Selçuk J. Appl. Math. Selçuk Journal of Vol. 10. No. 1. pp. 19-32, 2009 Applied Mathematics
A Method on The General Solution of Inhomogeneous Euler Differ-ential Equations
Ba¸sak Karpuz2, Hüseyin Yıldırım1
Department of Mathematics, Faculty of Sciences and Arts, ANS Campus, 03200, Afy-onkarahisar, Turkey.
e-mail: 2bkarpuz@ gm ail.com,1hyildir@ aku.edu.tr
Received: October 10, 2007
Abstract. In this article, an -th order Euler differential equation will be writ-ten as differential of multiplication of an -th order Euler differential equation and a function by multiplying the equation with a suitable integration multiplier, after making some arrangements the equation will be reduced to an ( − 1)-th order Euler differential equation and continuation will give the general solution. Key words: Euler differential equation, general solution.
2000 Mathematical Subject Classification: 34A30. 1. Introduction
An -th order Euler differential equation is given by (1.1) [] = () for ≥ 1 where ∈ C for = 0 1 ∈ N and
(1.2) := X =0 µ ¶ 6= 0
Throughout the paper, we shall assume that is a suitable function, i.e., is integrable over [1 ∞) for any ∈ C, and also its integrals satisfy the same property.
As it is well-known, an Euler differential equation can be transformed into an autonomous linear differential equation by the independent variable transform () = e for ≥ 0, then solution of homogeneous part can be obtained by characteristic equation and inhomogeneous part can be obtained by the variation
parameters, the method of undeterminate coefficients or the inverse operator method. On the other hand, the solution of homogeneous part can be obtained by plugging () = , where ∈ C, into equation then particular solution can be obtained by variation parameters or undeterminate coefficients method as in [1,2,3]. Also its order can be reduced under some certain conditions given in [4]. In this paper, we will show that order of the equation can be reduced even the condition in [4] is not satisfied.
2 Reduction of Order
Let for = 0 1 , the coefficients in (1.2) hold the following relation (2.1)
= ∗−1
= ∗−1+ ( − ) ∗ for = 1 2 − 1 0= −∗0
where ∈ C and ∗
∈ C for = 0 1 − 1 are undeterminate for now. It will be shown that the system above is indeed consistent, and and for = 0 1 − 1 are solvable.
Thus, for all ≥ 1, we have −(+1)[] =−(+1) X =0 () = X =0 −(+1)() =−(+1)()+ Ã−1 X =1 −(+1)() ! + 0−(+1) and by using (1.1), (1.2) and (2.1), we get
−(+1)[] =∗−1−(+1)()+ Ã−1 X =1 (∗−1+ ( − )∗)−(+1)() ! − ∗0−(+1) =∗−1−(+1)()+ Ã−1 X =1 ∗−1−(+1)() ! + Ã−1 X =1 ( − )∗−(+1)() ! − ∗0−(+1) = Ã X =1 ∗−1−(+1)() ! + Ã−1 X =0 ( − )∗−(+1)() ! = Ã−1 X =0 ∗−(+1) ! + Ã−1 X =0 ( − )∗−(+1)() ! = X−1 =0 ³ ∗−(+1)+ ( − )∗−(+1)()´
or −(+1)[] = −1 X =0 ∗³−(+1)+ ( − )−(+1)()´ = −1 X =0 ∗ ³ −()´ = −1 X =0 ³ ∗−() ´ = Ã−1 X =0 ∗−() ! = Ã − −1 X =0 ∗() ! Here, let us define
(2.2) ∗−1[] := −1 X =0 ∗() so that we get (2.3) −(+1)[] = ¡ −∗−1[]¢ Now, ∈ C and ∗
∈ C for = 0 1 − 1 will be calculated from (2.1). First, using the second and the third equalities in (2.1), we find that (2.4) ∗−1= + ( − )∗ for = 1 2 − 1 Letting = − 1 in (2.4), we obtain
∗−2= −1+ ( + 1 − )∗−1 and by using the first equality in (2.1), we have (2.5) ∗−2 = −1+ ( + 2 − ) Once again, by letting = − 2 in (2.4), we have
∗−3= −2+ ( + 2 − )∗−2 and by using (2.5), we get
∗−3=−2+ ( + 2 − )(−1+ ( + 1 − )) =−2+ ( + 2 − )−1+ ( + 2 − )( + 1 − ) =−2+Γ( + 3 − ) Γ( + 2 − )−1+ Γ( + 3 − ) Γ( + 1 − )
By the emerging pattern, it is not hard to see that ∗−1=+ Γ( + 1 − ) X =+1 1 Γ( + 1 − ) =Γ( + 1 − ) X = 1 Γ( + 1 − )
holds for = 1 2 . By taking into account the last equality in (2.1), the resulting above can be written in the following form
(2.6) ∗−1= Γ( + 1 − ) X = 1 Γ( + 1 − ) for = 1 2 Therefore, by letting = 1 in (2.6), we get
∗0= Γ() X =1 1 Γ( + 1 − ) which yields by using the last equality in (2.1) that
0= −Γ() X =1 1 Γ( + 1 − ) is true. Thus, we obtain
0 =0+ Γ() X =1 1 Γ( + 1 − ) =0+ Γ( + 1) X =1 1 Γ( + 1 − ) =Γ( + 1) X =0 1 Γ( + 1 − ) We now define (2.7) () := Γ( + 1) X =0 1 Γ( + 1 − )
Clearly, the characteristic equation given in (2.7) is an -th order polynomial of . Let ∗ ∈ C be an arbitrary root of (2.7), then by substituting ∗ in (2.6) one can calculate ∗
for = 0 1 − 1. Note that if ∗ = −1, then (1.1) already satisfies the condition in [4]. Otherwise, the equation below satisfies the condition in [4]. From (1.1) and (2.1), we obtain
−(∗+1)[] =
³
or the reduced equation
(2.8) ∗−1[] = ∗ Z
−(∗+1) () + ∗
where is the integration constant and coefficients of (2.2) are determined by (2.6). Obviously, (2.8) is an ( − 1)-th order Euler differential equation. An application of this method is given below.
Example 1. Let the operator 2be defined as follows 2:= 2
2 2 +
+ 1
and consider the following second-order Euler differential equation 2[] = for ≥ 1
Here, the coefficients of the operator 2 are given by = 1 for = 0 1 2. It follows from (2.7) that the characteristic equation for 2 is
2() =Γ( + 1) µ 1 Γ( + 1)+ 1 Γ()+ 1 Γ( − 1) ¶ =1 + + ( − 1) =2+ 1
and the roots of this parabola are 1= −i and 2= i. Let us reduce the equation by the root 2 = . Due to (2.6), the coefficients of the reduced operator are calculated as ∗
1= 1 and ∗0= i. Hence the reduced operator is ∗1:= + i so that we have ∗1[] =i µZ −(i+1) + 2 ¶ =i Z −i + 2i =1 + i 2 + 2 i
by using (2.8). This equation is a first-order Euler differential equation. Ap-plying the order reduction again for ∗
1[], we have the characteristic equation below
∗
whose root is 1= −i. By (2.6), we find that ∗0= 1. For the arbitrary constants 1 and 2, we have the following general solution
=−i µZ i−1 µ 1 + i 2 + 2 i ¶ + 1 ¶ =−i Z µ1 + i 2 i+ 22i−1 ¶ + 1−i =1 + i 2 −i Z i + 2−i Z 2i−1 + 1−i =1 2 − i 22 i+ 1−i As is well-known, for all ∈ R+ and ∈ R
±i=eln ±i =e±i ln =(cos( ln ) ± i sin( ln )) Thus, we have =1 2 − i
21(cos(ln ) + i sin(ln )) + 2(cos(ln ) − i sin(ln )) =1 2 + µ −i 21+ 2 ¶ cos(ln ) − µ 1 21+ i2 ¶ sin(ln ) and by rearranging the arbitrary constants we get the general solution
= 1cos(ln ) + 2sin(ln ) + 1 2 3 General Solution by Order Reduction
In this section, we give the general solution of (1.1) by repeatedly applying the method in § 2. Define the order reduction operator ∆ by
∆[] := ∗−1 where ∗
−1 is defined with respect to the discussion given in § 2, and for = 0 1 , we define (3.1) ∆[] := ⎧ ⎪ ⎨ ⎪ ⎩ = 0 ∗−1 = 1 ∆−1[∆[]] = ∆−1[∗ −1] = 2 3
As we see from (2.6) that the leading coefficients of the operators and ∆[] = ∗−1 are the same. Also by (3.1) leading coefficients of the opera-tor ∆[
] is the same for = 1 2 . Thus, we see from (1.1) and (3.1) that
(3.2) ∆[
Let −for = 0 1 −1 be the root of the characteristic equation ∆[ ],
which is used to reduce ∆[
] to ∆+1[]. So that by using (2.6), for the arbitrary constant −, one can obtain
¡ −−∆+1[ []] ¢ = −(−+1)∆[ []] or (3.3) ∆+1[[]] = − Z −(−+1)∆[ []] + −− by using (2.8). It is clear that ∆+1[
[]] is an ( − ( + 1))-th order Euler differential equation. The right-hand side of (3.3) consists of the function () and the arbitrary constants for = − −+1 . By taking = −1 in (3.3), we get ∆[ []] = 1 Z −(1+1)∆−1[ []] + 11 It follows from (3.2) that
= 1 Z −(1+1)∆−1[ []] + 11 or simply (3.4) () = 1 1 Z −(1+1)∆−1[ []] + 1 1
Thus, the function 1 is a solution of the homogeneous part of the equation
(1.1) and let us define this solution by
(3.5) 1() := 1
Again by using (3.3), we obtain = 1 1R−(1+1) 1 ³ 2 1 R −(2+1) 2 ∆−2[[]]2+ 212 ´ 1+11 = 1 1R2−(1+1) 1 ³R −(2+1) 2 ∆−2[[]]2 ´ 1 +2 1R2−(1+1) 1 1+1 1
We therefore see that the function 2() := 1
Z
2−(1+1)
1 1
is a solution of the homogeneous part of the equation (1.1) too, which is linearly independent from 1. By continuation, for = 2 3 , one may write (3.6) () := 1 µZ 2−(1+1) 1 µZ 3−(2+1) 2 · · · µZ −(−1+1) −1 −1 ¶ · · · 2 ¶ 1 ¶
Here, are linearly independent solutions of the homogeneous part of (1.1). On the other hand,
(3.7) () := 1 1 µZ 2−(1+1) 1 · · · µZ −(+1) () ¶ · · · 1 ¶
is a particular solution for (1.1). Finally, using (3.6) and (3.7), the general solution of (1.1) is given by (3.8) () = () + X =1 ()
where for = 1 2 are arbitrary constants. It is easy to see that () and () coincide when () ≡ 0.
4 Properties of Order Reduction and Rewriting the General Solution In this section, we give some properties related to the discussion given in § 2 and § 3.
Lemma 1 (Relation of the roots). Let be reduced to ∆[] = ∗−1 by the root ∗of the characteristic equation
(), then
() = ( −
∗) ∗
−1()
Proof . By (3.6), the characteristic equation of is
() = Γ ( + 1) P =0 1 Γ(+1−) = Γ ( + 1) Ã 1 Γ(+1−)+ P−1 =1 1 Γ(+1−)+ 1 Γ(+1)0 ! By substituting (3.1) into the resulting above, we get
() = Γ ( + 1) Ã 1 Γ(+1−)∗−1+ P−1 =1 1 Γ(+1−) ¡ ∗−1+ ( − ∗) ∗¢+Γ()1 (−∗) ∗0 ! = Γ ( + 1) Ã P =1 −∗ Γ(+1−)∗−1+ P−1 =0 (−∗) Γ(+1−)∗ ! = Γ ( + 1) Ã P−1 =0 1 Γ(−)∗−1+ P−1 =0 −∗ Γ(+1−)∗ ! = Γ ( + 1) Ã P−1 =0 − Γ(+1−)∗+ P−1 =1 −∗ Γ(+1−)∗ ! = Γ ( + 1) P−1 =0 −∗ Γ(+1−)∗
and thus, we have () = ( − ∗) Γ ( + 1)P−1 =0 1 Γ(+1−)∗ = ( − ∗) ∗ −1()
which completes the proof.
Lemma 1 shows that the roots, which were not used in reduction of can be used to reduce ∆[]. This fact can also be seen in Example 1.
Lemma 2. For = 0 1 − 1, let ∆+1[] be reduced from ∆[] by the root −of the characteristic equation ∆[](), then we have
() = Y =1
( − )
Proof .By Lemma 1, we can write
() = ( − ) ∆[]()
Again by Lemma 1, we have
() = ( − ) ∆[]() = ( − ) ( − −1) ∆2[ ]() = ( − ) ( − −1) ( − −2) ∆3[]() = " Q =−2( − ) # ∆3[ ]()
Repeating this procedure, we obtain
() = " Q =2( − ) # ∆−1[]()
We know from (2.6) that leading coefficient of ∆−1[] is. So, ∆−1[]can
be written as follows ∆−1[]() = ( − 1) Thus, we get () = " Q =2( − ) # ( − 1) = Q =1( − ) ·
Hence, we see that for = 1 2 , the roots of ∆−1[], are also roots of
As an immediate consequence of Lemma 2, we have the following corollary. Corollary 1 (Distinct roots). Let ∈ C for = 1 2 be distinct roots of (). Then, the solution of the homogeneous part of (1.1) is
= X =1
where are the arbitrary constants.
Lemma 3. Let ∈ C for = 1 2 be the roots of , then for each ,
the corresponding function is a solution of the homogeneous part of (1.1).
Proof . We see from Corollary 1 that for = 1 2 , the roots of ∆[]are
roots of at the same time. Using the arbitrariness of the roots in the order
reduction and letting = 1for any index in (3.5) complete the proof. Lemma 4 (Repeated roots). Let be (1 ≤ ≤ )-fold root of the polynomial , then the function
= X =1
(ln )−1
where for = 1 2 , are arbitrary constants, is a solution of the homo-geneous part of (1.1).
Proof . This proof will be considered in two cases.
(i) = 1. Then, is the unique root, hence = 11 is a solution of the homogeneous part of (1.1) by Lemma 2.
(ii) Let be -fold root for some such that 2 ≤ ≤ . Then, one can write () = ∙ Q =+1( − ) ¸ ∆−[]()
where 6= for = + 1 + 2 . For = 1 2 , let = be the roots of the characteristic equation ∆−[]. By (3.5), the function 1 = is a solution of the homogeneous part of (1.1). By (3.6), the function
() = ¡R 1− −1¡R2− −1¡R− −1−1 −1 ¢ 2 ¢ 1 ¢ = ¡R−1 1 ¡R −12 ¡R−1−1−1 ¢ 2 ¢ 1 ¢ = Γ()1 (ln )−1
which consists of − 1 integrals, are solutions of the homogeneous part of (1.1). Thus, for the arbitrary constants, the function
= X =1 = X =1 (ln )−1= X =1 (ln )−1
forms the linear combination of linear independent solutions of the homogeneous part of (1.1).
Next, we give the following theorem of which proof follows by making use of the series of lemmas above.
Theorem 1 (General solution). Let ∈ C be -fold roots of the char-acteristic equation , where P=1 = for 1 ≤ ≤ . Then, for
arbitrary constants, the function := X =1 X =1 (ln ) −1
is the solution of the homogeneous part of (1.1). The particular solution of (1.1) is given by () := 1 1 µZ 2−(1+1) 1 · · · µZ −(−1+1) −1 µZ −(+1) () ¶ −1 ¶ · · · 1 ¶ where for = 1 2 , the root is assumed to hold the following relation:
1 2 1 = 1 1+1 1+2 1+2 = 2 (−1=1)+1 (−1=1)+2 (
=1)
= −+1 −+2 = . Then := + is the general
so-lution of (1.1).
We give the following example as an application of Theorem 1. Example 2. Denote 4:= 4 4 4 + 8 3 3 3+ 15 2 2 2+ 7 + 1 To obtain the general solution of
4[] =
we consider the following characteristic equation of 4 4() =4+ 23+ 22+ 2 + 1
=( + 1)2(2+ 1)
whose roots are 1 2= −1, 3= i and 4= −i. Thus, by Theorem 1 = 11−1+ 12−1ln + 21−i+ 31
or equivalently
=11−1+ 12−1ln + 21(cos(ln ) − i sin(ln )) + 31(cos(ln ) + i sin(ln ))
=11−1+ 12−1ln + (21+ 31) cos(ln ) + i(31− 21) sin(ln )
is the solution of the homogeneous part of the equation, and = −1 µZ −11 µZ i2 µZ −(2i+1)3 µZ i44 ¶ 3 ¶ 2 ¶ 1 ¶ =1 8 is the particular solution. By rearranging arbitrary constants in , the general solution = 1 1 + 2 ln + 3cos(ln ) + 4sin(ln ) + 1 8 is obtained.
5 Maple Codes for the Method
In this section, we give the Maple programming codes for the method given above.2 >#Order Reduction OR:=proc(LCoef,f) local DeltaLCoef,InHomFunc,CharEq,Root,LNOrder: print("Given Equation"): print(sum(LCoef[k+1]*(x^k)*‘if‘(k>0,diff(y(x),x$k),y(x)),k=0..nops(LCoef) -1)=f): LNOrder:=nops(LCoef)-1: if LNOrder=0 then print("General Solution"): print(y(x)=f/LCoef[1]): else CharEq:=simplify(GAMMA(r+1)*sum(LCoef[k1+1]/GAMMA(r+1-k1),k1=0..LNOrder)): print("Characteristic Equation"): print(phi(r)=CharEq): Root:=[solve(CharEq=0,r)][1]:
print("Root of the Characteristic Equation"): print(tau=Root): DeltaLCoef:=[seq(subs(r=Root,simplify(GAMMA(r+1-k1) *sum(LCoef[k2]/GAMMA(r+1-k2),k2=k1..LNOrder))),k1=1..LNOrder)]: InHomFunc:=simplify(x^Root*(int(x^(-(Root+1))*f,x)+C[LNOrder])): if LNOrder=1 then print("General Solution"): print(y(x)=InHomFunc/DeltaLCoef[1]):
else print("Reduced Equation"): print(sum(DeltaLCoef[k+1]*(x^k)*‘if‘(k>0,diff(y(x),x$k),y(x)), k=0..LNOrder-1)=InHomFunc): end if: end if: end: >#General Solution GS:=proc(LCoef,f) local EqOrder,CharEq,Roots,UniqueRoots,OrderOfRoots,k1,k2,k3,HomSol,ParSol; print("Given Equation"): EqOrder:=nops(LCoef)-1: print(sum(LCoef[k+1]*(x^k)*‘if‘(k>0,diff(y(x),x$k),y(x)),k=0..EqOrder)=f): CharEq:=simplify(GAMMA(r+1)*sum(LCoef[k1+1]/GAMMA(r+1-k1),k1=0..EqOrder)): print("Characteristic Equation"): print(phi(r)=CharEq): Roots:=solve(CharEq=0,r):
print("Roots of the Characteristic Equation"): print(tau=[Roots]):
UniqueRoots:={Roots}:
for k1 from 1 to nops(UniqueRoots) do: k3:=0:
for k2 from 1 to nops([Roots]) do if UniqueRoots[k1]=[Roots][k2] then k3:=k3+1: end if: end do: OrderOfRoots[k1]:=k3: end do: HomSol:=sum(sum(C[k4,k5]*x^(UniqueRoots[k5]) *(ln(x))^(k4-1),k4=1..OrderOfRoots[k5]),k5=1..nops(UniqueRoots)): ParSol:=x^([Roots][EqOrder])*int(x^(-([Roots][EqOrder]+1))*f,x): if EqOrder>1 then
for k1 from 2 to EqOrder do
ParSol:=x^([Roots][EqOrder-k1+1])*int(x^(-([Roots][EqOrder-k1+1]+1)) *ParSol,x): end do: end if: ParSol:=simplify(ParSol/LCoef[nops(LCoef)]): print("General Solution"): print(y(x)=HomSol+ParSol): end: >#Example 1 OR([1,1,1],x):
"Given Equation" () + µ () ¶ + 2 µ 2 2() ¶ = "Characteristic Polynomial" () = 1 + 2
"Root of the Characteristic Polynomial" = i "Reduced Equation" i() + µ () ¶ = 1 2 + 1 2i + 2 i >#Example 2 GS([1,7,15,8,1],x): "Given Equation" () + 7 µ () ¶ + 152 µ 2 2() ¶ + 83 µ 3 3() ¶ + 4 µ 4 4() ¶ = "Characteristic Polynomial" () = 1 + 2 + 22+ 23+ 4 "Root of the Characteristic Polynomial"
= [−1 −1 i −i] "General Solution" () = 11 + 21ln() + 12 i+ 13(−i)+ 1 8 References
1. Weisstein, E. W., 2004, Euler differential equation, MathWorld - A Wolfram Web Resource
http://mathworld.wolfram.com/EulerDifferentialEquation.html
2. Ahmadi, G., 2006, Review of engineering Mathematics, Clarkson University, Pots-dam, New York-USA
http://www. clarkson.edu/projects/crcd/me537/downloads/0016_ReviewEngineeringMath.pdf 3. Sean Mauch, 2004, Introduction to methods of applied mathematics or advanced
mathematical methods for scientists and engineers, California Institue of Technology, California-USA
http: //www.ama.caltech.edu/∼seanm/applied_math.pdf
4. Exact differential equations, eFunda Engineering Fundementals - Web Source http://www. efunda.com/math/ode/linear_ode_varcoeff.cfm
