2nd Order De

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304 J. DAS (N EE CHAUDHURI)

4. The unique solution of the DE (1.1) determined by a BC U [y] = 0that is not natural with respect to (1.1)

We rst note that, as the DE (1.1) is homogeneous, the uniqueness of its solutionis to be understood up to a constant multiplier. That such a unique solution existshas been proved in [1].

As the BC U [y] = 0 is not natural w.r. to the DE (1.1), we can not haveU [] = 0 = U [].

If U [] = 0, then U [] = 0 and the required unique solution of (1.1) satysingU [y] = 0 is .

If U [] = 0, the required unique solution of (1.1) should be of the form = + ( u + iv )), where u , v are real numbers. Then U [] = 0 if and only if

1 + 2 u + 3 (1 + u 1 v2) + 4 (1 + u 1 v2 ) = 0(4.1)

and

2v + 3 (2 + u 2 + v1) + 4(2 + u 2 + v1 ) = 0 .(4.2)The equations (4.1)(4.2) determine u , v uniquely (and so, uniquely), sinceU [] = 0 implies ( 2 + 3 1 + 4 1)2 + ( 32 + 42)2 = 0. These observationslead to the following theorem:

Theorem II. Let the BC U [y] = 0 be not natural with respect to the DE (1.1) .(i) If U [] = 0, then the required unique solution of (1.1) , that satises U [y] =

0 is .(ii) If U [] = 0 , then the required unique solution of (1.1) , that satises U [y] =

0 is = + ( u + iv ) (u, v : real numbers ), where u and v are uniquely determined by the equations (4.1) (4.2) .

5. Compatibility of boundary conditionsLet U [y] = 0, U [y] = 0, be two linearly independent BCs, none of which is a

NBC with respect to the DE (1.1). Then, each of these BCs will determine a non-trivial solution of the DE (1.1), uniquely up to a constant multiplier; in general,the solutions determined by them will be different. In case the two BCs determinethe same non-trivial solution of the DE (1.1), then they are said to be compatiblewith respect to the DE (1.1). In this section we shall determine the condition thatwill guarantee that the two BCs U [y] = 0 = U [y] are compatible with respect tothe DE (1.1). Equivalently we determine the condition under which the Sturm-Liouville problem (SLP) : L [y] = 0 = U [y] = U [y] has a non-trivial solution,unique up to a constant multiplier.

In this connection, we need to use the following result which has been proved

in [1].

5.1. Proposition. For any solution of the SLP (1.1) , (1.2) , (1.3) , let B ()denote the set of vectors = ( 1 , 2 , 3 , 4 ) of R 4 such that

U [] = 1 (a) + 2 (1) (a ) + 3 (b) + 4 (1) (b) = 0 .(5.1)


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THE BCS ASSOCIATED WITH SECOND-ORDER DES 305

Then (A) for = + ( u + iv ),

dim B () = 2 if either (A 1 ) 2 2 2 2 = 0 ,or (A 2 ) 2 2 2 2 = 0 ,

(2 , 2 , 2 , 2) = (0 , 0, 0, 0) ,v = 0 ,

or (A 3 ) 2 2 2 2 = 0 ,

(2 , 2 , 2 , 2) = (0 , 0, 0, 0) ,v = 0 , 2 + u 2 = 0 , or , 2 + u 2 = 0 ,

or (A 4 ) 2 = 2 = 2 = 2 = 0 , v = 0 ;

and

dim B () = 3 if either (A 5 ) 2 2 2 2 = 0 ,

(2 , 2 , 2 , 2) = (0 , 0, 0, 0) ,v = 0 , 2 + u 2 = 0 = 2 + u 2

or (A 6 ) 2 = 2 = 2 = 2 = 0 , v = 0 ;

(B) for = ,

dim B () = 2 if (B 1 ) 12 1 2 = 0 ,dim B () = 3 if (B 2 ) 12 1 2 = 0 ,

5.2. Compatibility of two boundary conditionsLet U [y] = 0 = U [y] be two linearly independent BCs. We are to nd

conditions under which they are compatible with respect to the DE (1.1), assumingthat none of them is a NBC for the DE (1.1). We shall actually prove the following:

Theorem III. The two linearly independent BCs U [y] = 0 = U [y], none of which is a NBC for the DE (1.1) , are compatible with respect to the DE (1.1) if and only if

A34 (1 2 1 2 + 21 2 1) A13 2 A14 2 + A23 2 + A24 2 = 0 ,(5.2)

where A ij = i j j i (i, j = 1 , 2, 3, 4).

Proof. First suppose that the BCs

U [y] = 0 = U [y](5.3)


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are linearly independent and are compatible with respect to the DE (1.1). Then,both the BCs determine the same solution of the DE (1.1). We are to consider

the following two cases :(a) = + ( u + iv ) , (b) = .

Case (a):From the proposition given in 5.1, it is clear that the following six cases are to

be considered separately:(a 1) 22 2 2 = 0 ,(a 2) 22 2 2 = 0 , (2 , 2 , 2 , 2) = (0 , 0, 0, 0), v = 0 ,(a 3) 22 2 2 = 0 , (2 , 2 , 2 , 2) = (0 , 0, 0, 0), v = 0 , 2 + u 2 = 0 ,(a 4) 22 2 2 = 0 , (2 , 2 , 2 , 2) = (0 , 0, 0, 0), v = 0 , 2 + u 2 = 0 ,(a 5) 2 = 2 = 2 = 2 = 0 , v = 0 ,

(a 6) 2 = 2 = 2 = 2 = 0 , v = 0 ,

subcase (a 1) : In this case we know that dim B () = 2.Therefore, the equation U [] = 0 being equivalent to

1 + 2u + 3 (1 + u 1 v2) + 4(1 + u 1 v2) = 0 ,(5.4) 2 v + 3(2 + u 2 + v1 ) + 4(2 + u 2 v1) = 0 ,(5.5)

the system of equations (5.4)(5.5) in = ( 1 , 2 , 3 , 4) have exactly two linearlyindependent solutions, and = ( 1 , 2 , 3 , 4) and = ( 1 , 2 , 3 , 4) form onesuch pair. This requires that the rank of the coefficient matrix of (5.4)(5.5) istwo. Therefore, we are to further subdivide the subcase ( a 1) into the following:

(a 1 i) v = 0 ,(a 1 ii ) v = 0 , 2 + u 2 = 0 ,(a 1 iii ) v = 0 , 2 + u 2 = 0 .

subcase (a 1 i): 2 2 2 2 = 0 , v = 0 .Here, two linearly independent solution vectors of (5.4)(5.5) can be obtained

by taking ( 3 , 4) = (1 , 0) and (0,1)as

= u 2 v1 + ( u 2 + v2)2 , (2 + u 2 + v1), v, 0 ,(5.6)

= u 2 v1 + ( u2 + v2)2 , (2 + u 2 + v1), 0, v .(5.7)

If = + ( u + iv ) is the non-trivial solution of the SLP (1.1), (1.2), (1.3), thevectors = ( 1 , 2 , 3 , 4) and = ( 1 , 2 , 3 , 4) must belong to B (). As , are linearly independent, the vectors , (given in (5.6), (5.7)) of B () should beexpressible in terms of , . Hence there exist real numbers A,B,C,D (A, B ) =(0, 0), (C, D ) = (0 , 0) such that

= A + B and = C + D .(5.8)


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Eliminating u , v, u 2 + v2 , A , B , C , D from the eight equations of (5.8) we have,if the BCs U [y] = 0 = U [y] are compatible with respect to the DE (1.1), then

(5.2) holds.subcase (a 1 ii ): 2 2 2 2 = 0 , v = 0 , 2 + u 2 = 0 .

In this case equations (5.4)(5.5) become 1 + 2u + 3(1 + u 1) + 4(1 + u 1) = 0 ,

3(2 + u 2) + 4(2 + u 2) = 0 .(5.9)

Since B () = B ( + u ) = 2, equations (5.9) yield two linearly independentsolutions, which can be taken to be

1 = K, 0, (2 + u 2), 2 + u 2 , 1 = ( u, 1, 0, 0) ,

whereK = ( 1 + u 1)(2 + u 2) (1 + u 1)(2 + u 2) .

If the BCs U [y] = 0 = U [y] are compatible with respect to the DE (1.1), theremust exist real numbers A , B , C , D (in general, different from those used in (5.8)),(A, B ) = (0 , 0), (C, D ) = (0 , 0) , such that

1 = A + B , 1 = C + D .(5.10)

Then ( C, D ) = (0 , 0) implies A34 = 0 from the second set of equations in (5.10).Hence

uA 23 + A13 = 0 , (2 + u 2)A23 + ( 2 + u 2)A24 = 0 ,and

uA 24 + A14 = 0 .These imply

A23 A14 = A24 A13

andA23 (A13 2 A23 2 A24 2) + A13 A24 2 = 0 .

HenceA23 (A13 2 A23 2 A24 2 + A14 2) = 0 .(5.11)

Now, if A23 = 0, it follows that A13 = 0.Then A13 = 0 = A23 = A34 will lead to the fact that and are linearlydependent, which is contrary to our hypothesis. Hence A23 = 0. So, A34 = 0, and(5.11) then implies that (5.2) holds.

subcase (a 1 iii ): 2 2 2 2 = 0 , v = 0 , 2 + u 2 = 0 .

In this case we have 2 + u 2 = 0. Then (5.9) implies that 4 = 0. So 4 = 4 = 0 .(5.12)

As = ( 1 2 , 3 , 4), = ( 1 , 2 , 3 , 4 ) satisfy the rst equation of (5.9), and2 + u 2 = 0, we get

2 A23 2A13 = 0 .(5.13)


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(5.12)(5.13) will then imply that (5.2) is satised.

subcase (a 2 ): 2 2 2 2 = 0 , ( 2 , 2 , 2 , 2 ) = (0 , 0, 0 , 0) , v = 0 .

In this case we know from Theorem I that there is a unique NBC for DE (1.1).Proceeding as in subcase ( a 1 i), we obtain real numbers A , B , C , D such that(5.8) hold.

Using 22 22 = 0, the eight equations in (5.8) can be reduced to eightequations in A2 C 2 , B 2 D 2 , A2 C 2 , B 2 D 2 and v. As v = 0, itwill then follow that

A34 (12 12) A14 2 A13 2 = 0(5.14)

and

A34 (2 1 2 1) + A24 2 + A23 2 = 0(5.15)

(5.14) and (5.15) then imply that (5.2) is satised.

subcase (a 3): 2 2 2 2 = 0 , ( 2 , 2 , 2 , 2 ) = (0 , 0, 0 , 0) , v = 0 , 2 + u 2 = 0 .

As v = 0, = ( 1 , 2 , 3 , 4), = ( 1 , 2 , 3 , 4) are two linearly independentsolutions of (5.9). From the four equations so determined, as 2 + u 2 = 0 we canprove that

A34 = 0 ,(5.16)A13 + A23 u = 0 ,(5.17)A14 + A24 u = 0 .(5.18)

Using (5.17)(5.18), it is then easy to show thatA23 2 + A24 2 A13 2 A14 2 = 0 .(5.19)

Then (5.16) and (5.19) imply that (5.2) is satisied.

subcase (a 4): 2 2 2 2 = 0 , ( 2 , 2 , 2 , 2 ) = (0 , 0, 0 , 0) , v = 0 , 2 + u 2 = 0 .

Two cases arise: (i) 2 + u 2 = 0 (ii) 2 + u 2 = 0.In both cases, proceeding as before, it may be shown that (5.2) holds.

subcase (a 5): 2 = 2 = 2 = 2 = 0 , v = 0 .

In this case we shall show that there is no non-trivial BC, other than the NBCs,which is satised by = + ( u + iv ). If possible, suppose that the solution = + ( u + iv ), (v = 0) of the DE (1.1) satises the BC U [y] = 0 which is nota NBC of the DE (1.1). Then we get

2 + 3 1 + 41 = 0 ,(5.20) 1 + 3 1 + 41 = 0 .(5.21)


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As U [y] = 0 is not a NBC of the DE (1.1), ( 1 , 2 , 3 , 4) must be orthogonal tothe vectors ( 1 , 1 , 1, 0) and ( 1 , 1 , 0, 1), see Theorem I. So

11 + 21 3 = 0 ,(5.22) 11 + 21 4 = 0 .(5.23)

We nd that the determinant of the coefficient matrix of the linear equations(5.20)(5.21)(5.22)(5.23) is non-zero.

Hence 1 = 2 = 3 = 4 = 0.

subcase (a 6): 2 = 2 = 2 = 2 = 0 , v = 0 .

In this case we know that dim B () = 3 and there are two linearly independentNBCs for DE (1.1), which may be given by (3.5), (3.6).

Hence, if U [y] = 0 is a BC satised by , and if it is not a NBC for the DE(1.1), we have

1 + 2u + 3 (1 + u 1) + 4 (1 + u 1) = 0 11 + 2 1 3 = 0 11 + 2 1 4 = 0 .

It can be easily proved that the above system of linear equations in 1 , 2 , 3 , 4has a unique solution (up to a constant multiplier).

Hence the BCs U [y] = 0 = U [y] can not be compatible unless either theyare linearly dependent or at least one of them is a NBC for the DE (1.1), both of which are contrary to our hypothesis. So this case does not arise.

Case (b) : =

We note that U [] = 0 implies

2 + 31 + 41 = 0 , 32 + 42 = 0 .

Clearly two cases are to be considered :

(b1 ) 12 1 2 = 0 , (b2 ) 12 12 = 0 .

subcase (b1) : 1 2 1 2 = 0 .

From the above system of equations we can determine 2 : 3 : 4 . So, if theBCs U [y] = 0 = U [y] are compatible, we must have

2 2

= 3 3

= 4 4

.

Hence,

A23 = A24 = A34 = 0 .(5.24)


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Also,A13 2 + A14 2 = ( 1 3 3 1)2 + ( 1 4 4 1)2

= 1( 32 + 42) 1 ( 32 + 42)= 0 ,

since

U [y] = 0 = U [y] .(5.25)

These imply that (5.2) is satised.

subcase (b2) : 1 2 1 2 = 0 .

We rst note that ( 1 , 2 , 1 , 2 ) = (0 , 0, 0, 0), for, otherwise, we have (b) = (b) = 0, which is contrary to our hypothesis, as is a non-trivial solution.

Two further subcases are to be considered :

(i) (2 , 2) = (0 , 0) , (ii ) (2 , 2) = (0 , 0) .

subcase (b2 i) : Here U [] = 0 will imply 2 = 0. Hence U [y] = 0 = U [y]will imply 2 = 2 = 0, A34 = 0,

A13 2 + A14 2 = 0 .(5.26)

Then it follows that (5.2) holds.

subcase (b2 ii ) : 2 = 2 = 0 .Here U [y] = 0 = U [y] imply

A23 1A34 = 0 = A24 + 1 A34 .

It then follows that (5.2) is satised.

Now we prove the necessity part of Theorem III. We suppose that (5.2) holds,where U [y] = 0 = U [y] are two linearly independent BCs, none of which is aNBC for the DE (1.1). We are to show that the two BCs are compatible, i.e.,

L [] = 0 , U [] = 0 and (5.2) must imply U [] = 0 .

Once again we are to consider separately the following cases :

(I) = + ( u + iv ), (II) = .

Case (I) : = + ( u + iv ) .

Here U [] = 0 implies 1 + 2u + 3 (1 + u 1 v2) + 4(1 + u 1 v2 ) = 0

and

2 v + 3 (2 + u 2 + v1) + 4(2 + u 2 + v1 ) = 0


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or

( 1 + 31 + 41 ) + u( 2 + 3 1 + 41) v( 32 + 42 ) = 0

and

( 3 2 + 42 ) + u( 32 + 42 ) + v( 2 + 31 + 41 ) = 0

or

A + Bu Dv = 0 = C + Du + Bv ,(5.27)

whereA = 1 + 31 + 41 , C = 32 + 42 ,B = 2 + 31 + 41 , D = 3 2 + 42 .

(5.28)

Now, (5.2) can be rewritten in the formDP CQ + BR AS = 0 ,(5.29)

whereP = 1 + 31 + 41 , R = 32 + 42 ,Q = 2 + 31 + 41 , S = 32 + 42 .

(5.30)

As dim B () = 2 or 3, these three equations (5.27) and (5.29) in A, B , C ,D will have two / three linearly independent solutions. Hence the rank of thecorresponding coefficient matrix is not greater than two.

Hence we haveR + Su + Qv = 0 = P + Qu Sv .(5.31)

Now U [] = [P + Qu Sv ] + i[R + Su + Qv ]. Hence L [] = 0, U [] = 0 implyU [] = 0, by (5.31).

Case (II) : = .

We note that U [] = B + iD = 0 implies B = 0 = D . Then (5.2) impliesAS + CQ = 0.The three equations B = 0 = D = AS + CQ treated as linear equations in 1 , 2 , 3 , 4 must yield at least two solutions. This leads to 2 = 2 = 0. Thendim B () = 3. So, all second order minors of the coefficient matrix of the abovesystem of linear equations must also vanish, from which we have

2Q = 0 = 2 Q .

If (2 , 2 ) = (0 , 0), we get Q = 0. Hence U [] = Q + iS = 0. If 2 = 2 = 2 =2 = 0, there are two linearly independent NBCs given by (3.5), (3.6).

As U [y] = 0 is not a NBC, we have 11 + 2 1 3 = 0 = 11 + 21 4 .

In this case AS + CQ = 0 implies 1 = 0. Then B = 0 implies 2 + 31 + 41 = 2(1 + 21 + 21 ) = 0 or 2 = 0, so that we get 1 = 2 = 3 = 4 = 0, contraryto our hypothesis. Hence this case cannot arise.


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The proof of Theorem III is now complete.

5.3. Compatibility of three boundary conditions

We are to nd conditions under which three linearly independent BCs U [y] =0 = U [y] = U [y] are compatible.

Theorem IV. Three linearly independent BCs U [y] = 0 = U [y] = U [y] arecompatible with respect to the DE L [y] = 0 if and only if

AB

=P Q

=LM

,(5.32)

where A, B , P , Q are as dened in (5.28) and (5.30) , and

L = 1 + 31 + 41 , M = 2 + 3 1 + 41 .(5.33)

Proof. A solution of the DE L [y] = 0 will satisfy three linearly independentBCs, i.e., dim B () = 3, if

either (1) 22 22 = 0 , (2 , 2 , 2 , 2 ) = (0 , 0, 0, 0), v = 02 + u 2 = 0 = 2 + u 2 , = + u

or (2) 2 = 2 = 2 = 2 = 0 , v = 0 , = + u ,or (3) 1 2 12 = 0 , = .

Suppose that the three given BCs are compatible. Then,

U [] = 0 = U [] = U [] ,

which leads to

A + Bu = 0 = P + Qu = L + M u ,(5.34)

if = + u . In other words, (5.32) is satised.If = , we have U [] = 0 = U [] = U [], and 12 12 = 0. It will then

follow that 2 = 2 = 2 = 0, whence A = P = L = 0. Hence (5.32) holds.Conversely suppose (5.32) holds. We are to show that L [] = 0 = U [] should

imply U [] = 0 = U [].If = + u , U [] = 0 implies A + Bu = 0.Using (5.32), we can immediately show that P + Qu = 0 = L + M u , i.e.,

U [] = 0 = U [].If = , U [] = 0 implies B = 0 = D .Using (5.32) again, we deduce that Q = M = 0; in other words U [] = 0 =

U [].

6. Remarks

The present paper takes notice of the following:


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(A) All the solutions of a DE of the form (1.1) may satisfy one or more non-trivial boundary condition of the form U [y] = 0. Such a boundary condition has

been named a natural boundary condition (NBC).(i) A necessary and sufficient condition for the nonexistenceof such NBC has been derived.

(ii) The number of NBCs for a given DE has been determined.(iii) The NBC/s for a given DE have been presented.

(B) Every real second-order linear homogeneous DE possesses two linearly in-dependent NBCs.(C) Each boundary condition of the form U [y] = 0, which is not a NBC for DE (1.1) , determines a solution of DE (1.1) uniquely up to a constant multiplier.(D) Two boundary conditions U [y] = 0 = U [y] are said to be compatible withrespect to DE (1.1) if both of them determine the same solution of DE (1.1) .

(i) The necessary and sufficient condition for the boundary conditionsU [y] = 0 = U [y] to be compatible with respect to DE (1.1)has been obtained.

(ii) The necessary and sufficient condition for the boundary conditionsU [y] = 0 = U [y] = U [y] to be compatible with respect to DE (1.1)has also been derived.

These observations urge one to rethink about Sturm-Liouville Problems regardingthe number of boundary conditions to be taken into account.

References[1] Das, J. (nee Chaudhuri), On the solution spaces of linear second-order homogeneous ordinary

differential equations and associated boundary conditions , J. Math. Anal. Appl. 200 , (1996),

4252.[2] Ince, E. L., Ordinary Differential Equations , Dover, New York, 1956.[3] Eastham, M. S. P., Theory of Ordinary Differential Equations , Van Nostrand Reinhold,

London, 1970.

Department of Pure Mathematics, University of Calcutta35, Ballygunge Circular Road, Kolkata-700019, India

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