EXPLICIT REPRESENTATION OF THE SOLUTION TO SOME BOUNDARY VALUE PROBLEM

EXPLICIT REPRESENTATION OF THE SOLUTION TO SOME BOUNDARY VALUE PROBLEM

S. MAZOUZI AND R. N. PEDERSON

Abstract.

In the half ray the unique solution to the boundary value problem L…Dt†u…t† ˆf…t†;t>0 Bj…Dt†u…0† ˆj;forjˆ0; 1; :::;pÿ1

rapidly decreasing at in¢nity is shown to be explicitly represented in terms of Green's function and some boundary kernels, namely,

u…t† ˆX^pÿ1

kˆ0hk…t†k‡Z ₁

0g…t;s†f…s†ds

1. Introduction.

LetL…z†be a polynomial of degreem1, where the coe¤cient ofz^mis equal to 1, and letfB_j…z†g^pÿ1_jˆ0 beppolynomials of degreesfm_jg^pÿ1_jˆ0 respectively, so that mj<m for 0jpÿ1: We assume that L…z† has at most p roots having positive imaginary parts (counting multiplicities).

Throughout this paper we denote byD_t the di¡erential operatorÿi_dt^d;and ifJ Rthen we denote bys…J†the subspace ofC¹ …J†containing all the functionsu…t†such that

…1‡ jtj†^pju^q†…t†j are bounded for allpandqinN.

Consider the following boundary value problem L…Dt†u…t† ˆf…t†;t>0 …2:1†

Bj…Dt†u…0† ˆj; forjˆ0; 1; :::;pÿ1 …2:2†

Received October 19, 1995.

where the_j are complex constants. Set

L_p…z† ˆ …zÿ₀†…zÿ₁†:::…zÿ_pÿ1† …2:3†

where the_j's are theproots ofL…z†having positive imaginary parts.

Firstly, we assume thatmj<pforjˆ0; 1; :::;pÿ1, then B_j…z† ˆ

Xpÿ1 kˆ0

b_jkz^k …2:4†

for jˆ0; 1; :::;pÿ1:Now, if the matrix…bjk†is nonsingular, one can solve Eq. (2.4) for the unknown variablesz^k(for ¢xedz).

Hence, if we denote by…b^jk†the inverse matrix of…b_jk†, we get z^jˆ

Xpÿ1 kˆ0

b^jkBk…z†

…2:5†

forjˆ0; 1; :::;pÿ1:So that, Eq. (2.2) is equivalent to D^k_tu…0† ˆ

Xpÿ1 jˆ0

b^kj_j …2:6†

forkˆ0; 1; :::;pÿ1:Thus, any solution of the Cauchy problem (2.1), (2.2) is a solution of the problem

L…Dt†u…t† ˆf…t†;t>0 …2:7†

D^k_tu…0† ˆ Xpÿ1

jˆ0

b^kjj

…2:8†

forkˆ0; 1; :::;pÿ1:Conversely, any solution of (2.7), (2.8) is a solution of (2.1), (2.2).

Remarks. If the matrix…b_jk†is singular, then there are constants cj not all zero so that

Xpÿ1 jˆ0

cjBj…z† ˆ0 …2:9†

Consequently, a necessary condition for the problem (2.1), (2.2) to have a solution is the following

Xpÿ1 jˆ0

c_jjˆ0 …2:10†

This shows that we can not hope to get a solution for all choices of the_j's.

Furthermore, even when we can solve, the solution is generally not unique.

Now if we allow to themj's to be greater thanp, then, by partial fractions we can write

Bj…z† ˆQj…z†L^‡_p…z† ‡B⁰_j…z†

…2:11†

Where the degree ofB⁰_jis less thanp.

Whenp<mwe de¢ne the polynomial L^ÿ…z† ˆ L…z†

L^‡_p…z†

…2:12†

We conclude by the following Lemma [5]:

Lemma. Let L…D_t†be any constant coefficient differential operator, and let f…t†be any function ins(R).Then, there exists a function u…t† 2s…R†which satisfies the differential equation

L…D_t†u…t† ˆf…t†; for t>0 …2:13†

that the di¡erential equation

L^ÿ…Dt†v…t† ˆf…t†;t>0 …2:14†

has always a solutionv…t†belonging tos(R) for any choice off…t†ins…R†.

Thus, the boundary value problem (2.1), (2.2) is equivalent to the following L^‡_P…D_t†u…t† ˆv…t†; t>0

…2:15†

B_j⁰…Dt†u…0† ˆjÿQj…Dt†v…0†; jˆ0; 1; :::;pÿ1 …2:16†

which is the form just treated. If we write B_j⁰…z† ˆ

Xpÿ1 kˆ0

b⁰_jkz^k; 0jpÿ1 …2:17†

then the problem (2.15), (2.16) has a unique solution for any given v…t† in s(R) and {_j}Cif and only if the matrix…b⁰_jk†is nonsingular.

We observe from Eq. (2.9) that the matrix…b_ij†is nonsingular if and only if the polynomials Bj…z†are linearly independent. Similarly, the matrix …b⁰_ij† is nonsingular if and only if the polynomialsB⁰_j…z†are linearly independent.

We say that thefBj…z†g are linearly independent modulo L^‡_p…z† if the fB_j⁰g are linearly independent.

2. Main results.

Now we are able to give explicitly the solution to the boundary value pro- blem (2.1), (2.2) in terms of Green's function; we shall follow closely Pederson's work after rede¢ning in a convenient manner the functionsu^‡_j …t†

(see [3], [4]).

Theorem 1. Let L…z†be a polynomial of degree m, having at most p roots 0,1,...,pÿ1with positive imaginary parts and no real roots. LetfBj…z†g^pÿ1_jˆ0be p polynomials of degrees fmjg with mj<m, which are linearly independent modulo

L^‡_p…z† ˆ …zÿ0†…zÿ1†:::…zÿpÿ1†

Then, for any f 2s…R^‡†and for any choice of the constants₀; ₁; :::; _pÿ1; there exists a unique solution u…t† 2s…R^‡†satisfying the boundary value pro- blem(2.1), (2.2).Furthermore, this solution can be represented as follows

u…t† ˆ Xpÿ1

jˆ0

h_j…t†_j‡ Z ₁

0 g…t;s†f…s†ds …2:18†

Proof. Let us first consider the casef ˆ0; then, the general solution of the equation

L…D_t†u…t† ˆ0 …2:19†

has the form

u…t† ˆ^mÿ1X

kˆ0

_kexp…i_k†

where thek's are the roots ofL…z† ˆ0. It is worth to recall that the coe¤- cients_kbecome polynomials in twhenever there are multiple roots. Now, in order foru…t†to be inL²…0; 1†;the coe¤cients_kmust vanish for any k such thatIm _k0 otherwiseu…t†could not be inL²…0;1†. Therefore, the solution of (2.19) which belongs to the spaceL²…0;1†is

u…t† ˆ Xpÿ1

kˆ0

_kexp…i_k† …2:20†

withImk>0.

Following AGMON, DOUGLIS and NIRENBERG [1], we de¢ne L^‡_k…† ˆ

Xk jˆ0

a^‡_j ^kÿj; kˆ0; 1; :::;pÿ1 …2:21†

where the constantsa^‡_j are de¢ned through the expression L^‡_p…† ˆ

Xp jˆ0

a^‡_j ^pÿj …2:22†

Letÿ^‡andÿ^ÿbe recti¢able Jordan contours in the upper and the lower half plane enclosing the roots ofL^‡_p…z†andL^ÿ_p…z†respectively.

De¢ne the functions u^‡_j …t† ˆ 1

2i I

ÿ^‡

L^‡_pÿjÿ1…†

L^‡_p…† exp…it†d …2:23†

We claim that

L…Dt†u^‡_j …t† ˆ0; t>0 …2:24†

and

D^k_tu^‡_j …0† ˆ_jk …2:25†

for j; kˆ0; 1; :::;pÿ1, where _jk is the Kronecker Delta. Indeed, by dif- ferentiation under the integral sign (which is of course allowed), we get

D^k_t u^‡_j …t† ˆ 1 2i

I

ÿ^‡

L^‡_pÿjÿ1…†

L^‡_p…† ^kexp…it†d …2:26†

Now, if we takeÿ^‡ to be a large circle about the origin with radiusn 2N*

so thatÿ^‡ encloses₀; ₁; :::; _pÿ1;then, D^k_t u^‡_j …0† ˆ 1

2 Z ₂

0

L^‡_pÿjÿ1…ne^i!†

L^‡_p…ne^i!† n^k‡1 e^i!…k‡1†d!

ˆ 1 2

Z ₂

0

Q…ne^i!† L^‡_p…ne^i!†d!

Where Q is a polynomial of degreep‡kÿj in n. Since L^‡_p is of degree p, then, by lettingngo to in¢nity, we obviously get

D^k_tu^‡_j …0† ˆjk; for kÿj0

For the case kÿj>0, we note that the polynomial ^kL^‡_pÿjÿ1 di¡ers from ^kÿjÿ1L^‡_p by a polynomialQof degree at most equals tokÿ1. Thus,

D^k_tu^‡_j …0† ˆ 1 2i

I

ÿ^‡

L^‡_pÿjÿi…† L^‡_p…† ^kd

ˆ 1 2i

I

ÿ^‡

^kÿjÿ1L^‡_p…† ‡Q…† L^‡_p…† d

ˆ 1 2i

I

ÿ^‡kÿjÿid 1 2i

I

ÿ^‡

Q…† L^‡_p…†d

ˆ 1 2i

I

ÿ^‡

Q…† L^‡_p…†d

By the same argument as before, since the degree of Q…z† is equal to kÿ1<pÿ1, we can observe that the last integral is zero. As a con- sequence, we obtain

D^k_tu^‡_j …0† ˆjk; j;kˆ0;1; :::;pÿ1 On the other hand we have

L…Dt†u^‡_j …t† ˆ 1 2i

I

ÿ^‡

L^‡_pÿjÿi…†L^‡_p…†L^ÿ…†

L^‡_P…† e^itd ˆ0:

It follows that the setfu^‡_j …t†gspans the negative exponential solutions of the homogeneous boundary value problem associated to (2.1), (2.2). Now, in order to obtain a solution to the inhomogeneous boundary value problem (2.1), (2.2), we de¢ne the functions

v…t† ˆ 1 2

I

ÿ

e^it L…†d …2:27†

It follows from the fact that the contour ÿ^‡[ÿ^ÿ can be deformed into a large circle that

D^k_t…v^‡…0† ‡v^ÿ…0†† ˆimÿ1;k

…2:28†

wherekˆ0; :::;mÿ1. As a consequence, the function Z _t

0 …v^‡…tÿs† ‡v^ÿ…tÿs††f…s†ds …2:29†

is a solution of the Eq. (2.1) with zero Cauchy Data.

Thus, the general solution of the inhomogeneous boundary value problem (2.1), (2.2) which is bounded must have the form:

u…t† ˆX^pÿ1

jˆ0

_ju^‡_j …t† ‡ Z _t

0 …v^‡…tÿs† ‡v^ÿ…tÿs† †f…s†ds …2:30†

ÿ Z ₁

0 v^ÿ…tÿs†f…s†ds

This is a consequence of the facts that v^‡…t† is a sum of the negative ex- ponentials when t>0, and v^ÿ…t† is a sum of negative exponentials when t<0. Now, by virtue of the complementing condition (linear independence of theB⁰_js†, we conclude that

Bk…Dt† Z _t

0 …v^‡…tÿs† ‡v^ÿ…tÿs††f…s†ds ˆ0

tˆ0

Since the function (2.30) is a formal solution of (2.1), (2.2) then, it must sa- tisfy the following

kˆBk…Dt†u…0† ˆX^pÿ1

jˆ0

jBk…Dt†u^‡_j …0†‡

‡ B_k…D_t† Z _t

o …v^‡…tÿs† ‡v^ÿ…tÿs††f…s†ds ÿ

tˆ0

ÿ Z ₁

0 f…s†Bk…Dt†v^ÿ…tÿs†ds

tˆ0

ˆX^pÿ1

jˆ0

_jfQ_k…D_t†L^‡_p…D_t†u^‡_j …0† ‡B⁰_k…D_t†u^‡_j …0†gÿ

ÿ Z ₁

0 f…s†Bk…Dt†v^ÿ…tÿs†ds

tˆ0

ˆX^pÿ1

jˆ0

_j b^‡_kjÿ Z ₁

0 f…s†B_k…D_t†v^ÿ…tÿs†ds

tˆ0

whereB_k…z† ˆQ_k…z†L^‡_p…z† ‡B⁰_k…z† ˆB_k⁰…z†mod…L^‡_p†and

B⁰_k…z† ˆX^pÿ1

jˆ0

b^‡_kjz^j

We get an algebraic system of equations with unknown variablesj

ˆX^pÿ1

jˆ0

b^‡_kljˆ_k‡ Z ₁

0 f…s†B_k…D_t†v^ÿ…tÿs†ds

tˆ0

; kˆ0;1; :::;pÿ1

We deduce from the complementing condition that the determinant of the matrix…b^‡_kj†is not zero; so, as a consequence, the above set of equations has a unique solutionf0; ::: ; pÿ1g:

De¢ne the inverse matrix

…b^kj_‡† ˆ …b^‡_kj†^ÿ1 Hence,

_jˆX^pÿ1

kˆ0

b^jk_‡k‡ Z ₁

0

X^pÿ1

kˆ0

b^jk_‡f…s†B_k…D_t†v^ÿ…tÿs†

tˆ0

; jˆ0;1; :::;pÿ1

and upon substitution of the_j's into (2.30) we obtain the bounded solution of the given BVP,

u…t† ˆX^pÿ1

jˆ0

X^pÿ1

kˆ0

b^jk_‡ku^‡_j …t†‡

‡ Z ₁

0

X^pÿ1

jˆ0

X^pÿ1

kˆ0

b^jk_‡B_k…D_t†v^ÿ…tÿs† f…s†ds

( )

u^‡_j …t†‡

tˆ0

‡ Z _t

0 v^‡…tÿs†f…s†dsÿ Z ₁

t v^ÿ…tÿs†f…s†ds If we set

h_k…t† ˆ 1 2i

I

ÿ^‡

X^pÿ1

jˆ0

b^jk_‡L^‡_pÿjÿ1…† L^‡_p…† e^itd forkˆ0; ::: ;pÿ1

g1…t† ˆ 1 2

I

ÿ^‡

e^it

L…†d; if t>0

ˆÿ1 2 I

ÿ^ÿ

e^it

L…†d; if t<0

g₂…t;s† ˆ ÿi 4²

I

ÿ^‡

I

ÿ^ÿ

X^pÿ1

jˆ0

X^pÿ1

kˆ0

L^‡_pÿjÿi…†B_k…†

L^‡_p…†L…† e^i…tÿs†d d and

g…t;s† ˆg₁…tÿs† ‡g₂…t;s†

then, the solution of the boundary value problem (2.1), (2.2) takes the ¢nal form

u…t† ˆX^pÿ1

jˆ0

h_j…t†_j‡ Z ₁

0 g…t;s†f…s†ds

An immediate computation shows that the above kernels satisfy the follow- ing estimates

jD^k_th_j…t†j C₀exp…ÿr₀t†; 8t>0;8kˆ0; 1; :::

jD^k_tg1…t†j C1 exp…ÿr1jtj†; 8t2R;8kˆ0; 1; :::

jD^k_tg₂…t;s†j C₂ exp…ÿr₂…t‡s††; 8t>0;8s>0;8kˆ0; 1; :::

for some positive constantsC₀,C₁,C₂,r₀,r₁, andr₂, depending only onL…z†

andfB_jg.

Now to see that the expression (2.18) is rapidly decreasing at in¢nity it su¤ces to show that the function

v_j…t† ˆt^j Z ₁

0 jf…s†j …C₁ exp…ÿr₁jtÿsj† ‡C₂ exp…ÿr₂…t‡s††ds is bounded for each nonnegative integerj. Since f 2s…R^‡†there is a con- stantC>0 such that

jf…s†j C

…1‡s†^j‡2;8s>0 it then follows that

vj…t† C⁰t^j 2 t

jZ t 2

0

ds

…1‡s†^j‡2‡C1 …2t†^j …2‡t†^j

Z _t

t2

ds …1‡s†²

‡C₁ t 1‡t _jZ ₁

t

ds

…1‡s†²‡C₂⁰t^j Z ₁

0 exp…ÿr₂…t‡s††ds C…j†

Z ₁

0

ds

…1‡s†²‡C3t^jexp…ÿr2t†<‡1

Hencevj…t†is bounded inR^‡and consequentlyu…t† 2s…R^‡†:

Finally, using classical techniques we can easily prove the uniqueness of this solution. This establishes the proof of the given theorem.

Let us denote byH^k…R^‡†;k0 the completion of the space s…R^‡†with respect to the norm

kuk²_kˆX^k

jˆ0

Z ₁

0 ju^k†…t†j²dt …2:31†

and we de¢ne the subspace

V^kˆH^k…R^‡† \C^k‰0;1Š

As a consequence of the previous representation theorem and Theorem 6^9 [5] we obtain the estimate of the solution to the problem (2.1)^(2.2) in terms of the Dataf and₀; :::; _pÿ1:

Theorem 2. Under the same assumptions of Theorem 1, we conclude that for each k2N;there is a constant C>0(depending only on L…z†;B_j…z†and k) such that, for each f 2V^kand₀; ::: ; _pÿ12C^p;the solution u2V^m‡kto the BVP…2:1† ÿ …2:2†satisfies the estimate

kukm‡kC X^pÿ1

jˆ0

jjj ‡ kf kk

!

and has the representation u…t† ˆX^pÿ1

jˆ0

h_j…t†_j‡ Z ₁

0 g…t;s†f…s†ds …2:32†

(whereh_j…t†andg…t;s†are the same as in Theorem 1.).

Proof. We deduce form the density of s…R^‡† in V^k that there is a se- quence …fn† s…R^‡† converging to f in V^k: On the other hand there corresponds to eachfn at most one solutionun 2s…R^‡†satisfying:

L…Dt†un…t† ˆfn…t†; …t>0†

Bj…Dt†un…0† ˆj; jˆ0; ::: ;pÿ1 and given by

u_n…t† ˆX^pÿ1

jˆ0

h_j…t†_j‡ Z ₁

0 g…t;s†f_n…s†ds …2:33†

We conclude by Theorem 6ÿ9 [5] that there is a constantC₀>0 depending only onL…z†andksuch that

kunÿX^pÿ1

jˆ0

hj…t†jkm‡kC0kfnkk

…2:34†

and

kunÿukm‡kC0kfnÿfkk

…2:35†

Hence,

kuk_m‡kX^pÿ1

jˆ0

kh_jk_m‡kj_jj ‡C₀kfk_kC X^pÿ1

jˆ0

j_jj‡kfk_k

! …2:36†

whereCˆmaxC₀;kh_jk_m‡k; jˆ0; ::: ; pÿ1 : The estimate (2.36) shows that the isomorphism

P:ÿ0; ::: ; pÿ1; f

!u

is continuous fromC^pV^kontoV^m‡k:Consequently, by lettingn! ‡1in (2.33) we obtain

u…t† ˆX^pÿ1

jˆ0

h_j…t†_j‡ Z ₁

0 g…t;s†f…s†ds; …t>0†

ˆPÿ_o; ::: ; _pÿ1;f This proves the theorem.

Remarks. 1) If L…z† admits a real root then we cannot hope to get an

estimate of the form (2.36) even under smooth Data as shows the following example:

du dtˆ 1

t‡12L²…R^‡† \C¹…R^‡†;

u…0† ˆ0

whose unique solution is

u…t† ˆ₀‡ln…1‡t†; …t>0†

which is not inL²…0;1†whatsoever the value of the constant0:

2) The best constantC in (2.36) is equal to the norm of the isomorphism Pde¢ned by

SupjX^pÿ1

jˆ0

h_j…t†_j‡ Z ₁

0 g…t;s†h…s†dsj

where the supremum is taken over allh2V^kand₀; ::: ; _pÿ12Csuch that X^pÿ1

jˆ0

jjj ‡ khkkˆ1

REFERENCES

1. S. Agmon, A. Douglis and L. Nirenberg,Estimate near the boundary for solutions of elliptic P.D.E., Comm. Pure Appl. Math. 12 (1959).

2. S. Mazouzi,Linear elliptic Boundary Value Problem in the half space, Master's Thesis, Car- negie Mellon Univ., PA., USA (1985).

3. R. N. Pederson,Explicit formulae for complementary boundary operators of linear elliptic problems, J. Differential Equations 46 (1982).

4. R. N. Pederson,Green's function and boundary kernels for elliptic boundary value problems, Unpublished class notes.

5. M. Schechter,Modern Methods in Partial Diff. Eq., McGraw Hill (1977).

INSTITUTE OF MATHEMATICS P.O. BOX 12, EL-HADJAR ANNABA

ALGERIA

DEPARTMENT OF MATHEMATICS CARNEGIE MELLON UNIVERSITY PITTSBURGH, PA. 15213 USA