρ 2 ln 1 x+µ 1 y +ν 1 z ρ 1 z 2, ρ 2 z 2 ρ 2 z 1 = 0, 17 where z 1, z 2 are given by 13 and 14 respectively, and as in Case I and Example 1, F is an arbitrary real-valued function of two real variables. Example 2. yp+zq = x. 18 The simultaneous ordinary differential equations corresponding to this PDE are dx y = dy z = dz x. 19 It is noted that the fourth equation of 4 is automatically satisfied.

8 72 S.M.H. Islam, J. Das N. Chaudhuri In this case, the equation 5 becomes ρ ρ ρ = 0 20 or, ρ 3 1 = 0. Roots of this equation are 1, ω, ω 2 ; where ω = 1±i 3 2. For ρ = 1, For ρ = ω, For ρ = ω 2, So each ratio of 19 is equal to 1 = µ 1 = ν 1. ω = µ 1 = ν ω 2. ω 2 = µ 1 = ν ω. dx+dy +dz y +z +x = ωdx+dy +ω2 dz ωy +z +ω 2 x = ω2 dx+dy +ωdz ω 2 y +z +ωx = ωdx+dy +ω2 dz ωωx+y +ω 2 z = ω2 dx+dy +ωdz ω 2 ω 2 x+y +ωz. The required solutions of the PDE 18 are then obtained as F x+y +z ω ωx+y +ω 2 z 1, x+y +z ω2 ω 2 x+y +ωz 1 = 0, where F is an arbitrary real-valued function of two real variables. 5. Case III: ρ 1 = ρ 2 ρ 3 In this case, the method described above for the cases Case I and II does not work. This is exhibited by citing the following example. Example 3. y zp+z +xq = x+ 3 y. 21 4

9 A METHOD OF SOLVING LAGRANGE S FIRST-ORDER The simultaneous ordinary differential equations corresponding to this PDE are dx y z = dy z +x = dz x y. For this PDE the equation 5 becomes ρ ρ 4 = 0, ρ which leads to Roots of this equation are 1, 1 2, 1 2. ρ 12ρ+1 2 = 0. For ρ = 1, 1 = µ 1 = ν 0. For ρ = 1 2, 2 = µ 5 = ν 6. So each ratio of 22 is equal to dx+dy y +x 2dx+5dy 6dz = 1 2 2x+5y 6z. This relation gives us the single solution of the equation 2, in the present case, as u = x+y2x+5y 6z 2 = constant. Being unable to find another solution of the simultaneous equations 2, in the present case, the method described above fails to derive solutions of the PDE Case IV: ρ 1 = ρ 2 = ρ 3 In this case, no solution of the simultaneous equations 2 can be derived and hence the method presented fails to find any solution of the PDE under consideration. The following example exhibits the difficulty. Example 4. x 2y +zp+2x+y +zq = 2x+2y +z. 24

10 74 S.M.H. Islam, J. Das N. Chaudhuri The simultaneous ordinary differential equations corresponding to this PDE are dx x 2y +z = dy 2x+y +z = dz 2x+2y +z. 25 For this PDE the equation 5 becomes 1 ρ ρ 2 = 0, ρ which leads to the equation Roots of this equation are 1, 1, 1. 1 ρ 3 = 0. For ρ = 1, 1 = µ 1 = ν 1. So, in this case, the equation 3 gives us the only ratio dx dy +dz. 27 x y +z Hence the method described above fails to give us the required solutions of the PDE Alternative Approach to Solve the Problems in Case III and Case IV To find the solutions of the PDEs 21 and 24 the following approch has been found to be suitable in case of the examples cited Solution of the PDE 21 The simultaneous ordinary differential equations corresponding to the PDE21 are dx y z = dy z +x = dz x y. Suppose it is possible to find numbers α i, β i, γ i i = 1,2,3 and ρ such that each ratio of 28 is equal to α 1 x+β 1 y +γ 1 zdx+α 2 x+β 2 y +γ 2 zdy +α 3 x+β 3 y +γ 3 zdz α 1 x+β 1 y +γ 1 zy z+α 2 x+β 2 y +γ 2 zz +x+α 3 x+β 3 y +γ 3 zx+ 3 4 y

11 A METHOD OF SOLVING LAGRANGE S FIRST-ORDER = α 1 x+β 1 y +γ 1 zdx+α 2 x+β 2 y +γ 2 zdy +α 3 x+β 3 y +γ 3 zdz α 2 +α 3 x 2 +β β 3y 2 + γ 1 +γ 2 z 2 +α 1 +β α 3 +β 3 xy = dd ρd, where + β 1 +γ 1 +β γ 3yz + α 1 +α 2 +γ 2 +γ 3 zx 29 D = α 2 +α 3 x 2 +β β 3y 2 + γ 1 +γ 2 z 2 +α 1 +β α 3+β 3 xy+ β 1 +γ 1 +β γ 3yz+ α 1 +α 2 +γ 2 +γ 3 zx, and dd denotes the total derivative of D. We see that 29 holds if ρα 1 = 2α 2 +2α 3, ρβ 1 = α 1 +β α 3 +β 3, ργ 1 = α 1 +α 2 +γ 2 +γ 3, ρα 2 = α 1 +β α 3 +β 3, ρβ 2 = 2β β 3, ργ 2 = β 1 +γ 1 +β γ 3, ρα 3 = α 1 +α 2 +γ 2 +γ 3, ρβ 3 = β 1 +γ 1 +β γ 3, ργ 3 = 2γ 1 +2γ 2. The above equations give us a linear homogeneous system of equations ρα 1 +2β 1 +2γ 1 = 0, α 1 ρβ 1 +β γ 1 +γ 2 = 0, 2β 1 ρβ γ 2 = 0, α 1 +β 1 ργ 1 +γ 2 +γ 3 = 0, 30 β 1 +β 2 +γ 1 ργ γ 3 = 0, 2γ 1 +2γ 2 ργ 3 = 0,

12 76 S.M.H. Islam, J. Das N. Chaudhuri where α 2 = β 1, α 3 = γ 1, β 3 = γ 2. The linear homogeneous system 30 will have a non-trivial solution for α 1,β 1,β 2,γ 1,γ 2,γ 3 if the determinant of the coefficient matrix of the system 30 is zero, i.e. This leads to the equation ρ ρ ρ ρ 1 1 = ρ ρ ρ ρ4 7 4 ρ ρ ρ 1 2 = 0. Roots of this equation are found to be 2, 1, 1, 1, 1 2, 1 2. Using these values of ρ in 30 solutions for α 1,β 1,β 2,γ 1,γ 2,γ 3 are found to be as follows: For ρ = 2, α 1 = β 1 = β 2 = 1, γ 1 = γ 2 = γ 3 = 0. For ρ = 1, α 1 = 4, β 1 = 10, β 2 = 25,γ 1 = 12, γ 2 = 30, γ 3 = 36. For ρ = 1 2, α 1 = 4, β 1 = 7, β 2 = 10, γ 1 = 6, γ 2 = 6, γ 3 = 0. Again we have the relations α 2 = β 1, α 3 = γ 1, β 3 = γ 2. So using the above values of ρ, α i, β i, γ i i = 1,2,3 in 29 we have the ratios: dx 2 +y 2 +2xy 2x 2 +y 2 +2xy = d2x y2 +18z 2 +10xy 30yz 12zx 12x y2 +18z 2 +10xy 30yz 12zx = d2x2 +5y 2 +7xy 6yz 6zx 22x 2 +5y 2 +7xy 6yz 6zx. These ratios give us the required solution of the PDE 21 as F x+y2x y2 +18z 2 +10xy 30yz 12zx, 2x 2 +5y 2 +7xy 6yz 6zxx+y 2 = 0, where F is an arbitrary real-valued function of two real variables.

13 A METHOD OF SOLVING LAGRANGE S FIRST-ORDER Solution of the PDE 24 The simultaneous ordinary differential equations corresponding to the PDE24 are dx x 2y +z = dy 2x+y +z = dz 2x+2y +z. 31 Suppose it is possible to find numbers α i, β i, γ i i = 1,2,3 and ρ such that each ratio of 31 is equal to α 1 x + β 1 y + γ 1 zdx + α 2 x + β 2 y + γ 2 zdy + α 3 x + β 3 y + γ 3 zdz α 1 x + β 1 y + γ 1 zx 2y + z + α 2 x + β 2 y + γ 2 z2x + y + z + α 3 x + β 3 y + γ 3 z2x + 2y + z = = dd ρd, where α 1 x + β 1 y + γ 1 zdx + α 2 x + β 2 y + γ 2 zdy + α 3 x + β 3 y + γ 3 zdz α 1 + 2α 2 + 2α 3 x 2 + 2β 1 + β 2 + 2β 3 y 2 + γ 1 + γ 2 + γ 3 z 2 + 2α 1 + β 1 + α 2 + 2β 2 + 2α 3 + 2β 3 xy + β 1 2γ 1 + β 2 + γ 2 + β 3 + 2γ 3 yz + α 1 + γ 1 + α 2 + 2γ 2 + α 3 + 2γ 3 zx D = α 1 +2α 2 +2α 3 x 2 + 2β 1 +β 2 +2β 3 y 2 +γ 1 +γ 2 +γ 3 z 2 + 2α 1 +β 1 +α 2 +2β 2 +2α 3 +2β 3 xy +β 1 2γ 1 +β 2 +γ 2 +β 3 +2γ 3 yz +α 1 +γ 1 +α 2 +2γ 2 +α 3 +2γ 3 zx, and dd denotes the total derivative of D. We see that 32 holds if ρα 1 = 2α 1 +4α 2 +4α 3, ρβ 1 = 2α 1 +β 1 +α 2 +2β 2 +2α 3 +2β 3, ργ 1 = α 1 +γ 1 +α 2 +2γ 2 +α 3 +2γ 3, ρα 2 = 2α 1 +β 1 +α 2 +2β 2 +2α 3 +2β 3, ρβ 2 = 4β 1 +2β 2 +4β 3, ργ 2 = β 1 2γ 1 +β 2 +γ 2 +β 3 +2γ 3, ρα 3 = α 1 +γ 1 +α 2 +2γ 2 +α 3 +2γ 3, ρβ 3 = β 1 2γ 1 +β 2 +γ 2 +β 3 +2γ 3, ργ 3 = 2γ 1 +2γ 2 +2γ 3. The above equations give us a linear homogeneous system of equations 2 ρα 1 +4β 1 +4γ 1 = 0, 2α 1 +2 ρβ 1 +2β 2 +2γ 1 +2γ 2 = 0, 4β 1 +2 ρβ 2 +4γ 2 = 0, α 1 +β 1 +2 ργ 1 +2γ 2 +2γ 3 = 0, β 1 +β 2 2γ 1 +2 ργ 2 +2γ 3 = 0, 2γ 1 +2γ 2 +2 ργ 3 = 0, 32 33

14 78 S.M.H. Islam, J. Das N. Chaudhuri where α 2 = β 1, α 3 = γ 1, β 3 = γ 2. The linear homogeneous system 33 will have a non-trivial solution for α 1,β 1,β 2,γ 1,γ 2,γ 3 if the determinant of the coefficient matrix of the system 33 is zero, i.e. 2 ρ ρ ρ ρ 2 2 = ρ ρ This leads to the equation 2 ρ 6 = 0. So all the roots of this equation are equal to 2. For ρ = 2 the system 33 has two linearly independent solutions, namely 3, 1,3,1, 1,0, 2,0,2,0,0, 1 for α 1,β 1,β 2,γ 1,γ 2,γ 3. Using the values of ρ, α i, β i, γ i i = 1,2,3 in 32 we see that each of the ratios in 32 is equal to d3x 2 +3y 2 2xy 2yz +2zx 23x 2 +3y 2 2xy 2yz +2zx = d2x2 +2y 2 z 2 22x 2 +2y 2 z 2. Again by 6.2 we have found that each ratios in 6.2 is equal to dx y +z x y +z. From the above three equal ratios we get the solution of the PDE 24 as 2x 2 +2y 2 z 2 F x y +z 2, 3x 2 +3y 2 2xy 2yz +2zx x y +z 2 = 0, where F is an arbitrary real-valued function of two real variables. 8. Remarks The genesis of the method described in 3, 4, 7, lies in finding an equivalent ratio of the ratios in 2.1, in which the numerator is the total differential of the denominetor. The idea may be employed to other types of equations. Further works in this direction are under progress, to be reported in future.

15 A METHOD OF SOLVING LAGRANGE S FIRST-ORDER References [1] E.L. Ince, Ordinary Differential Equations, Dover Publications, New York, 1956.

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