Inverse Problems for Kinetic Equations
Theorem 2.8. Assume that the functions
2.6 THE GENERAL UNIQUENESS THEOREM
We consider the simultaneous determination of three functions (F, Η, λ) from (2.1) where F is the distribution function, H is the Hamilton function, and λ = St F + q. The uniqueness of the solution to the inverse problem can be proved under special restriction concerning both the data and the sought functions (F, H, A). Let D C Rn be a domain of variables ρ with a smooth boundary; Q = D χ D\, χ £ D, and dQ be the boundary of Q.
The data for the inverse problems are the functions F0, A, B, C, H0 defined as follows:
F0(s,p,t) = F |r, s e r = 3D, ρ £ R " , a < t < ß ,
(2.11)
A(x,p) = F \t=a, B(x,P) = F \t=0, C(x,p) = f |( = a, H0 = H \dQ .
Suppose that the following conditions are satisfied.
1. The sought Hamilton function H(x,p) is analytic in ρ € Rn and H(x,p) £ C2(DxRn).
" d2 H n d2 Η
2. Σ ö - ä - 6 6 < 0, Σ ä - ä - f i f i ^ α Iii2 - € R n' w i t h constant α > 0.
i,j=i öxiöxj opidpj
3. \ΌαΗ(χ,ρ)\ < (Ν + |p|)m, \a\ < 2, Ν > 0, τη > 0 are constants, Da being a differentiation operator with respect to variables (x,p).
4. The known potential Φ(χ,ί) belongs to the class C2(R" Χ [a, 6]), ρ £ R", a <t < β.
5. |££Φ| < (Ni + |p|)m i , |7| < 2, Νλ > 0, m, > 0 are constants.
Θ2Φ
t,j=i dxidxj
7. The given function obeys the conditions
8. The sought quantum distribution function F(x,p,t) e C2(D χ R" χ [α, 6]) together with its first and second derivatives, rapidly vanishes by
lim \DaF\ |p|m = 0, Vm > 0, \a\ < 2.
IPHoo
9. The unknown function q(x,p,t) is twice and continuously differentiable and satisfies the equation
fri dxjdpj 10. The operator
η Q2 dxjdpj
is supposed to be known, defined on the set of distribution functions and to map this set onto itself.
11. For any two distribution functions Fi(x,p,tf) and F2(a;,p, t), F\ φ Fj, the following inequality holds
JI / {?
b|grad * (fi ~ Fî)|2+{Fì ~ f2) (MFi - MFî] ) dxdpdt >
a D R"
Let us make some remarks concerning the conditions stated. Conditions 2, 6 and 7 imply the convexity of the functions H(x,p), A(x,p) and Φ (χ, ρ) with respect to the corresponding variables. In the case η = 1 these functions may be as follows
H=\p\ ρ 6 R \ A = e - p 2, W < - L
Φ(χ, t) = - 7 ( i ) x2 + ß(t)x + c{t), 7 > 0.
Conditions 3, 5 and 8 characterize an increase and a decrease, respectively, in functions Η(χ,ρ), Φ ( χ , ρ ) and F(x,p,t) at infinity. Condition 8 is usually considered to be satisfied.
Sometimes the sources q(x,p, t) are independent of ρ or x. These cases are embraced by condition 9. Items 10 and 11 state the restrictions concerning the aprioriunknown operator St. Condition 11 is connected with the monotonicity of the operator
η Q2
M ~ £ dxjdPi S t
which, unlike operator St is considered as given.
The inverse problem is: find the functions ( F , H , \ ) , given (2.11) for equation (2.1) and provided that conditions 1-11 are satisfied.
T h e o r e m 2.10. The inverse problem has no more than one solution.
PROOF. We first prove that the Hamiltonian Η(χ,ρ), χ € D, ρ € Rn is uniquely deter-mined.
Setting t = 0 in (2.1), we have
C(x,p) + {A,H} = — ^ J [Φ]οe'!,','-p''i4(x,p/)dp'dy + S t F + q0(x,p). (2.12)
^ '
Ri-go = q{x,p, 0), [Φ]ο = φ ( χ - \ hy, o ) - Φ ( χ + ì hy, θ ) ] .
Inverse Problems for Kinetic Equations 39 If Hi and H2 are two solutions of (2.12) satisfying H} \dQ = H0, we consider the difference φ = Hi — Hi. We have
{ιψ, A} = St2 A - St2 A + qoi - qo2 - λ0, vlaç = 0.
The right-hand side St2 A - St2 A + q01 - q02 = λ0(ζ,ρ) can be shown to satisfy
% dxjdpj Indeed, by conditions 9 and 10
Thus,
Σ 7Γ7Γ t(St2 A - St2 A ) + (901 - 902)1 = M A * M A =
j=l OXjOPj
η β2\
{φ, A} = λ0, Σ < Γ 7 Γ = 0> ψ I9« " °· (2·1 3 )
j= 0 OXjOPj Using (2.13) one can verify directly that
y r È L J L i a\ - 1 y ( d*A 9 φ 9 ψ d*A d í p 9 φ\ di, dpj ' 2 \dpidpj dxi dxj dx,dxj dp, dp3 J
, 1 A dip d θ 1 " d d ,x
1 ί£ΙΑδφ_άφ\ _ 1 A d_ ÍdA δφ 3ψ\
2 ^ dx, \dp, dxj dp,/ 2 öp, V^a;, dx3 dp,)
¿ í ^ i r ^ p j
-Recalling the condition = 0 and making use of the essential fact that the domain Q is a Cartesian product D χ Di, χ € D, ρ € Di, we then obtain by integration
/ y ^ Q2A 9 φ 9 φ \ ^ = o J \dpidpj dxi dxj dxidxj dpidpj J
From this and condition 7 we obtain d<p/dxj for j = l , . . . , n , which in turn leads to the relation φ(χ,ρ) = 0, (χ, ρ) 6 Q. Thus, Hi = H2 in Q. Since by condition 1 the Hamiltonian is analytic, it has a unique continuation from the domain Q to D χ Rn, on which it will henceforth be taken as known.
Let (Fi, Ai) and (F2,X2) be two solutions of the inverse problem satisfying all the hypotheses of the theorem. Writing F = Fi — F2 and q = qi — <72, from (2.1) and (2.12) we obtain
dF
dt ' ' "+ =J - (2tt) (¿Wi J We'y{r~p,)F(x>P>t)dP'áy + Sti Fl + St, F2 + q, (2.14) F\t=a = 0, F |( = 6 = 0, F |Ι€Γ = 0.
Differentiating (2.14) with respect to pj, multiplying by dF/ dxj, and summing over j, we get the equality
The following relations can be verified directly:
— \ 1 d ( d F d F \ _ A d F d2F
2 dt \<9j:; d p j ) 2 dxj \ dt dp J + 2 dp, \ dt dx, / ^ dx, dpjdt' ( ' }
r d F d l F m - 1 ν d2H dF dF d2H dF dF
dpidpj dxi dxj dxidxj dp, dp,
1
+ 2 . f e dp,
dF (dF^dH_ d F d H \ dxj \dxi dp, dp, dxi J
_ 1 A _d_
2 d x>
dF ( W < W _ d F d H \ dpj \dx{ dpi dpi dx, J
1 " d_ (dH_ dF_dF\ _ 1 " I d t i dF d F \
2 t j ^ ! dx, V dp, dx, dpj J 2 ^ dpi \ dx, dx, dpj )
Since by hypothesis
(2.17)
J^ dxjdpj dxjdpj
we have the equality
η f t p β
E x i s t í · . - « « + , )
- ± TT Fi t ' <S l f i - S' f i + «j - IF, - F,, (MF, - MF,)
• j ax j apj ( 2 . 1 8 )
We now establish that
b
α D R" 1- R2n
where
We have
A dF d i
• C E ' l=l 3
F = { 2 ^ J F ^ e ì y P ' d P ' ·
(2.19)
ξ ¿ 3 di 3 ( ¿ ñ / [* (* - 5 4 ) - * ( ' + 5 hy ' 01 eiy(P '" P) x
Inverse Problems for Kinetic Equations 41
J=1 J ν > R„ whence by integration we obtain
JJJtMjï^J^^^
1 D R" 3 R"
= { } J j £ d F'{ dy, t ]F ( * , y , m y > d y dxd t ,
" D R " ·*=1 J
where
R"
We prove next that
ι hr r r n dF* ~ \ r r r n d \ - \2
= ^ d ^ d , (2.20)
o D R" J -1 J a D Rn 3=1
Let
-1 Rn
Since the function F is real-valued, Ρ and Q are even and odd in the variable y, respec-tively. We have
" dF* ^
t t t il r
/ / / E ^ W w d y d x d i α D R" ·>=1 J
/Ι η Un J=l \ J J / a D R"
L>-Since the function [Φ] is odd in y, the same is true of the function
for all j, 1 < j <n, and we have This establishes (2.20).
We now show that
+ > 0 . (2.21)
Indeed,
η g η j=1 ux] j = l
d$(x-\hy,t) d$(x + \hy,t) dxj dxj
We fix χ, y and t, and consider the function
d$(x-\zy,t) dΦ (x + \zy,t}
9{z) = Σ&
7=1 dx, dxj ζ > 0.
Since jr(O) = 0, we have g(z) = g'(0z)z, 0 < θ < 1, and so
<K*) = - Σ fi
dxjdxk dxjdxk Vkz- (2.22)
Setting ζ = h we obtain from (2.22)
= -h Σ
*¿=i
0 Φ2( χ - 0 Φ2( χ + ^0/ΐ2/,ί) dxjdxk -VjVit +
dxjdxk VjVk By condition 6
*,¿=i Σ
0Φ 2 (χ - \ 9hy,t) 5Φ 2 (x + £ ehy, t)
dxjdxk dxjdxk yflk < 0.
Therefore, g(h) > 0, which corresponds to (2.21).
Since F = Fi- F2, we have F|(=a = 0, F |t=6 = 0, F|r = 0.
Integrating (2.15) over x, p, and i, recalling (2.16)—(2.20) and using the specific prop-erties of the divergence terms and the Cartesian product representation D χ Rn χ [α, 6]
of the domain, we obtain the fundamental identity
III f t ,
a D R" ''J-1 dxidxj dpi dpj dpidpj dx, dxd2H dFdF d+
2H dF dF } dpdxdt +iJJJ\p\2£¿rWy>áydxdta D Rn 1
+ J J j (Fi- Fi) (MFt - MF
b2 )dxdpdt = 0.
(2.23)α D R"
Since
" d2H dF dF > " f d F \ 2 " d2H dF dF >
dpidpj du dx, ~ ¿Í [ d x j ' dxßxj dp, dp3 ~
n d
and (as proved above) Σ J,— ^ 0 (2.21) implies that j—i °x5
b
J J J { f lgrad^ ~ F2)12 + ( F l ~ Ρ ί ) ( M F l ~ MF2)} d l d p d < -α D R"
Inverse Problems for Kinetic Equations 43 which for Fi φ Fi contradicts condition 11. Therefore we must have F1 = F2. From (2.1) it follows in addition that Αι = λ2. The theorem is proved.
Some conditions of the theorem are essential. We give an example of the nonuniqueness for the case when H(x,p) is specified.
Assume that Q = {(ζ,ρ) : χ2 + ρ2 < 1}, i.e. Q is not a Cartesian product, and F(x,p) = 1 - e1- ^ "2) , Η = \ (x2 + p2). We have F\dQ = 0, {F, H} = 0, F φ 0.