On the Cauchy problem for the polyparabolic equation

ANNALES SOCIETATIS MATHEMATICAE POLONAE Series I: COMMENTATIONES MATHEMATICAE XXII (1981) ROCZNIK1 POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO

Séria I: PRACE MATEMATYCZNE XXII (1981)

J. M i l e w s k i (Krakôw)

On the Cauchy problem for the polyparabolic equation In [3] M. Nicolescu solved some limit problems for the polycaloric equation. In the present paper we shall solve the Cauchy problem for the polyparabolic equation of degree 3n. The solution will be given in the form of a sum of polyparabolic potentials.

1. Let us consider the half plane Û = {(x, t): x gR, t > 0} and the equation

(1) И и (x ,t) = 0,

where L = Dbx — Dt — c\ c = const ^ 0, L" = L (L "_1), nsN.

Denote by Я the set of all functions u which are continuous together with the derivatives D? Dpu, p + 3q ^ 3n in Q and satisfying there equation (1).

We shall construct the function we Я satisfying the initial conditions (2) D*w(x, 0) = gk(x) for x e R , к = 0, 1,..., n — 1.

The problem of constructing of the function we Я satisfying the initial conditions (2) will briefly be called the (L-C ) problem.

2. Let

j (t - s )~ 1/3f ( ( x - y ) ( t - s ) ~ 1/3) for s < t,

U (x, t; y, s) = < '

where

/ ( 0 =

3 V 3 Г ' Ч з у з

,3/2 + J. _1/3 з у з

,3/2

and JY(z) denote the Bessel functions of the first kind. It is well known [1]

that the function U is the fundamental solution of the equation (D l ~ D t)u = 0.

Let us consider the function

V( x, t; y,s) = e c(t s) U (x, t; y, s);

this function is the fundamental solution of the equation Lu = 0.

296 J. Mi l ews ki

3. Let

1 00

uk( x , t ) = ---J <Pk(y)V{x,t; y,0)dy, K — 00

where

<pM = T r ^ W - (^ - 11)!Y 7 il î - . , 0 » + - + ( - i y - 5L r *S(y) and

hk(y) = gk(y), * î - i( y ) = ( — c)^fc_ ! (y) -h c i ! (y ),..

hko(y) = C°k( - c ) kg0(y) + C l ( - c ) k- 1 g^(y)+, ... + < 2 й 34,Ы for к - 0, 1 , n — 1 ; СЦ = (J).

We shall prove that the function n-l

(4) u(x ,t ) = X **“ *(*» 0

k= 0

is the solution of the (L - C ) problem. We shall make use of the following Lemmas:

Lemma 1. Let the function v ( x, t ; y, s) satisfy the equation (D* — Dt — c)v = 0 for every (y, s)e(2; then

( D l ~ D t- c ) k( t h ) = \—l )kCjk\tJ~kv 0

/or к ^ j, for к > j, where j e N , k e N .

We omit an easy proof which can be carried by induction.

Let

(5) I ( x , t ) = — -

J

7t -00

Lemma 2. Let the function g as well as gik) be continuous, absolutely integrable over the real axis vanishing at ± c c for к = 1,2,..., 3n + 1; then the function I is continuous, bounded and has continuous and bounded derivatives D?D%I, p + 3q ^ 3n in Q.

P roof. Integral (5) can be represented as e~ci oo

I ( x , t ) = ---J g(y)Dyv ( x, t ; y,0)dy,

w h e re

Cauchy problem fo r the polyparabolic equation 297

(6) y(x, t; y, 0) = J о f(z)dz.

x-y

r 1/3

Integrating by parts, we obtain

l{ x , t) = Î g '(y )v(x, t; y, 0)dy,

therefore

e~ct 00

|/(x, oi < --- I \v\\g'\dy < Co,0,

Я — ocj

where C0>0 denotes a positive constant.

We shall prove now that the integrals appearing in /, differentiated under the sign of the integral are uniformly convergent. Since Dx v = — Dyv we get, integrating by parts

g - C I C O e ~ c t CO

--- J g’ (y)Dx v ( x, t ; y,0)dy = --- f g"(y)v(x, t; y, 0)dy.

Я — cc 7 1 —qo

Therefore

IDx I ( x , 01 < Clt0,

where C1>0 denotes a positive constant.

By similar estimations, using the identity Dx U — Dt U we can prove that

\Щ Щ 1(х, 01 < с м , where СР;<г denotes a positive constant, p + 3q ^ 3n.

Lemma 3. Let the functions g and g' be continuous and absolutely integrable over the real axis and vanishing at ± oo ; then

lim I ( x, 0 = g (x 0) as ( x , t ) - > ( x 0, +0), x0eR.

P roof. Let x £ (x b x 2), where Xi = x0 —<5andx2 — x0 + ^;^ = const > 0, Integral (5) may be written as

4

I ( x, 0 = £ h ( x , t),

Jlc=l

where.

j *i

h ( x , t ) = --- J g ( y ) V ( x , t ; y,0)dy,

Я — a n

{ \ x2

I i { x , t ) = - J L ^ s L J V(x,t; y,0)dy, я x i

298 J. Mi l e ws ki

1 ^{x 2}

h ( x , t) = ---f l g ( y ) - g ( x 0) } V ( x , t ; y,0)dy,

ft x j

1 00

U { x , t ) = ---- - j g { y ) V ( x , t ; y,0)dy.

f t x 2

The integral I 2 can be represented as

g(x0)e ^ct w<x>*2*‘) I 2(x, t) = —

f t V > ( x , X l , t )

where a ) ( x ,y ,t ) = (x — y)/t113. Therefore (see [1]) g(x0)

j f (z)dz,

lim I 2(x, t) as £2э(х, t) -> (x0, +0). Next

-cr x2

j f ( z )d z = g( x0)

J3(x ,t) = — S [ g ( y ) - g ( x 0) ] DyVi(x,t; y,0)dy xi

---e ctV! (x,t; у,0) 1д ( у ) - д ( х оУ\\хх21 +

1 *2

+ e J g'(y)v1(x,t; y,0)dy, xi

where

whence

c>(x,X2,t)

v i ( x , t ; y, 0) = J f (z)dz;

o>(x,y,t)

l / 3 ( x , t ) j ^ — l v t ( x , t ; y ^ j l l g t x j - g t x 0)| +

к

1 *2

+ — j W(y)\\vi(x,t; y,0)\dy ^ Cxd,

ft x!

Ci being a positive constant. If x-^Xq, we can choose Ô = |x — x0| and by the inequality just obtained

lim /3(x,r) = 0 as й э ( х , t) -> (x0, +0).

Similar argument shows that

lim I k(x, t) — 0 as й э(х, t) (xq, -4-0), k = 1,4.

Cauchy problem fo r the polyparabolic equation 299

Let now

1 00

K ( x , t ) = --- f g( y ) U ( x, c , y,0)dy.

К — oo

We shall need one lemma more

Lemma 4. Let the functions g and g{k) satisfy the assumptions as in Lemma 1, for k = 1,2,..., 3m + 1; me IS. Then

lim A mX (x ,t ) = g{3m)(x0) as £2э(х, t) -* (x0, +0), x 0eR.

P roof. Using Lemma 2 and the identity Dt U = D^U, we obtain 1 00

D™K(x, t) = ---f g ( y ) D? U ( x , t; y,0)dy

Я — 00

1 00

= ---1 g( y) Dl mDyv( x, t ; y, 0)dy, tl ^{— 00}

where the function v being defined by formula (6). Let us observe also that Dxv = —Dyv; hence integrating by parts 3m times gives

1 00

D ~ K ( x , t ) = ---f g<3" ' ( y ) U ( x , f , y,Q)dy.

Tt — oo

Therefore, by Lemma 3

lim D ? K ( x , t ) = 0(3m)(xo) as й э (х , t ) —►(xcq, +0).

Now we are able to prove the

Theorem. Let the functions gk(y) for k = 0 ,1 ,..., и— 1 as well as the functions # (y) for j = 1,2,..., 6n —3/c —2 be absolutely integrable over the real axis and vanishing at +oo; then the function и defined by formula (4) gives the solution of the (L - C ) problem.

P roof. By Lemma 2 the function и as well as its derivatives D?D%u are continuous in Q for p + 3q ^ 3n. By Lemma 1 и satisfies equation (1) in Q. It remains to prove that it satisfies the initial conditions (2). Indeed, by Lemmas 2 and 3

lim u{x,t) = g (x0) as £2э ( x , t ) —►(.Xq, ■+• 0).

Differentiating /с-times identity (4) with respect to t we obtain for 1 ^ k ^ n— 1

(7) Dku ( x , t ) — w0(x, t) + twl (x, t) + ... + tn~l wn^x(x, t),

w h ere

300 J. Mi l ews ki

(8) w 0 ( x , t) = D t U 0 ( x , t) + k D f 1 u 1(x, t) + ... + k\ u k ( x , t).

Moreover, the functions wk for к = l,2 ,...,n —1 are linear combinations of the integrals of the type (5). Hence by formula (7) and by Lemma 2 follows the condition

lim D*u(x, t) = lim w0(x, t) as й э ( х , t) -* (x0, +0), x0eR.

Substituting the functions u0, ul t uk defined by (3) into formula (8) we obtain

lim w0(x, t) = lim 1 00

— J 0*ООП**f; y , Q ) d y

^ — 00

as й;э(х, t) -* (x0, +0), x0eR.

9k (*o)

References

[1 ] L. C a t t a b r ig a , Un problem a al kontorno p er una equazione di ordine dispari, Analli del scuola normale superiore di Pisa fis. e mat. 13 (1959), p. 163-203.

[2 ] J. M ile w s k i, On a certain lim it problem fo r p o lyparabolic equation, Comment. Math.

20 (1977), p. 133-145.

[3 ] M. N ic o le s c u , E quatia ite ra ta a caldurii, Studii si cercetari mat., Acad. RSR 5 (1954), p. 3-A