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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