Application 8.3 Automating the Frobenius Series Method

Transcription

1 Application 8.3 Automating the Frobenius Series Method Here we illustrate the use of a computer algebra system to apply the method of Frobenius. In the paragraphs that follow, we consider the differential equation + + = () y 3 y ( ) y of Eample 4 in Section 8.3 of the tet, where we found the two indicial roots r = and r =. We carry through the formal Frobenius method starting with the larger indicial root r, and you can then apply the same process to derive the second Frobenius series solution (found manually in the tet) corresponding to r. In the following eamples, use this method to derive Frobenius series solutions that can be checked against the given known general solutions... y y y =, y y y =, y ( ) = Acos + y ( ) = Acos y + y + y =, y ( ) = Acos + 4. y y y + + =, y ( ) = ( Acos + ) + + =, y ( ) = ( Acos + ) 5. 4y 6 y y y y y (4 ) =, y ( ) = ( Acos + ) y y y =, y ( ) = ( Acos + ) 8. + =, y y y y ( ) = A( ) + B( + e ) The net three problems involve the arctangent series 8 Chapter 8

2 tan = =, y ( ) = ( A+ Btan ) 3 ( ) y ( 4 ) y y =, y ( ) = ( A+ Btan ) ( ) y (3 5 ) y y =, y ( ) = ( A+ Btan ) ( ) y (3 7 ) y 8 y Using Maple Beginning with the indicial root r := /: we first write the initial k+ terms of a proposed Frobenius series solution: k := 6: a := array(..k): y := ^(/)*sum( a[n]*^(n), n =..k); ( ) y: = a + a + a + a + a + a + a Then we substitute this series (partial sum) into the left-hand side of Equation (), de := *^*diff(y,$) + 3**diff(y,) - (^+)*y: eq := simplify(de); 3 ( deq : = 5a 4a 7a 44a 65a 9a a + a + a + a + a + a + a ) 3/ Noting the factor, we multiply by powers of. 3/ and then collect coefficients of like eq := collect( ^(-3/)*eq, ); deq : = a a + ( a + 9 a ) + ( a + 65 a ) ( a + 44 a ) + ( a + 7 a ) + (4 a a ) + 5a Application 8.3 9

3 Net we set up the equations the successive coefficients must satisfy. We do this by defining an array and then filling the elements of this array by equating (in turn) each of the series coefficients to zero. eqs := array(..5): for n from to 5 do eqs[n] := coeff(eq,,n) = : od: coeffeqs := convert(eqs, set); { coeffeqs : = 5 a =, a + 44 a4 =, a3+ 65 a5 =, 9 a a =, 4 a a =, a + 7 a = We have here a collection of si linear equations relating the seven coefficients a through a 6. Hence we should be able to solve for the successive coefficients in terms of a : succcoeffs := convert([seq(a[n], n=..6)], set); { } succcoeffs: = a, a, a, a, a, a ourcoeffs := solve(coeffeqs, succcoeffs); ourcoeffs : = a =, a3 =, a5 =, a = a, a4 = a, a6 = a Finally we substitute all these coefficients back into the original series. partsoln := subs(ourcoeffs, y); 4 6 partsoln : = a + a + a + a } Note that (after factoring out a ) this result agrees with the first particular solution found in the tet. y 4 6 = / ( ) a Using Mathematica Beginning with the indicial root 3 Chapter 8

4 r = /; we first write the initial k+ terms of a proposed Frobenius series solution: k = 6; y = ^r (Sum[ a[n] ^n, {n,,k} ] + O[]^(k+)) a() a() a() a(3) a(4) a(5) 3/ 5/ 7/ 9/ / a O 3/ 5/ (6) + ( ) Then we substitute this series into the differential equation in (), deq = ^ D[y,,] + 3 D[y, ] - (^ + )y == 3/ 5/ 7/ 5 a() + (4 a() a()) + (7 a(3) a()) + 9/ / (44 a(4) a()) + (65 a(5) a(3)) + 3/ 5/ (9 a(6) a(4)) + O( ) == Mathematica has automatically collected like powers for us, and we can use the LogicalEpand command to etract the equations that the successive coefficients satisfy. coeffeqns = LogicalEpand[ deq ] 5 a() == 4 a() a() == 7 a(3) a() == 44 a(4) a() == 65 a(5) a(3) == 9 a(6) a(4) == We have here a collection of si linear equations relating the seven coefficients a = a() through a 6 = a(6). Hence we should be able to solve for the successive coefficients succcoeffs = Table[ a[n], {n,, 6} ] { a(), a(), a(3), a(4), a(5), a (6)} in terms of a(): ourcoeffs = Solve[coeffEqns, succcoeffs] a() a() a() a(), a(), a(3), a(4), a(5), a(6) Finally we substitute all these coefficients back into the original series. Application 8.3 3

5 partsoln = y /. ourcoeffs // First 3/ 5/ 9/ a() 5/ a() + a() + a() + + O( ) Note that (after factoring out found in the tet. y a / / ( ) a ) this result agrees with the first particular solution 4 6 = Using MATLAB Beginning with the indicial root r =, we first write the initial seven terms of a proposed Frobenius series solution: syms y a a a a3 a4 a5 a6 y = ^(/)*(a+a*+a*^+a3*^3+a4*^4+a5*^5+a6*^6); pretty(y) / (a + a + a + a3 + a4 + a5 + a6 ) Then we substitute this series (partial sum) into the left-hand side of Equation (): de = *^*diff(y,) + 3**diff(y) - (^ + )*y; simple(de) ans = -^(3/)*(^6*a5+^7*a6+*a+^4*a3+^3*a+^*a +^5*a4-65*a5*^4-44*a4*^3-7*a3*^ -4*a*-5*a-9*a6*^5) Noting the like powers of. 3/ factor, we multiply by 3/ and then collect coefficients of de = collect(simplify(^(-3/)*de)) de = -^7*a6-^6*a5+ (9*a6-a4)*^5 + (-a3+65*a5)*^4 + (-a+44*a4)*^3 + (-a+7*a3)*^ + (4*a-a)* + 5*a 3 Chapter 8

6 Net we set up the equations the original coefficients in our Frobenius series must satisfy. By cutting and pasting it is a simple matter to pick out the successive coefficients in de and equate each of them to zero. eq = 5*a; eq = 4*a-a; eq3 = -a+7*a3; eq4 = -a+44*a4; eq5 = -a3+65*a5; eq6 = 9*a6-a4; We have here a collection of si linear equations relating the seven coefficients a through a6. Hence we should be able to solve for the successive coefficients a through a6 in terms of a: soln = solve(eq,eq,eq3,eq4,eq5,eq6, a,a,a3,a4,a5,a6); coeffs = [soln.a,soln.a,soln.a3,soln.a4,soln.a5,soln.a6] coeffs = [, /4*a,, /66*a,, /5544*a] Finally we substitute all these coefficients back into the original series y: y = subs(y, [a,a,a3,a4,a5,a6], coeffs); pretty(y) / 4 6 (a + /4 a + /66 a + /5544 a ) Note that (after factoring out a ) this result agrees with the first particular solution found in the tet. y 4 6 = a / ( ) Application