A R K I V F O R M A T E M A T I K B a n d 7 n r 26 1.67 07 1 Communicated 26 April 1967 by OTTO FROSTlgAN
Estimates o f the age o f a heat distribution
B y J o H A ~ PHILIP
A B S T R A C T
The paper deals with the possibility to solve the heat equation backwards in time. M o r e specifically, we treat the following problem. Given the temperature at a finite number of points of a homogeneous bar, how old can the heat distribution be? I n the case that the temperature is given at equidistant points xi, the problem is completely solved. I n the ease of nonequidistant x~ we find an upper bound for the age. Such a bound is also obtained when the information about the heat distribution is given by the value of a finite number of linear funetionals.
I. Introduction
W e c o n s i d e r t h e h e a t d i s t r i b u t i o n ( t e m p e r a t u r e d i s t r i b u t i o n ) in a h o m o g e n e o u s b a r of i n f i n i t e l e n g t h ( c o o r d i n a t e x) as a f u n c t i o n of t i m e (t). O u r h e a t d i s t r i b u t i o n s w i l l b e c o n s i d e r e d as p o s i t i v e m e a s u r e s ut(x).
T h e f u n d a m e n t a l s o l u t i o n of t h e h e a t e q u a t i o n (02u/~x 2= ~u/~t) is
wdx) =(1/2V2t) exp (-xV4t) (t>0). (1)
A n " i n i t i a l h e a t d i s t r i b u t i o n " % a t t - 0 g i v e s t h e f o l l o w i n g d i s t r i b u t i o n a t t h e t i m e t:
ua =Vt-~Uo. (2)
W e s h a l l b e c o n c e r n e d w i t h p r o b l e m s c o n n e c t e d w i t h s o l v i n g t h e h e a t e q u a t i o n b a c k w a r d s i n t i m e , v i z . w i t h t h e f o l l o w i n g p r o b l e m : I f v is a b o u n d e d p o s i t i v e m e a s u r e , f o r w h i c h t d o e s t h e r e e x i s t a b o u n d e d p o s i t i v e m e a s u r e u 0 s a t i s f y i n g
v = ~0~.u0? (3)
W h e n t - > 0 , % a p p r o a c h e s t h e D i r a c m e a s u r e a t t h e origin, so f o r t=O, (3) h a s t h e s o l u t i o n uo=v. W h e n t-+ co, V t - x - u 0 ~ 0 f o r e v e r y x, so (3) h a s n o s o l u t i o n f o r l a r g e t if v 4 0 . F u r t h e r m o r e , w e h a v e
~t~ * ~0t~ = ~t,+t~, (4)
so if (3) h a s a s o l u t i o n % for t=~, i t h a s t h e s o l u t i o n Uo~-y:~_. for a t i m e ~ K r . Thus, i t is m e a n i n g f u l t o a s k f o r t h e l a r g e s t i n t e r v a l (0, t) in w h i c h (3) h a s a solution.
J . PHILIP, Age o f a heat distribution
I n our p r o b l e m s , h o w e v e r , t h e i n f o r m a t i o n a b o u t v will be i n c o m p l e t e a n d g i v e n b y n r e a l n u m b e r s only. I n sections I I - I V t h e i n f o r m a t i o n is t h e v a l u e s v(xt) a t n p o i n t s x~. I n s e c t i o n V, we a s s u m e t h e v a l u e s of n l i n e a r f u n c t i o n a l s of v to b e k n o w n , w h i c h is a m o r e r e a l i s t i c s i t u a t i o n f r o m a p h y s i c a l p o i n t of view.
II. F o r m u l a t i o n o f Problem I
L e t t h e i n f o r m a t i o n a b o u t v be g i v e n b y t h e v a l u e s a~ = v ( x i ) a t x 1 < x 2 < x a < ... <x~, so t h a t we h a v e t h e e q u a t i o n s
a t -~vt~eUo(X,) (1 <~i<~n). (5)
I f u 0 - - 0 we h a v e a , - 0 (1 <~i<~n). I f u0~:0 a n d t > 0 we h a v e a t > O , (1 ~ i < ~ n ) , since
~ f t > 0 for t > 0 . Thus, if a t - 0 for some i b u t n o t for all we m u s t h a v e t = 0 . F u r t h e r , for n - 2 , (5) h a s a s o l u t i o n for a n y t~>0 consisting of a single D i r a c m e a s u r e of s u i t a b l e size a n d p o s i t i o n ( b o t h d e p e n d i n g on t). W e pose
P r o b l e m I. F o r given v(xi) - a~ > 0 (1 ~ i ~ n; n >~ 3 ) , / i n d the s u p r e m u m T o / a l l t / o r which there exists a positive bounded u o satis/ying (5).
F o r t fixed, our p r o b l e m is a f i n i t e m o m e n t p r o b l e m . W e t a k e t h e following condi- t i o n for e x i s t e n c e of a s o l u t i o n t o t h i s p r o b l e m f r o m R o g o s i n s k i (1958), T h e o r e m 1 a n d C o r o l l a r y 1:
T h e o r e m R. There exists a positive u o satis/ying (5) i / a n d only i / the p o i n t a = (ai, a 2 .. . . . an) E R ~ is i n the hull cone I P t o / t h e curve
pt(x) = (~)t(Xl x), ~ ) t ( x 2 - x ) . . . ~)t(Xn - x ) ) , - c~ < x < -k oo.
B y t h i s t h e o r e m we h a v e
(6)
T = s u p {t : a E P t } . W e shall i n v e s t i g a t e t h e p r o p e r t i e s of Pt.
(7)
L e m m a 1. P t 2 ~ P t ~ i / tl <~t 2. (8)
P r o @ I f y E P ~ 2 t h e r e e x i s t s a p o s i t i v e m e a s u r e # such t h a t Y = # * P t 2 . T h e n w e h a v e b y (4) y = f l y , p t 2 - # - ~ ) 4 _ t l b 6 p t r Since [A-)~t2--t, is a p o s i t i v e m e a s u r e t h e l e m m a is p r o v e d .
Since yzt(x)>~0, P t is a s u b s e t of t h e p o s i t i v e o r t h a n t in R ~. I f x # 0 , y4(x)-+0 w h e n t ~ 0 , i m p l y i n g t h a t t h e r a y f r o m t h e origin t h r o u g h t h e p o i n t pt(xt) a p p r o a c h e s t h e y~-axis of R n as t-*-O. Thus, P t is m o n o t o n i c a l l y i n c r e a s i n g t o t h e whole p o s i t i v e o r t h a n t w h e n t-+0. Since we h a v e a s s u m e d a ~ > 0 ( l ~ i ~ < n ) , t h e r e exists a n s > 0 such t h a t a EP~.
W h e n t-+ co, P t decreases to a subcone, s a y Poo of t h e p o s i t i v e o r t h a n t . The cone Poo is t h e s e t of p o i n t s a for w h i c h (5) h a s a s o l u t i o n for all t. Poo is d e s c r i b e d b y t h e o r e m 1 in t h e case of e q u i d i s t a n t x~.
B y L e m m a 1 we h a v e
1 The hull cone of a set A is defined as the smallest convex cone with vertex 0 that contains A.
ARKIV F6R MATEMATIK. B d 7 n r 26 sup {t: a EPt} : inf {t: a r
if we i n t e r p r e t t h e r i g h t - h a n d side of (9) as ~ w h e n a E P ~ .
(9)
III. Equidistant data
A s s u m e t h a t t h e v a l u e s a~=v(x~) are o b t a i n e d a t e q u i d i s t a n t p o i n t s , i.e. a s s u m e x ~ = b + i h (l~<i~<n), w h e r e b a n d h a r e c o n s t a n t s . Since t h e p o s i t i o n of t h e origin on t h e x - a x i s is i m m a t e r i a l , we p u t b = 0 .
T h e o r e m 1. A s s u m e a ~ > 0 (l~<i~<n) and define T = e x p (h2/4t) /or t > 0 so that t = h2/4 log T. Consider the quadratic/orms
QI(T) = E~E,,a~+mTa+m)'~m, 1 ~<i, m ~< [n/2], Q2(T) : ~i~mai+m_l T(~+m 1)z~i~m , 1 <~i, m <~[(n + l)/2].
Let T O be the smallest T >~ 1 such that t h e / o r m s Q1 and Q2 are both positive semidefinite /or T o <~ T < ~ . I / ~ o > 1, we have T = h2/4 log T 0. I] T o = 1, then T = c~, that is a EPoo and the e~uations (5) are solvable/or all t>~O.
P r o @ A s y m m e t r i c m a t r i x is p o s i t i v e d e f i n i t e if all its d i a g o n a l s u b d e t e r m i n a n t s a r e p o s i t i v e . F o r t h e m a t r i c e s of t h e f o r m s Q1 a n d Q2 t h e s e d e t e r m i n a n t s a r e p o l y - n o m i a l s in T a n d i t is e a s i l y s h o w n t h a t t h e i r l e a d i n g coefficients are p o s i t i v e . B y t h e d e f i n i t i o n of T 0 we t h e n k n o w t h a t t h e r e e x i s t s a T1 s u c h t h a t Qa(T) a n d Q2(T) b o t h a r e s t r i c t l y p o s i t i v e for To < T <T~.
Now, we w r i t e o u t t h e e q u a t i o n s (5) w i t h x~ = i h
at = ( 1 / 2 I / ~ ) e x p ( - (ih - x)2/4t) %(dx). (5') R e a r r a n g i n g (5'), we g e t
a~ = e x p ( -- i2h2/4t) e x p (ihx/2t) ( 1 / 2 ~ t ) e x p ( - xe/4t) uo(dx).
N o w , t h e m e a s u r e w = (1/21/~) e x p ( -x~/4t)Uo is p o s i t i v e if a n d o n l y if u 0 is p o s i t i v e , so we h a v e t h e question: for w h i c h t does t h e r e e x i s t a p o s i t i v e w s a t i s f y i n g
a~ = e x p ( - i2h2/4t) e x p (ihx/2t)w(dx).
W e m a k e a c h a n g e of v a r i a b l e in t h e i n t e g r a l b y p u t t i n g e x p (hx/2t)= 7' Since ~]
is a m o n o t o n i c f u n c t i o n of x, t h e p o s i t i v e m e a s u r e w(dx) changes to a p o s i t i v e m e a s u r e , s a y ~u(d~), a n d we g e t
F
at = e x p ( - i2h2/4t) ~ # ( d ~ ) ,
0
or at T ~' = ~ju(d~]) (1 ~ i ~ n). (10)
J.
PHILIP,Age of a heat distribution
This s e t t i n g of t h e p r o b l e m is k n o w n as Stieltjes' m o m e n t p r o b l e m . A sufficient c o n d i t i o n for t h e possibility of r e p r e s e n t i n g t h e q u a n t i t i e s a~T ~ b y a positive m e a s u r e as i n (10) is t h e strict p o s i t i v i t y of t h e forms Q1 a n d Qe. (See e.g. K r e i n , 1951.) Thus, if T 0 < T < T 1 we h a v e a r e p r e s e n t a t i o n (10), p r o v i n g t h e s o l v a b i l i t y of (5) for T i n t h e open i n t e r v a l (To, T1). B y L e m m a 1, however, (5) is t h e n solvable for e v e r y v > T o.
T h e o r e m 1% shows t h a t t h e s o l u t i o n u 0 c a n be t a k e n as a finite s u m of Dirac measures.
Conversely, if (5) has a solution fi for ~ > 1 we h a v e
Q~(~)= f : (~m~m~f~)2/~(@)~O,
a n d
Q2("~) = f : (Ym ~m ~m)2 (1/r~) fi(d~l)~
O.i m p l y i n g f >~T o. I n t h e case 3 o > 1 there is c o n s e q u e n t l y n o solution for T in t h e open i n t e r v a l (1, To), p r o v i n g t h a t
T=h2/4
log T 0. I n t h e case 3 0 = 1 the eqs. (5) are solv- able for e v e r y t >~ 0 i m p l y i n g T = ~ .Remark.
The p o s i t i v i t y of Q1 or Q2 is a c o n d i t i o n on a n odd n u m b e r of consecutive% F o r n odd, QI~>0 is a c o n d i t i o n on a2, a3, ..., an_ 1 a n d Q2>~0 a c o n d i t i o n on al, a z ... a n. F o r n even, Q1 ~> 0 is a c o n d i t i o n on a2, a s ... a~ a n d Q2 >~ 0 a c o n d i t i o n o n a l , a2, ..., an_ 1.
Example 1.
L e t n - 3 so t h a t al, a 2 a n d a s are given. T h e p o s i t i v i t y of Q1 a n d Q2 t h e n gives the i n e q u a l i t i e sa2v4~O
a n dalaaTl~
I fa~>ala3, we
get t h e e s t i m a t e 3 o-a2/~hla a
orT =h2/4
log(a2/~hla3).
I fa~ •ala3, T = c<).
Example 2.
This e x a m p l e shows t h a t there does n o t necessarily exist a solution of (5) for t = T. L e t n - 4 so t h a t al, ae, a a a n d a 4 are given. The p o s i t i v i t y of Q1 a n d Q2 givesala3Tl~
a n da2aaT2~
Suppose
aeo/aiaa>max
(1,a~/a2aa)
so t h a tTo=a2/ala 3.
2 2 K r e i n (1951) shows t h a t if t h e r e exists a r e p r e s e n t a t i o n (10), there exists one of t h e formi~ ~+ i (11)
a~T = ~ 1 ~ 2 ( 1 ~ < i ~ < 4 ) , ~ 1 > 0 , ~ 2 > 0 , ~ 1 ~ < ~ 2 9
T h e four eqs. (11) are a c t u a l l y j u s t sufficient to d e t e r m i n e t h e four q u a n t i t i e s ~1, ~z, r h a n d ~72. C o m b i n i n g t h e eqs. (11) for (1 ~<i<3) we get
Ql~O2~]l~]2(r]l --~/l) 2 = Ts(ala3 T2 --a22).
F o r T =3o these expressions e q u a l 0. Thus, either ~]1 = 0 or ?]1 =~2, so t h e r e p r e s e n t a - t i o n has o n l y one p o i n t m a s s a n d h a s the form Q2~. This r e p r e s e n t a t i o n does n o t satisfy t h e e q u a t i o n for i - 4.
IV. Nonequidistant data
W h e n t h e p o i n t s x~ are n o t e q u i d i s t a n t , we o n l y give a n e s t i m a t e t h a t t a k e s i n t o a c c o u n t three p o i n t s a s =v(xi). Of course, if more t h a n three v a l u e s are k n o w n , those three t h a t give t h e smallest e s t i m a t e T should be used.
ARKIV FOR MATEMATIK. Bd 7 nr 26 Theorem 2. Let a~=v(x~)>O (1 ~<i~<3), be the values o] v at three points xl <x~ < x z.
Then,
T - (1/4) (x 3 - x i ) ( x 3 - x 2 ) (x 2 - x i )
(x a - x l ) log a 2 - ( x a - x 2 ) log a ~ - (x~-x~) log a a i / t h i s expression is positive. I n other cases T = ~ .
Proo]. B y (7) a n d (9) T can be characterized b y the fact t h a t if t > T, t h e n there exists a plane in/~a separating a f r o m Pt- L e t
or
= ~lYl + ~2Y2 -{- ~3Y3 = 0be a plane in R 3. We normalize it b y p u t t i n g as = - 1 . We shall t r y to find a~ a n d
~3 so t h a t
/(x) : aTpt(x) >~O for all x
a n d o~ra ~ O.
I f this is possible, we obtain a plane t h a t separates (not strictly) a a n d Pt. F u r t h e r , we shall choose al a n d a3 so t h a t the plane supports the cone P t along the r a y t h r o u g h pt(O) for some O, giving t h e e q u a t i o n
/ ( 0 ) = aTpt(O ) = 0. (12)
Since we r e q u i r e / ( x ) >~ O, we m u s t also h a v e
/'(O) = arp~(O) = 0. (13)
The equations (12) a n d (13) determine al a n d a3 as functions of t a n d @:
al(t, @) = [(x 3-x2)/(x ~ - x l ) ] exp ((x I - x 2 ) ( x 1 + x 2-2@)/4t), a3(t, | = [(x2 - X l ) / ( x 8 -x~)] exp ((x z - x 2 ) ( x 3 +x2 - 2 | To see if there exist t a n d @ so t h a t
ara = al(t, O ) a ~ - a 2 +a3(t, O ) a 3 ~ 0
we seek m i n 0 a r a b y solving (~/~O)aTa=O. W e find a unique solution O0 = 89 + 89 + (2t log a3/al)/(x 8 - x l )
a n d get
mino ra X" X"a exp ((x2- (x3- ]
(14)
F o r some al, a 2 a n d a3, this expression is n o t negative even when t--> co, m e a n i n g t h a t there does n o t exist a n y plane separating a a n d P t even for large t. F r o m such a,, our t h e o r e m gives no estimate. F o r other a~, t h e expression is negative for large t.
T h e n T is the value of t for which m i n o a r a = 0.
J . PHILIP, Age of a heat distribution
Remark. F o r three equidistant points, Theorem 2 gives the same estimate as T h e o r e m 1. Cf. example 1.
Remark. F o r equidistant xi, t h e m e t h o d of proof for t h e o r e m 2 can be generalized to handle 5, 7, etc. points b y introducing 2, 3, etc. points of c o n t a c t like @. Such a generalization gives e x a c t l y the estimate of theorem 1. Also for n o n e q u i d i s t a n t x~, t h e generalization of t h e above m e t h o d is usable, b u t explicit expressions correspond- ing to (14) are n o t obtained.
V. P r a c t i c a l m e a s u r e m e n t s
Since it is h a r d to construct a t h e r m o m e t e r t h a t gives the t e m p e r a t u r e v(x~) at a single point, we shall consider t h e situation t h a t the information a b o u t v is given b y the values of n ( ~ 3) linear functionals
d~ = S v ( x i - x ) v ( d x ) (1 <~i<~n), (15) where v is a positive measure with total mass equal to 1. The measure v is t h o u g h t t o describe the measuring i n s t r u m e n t a n d the values d i are obtained when the in- s t r u m e n t is " c e n t e r e d " at the points x~. Physicist often assume
v(dx) = ( 1 / 2 ~ a ) e x p ( - x2/2aZ)dx = ygo,12(x)dx (16) for some a > 0 . The limit case w h e n a - + 0 corresponds to v-~(~ a n d d~->v(x~)=ai, t h a t is the case in secs I I - I V . W e pose
Problem II. For given d i > 0 (1 ~ i <~n, n>~3), ]ind an upper b o u n d / o r the supre- m u m T o / a l l t / o r which there exists a positive bounded u o satis/ying (15).
Remark. This problem has n o t always a solution since there m a y n o t exist a n y t for which the d~ (1 ~<i ~<n) is a conceivable set of data. Cf. the corollary of L e m m a 3.
The equations (15) can also be written
d~=v~(x~) ( l < i < n ) (15')
or b y (3) d~ =uo-)~t-)~(x~) (1 <~i<~n). (15")
L e t ut = Y3t~v (17)
so t h a t di = ut~eUo(X~) (1 ~<i~n). (18)
B y l%ogosinski (1958), t h e o r e m 1 a n d corollary 1 we get
Theorem R'. There exists a positive bounded u o satis]ying (18) i/ and only i/ the point d = (dl, d 2 ... d~) C R ~ is in the hull cone K t o] the curve
k~(x)=(~(xl x),z~(x2-x),...,z~(x~-x)), - ~ < x < + ~ .
B y this theorem we have
A R K I V F O R M A T E M A T I K . Bd 7 n r 26
= s u p {t : dEKt}.
Most of the results of section I I h a v e their c o u n t e r p a r t s here.
L e m m a 2. K t ~ Kt~ i/ t l <t 2.
(19)
(20)
Pro@ If y E Kt~ there exists a positive measure/~ such t h a t y = # ~-kt, = # %Pt~ % v =
#-)~t2_t~-)~pt~%v--#~-~)t2_t~-~kt~. Since/~%yJt~_t~ is a positive measure t h e l e m m a is proved.
B y this l e m m a we have, as in section I I , t h a t
= s u p {t: d E K t } = i n f {t: d ~ K t }
(21)
if we interpret the i n f i m u m as c~ w h e n d E K t for all t ~> O.
Define
0 = sup {t :v =~t~-~, ~ positive measure}.
T h e total mass of ~ equals t h a t of v so it is equal to 1. We can t h i n k of Y~t, v a n d )t as probability distributions corresponding to r a n d o m variables Xv, X , a n d Xx.
T a k i n g variances, we get
Var (X~) = 2v~+Var (Xx).
Since yh-~8 when t - ~ 0 a n d Var (Xx)>~0 we h a v e 0 < v~ < 89 Var (X,).
Of course, v~ > 0 only when v is infinitely differentiable. F o r the particular v in (16) we get ~ = 89 Var (X,) = 89 2.
L e m m a 3. K t c Pt+ ~.
Pro@ I f y E K t there exists a positive measure # such t h a t y = k t ~ e / ~ = p t ~ e r ~ e # = p t ~ o o ~ t % / ~ - p t + o ~ # . Since ~ # is a positive measure, the l e m m a is proved.
Corollary. I n general, K o = limt_~o K t is not the whole positive orthant. Since Problem 11 has a solution i / a n d only i / d E Ko, the condition d~ > 0 (1 <~ i <~n) is not in general su//ieient ]or Problem I I to have a solution.
Theorem 3. Assume d E K o and apply Theorem 1 or 2 with a = d so that an estimate T is obtained. Then,
~< T - v ~.
Pro@ F o r every t < T we h a v e d E K t c P t + ~ b y L e m m a 3. Then, t + ~ < ~ T = sup {t : d E P t } p r o v i n g t h e theorem.
The physical interpretation of t h e last formula is t h a t the dispersion of u 0 caused b y h e a t c o n d u c t i o n during the time T a n d the dispersion caused b y the measuring instrument, which is larger t h a n if it h a d been aged ~, together f o r m a dispersion corresponding to h e a t conduction during a time T.
Z. PHILIP, Age o f a heat distribution
A C K N O W L E D G E M E N T
T h e problem here t r e a t e d was s u g g e s t e d by I-Ians R h d s t r S m . H e h a s also m a d e m a n y v a l u a b l e r e m a r k s during t h e p r e p a r a t i o n of this paper.
Department o] Mathematical Statistics, Royal Institute o/ Technology, Stockholm, Sweden
R E F E R E N C E S
KRE~N, M. G., T h e ideas of P. L. C h e b y s h e v a n d A. A. Markov in t h e t h e o r y of limiting v a l u e s of integrals a n d their f u r t h e r development, Amer. Math. Soc. Translations, ser. 2, n o 12, 3-120 (1951).
ROGOSI~SKI, W. W., M o m e n t s of n o n - n e g a t i v e mass, Proc. Roy. Soc. (London), A245, 1-27 (1958)
T r y c k t den 8 april 1968
Uppsala 19ff8. Almqvist & \Viksells Boktryckeri AB
