non-linear parabolic problem associated with the penetration of a magnetic field into a substance
TRANSCRIPT
Mathematical Methods in the Applied Sciences, Vol. 16, 281-295 (1993) MOS subject classification: 35 M. 35 Q, 65 N
Non-linear Parabolic Problem Associated with the Penetration of a Magnetic Field into a Substance
Nguyen Thanh Long
Ecole Polytechnique. Dkpartement de MalhCmaliques, Ho-Ch-Minh Ville, Vietnam
and
Pham Ngoc Dinh Alain
Dkpartement de Mathkmatiques. Universitk d’OrlCans-BP 6759, 45067-Orleans Cedex. France
Communicated by W. Tornig
We study the following initial and boundary value problem:
u, - V * ( a ( j: (Vu1”dr)Vu) +f(u) = 0.
u = 0 on aQ U(X, 0) = u,,(x).
In section 1. with ug in L2(12), fcontinuous such thatf(u) + EU non-decreasing for E positive, we prove the existence of a unique solution on (0, T), for each T > 0. in section 2 it is proved that the unique solution u belongs to Lz(O, T; HA n H’) n L“(0, T; HA) if we assume uo in HA andf in C ’ ( R . W). Numerical results are given for these two cases.
0. Introduction
In this paper, we study the following initial and boundary value problem:
u, + A(u) + j ( u ) = 0, (X, t )ER x (0, T ) , (0.1)
u = 0 on an x (0, T ) , (0.2)
4x9 0) = uo(x), (0.3) where
01 70-42 1 4/93/0428 1 - 15s 1 2.50 Q 1993 by B. G. Teubner Stuttgart-John Wiley & Sons, Ltd.
Received 29 Augusl 1991 Revised 5 February 5 1992
282 N. T. Long and P. N. D. Alain
R being a bounded domain in RN with a sufficiently regular boundary XI . The assumptions on the functions a and f needed for our purpose will be specified later. This problem was considered by Laptev [2] with f = 0; its physical interpretation is that of the penetration of a magnetic fied into a substance. To describe the phenomena that are produced in massive conductors placed in an externally varying magnetic field H, in the case where the temperature f3 rises and has an effect on the resistivity p(0) , the following system of equations was first proposed in [l]:
aH/at = - (c2/47t) rot[p(8) rot HI,
c,(e)ae/at = p(e) 1(c/4n)rot HI^, div H = 0, (0.5)
(0.6) where c is a constant and C,(f3) is the specific heat capacity of the substance, which, in general, also depends on temperature.
Laptev transforms (0.5) and (0.6) into one equation by introducing the function s(f3):
Takiqg into account that the function s(0) is monotonically increasing, system (0.5) and (0.6) can be rewritten as
div w = 0, (0.9) where a(s) = (c2/4n) p[&s)], w = ( c / ~ A ) H and #(s) is the inverse of the function s(0). Assuming that the field w has the form
w = (O,O, 4, I4 = 4% y, 0, (0.10)
equation (0.8) takes the form (0.1) in two dimensions while the second equation of (0.9) is automatically satisfied. If the function p ( 0 ) is uniformly bounded, then a(s) possesses an analogous property. This phenomenon is observed in both semiconductors and plasmas.
In [2] Laptev establishes existence and uniqueness theorems for equation (0.1) with f = 0, uo in HA (Q), and a and a’ continuous under the conditions
(0.12)
The paper consists of three sections. In section 1, under conditions (0.5) and (0.6) with uo in L2(R),fcontinuous,j(0) = O,f(u) + E u non-decreasing for E > 0 ‘small’ and taking bounded sets of L2(Q) into bounded sets of L2(R), we prove the existence of a unique solution on (0, T), for each T > 0. We also prove an exponential decay of the solution to zero as t tends to infinity.
In section 2 we strengthen the hypotheses and assume, as in [2], uo in Hl(Q); then with f in %“(R, R), J(0) = 0, f’ 2 E, E > 0 and, under a certain local Lipschitzian condition onf , an existence and uniqueness theorem is proved on (0, T), for each T > 0, generalizing the results of Laptev.
In section 3 we present the numerical results.
Non-linear Parabolic Problem 283
1. Solution of the problem
Let
L2 = L2(Cl), HA = HA@), H z = H2(Q).
Here HA@) and H 2 ( Q ) denote the usual Sobolev spaces on R. Let (.;) denote either the L2 inner product or the pairing of a continuous linear functional with an element of a function space. Let 11 * \ I x be a norm on a Banach space X and let X * be its dual. We denote by Lp(O, r X), 1 < p d co , the space of functions f o n (0, T ) to X such that
IlfIILP,O.T.XI = (s,' l l f~f) l l~df) l 'p < 00 for 1 d P < 00.
Ilf IIL",O. T : X ) = ess SUP(0. T) Ilf(t) Ilx
We make the following assumptions:
Al : uo in L2. A2: a in %" (R, , R) satisfies the following conditions:
for p = co.
We denote the norm in LZ(Cl) by I1 * 11.
(i) b = (1; slaf(s)i2ds)li2 < + 00, (ii) there exist ao, a , > 0 such that
26 < < U ( S ) < (11 V, 2 0.
A3: f i n %(R, R) satisfies the following conditions:
(9 f ( 0 ) = 0, (ii) f(u) + EU is non-decreasing, with E such that
0 < E < (a0 - 2 b ) / C i ,
where Co > 0 such that
l l u l l ~ 2 d Co IIVuIILz VUEHA.
(iii) f(u) = f o u, for u in L2, takes bounded sets into bounded sets.
We then have the following theorem.
Theorem 1. Let assumptions A l , A2 and A3 hold. Then the problem dejned by (0.1)-(0.3) has a unique solution in L"(0, T; Lz) n L2(0, Z HA), for each T > 0.
The proof of Theorem 1 is a combination of compactness and monotonicity arguments and consists of several steps.
First we shall require the following two Lemmas:
Lemma 1. Let A be the operator dejned by (0.4). Then we haue the following inequality:
(ao - 2b) joT I1 Vu - Vu 11' dt < ( A u - Au, u - u ) dt
T loT
G (al + 2 b ) j 0 IIVu - Vull'dt, (1.1)
Vu, U E L2(0, T; HA), for each T > 0.
284 N. T. Long and P. N. D. Alain
Proof. Let
Using the formula
F(u) - F(u) = d/do(F(u + Ow))do, w = u - u sd
s:
and assumption A2 we obtain
J = j: a(s) Vw do + a’(s) (ds/da) * (VV + oVw) do, lo1 where
q = S(X, t, o) = IVU + oVwlZ d7.
Consider jOr ( J , w)dt. By (1.3),
IoT (5, Vw)dt = IOT 1’ a(~)IVw1~dadxdt + K,
K = IoT jn l: a+) (dslda) - (VV + OVW) . Vw do dx dt.
n o
where
We claim that
1K1 < 2b 11 VW l l i 2 (p), with Q = R x (0, T ) .
We have
I K I 6 I I v W I I L ~ ( Q ) Jb’ llJi 112L2(Q)dgr
where
IJJl ( I i 2 ( Q ) = Jn IoT la’(s)12 I(ds/do)12 IVV + oVw12dxdt.
Inserting (1.4) into (1.9) we obtain, after inverting variables z and t.
l l J l 1 1 2 2 ( Q , 641ndx]or IVwl2dz[ s(x.r,d S(X, T,a)la’(rl12rldtl
6 4 jn dx loT IW2 ds [y la’(rl)12 rl drl,
from which (1.7) follows. Consequently, from (1.7) and assumption A2 we have
a0 IlVwl12z(Q) - l K I ( J , Vw)dt a1 llvw IIL2(Q) + IKl,
(1.10)
(1.11)
Non-linear Parabolic Problem 285
which implies inequality (1.1) since
( A u - Av, u - v)dt = (J,Vw)dt, w = u - ti jOT
vu, V E L2(0, T; HA). j O T
Lemma 2. For each u, w E Lz(O, T; HA) we have the following:
lim joT ( A ( u + Aw), w) dt = ( loT Au, w) dt. ,Id m
and
(1.12)
there exists a sequence (A,}, A, 2 0, lim A, = 0, such that
lim joT (f(u + Amw), w) dt =
r n d + m
(f(u), w ) dt. (1.13) m-tm !OT
Proof: Equality (1.12) is immediate, thanks to Lemma 1. Let {A,}, 0 < 1, < 1, be a sequence such that lim A, = 0. By using the continuity of
I, we can extract from the sequence {A,} a subsequence, still denoted by {A,,,} such that
(1.14) f(u + A, w ) +f(u) a.e. in Q, for each u, w E L2(0, T; HA). Let g,(u) =f(u) + EU, E > 0. Since g,(u) is non-decreasing, we have
Igc(u + Amw)l g,(lul + Iwl) - gc( - IIO - Iwl). (1.15)
Therefore,
If(u + Amw)l G f(lul + Iwl) - f( - Iu1 - Iwl) + 3&(lul + Iwl) = F(lul + Iwl). (1.16)
Since u, w E L2 (0, T; HA), it follows from (1.16) that
Ilf(u -k 1mW)IIL2(Q) G 11 F(Iul + IwI)llL2(Q) = c1, (1.17)
where C1 is a constant independent of m. Equations (1.14) and (1.17) enables us to apply Lemma 3 (see section 2), which shows that
f(u + A, w) +f(u) in L2(Q) weakly for m + 00 .
We turn now to the Proof of Theorem 1.
Proofof Theorem 1. Consider a special basis of H;, w l , w2, . . . , w,, . . . , formed by the eigenfunctions of the Laplacian A. Let (wl, w2, . . . , w,) be the linear space generated by w l , . . . , w,.
QED
n
U y t ) = c C j , ( t ) wj j= 1
be a solution of the following system:
(1.18)
(u?', wj) + (a( f: IVu(n)/2dr) VU("), Vwj) + (f(u'")) wj) = 0
f o r j = 1,2,. . . ,n, (1.19)
u(")(O) = uOn + uo in L2 strongly. ( 1.20)
286 N. T. Long and P. N. D. Alain
Note that such a solution clearly exists on a sufficiently small interval [0, T,,]. The a priori estimates which follow allow us to take T,, equal to T.
1.1. A priori estimates
j = 1, 2, . . . , n and integrating with respect to the time variable, we have 1. Multiplying thejth equation of system (1.19) by cjn(t), adding these equations for
+ j: (f(u(")), u(") ) ds = f II uOn II '.
Assumptions A2(i) and A3(i), (ii) imply that
f (I d")( t ) ( 1 + a. 1: ( 1 VU(") I( ds
,< 1 1 ~ 0 ~ [ I 2 + (00 - 26) 11 Vd") 11 ds. 1: Finally, from (1.22) we obtain the first a priori estimate
I I ~ ( " ) ( t ) 1 1 ~ + 46 (IVU(")I12ds G C2, s: where C2 is a constant independent of n.
2. Let A be the operator defined by (0.4),
A:L2(0, T; H i ) -+ L2(0, T; H - I ) ,
and satisfying
(Au,u)dt = I O T ( a ( J:lBu12d~)Vu,Vv)dt
for dl u, u in ~' (0, T; HA). loT
From (1.24) and assumption A2 we easily deduce that
(1.21)
(1.22)
(1.23)
( 1.24)
G a1 IIU IIL'(0,T:H;) V u E L2 (0, T; HA). (1.25)
Combining (1.23) and (1.25) we obtain
11 Ad") I ~ L Z ( O , T ; H ; ) ai J(Cd46)
3. By assumption A3 (iii) and (1.23) it follows that
( 1.26)
ll.w"))Il G c3, (1.27)
By (1.23), (1.26) and (1.27), we can extract a subsequence of {u(n)} , still denoted by
u(") -+ u in Lm(O, T; L 2 ) weakly*, (1.28)
where C3 is a constant independent of n.
{u(n)}, such that
Non-linear Parabolic Problem 287
u'") -+ u in L2(0, T; H A ) weakly*, (1.29)
Ad") -+ x in L2 (0, T; H - I ) weakly,
f(u'")) -+ x1 in L"(0, T; L2) weakly*,
u(")(T) -+ 5 in L2 weakly.
(1.30)
(1.31)
(1.32)
1.2. Limiting process
integrating with respect to the time variable, Passing to the limit in (1.19), we obtain, after multiplying by q ~ . 9 ( [ 0 , T I ) and
( 5 , W j > V ( T ) - (UO, wj> d o ) - PT PT
by virtue of (1.24). Equation (1.33) implies that
u, + x + x1 = 0 in L2 (0, T; H - I).
(1.33)
(1.34)
Taking the inner product of (1.33) with q ~ . 9 ( [ O , T I ) , we get
u(T)cp(T) - u(0) q(0) - (x + Xl)qdt = 0. (1.35)
It follows from (1.33) and (1.35) that
(1.36)
(1.37)
Let
x', = loT (A;(") - Ju, u(") - u ) dt for all u in L2(0, T; HA), (1.38)
where
i u = Au + f(u). (1.39)
f ( u ) + EU being a non-decreasing map, it follows that
From (1.40) and assumption A3 (ii) we deduce that
The relation
(1.41)
( 1.42)
288 N. T. Long and P. N. D. Alain
implies, by passing to the limit
= joT (x + x I , u)dt.
Then, by virtue of (1.43), (1.29) and (1.31) we obtain
(1.43)
(x + x1 - z u , u - u ) dt for all u in L2(0, HA). (1.44)
In (1.44), if we choose u = u - A,,, w, A,,, being the sequence defined in Lemma 2, and use Lemma 2, we conclude that
n. -a ,
fT J (x + x1 - Au - f (u) , w ) d t > O for all w in ~ ~ ( 0 , T; HA), 0
( 1.45)
i.e.
1.3. Proof of uniqueness
w = u - u is a weak solution of the problem Let u and u be two weak solutions of the problem defined by (0.1)-(0.3). Then
( 1.46) w, + z u - z u = 0,
w = 0 on aRx(0, T ) , (1.47)
w(0) = 0. (1.48)
Taking the inner product of (1.46) by w and integrating with respect to the time variable, we get
P T
(1.49)
which implies that w = 0, by virtue of (1.41).
Now we consider asymptotic behavior for t tending to infinity. More precisely, we have the following proposition.
Proposition 1. Under the hypotheses ofthe Theorem 1, ifu(t) is the unique solution ofthe problem defned by (0.1)-(0.3), then we haue
I1 u(t) 11 G 11 uo 11 e-kl jor each t > 0, where
k = (ao/C;) - E > 0.
Pro05 From (1.19) we deduce that
Non-linear Parabolic Problem 289
2. Existence and uniqueness theorem
In this section we strengthen the hypotheses and assume, as in [ l ] , that
B1 u o € H ; . B2 a in %'(R+, R) satisfies
(i) b = (s(a'(s)("s)li2 -= + a, (ii) there exist a,, a, > 0 such that
j: 2b max(1, c) < a, d a(s) d at Vs 2 0,
where c is a positive constant satisfying
la2u/axiaxj12dx < c2 jlAu(12 VUE Hh n H 2 , s" fl i . j = 1
while f satisfies the following condition:
B3 (i) f E%"(R, R),f(O) = 0,f' 2 - E, E > 0 not necessarily small, (ii) for each bounded subset B of HA, there exists K B > 0 such that
M Y ) - f(z)II d KBII VY - VZII V Y , z E B.
Theorem 2. Let assumptions Bl-B3 hold; then the problem dejned by (0.1)-(0.3) has exactly one solution, u EL'(O, T ; HA n H 2 ) n L"(0, T ; HA) such that u, E L2(Q), Q = x (0, T ) f o r each T > 0.
ProoJ: Proceeding as in the proof of Theorem 1 , we consider equation (1.21) and get, after using assumptions B2(i) and B3(i),
11 u(")(c) /I + 2ao joT 11 VU(") (s) /I ds d C, + 2~ 11 u(")(s) I( ds, joT (2.2)
where C4 is a constant independent of n and t .
form a bounded set in the space It follows from (2.2) and Gronwall's lemma that the elements of the sequence {u"(t ) ]
(2.3) ~"(0, T; L2) n ~ ~ ( 0 , T; HA).
290 N. T. Lung and P. N. D. Alain
2. Let Q1 = R x(0, Tl) with 0 < TI < T. Using the fact that the wj(x) are the eigenfunctions of the Laplacian A, equation (1.19) can be rewritten as
(~f", - Awj) + ( A d " ) , - Awj) + ( f ( d " ' ) , - Awj) = 0
f o r j = 1 , 2 , . . , n . (2.4)
Multiplying each equation in (2.4) by cj,,(t) and summing up with respect to j, we have, after integration from 0 to T1,
loT' (up' - Ad"') + (Ad") , - Ad")) + (f(u'"I), - Au("') = 0. (2.5) s:' l' We have
N AU(") = - C @/axi) [ u ( ~ ( n ) ) (aP/axi ) ] = - ~ d " ) - u ' ( s ( " ) ) ( ~ s ( n ) . vd"))
(2.6) i s 1
and (f(~'"'), - Ad"') = (f'(~'"') VU'"), VU'"')
since u'") = 0 on aR x (0, T). Hence, by (2.6) and (2.7) we get from (2.5)
(2L)IIV~'")(Tl)1I2 + jar' jn u ( ~ ( ~ ) ) l A d ~ ) l ~ dxdt + lor' jn f ' (un) IVu(")12 dx dt
= lor' (P,, Ad") dx dt + f (IVuOn (I2, (2.8)
where
and
pn(x, t ) = - a'(.$'"') Vs("' * V P .
)I q n IIL+Q,, G 2hc IIAu'") IILz(Q,,.
(2.10)
Define R ' c R such that iz' is compact and let Q2 = R' x (0, Tl). We claim that
(2.1 1)
Indeed, from (2.10) we have
From (2.9) we obtain
I V S ( " ) I ~ G 4s(") 1; ( la2u(")/axi i)xjI2 d7. i. j = I )
Combining (2.12) and (2.13) and after inverting variables 7 and t, we get
(2.12)
(2.13)
Non-linear Parabolic Problem 291
G 4b2 loT1 ( l a w / a x , axjt2 d.x dr, (2.14)
with sl = s(")(x, t) and s2 = s(")(x, T I ) . Inequality (2.14) implies (2.1 l), c being the constant defined by (2.1).
i, j = 1 )
joT1 From (2.8), (2.11) and assumptions B1 and B3(i) we deduce that
I( Vu(")( TI) 11 + 2(ao - 2bc) IIAu(") 11' dr
< C5 + 2~ 11 Vu(") 11' dr for each TI, 0 G T1 < T, (2.15) s:' where C5 is a constant independent of n and t . Inequality (2.15) implies that
II Au'") )I Lz(Q) + I\ VU'") 11 < CT, Q = SZ x (0, T ) , (2.16)
CT indicating a constant depending on T. Since we have
l IAu(") l lLz(Q) < ( a l + 2bc)llAu(")llLz(Q),
Ad") is bounded in LZ(Q).
Taking pz = CT implies, with assumption B3(ii), that
II .f(u'"') II L2 G P * K,
since JJW") 1) < p . The last a priori estimate follows from the inequality
(2.17)
(2.18)
IIUI") IlL'(Q) 11 llL'(Q) + llL'(Q) d c T , (2.19)
Consequently, by (2.16) and (2.19), we can extract from {u(")} a subsequence, still C7. always indicating a constant depending on T.
denoted by { u(")}, such that
u(") + u in L"(0, T; HA) weakly*,
u(") + u in L2(0, T; HA n H,) weakly*,
(2.20)
(2.2 1 )
u?) + u, in L2(Q) weakly*. (2.22)
Using a lemma on compactness [2] applied to (2.21) and (2.22, we deduce that
the set {u (" ) } contains a subsequence, still denoted by {u'")}, strongly convergent to u in L2(0, T; HA). (2.23)
We shall now require the following lemma [2].
Lemma 3. Let 0 be an open bounded set c RN and g,,, g E Lq(0) (1 < 4 < co ) such chat 11 g, IILq(B) < C, where C is a constant independent of p, and g,, -+ g a.e. in 0. Then, g,, + g in Lq(0) weakly.
By the Riesz-Fischer theorem, we can extract from the sequence {u(" ) } a subsequ- ence {u(")}, such that
dn) + u a.e. in Q = 0 x (0, T ) . (2.24)
292 N. T. Long and P. N. D. Alain
I t follows from (2.18) and Lemma 3 applied to the functionfthat
f(u'")) + f(u) in L2(Q) weakly. (2.25)
We note from the strong convergence of the sequence {d")} to u in L2(0, T: HA) that rr PI
J (Vu'")l2 dr -, J 1Vu12 dr in L'(Q) strongly 0 0
since we have
< T 1) V I P ) + VU IILz(Q) * ( 1 VU'") - VU IILz(Q).
By assumption B2, (2.23) and (2.26) we deduce that
a ( 1: [Vu(")I2 dr) Vu'") + a( 1: IVuI2 dr) * Vu a.e. in Q.
On the other hand, we have
Taking (2.27) and (2.28) into account, and by virtue of Lemma 3, we have
(2.26)
(2.27)
(2.28)
(2.29)
Integrating (1.19) on (0, T) and using (2.22), (2.25) and (2.29), we finally obtain
jOT ( u l , u ) dr + (a@) Vu, V v ) dr + ( f ( u ) , v) dt = 0 VUEL'(O, T ; Hh),
(2.30)
u, + Au + f ( u ) = 0 in L2(Q). (2.3 1)
JOT ]Or
i.e.
Let w ~ L ~ ( i z ) . Since u(") and u are in W(0, T: L2), we have
loT (@), ( T - t) w) dr = - (uOn, w) T + (d"), w ) dr, (2.32) I: which implies, passing to the limit for n + co ,
f T PT
J ( u ~ , ( T - t ) W ) dt = - ( ~ 0 , W ) T + J (u, W ) dt, 0 0
from which the initial condition
u(0) = uo
follows since
loT ( t i l , ( T - t) w) dt = - T(u(O), W ) + (u , W) dt. 10T
QED
Non-linear Parabolic Problem 293
2.1. Uniqueness of the solution
of the problem defined by (0.1)-(0.3) since Proceeding as in the proof of Theorem 1, we deduce the uniqueness of the solution
II w( Tl [ I 2 + 2(ao - 2b) loT’ I( V w (1 dt G 2.5 loT’ 11 w 11 dt,
W = U - V , O < T , < T .
3. Numerical results
Consider the problem defined by
U, - a / a ~ [ a ( J : u:(x, r)dr)uX] + f ( u ) = F(x, t), 0 < x < 1, (3.1)
u(0, t ) = u(1, t ) = 0,
u(x , 0 ) = sin nx, (3.2)
(3.3) where a(s) = (s + 2)/(s + 1) and F(x, t ) €L2(Q), Q = (0, T ) x (0, 1).
We have
b = (1; sIat(s)”dg)lil = 1/J6
and
1 < a(s) < 2.
In (3.1) the functionf(u) is taken such that assumptions Al-A3 and Bl-B3 are fulfilled.
1 . We choose in equation (3.1)
F ( x , t ) = e-‘sin nx [(27r2 - 1 ) + (n2 - 2 ) cp + (n2 - 1) c p 2 ] / ( 1 + ( P ) ~
+ (e- sin K X ) ” ~ + 1/3 (e-lsin nx), (3.4) where
cp = ( 1 - e-”) (n2 cos’ nx)/2,
f(u) = u1I4 sgnu + EU, E = 1/3. (3.5) As a matter of fact, in assumption A3 we have a. = 1, b = 1/,/6 and Co = 1J2, which implies that we must choose E in the interval (0,0.368). The exact solution of the problem defined by (3.1)-(3.3), with F ( x , t) andf(u) defined in (3.4) and (3.9, respec- tively, is u(x, t) = e-‘ sin I I X . To solve numerically this problem we consider the non-linear differential system for the unknowns uk(t) = u(&, t ) , xk = kh, h = 1/N:
duk/dt - ( g ( x k , t) + 2) / (g(xk , t ) + l ) Uxx(Xk,t)
+ ( l / (g(Xk, t ) + (%/ax) (xk, t ) U x ( X k r t ) + f ( U k ) = F ( x k , t),
uk(0) = sinnkh, k = 1 , . . . , N - 1, (3.6)
294 N. T. Long and P. N. D. Alain
0.6
0.5
0.4
0.3
0.2
0. L
0.
0. 0.2 0.4 0.6 0.8 1.0
T = I / 2 __ hPPROX.SOL - E Y AC. SOL
Fig. 1
0.0 .
0. 0.2 0.4 0.6 0.8 1.0
(3.7)
X
X
Non-linear Parabolic Problem 295
We start with the initial conditions
to = 0, a k . 0 = sinakh, h = 1/N, (3.8) U x ( X k , 0) = XCOS nkh, U , , ( X k , 0) = - 7c2 Sin nkh
fork = 1 , 2 , . . . , N - 1. (3.9)
The linear differential system (3.7)-(3.9) is solved by searching its eigenvalues and eigenfunctions. For a step h = 1/11, we obtain the curves in Fig. 1 for the approximate solution U k ( t ) and the exact solution s o l k ( t ) , k = 0, 1,. . . , 11 and for the time T = SO/lOO. Always with a step h = 1/11 and for times T = 1/50, 1/10, 1, 2, 3, we obtain the curves in Fig. 2 for the approximate solution u k ( t ) , k = 0, 1, . . . , 11, which show that the numerial solutions tend rapidly to zero when the time T is ‘great’ enough.
2. The functionf(u) = u3 satisfies the assumption B3 of Section 2. We take
F(x, t ) = e-‘sin ax [2n2 - 1 + (aZ - 1) cp + (n2 - t)cp2]/(1 + cp)’
+ (e-’sin
where cp = (1 - e-”) a2 cos2 ax/2 . The exact solution of the problem defined by (3.1)-(3.3), with F(x, t) in the form
(3.10) andf(u) = u3, is u(x, t ) = e-lsin ax. We solve in the same way as previously the linear differential system associated with the non-iinear differential system and obtain similar curves as under 1 above.
Acknowledgement
The authors wish to thank Professor W. Tomig for his constructive and useful remarks.
References
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