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QUARTERLY OF APPLIED MATHEMATICSVOLUME LIX, NUMBER 3SEPTEMBER 2001, PAGES 507-528
STEADY STATES FOR A ONE-DIMENSIONAL MODELOF THE SOLAR WIND
By
JACK SCHAEFFER
Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA
Abstract. A one (space) dimensional Vlasov equation is used to model the solarwind (a collisionless plasma) as it moves past an applied magnetic field (an obstacle).The goal is to understand physically reasonable steady states for this situation. Whenthe applied magnetic field is sufficiently small, appropriate states are constructed.
1. Introduction. Consider the following simplified model of the solar wind:
dtf + vidxf - (E + v2B)dVlf + v1BdV2f = 0,p(t,x) = ff(F(vi,v2) - f(t,x,v1,v2))dv2dv1, (1.1)E(t,x) = pdx.
Here / gives the density in phase space of mobile negative ions with mass one and chargeminus one. This is assumed to depend only on time, scalar position x = x\, and velocityv = (i>i,i>2)• The positive ions have charge positive one, but are taken to have infinitemass so that they form a fixed background density given by F(v\,v2) (see [13]). We takeF to be smooth and compactly supported with
F(vi,v2) =0 if Vi < 0.
B = B(x) is a given magnetic field which we assume is compactly supported. The speedof the solar wind is a few tenths of one percent of the speed of light; so the omission ofthe self-consistent magnetic field due to the plasma is reasonable. We consider (1.1) withthe initial condition
f(0,x,v) = F(v) + g{x,v) (1.2)
where g has compact support and zero average value.
Received September 29, 1999.2000 Mathematics Subject Classification. Primary 35Q60; Secondary 86A25.Supported in part by NSF DMS-9731956.
©2001 Brown University507
508 JACK SCHAEFFER
The goal is to understand the possible steady behavior of (1.1), (1-2). Thus, weconsider the steady state problem
vidxf - (E + v2B)dVlf + viBdV2f = 0,Pix) = f(F ~ f)dv, (1.3)
,E(X) = U-ocPdx~ 1ST Pdx-We seek physically appropriate conditions at x = — oo. We consider
lim E(x) = 0 (1.4)x—►— OO
and
lim / = F uniformly on v\ > 0. (1.5)x—► —oo
Note that (1.5) imposes no restriction for v\ < 0. We will show that solutions of thedynamic problem (1.1), (1.2) satisfy (1.4) and (1.5) for any fixed t > 0. We will also seethat conditions (1.4) and (1.5) are appropriate when B is sufficiently small, but it is notclear that they are for B large.
The plasma physics literature on collisionless shocks is extensive, for example, see [3,16]. Many mathematical works consider the existence of steady states ([2, 6, 15]) andstability of steady states ([5, 7, 8, 14]) in collisionless plasmas. This work differs fromthose mentioned above in that the applied field, B, is included to model flow past anobstacle. Then we seek appropriate conditions at infinity to capture the steady behaviorof the dynamic problem. We mention also two related references [9] and [12].
Section 2 sketches the existence of smooth solutions to the dynamic problem (1.1),(1.2). Section 3 shows some consequences of (1.3), (1.4), and (1.5) and also constructssolutions when B is sufficiently small. Section 4 presents numerical results where thesolution of (1.1), (1.2) tends as t —> +oo to a solution of (1.3), (1.4), (1.5). Finally, theappendix contains some of the details from Sec. 3.
2. The dynamic problem. Define
A(x) = f B{x)dx.J — OO
For a smooth solution of (1.1) define (X(s,t,x,v), V(s,t,x,v)) by
Vi, X(t,t,x,v) = x,
= -(E(s,X) + V2B(X)), Vl(t,t,x,v)=v1, (2.1)
dXds
dVids
Then
C^- = V1B(X), V2(t,t,x,v) = v2.as
JL(V2 - A(X)) = 0 and ±{f{x,X,V))= 0.
ONE-DIMENSIONAL MODEL OF THE SOLAR WIND 509
Theorem 2.1. Let B £ Cq(IR), F e Cq(K2) with F > 0, and g € Cg(K3) with
gdvdx = 0II'and
+ 9 > 0.Then (1.1), (1-2) has a unique C1 solution for all £ > 0 with
(x,v) ' ^ f(t,x,v) - F(v)
compactly supported for every £ > 0.The proof of existence is not immediate because /(0, x, v) does not decay as \x\ —+ oo
and hence the mass and energy are infinite. Uniqueness and local-in-time existence maybe handled as in [1], where the iteration
'dtfn+1 + vidxfn+1 - (En + v2B)dVlfn+1 + VlBdV2fn+1 = 0,pn+i =ff(F-fn+1)dv2dv1,En+i = 11- oo pn+idi - \ r pn+idi
is used. We omit this here since it has become standard. To continue the local solutionin time it is sufficient that
R(t) = sup{ |a;|: f(s, x, v) — F(v) ^ 0 for some »eR2 and s £ [0, £]}
-I- sup{|a:|: B(x) ^ 0}
remains finite. Thus, we will focus on obtaining an a priori bound on R(t).
Note thatd
and
so
Hence
and
Also, if we take
then
dJJU-F)dvdx = 0
J J (/(0, x, v) - F(v))dv dx = 0;
J J (/ — F)dv dx = — J pdx — 0.
■y rX /"OO pX
E=2LPdi~ 2 J, "di'L
|x| > R(t) => E(t,x) = 0.
j = J(F~ f)vidv,
dtP = - J dtfdv = J vxdxfdv = -dxjx\
pdxOO
510 JACK SCHAEFFER
SO
Next
so
Let
Then
Hence
d,E/X pX
dtpdx = - dxjidx = -j i--OO J — OO
rR(t) i-
= / / \v\2dtfdv dxJ-R(t) J
rR(t) r
= / \v\2(-vidxf+ [E + v2B]dVlf-viBdV2f)dvdxJ-R{t) J
/pR(t) nf\v\2v1dv\R_{^)(t) - / 2v1EfdvdxJ — R(t) J
rR(t)
-R(t) .
<j\dv 1 dx
= -jt (^j E2dx^j -2 J Fvidv J Edx;
d_ ( rRwdt / / f\v\2dvdx+ / E2dx
J-R(t) J J }
= —2 J Fv\dv J Edx.
rR(t) p r
Q(t) = / / /|f|2rft; dx + / E2dx.J-R(t) J J
,-R(t)\Q'(t)\<C \E{t, x)\dx
J-R(t)
< C\ 2R(t) / E2{t,x)d.V
S«!C>< Ci?2(i)
ONE-DIMENSIONAL MODEL OF THE SOLAR WIND 511
and2
It
/-//: c
<-JX C dv2 dv i + r 2k< Cr + r~2k.
Next using (2.4) we have
rR(t)/n(t) p/ \F-f\dvdx
■R(t) J-R(t)rR(t) r
<CR(t)+ / / fdvdxJ-R(t) Jfm i
< CR(t) + / Ck*dxJ—R(t)
R(t)< CR(t) + CR$(t) IJ kdxR(t) ,
(2.2)
Q(t) < (qH0) + J CRk*{r)dT
< C{Q(0) + t2R(t))< C{C + t2R(t)) < C( 1 + t2)R(t).
Next, we bound the V2 support of /. Suppose f(t,x,v) ^ 0. Then
f(t,x,v) = f(0,X(0,t,x,v),V(0,t,x,v)) / 0.
So
\V2{®,t,x,v)\ < C.
But
V2 - A(x) = V2(0,t,x,v) - A(X(0,t,x,v)).
So
\v21 = 1^(0,t,x,v) + A(x) - ^4(X(0,£,:e,u))| < C. (2.3)
We now use an energy estimate from [10, 11]. Let
k = J f\v\2dv.Note that for any r > 0
rcf dv2 dv 1
Taking r = k1'3 yields
J fdv<Ck5. (2.4)
512 JACK SCHAEFFER
But since
kdx < Q(t),R(t)
using (2.2) yields
\E{t,x)\ <C{l + t2)R(t). (2.5)
Consider (t,x,v) with F(v) — f(t,x,v) ^ 0. If f(t,x,v) ^ 0, then
f(0,X(0,t,x,v),V(0,t,x,v)) ^ 0.
So
|Vx(0 ,t,x,v)\<C
and hence by (2.5) and (2.3)
\Vi(s,t,x,v)\ <C+ [ \(E+ V2B)\(TiX(T,t,x,v))\dTJo
<C + Cs + C{l + s2) [ R(r)dTJo
for 0 < s < t. If f(t,x,v) = 0, then F(v) ^ 0. So |w| < C and hence by (2.5) and (2.3)
\Vi(s,t,x,v)\ = Vl - [ {E + V2B)J S
dr(t,X (T,t,X,V))
t
<
Thus, in both cases,
C + C(t- s) + C(l + t2) f R(t) dr.J S
|Vi(s,t,x,t;)| <C+ Ct + C{l + t2) f R(r)dr = S(t) (2.6)Jo
for 0 < s < t. S(t) is defined by (2.6).Define
R(t) = R(0) + I S(r)dTJo
and claim that R(t) < R(t). Define
H(t) = sup{|F(t;) - f(t, x, »)|: \x\ > R(t) and v € R2}
and note thatF(v) - f(t,x,v) = F(v) - f(0,X(0,t,x,v),V(0,t,x,v))
= F(V(0, t, x,v)) — /(0, X(0, t, x, v), V(0, t, x, v))
- [ dVlF(V)(E(s,X) + V2B(X))ds (2-7)J 0
+ [ dV2F(V)VlB(X)dsJo
ONE-DIMENSIONAL MODEL OF THE SOLAR WIND 513
where we abbreviate X = X(s,t,x,v). Now if f(t,x,v) - F(v) ^ 0 and |a?| > R(t), thenby (2.6) we have
\X(s,t,x,v)\ =
Hence, since R(s) > R(0),
f Vi(r, t,x, v)dr > R(t) — j S(t)<It = R(s).J S J S
B(X(s,t,x,v)) = 0.
Also
|X(0, x, u)| > i?(0) = il(0);
so
F(V(0,t,x,v)) - f(0,X(0,t,x,v),V(0,t,x,v)) = 0.
Thus, (2.7) yields
\F(V) ~ f(t,x,v)\ < C [ \E(s,X(s,t,x,v))\ds. (2.8)J 0
If X > R(s), then using (2.3) and (2.6) we have
|£(S,X)| = / p(s,y)dyJ — OO
-J J(F{v) - f{s,y,v))dvdy< R(s)H(s)CS(s).
Similarly, if X < —R(s), then
|E(s,X)| < R(s)H(s)CS(s);
so by (2.8)
It follows that
|F(v)-f(t,x,v)\<C [ R(s)H(s)S(s)ds.Jo
H(t) < C f R(s)H(s)S(s)ds.Jo
By Gronwall's inequality, H(t) = 0 for all t > 0. The claim that R{t) < R(t) now follows.Finally, by (2.6)
R(t) < R(t) = R(0) + f S(r)dTJo
= i?(0)+jf (c + Ct + C(1 + t2) R(s)d^jdr
< C(1 + t + t2 + t(l + t2)) I R(s)ds.Jo
So by Gronwall's inequality, a bound on R(t) follows.
514 JACK SCHAEFFER
3. The static problem. As before we assume
Fe^(t2), F > 0, VI <0=> F{v) = 0. (3.1)
We define G by
F(v) = G(±\v\2,v2)IVl>0
(where IVl>o = 1 if ui > 0, 0 else) and assume
F(v) = 0 if ^\v\2 < ^min or |u2|>S (3.2)
with £m;n — \S2 > 0. We also assume B £ Cq(K) with
B(x) = 0 if x £ (0, L). (3.3)
Now we seek solutions of (1.3) with
/ > 0, (3.4)
p bounded, (3.5)
E integrable on (—oo,x) for all ieR, (3.6)
and with (1.4) and (1.5). By (3.5) E is Lipschitz continuous. So we may define (X(s, x, v),V{s,x,v)) by
= vi, X(0,x,v)=x,
dVi-± = -(E(X) + V2B(X)), V1(0,x,v) = v1, (3.7)as
dVo~^=V1B(X), V2(0,x,v) = v2.as
We will interpret the Vlasov equation as the requirement that f(X(s,x,v), V(s,x,v))is constant in s and not assume that / is continuous. To motivate this, consider thefollowing example. Let U(x) be a smooth nondecreasing function with
' U(x) = 0 if x < 0,0 < I/(x) < 1 if 0 < x < 1,1 = U(x) if 1 < x.
Now we take f(x,vi) = 1 (drop v2) on the set
{(x,vi): vi > 0} U {(x,ui): \v\ + U(x) < 1}
and f(x,v\) = 0 at all other points. Then / is constant on characteristics, i.e.,f(X(s,x,vi), V(s,x,vi)) is constant in s where
as
Also
F(v\) := lim f(x,vi) = 1 for V\ > 0
ONE-DIMENSIONAL MODEL OF THE SOLAR WIND 515
is smooth for v\ > 0; so the "upstream condition" (1.5) is smooth. However, / isdiscontinuous at each point (x, 0) for x > 1.
Define
and
Then
and
Let us denote
and
d_ds
U(x) = — f E(x)dxJ —OO
A(x) — f B{x)dx.J —OO
)^\V(s,x,v)\2 - U(X(s,x,v)) = 0 (3.8)
■^[V2{s,x,v) - ^4(X(s, x, v))] = 0. (3.9)
£{x,v) = i|^(s,x,w)|2 -U(X(s,x,v))
£(x,v) = V2 (s,x,v) — A(X(s,x,v)).
If we further define
r(y) = l(£ + A(y))2-U(y),then for (x,v) fixed we have
d2 X— = -{E{X) + V2{B{X))
= -{-U'{X) + {C + A{X))A'{X)) (3-10)= -V'{X).
Since (3.10) is reversible in s it follows that if V\ = 0 for two values of s, then X isperiodic. If
lim X(s, x, v) = — oo,s—►—OO
then Vi(s, x, v) = 0 for at most one s. Let si be that value of s if it exists and «] — +oootherwise. In both cases
Vi(s,x,?;) > 0 for s < si. (3-11)
If Si is finite, then by reversibility
V\(s,x,v) < 0 for s > si (3-12)
and
lim X(s,x,v) — — oo. (3.13)s—>+oo
516 JACK SCHAEFFER
Lemma 3.1. If
lim X(s,x,v) = —oos—> — oc
for some (x,v), then
lim V\ (s,x,v) exists and is nonnegatives —> — OC
and
f(x,v) = G(\\v\2 - U(x),v2 - A(x)).
The proof is deferred to the appendix.Note that lim^^-oo U{x) = 0; so there exists xq < 0 such that
x < x0 => A(x) = 0 and |f7(:c)| < |(£min - |<S2). (3.14)
Lemma 3.2. If X(so,x,v) < xo, Vi(so,x,v) > 0,
±\V(s0,x,v)\2 - U(X{s0,x,v)) > £min,
and
|V2(so,a;,w)| < <S,
then
Vi (s,x,v) > yj£min - \S2
for s < so and
lim X(s,x,v) = — oo.s—► — oo
The proof of Lemma 3.2 is also deferred.
Lemma 3.3. For x < xq and v\ > 0,
f(x,v) > G{\\v\2 -U(x),v2 - A(x)).
If £(x,v) > £„nn and \v21 < S also, then
f(x,v) = G(|\v\2 - U{x),v2 - A(x)).
Proof. Consider x < xq and V\ > 0. If £(x,v) > £min and \v2\ < S, then by Lemma3.2 we have lim5^_00X(s,x,v) = —oo. Hence
/(x,v) = G(±|u|2 - U(x),v2 - A{x))
by Lemma 3.1. If either £(x,v) < £min or \v2\ > S, then
0 = G(|\v\2 - V(x),v2 - A(x)) < f(x,v)
since f(x,v) > 0. This completes the proof. □
ONE-DIMENSIONAL MODEL OF THE SOLAR WIND 517
Lemma 3.4. If
f(x,v)^0 and lim X(s,x,v) = — oo,s—► — oo
then
V\ (s,x,v) > 0 for all s
and
lim X(s, x, v) = +00.s—►+OO
The proof is deferred, but we mention that the condition (1.4) is central to the proof.Consider the following assumption:
iff(x,v)^0, then lim X(s,x,v) = —oo. (3.15)s—►—OO
This would fail to be true if / were nonzero on some bound periodic orbits.
Theorem 3.5. Assume (1.3), (1.4), (1.5), and (3.1) through (3.6) hold. If (3.15) alsoholds, then
f(x,v) = G(±H2 - U(x),v2 - A(x))
for V\ > 0 and f(x, v) = 0 for v\ < 0.
Proof. Consider (x,v) with f(x,v) ^ 0. By assumption linis-.-oo JT(s, x, v) = -ooand so by Lemma 3.1
f(x,v) = G{\|u|2 - U(x),v2 - A(x)).
By Lemma 3.4, > 0. It follows that / = 0 on I x (—oo, 0] x R.Now consider (x,v) with vi > 0 and
G(\\v\2 - U(x),V2 - A{x)) ± 0.
Choose (y,w) with y < xo,wi > 0,
\\™\2 ~U{y) = \\v\2 -U(x), (3.16)
and
w2 - A(y) = v2 - A(x). (3.17)
We may do this since
\\v\2 - U(x) > £min, \v2 - A(x)\ < S,
and
\U(y)\ < i(fmin - |52).
Now by Lemma 3.3
f(y,w) > G(±\w\2 - U{y),w2 - A(y))
= G(\\v\2 - U(x),v2 - A(x)) > 0.
Since f(y, w) ^ 0, from before we have
f{y,w) = G{\\w\2 - U{y),w2 - A{y)).
518 JACK SCHAEFFER
By Lemma 3.4, lims^+00 X(s, y, w) = +00. So there exists s0 such that x = X(so5 V, w).It follows by (3.16) and (3.17) that
vi = Vi(s0,y,w)
and hencef(x,v) = f{y,w)
- G(£(y, w),C(y, w))= G(S(x,v),£(x,v)).
This completes the proof. □Next we seek to construct solutions that have the form stated in Theorem 3.5. Note
that
G(|\v\2 - U(x),W2 - A(x)) 0
=> \v\ > £min + U(x) — \v\ and \v2 - A(x)\ < 5
=> \v\ > £min + U{x) - |(5 + |v4(a;)|)2.
So if
£min + U(x) - 1(5 + \A(x) I)2 > C > 0, (3.18)
then G(|\v\2 — U(x),v2 — A(x)) vanishes for v\ near zero and hence
G(i|v|2 -U(x),v2 - A(x))IVl>0
is smooth.
Theorem 3.6. Consider B(x) = eB\(x) and assume (3.1) through (3.3). There exists£0 = Eo(-Bi) such that for |e| < sq there is a solution of (1.3) that satisfies (1.4), (1-5),(3.4) through (3.6), (3.15) and is of the form
f(x,v) = G{\\v\2 -U(x),v2 - A(x))IVl>0.
Proof. Define
pc{u,a) = J J G(^\v\2 - u,v2 - a)dv2dvi
and note that
/OC /»/ G{£ + \vl,v2 - a)dv2-u J y 2(c + u)
If
£min + U — 5(5 + |d|)2 > 0,
then
G(— u + \v\, v2 — a) = 0 for all ^2,
and hence 2 f°° f 3duPG(u,a) = J J G{£ + \v\,v2 - a)dv2{£ + u)~^d£ < 0.
ONE-DIMENSIONAL MODEL OF THE SOLAR WIND 519
U must satisfy
U"(x) = —p(x) = J f dv — J F dv
= [ ( G(\\v\2-U{x),v2-A(x))dv2dviJo J (3.19)
-J J G(±\v\2,v2)dv2dvi= pG(U(x), >l(a;)) - pG(0,0).
Define
cr(u, a) = - (pg(u, a) - pG{0,0))duJo
and note that A(x) = constant on (—oo, 0) and on (L, oo). So
W'{x))2 + a(U(x),A(x))\ = 0
on (—oo, 0) U (L, oo). Since U(x) —> 0 and U'(x) —► 0 as x —» —oo
(3.20)
\{U'(x))2 + <t(U(x), 0) = 0
on (—oo, 0). Since ct(0,0) = 0, <9„<7(0,0) — 0, and
^wcr(0,0) = —3„pg(0,0) > 0
it follows that U(x) = 0 for all x < 0.Next choose C\ and C2 such that (it, a) € N := (—C\,C{) x (—C2, C2) => (3.18) holds
and
d2ua = -dupG > ~\dupG{0,0) := C3 > 0.
Then for (u,A(L)) € N we have
er(u, >1(L)) > <r(0, ̂ 4(£)) + 3Mcr(0, A(L))u + \C3u2
= -(pg(0, A(L)) - pG{0,0))u + 5C3U2.
For £ sufficiently small we have
|X(L)| < |e| [ \Bi{x)\dx < C2.Jo
By further restricting e we may force U, the solution of (3.19), to satisfy
\U(x)\ <Ci fovx<L
and hence
(U(x),A(x)) € N for x < L.
Define
L* = sup{x > L: \U\ < C\ on [L,x]}.
520 JACK SCHAEFFER
Then for L < x < L* we have by (3.20)
\C3U\x) < \(U\x))2 + a(U(x), A(L))+ (pG(0,A(L))-pG(0,0))U(x)
< \{U\L))2 + o(U(L),A(L))+ (Pg(0,A(L))-Pg(0,0))C1.
(3.21)
The right-hand side of (3.21) tends to zero as e —> 0. So we may force |J7(x)| < \C\to hold on (L, L*) by restricting e. In this case it follows that L* = oo, (U(x), A{x)) G Nfor all x, and hence (3.18) holds for all x. The proof is now complete. □
We comment that for x > L,U(x) is periodic and will not approach zero asu +ooin general.
4. Numerical results. In this section we describe the results of approximating so-lutions of (1.1), (1.2) with a particle method. The case considered is
and
B(x) = 9(4(x - l)(x - 2))2/(1,2}(x)
g{x,v) = 0,
F(v) = 10<5(t>i — 5)6(v2)-
Note that previously F was assumed to be smooth and because of this Theorem 2.1 doesnot apply. In fact, the proof of Theorem 3.6 can be adapted to this case and the solutionsconstructed there may be computed using the equation
U"(x) = pG(U(x),A(x)) - pg(0,0)
50 50 (4.1)~ t/25+W(^P^x) y/25'
The solution to (1.1) was computed on the interval 0 < x < 10 for t small enough thatf(t, x,v) = F(v) for x near 0 and 10. The spatial step dx — 3.125 x 10-3 was used tocompute E with 100 particles per spatial step.
Figures 1 and 2 show the graphs of
J f dv = p+ 10
and
fv 1 dv - 50 - ji1at the time t = 0.8, t = 1.0, and t = 1.2. Note that all three graphs in each figure agreefor x < 4. For x < 4 steady conditions have been attained by t = 0.8. Note also that for
ONE-DIMENSIONAL MODEL OF THE SOLAR WIND 521
Fig. 1. Time series of rho
a steady state we must have
fvidv = 50,/■
which is the case in Fig. 2 for x < 4.Note also the sharp spike which occurs in both figures at t = 1.2. To understand this
we graph
/ fvidvvelocity = J fdv
at the same three times in Fig. 3. Again the curves match for x < 4. For x > 4 wesee faster velocity particles catching up with slower particles downstream, resulting in asteepening curve.
Figure 4 shows f fdv at time 1.2 computed with the particle code as a dotted curve.The solid curve is the solution of (4.1). Steady conditions hold for x < 7 but the sharpspike is a transient feature. Moreover, the solution of (4.1) correctly predicts the steadybehavior of the dynamic problem in this case. If the amplitude of B is increased by 20%,then the solution of (4.1) is not globally defined and particles are reflected by B.
522 JACK SCHAEFFER
350 i ; i i i i i i r*
300 -
t=0.8t=1.0t=1.2
fi|
250 [- !
1200 h
150
100 II I
' \50 ■
_l I I I L_01 23456789 10
Fig. 2. Time series of ji
5. The appendix.The Proof of Lemma 3.1. We have
A(:r) = 0 for x < 0.
So by (3.9)
lim V^s, x, v) = i>2 — A(x). (5-1)s—* — oo
Similarly
lim U(x) = 0.x—> — OO
So by (3.8) and (5.1)
5 lim Vi(s,x,v) = h\v\2 - U(x) - |(v2 - A{x))2.s—* — OO
It follows that lims-,-00 V\ (s, x, v) exists. Since limg^-oo X(s, x, v) = —oo it follows thatlims^-oo Vi(s, x, v) > 0 and
lim Vi(s,x,v) = \/\v\2 — 2U(x) — (v2 — A(x))2.
t=0.8t=1.0t=1.2
ONE-DIMENSIONAL MODEL OF THE SOLAR WIND 523
\\i
7 -
O5CD>
3 -
2 -
_l I 1 I I L- I I L.
FIG. 3. Time series of velocity
By (3.11), Vi(s,x,v) > 0 for s sufficiently negative. So by (1.5)
lim f(X(s,x,v),V(s,x,v))=F\ lim V(s,x,v)s—► — oo \s—► —oo
and hence
f(x,v)= lim f(X(s,x,v),V(s,x,v))s—>— —
= g( lim Vs—>—OO
, lim V2 ] I lim V\ > 0+ —OO / 5—♦ —OO
8 9 10
= G{\\v\2 - U{x),v2 - A(x)).
Note that if lims^_oo V\ — 0, then G(\\v\2 — U(x),v2 — A{x)) = 0. This completes theproof. □
The Proof of Lemma 3.2. Define
s = inf{s < s0 : Vi(t, x, v) > 0 on r € (s, so)}.
For all s
£min < £(x,v) = i|y(s,X,v)|2 -U{X(s,X,v))
524 JACK SCHAEFFER
Fig. 4. Comparison with exact values
and
S > \£(x,v)\ = IV2(s,x,v) — A(X(s,x,u))|.
So for s £ (s, so) we have X(s, x, v) < xq. So A(X(s, x, v)) = 0,
IV2(s,x,v)j < <S,
and by (3.14)
\V?(s,x,v) > £min + U(X(s,x,v)) - 5V22(s, x, v)
^ f — Iff _ I <?2>i _ I c2^ '-'mm 21 mm 2 / 2
_ Iff . _ I2 V°min 2 ''
Thus
Vi(s,x,v) > - i<S2
for s G (s, so). It follows that s = —00 and the lemma follows. □Before proving Lemma 3.4 we consider the following:
Lemma 5.1. If
f(x,v)^ 0 and lim X(s,x,v) = — 00,
ONE-DIMENSIONAL MODEL OF THE SOLAR WIND 525
then there exists a neighborhood, N, of v such that
w £ N => lim X(s,x,w) — —oos —* — oo
and
w € N => f(x,w) = G(\\w\2 — U(x),w2).
Proof. By Lemma 3.1'<2
SoG(\\v\2 - U(x),v2 - A(x)) = f(x,v) / 0.
2 1^1 ^ ^min
and
\v2 - A{x) | < S.
There exists so < 0 such that
< xq and ^(so, x, v) > 0.
By continuity with respect to initial conditions there is a neighborhood, N, of v suchthat for w £ N we have
\\w\2 - U(x) > £min,
\w2 - A(x)\ < S,X(s0,x,w) < x0,
and
Vi («o, a:, w) > 0.
By Lemma 3.2, w € N => lims__oo X(s,x,w) = —oo. Now using Lemma 3.1 completesthe proof. □
The Proof of Lemma 3.4. We proceed by contradiction. First suppose there is (x,v)with /(x,v) / 0, Vi < 0, and lim^-oo X(s, x, v) = —oo. We will show this contradicts
(1.4).By (3.12) and (3.13),
s > 0 =>• V\ (s, x, v) <0
and
lim X(s, x, v) = — oo.s—►-f-oo
So without loss of generality we take V\ < 0 and x < x$ (by replacing (x,v) with(X(s,x,v), V(s,x,v)) with s large). Let N be the neighborhood of v from Lemma 5.1.
Next choose £,£,£,£ as follows. Denote
N = {w: wi < 0,£ < \\w\2 - U(x) < £,C <w2<C}
and choose £,£,£,£ so that TV c N, N ^ 0, and G(\\w\2 — U(x),w2) > C > 0 forw G N.
526 JACK SCHAEFFER
Now consider (y, z) £ I x I2 with
y < x, zi < 0,
£ < \\z\2-U{y) <£,
and
C < z2 < £.
We may choose w & N such that
\\w\2 - U{x) = \\z\2 — U(y)
and
W2 = Z2.
There is s such that
y = X(s,x,w) and z = V(s, x, w).
Then
f(y,z) = f{x,w) = G{\\w\2 - U(x),w2) > C.
So for y < x, z\ < 0, ||z|2 - U{y) £ (£,£), z2 € (£, C) we have
f(y,z) > C.Since U(y) —► 0 as y —> —oo this implies
[ [ f{y.*0> cJ — OO J
for y sufficiently negative.Finally, using Lemma 3.3,
U"{y) = J J f(y, z)dz dy- J F{z) dz
> J j G(\\z\2 -U{y),z2)dzdy + C
- J J G(\\z\2 ,z2)dzdy.
But since U(y) —> 0 as y —> —oo
f/"(y) > cfor y sufficiently negative. This contradicts the assumption that E — —U' 0 asy -> -oo.
Note that if Vi(s,a:,w) — 0 for some s, then by (3.12), V\(s,x,v) < 0 for s > s. So theabove argument shows that
f(x,v)y^ 0 and lim X(s, x, v) = — oo => Vi(s, x, v) > 0 for all s.
ONE-DIMENSIONAL MODEL OF THE SOLAR WIND 527
Next assume f(x,v) ^ 0, lims^_ooX(s,a;,u) = —oo, and X(s, x,v) is bounded fromabove. Let
x = lim X(s,x,v). (5.2)s—>+oo
Since vjI(s,i,d) is bounded for s > 0, it follows that
From (3.10)
lim Fi(s, x,v) = 0. (5.3)s—>+oo
\V?{s,x,v) + V{X(s, x,v))
is independent of s. From (5.2) and (5.3) it follows that
V(x) = S(x, v)
and
^-X(s,x,v) = \j2(P{x) - V(X(s,x,v))). (5.4)
Clearly V(x) > 0. If V'(x) > 0, then the solution of (5.4) would attain V\ = 0 in finitetime, contrary to assumption. Thus
V\x) = 0.Let N be the neighborhood of v from Lemma 5.1. Choose £ < £(x,v) such that
0 < w\ and £ < £(x, wi, v?) < £(x, v)
=> (w1,v2) £ N and f(x,wi,v2)^0.
By Sard's theorem (p. 61 of [4]) the set
{V(y)\ V'(y) = 0 and x <y <x}
has measure zero. Hence, we may choose £q G (S,S(x,v)) with
So & {V(y): V'(y) = 0 and x < y < x}. (5.5)Now choose Wi > 0 with £(x, wi, u2) = So and let W2 = V2■ Then w £ N and sof(x, w) 7^ 0 and lim,,^-,^ X(s, x, w) = —oo.
As above V\(s,x,w) > 0 for all s. Also note that C(x,w) = C{x,v). So V in (3.10) isunchanged. So
So = m2(s,x,w) + V{X(s,x,w))
— \w\ + V(x) < \v\ + V(x)= S(x, v) = V{x).
Hence, V(X(s,x,w)) < V(x) and X(s,x,w) < x for all s. As before it follows thatlims_,+00 Vi(s,x,w) = 0. But on the other hand,
S0 = V lim X(s,x,w) ) ;ys—>+oo J
so by (5.5)
V' ( lim X(s,x,w) ) / 0.ys—>+oo J
528 JACK SCHAEFFER
Since
■^-X(s1x,w) = y:2 (^P ̂ lim X(s,x, w)J - P(X(s, x, w))^,
it follows that V\ (s, x, w) = 0 for some s. This is a contradiction and completes theproof. □
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