Exercise 2.1.2 A Nonhyperbolic Cauchy Problem for the Wave Equation)
8.1 The one-dimensional case
and standard energy integrals.
To conclude this brief introduction, we recall a known relation between hyper- bolic problems and corresponding spectral problems.
Letumbe Neumann eigenfunctions of the elliptic operatorAwhenb=0,0≤c and the coefficients do not depend ont:
Aum=λ2mumin, a∇umãν=0 on∂.
It is known that this self-adjoint elliptic problem has an L2() complete and orthonormal system of eigenfunctionsumcorresponding to nondecreasing eigen- valuesλ1≤λ2≤ ã ã ã ≤λm≤. . .. For simplicity we will assume that 0< λ1. In fact, this inequality can be achieved by using substitutionu =veτt. So a solution to the hyperbolic problem (8.0.1)–(8.0.3) witha0=1,u0=u1=0,f =0 can be written as
u(x,t)=
g1m(t)um(x), g1m(t)=
∂×(0,t)
sin(λm(t−τ))/λmum(y)g1(y, τ)dγ(y)dτ,x,y∈∂ (see, e.g., the book of Lions [Li, Ch. 4, section 7, p. 320]). The series foru is convergent inL2(). From the last two formulas it is not difficult to see that the Neumann-to-Dirichlet operatorlis an integral operator with the kernel
sin(λm(t−τ))/λmum(x)um(y)(the sum overm=1,2, . . .).
The data of the inverse spectral problem are Neumann eigenvalues λm and the Dirichlet traces on∂theL2(∂) normalized Neumann eigenfunctions. The above formula shows that these data uniquely determine the Neumann-to-Dirichlet lateral operatorl. Therefore most of the uniqueness results for coefficients with the given l give uniqueness results for the inverse spectral problem. For more results on inverse spectral problems we refer to [KKL].
8.1 The one-dimensional case
We consider two simple inverse problems that illustrate limitations and possibilities of the multidimensional situation. These problems are one-dimensional, which makes them relatively easy, but still there are some fundamental and unanswered questions about them.
Letbe the half-axis (0,+∞) and Q=×(0,T). Let us consider the hy- perbolic equation
(8.1.1) (∂t2−∂x2+c)u =0 inQ
with the coefficientc=c(x)∈C[0,∞), the zero initial data (8.0.2) (u0 =u1= 0), and the Neumann datag1=δ(0) on the lateral surfaceγ1×[0,T), γ1=∂.
Hereδis the Dirac delta function. When the datag1∈C1[0,T] and satisfy the com- patibility conditiong1(0)=0, there is a unique solutionu ∈C2(Q). Letu0(x,t)
be 1 when x<t and 0 whent ≤x. This function is a generalized solution to the string equation satisfying the Neumann condition with the delta function. Let u=u0+U. ThenUsolves the initial boundary value problem
(∂t2−∂x2)U = −c(u0+U) inQ,U =∂tU=0 on× {0},
∂xU =0 on∂×(0,T).
We extend U,c, . . . onto {x<0} by the symmetric reflection U(−x,t)= U(x,t),c(−x)=c(x),u0(−x,t)=u0(x,t). By d’Alembert’s formula, the ex- tended initial value problem is equivalent to the integral equation
(8.1.2) U(x,t)= −1
2
con(x,t)
c(u0+U),
where con(x,t) is the backward characteristic triangle {(y,s) :|x−y|<|t− s|,s<t}. This Volterra-type equation has a unique solutionU∈C(Q). Indeed, let us define the operator
BU(x,t)= −1 2
con(x,t)
cU.
By using induction, it is not difficult to show that
BkU∞(Q)≤à2kT2k/(2k)!U∞, à= c1/2∞(Q).
This bound implies that the operator (I−B) is invertible (inC(Q) orL∞(Q)) with inverse (I−B)−1=I+B+ ã ã ã +Bk+ ã ã ã. From this formula and the bound onBkwe have(I−B)−1 ≤eàT. The Volterra integral equation (8.1.2) can be written as (I−B)U=Bu0, so its unique solutionUcan be obtained via the inverse operator. From the above bounds it follows that
U(;c)∞(Q)≤ 1
2T2eàTc∞(Q), (8.1.3)
U(;c2)−U( ;c1)∞(Q)≤ 1
2eà1T(1+1/2à22T2eà1T)T2c2−c1∞(Q).
To obtain the second bound, observe that by subtracting the integral equations (8.1.2) forU2,U1we will have
(U2−U1)(x,t)= −1 2
con(x,t)
(c2−c1)(u0+U2)−1 2
con(x,t)
c1(U2−U1), which can be considered as a Volterra integral equation with respect toU2−U1. Denoting by B1 the operator B withcreplaced by c1 as above, we obtain the bound onB1kwithàreplaced byà1= c11/2∞, and the corresponding bound on (I −B1)−1. Using the first inequality (8.1.3) to boundU2∞, we can conclude that theL∞-norm of the first term on the right side of the equation forU2−U1is not greater than
1
2T2c2−c1∞(Q)(1+1/2à22T2eà1T).
Thus, we obtain the second bound (8.1.3).
8.1. The one-dimensional case 223 Differentiating the integral (8.1.2) with respect to x,t, we conclude that
∇U ∈C(Q+), where Q+ is {x≤t} ∩Q. The differential equation forU can be understood in the generalized sense.
In the inverse problem, one is looking forc∈C[0,+∞) given the Dirichlet data (8.0.4) on∂×(0,T).
Lemma 8.1.1. The inverse problem is equivalent to the following nonlinear Volterra-type integral equation:
c(τ)+2 τ
0
c(s)∂tU(s,2τ −s;c)ds= −2g0(2τ), g0(0)=1,g0(0)=0,2τ <T,
(8.1.4)
where U is a solution to equation (8.1.2).
PROOF. We define the functionvas 1 whent+x<2τ <T and as 0 otherwise.
We approximatev by functionsvε(t+x) inC∞such that thevε are decreasing with respect toε, andvε=von∂×(0,2τ−ε). From the definition of a weak solution to the problem (8.1.1), (8.0.2), (8.0.3) we have
Q
u(∂t2−∂x2+c)vε=
∂×(0,T)
(−g0∂xvε−vεg1)dt.
Using thatvε solves the homogeneous wave equation, that thevε converge tov inL1(Q) and inL1(∂×(0,T)), and that the∂xvεconverge to the Dirac delta function concentrated at 2τ asε→0, in the limit we obtain
Q
cuv= −g(2τ)−1.
We denote the right side byF1(τ) and observe that the first integral is actually over Tr(τ) (the triangle with vertices at the points (0,0), (τ, τ), (0,2τ)), wherev=1.
Writing this integral as a multiple one, we obtain τ
0
c(s)
2τ−s s
(1+U)dt
ds=F1(τ).
We differentiate both sides with respect toτ to get τ
0
c(x)2(1+U(x,2τ −x))d x=F1(τ).
Differentiating once more and using that U(τ, τ)=0, we obtain the relation (8.1.4).
Letcbe any solution to equation (8.1.4) and letg• be the data of the inverse problem generated by this coefficient. Starting with the relation (8.1.4) and using the conditions ong0, return to the above relation containing g0(2τ)+1. On the other hand, repeating the previous argument, we will obtain the same relation with g•instead ofg0. Since both relations are valid when 0< τ <T/2, we conclude thatg=g•on (0,T).
The proof is complete.
By inspecting equation (8.1.4), one concludes that whenc∈C[0,T], the data g0must be inC2[0,T].
Corollary 8.1.2. A solution to c to the inverse problem is unique on(0,T/2).
Let g0∈C2[0,T].
Then there is T0≤T such that a solution c to the inverse problem exists on [0,T0/2].
PROOF. We will show that the nonlinear operator B•c(τ)=2
τ
0
c(x)∂tU(x,2τ −x;c)d x fromL∞(0,T) into itself admits the following bound:
(8.1.5) B•c2−B•c1∞(0,T)≤C(M)(T−T0)c2−c1∞(0,T), whereC(M)=(4Me+e2), provided thatcj∞≤ M2,T M ≤1,0≤T0≤T, andc1 =c2 on (0,T0). This implies that for smallT −T0 the operator B• is a contraction, and then uniqueness follows.
Observe that subtracting two integral relations (8.1.2) forU2andU1, we con- clude that for fixedc1the operatorU2−U1can be considered as linear with respect toc2−c1. Therefore, we can write the difference of values ofB• atc2andc1as B(c2−c1), whereBis a linear operator. We have
B•c2−B•c1=2 τ
0
(c2−c1)∂tU2+2 τ
0
c1∂t(U2−U1).
Differentiating the integral (8.1.2) with respect totand subtracting the equations forU2andU1, we have
∂t(U2−U1)= −2−3/2
∂con(x,t)
((c2−c1)U2+c1(U2−U1)).
Using that U1=0 when c1=0 and the first bound (8.1.3), we conclude that∂tU2∞(Q)≤ M/2eT M, and therefore the L∞-norm of the first integral is bounded by
(T −T0)c2−c1∞MeT M.
Using the second bound (8.1.3) withT replaced byT −T0(c1=c2 on (0,T0)), we similarly conclude that
∂t(U2−U1)∞(Q)≤ 3
2eM T +1 2e2M T
(T −T0)c2−c1∞. We conclude thatB•c2−B•c1∞is bounded by
(4MeM T+e2M T)(T −T0)c2−c1∞. Using the conditionT M≤1, we obtain (8.1.5).
As we noted above, global uniqueness of the solution of equation (8.1.4) follows from (8.1.5). Indeed, letT0be maximal with the property thatc1=c2on (0,T0).
8.1. The one-dimensional case 225 IfT0<T, then we can find (probably smaller)T such thatC(M)(T −T0)<1.
From (8.1.5) it follows that the operatorB•is a contraction inL∞(T0,T), which as in the Banach contraction theorem implies uniqueness of a solution of equation (8.1.4). Soc1=c2 on (T0,T), which contradicts the choice ofT0. Accordingly, T0=T, and we have global uniqueness.
Local existence also follows from the bound (8.1.5) and the Banach contraction theorem. Indeed, letg2(τ)= −2g0(2τ) andr = g2∞(0,T). LetM =√
2r. We can assume that T is so small that T M ≤1 andC(M)T < 12. Consider the set S⊂L∞(0,T) of functionscsuch thatc−g2∞(0,T)<r. Equation (8.1.4) can be written asc=g2−B•c. We claim that the operator on the right side of this equation maps S into S and is a contraction. Indeed,c∞(0,T)≤r+r=2r whenc∈ S, soc∞≤ M2, and we have the bound (8.1.5), which implies that B•c2−B•c1•≤ 12c2−c1∞. Now it is easy to check that the operator maps Sinto itself.
The proof is complete.
Observe that the existence of a solutioncto the inverse problem is guaranteed when
2r T2≤1, (4√
2r e+e2)T < 1 2, wherer =2g0∞(0,T).
The next problem about determination of the speed of propagation is more complicated, even in the one-dimensional case.
We consider the equation
(8.1.6) (a20∂t2−∂x2)u=0 inQ with the same initial and lateral boundary conditions.
Again, the inverse problem is to determinea0 fromg0. We will study it in the new (characteristic) variables
(8.1.7) y=y(x)=
x
0
a0(θ)dθ, τ =t.
Lemma 8.1.3. Equation(8.1.6)is equivalent to the following one:
(8.1.8) ∂τ2v−∂y2v+c(y)v=0in Q, where
c(y)=1 2∂y2l+1
4(∂yl)2,l(y)=lna0(x(y)) for the function
v(y, τ)=a01/2(x(y))u(x(y), τ).
PROOF. This lemma follows from the following elementary calculations.
From formula (8.1.7) according to the chain rule, we have∂xy=a0(x),∂x = a0(x(y))∂y,∂t =∂τ. So
∂xu = −1
2a−3/20 ∂xa0v+a01/2∂yv,
∂2xu =a3/20 ∂y2v− 1
2a0−3/2∂2xa0−3/4a−5/20 (∂xa0)2
v, and equation (8.1.6) is equivalent to equation (8.1.8) with
c(y)= 1
2a0−3∂x2a0−3
4(a0−2∂xa0)2,x=x(y).
Observe that this expression forccan be written as
−1
2a−10 ∂x2(a0−1)+1
4(∂x(a0−1))2,
or using that according to (8.1.7), ∂yl=a−01∂ya0 =a0−2∂xa0= −∂x(a0−1) and thereforea0−1∂x2(a0−1)= −a0−1∂x∂yl= −∂y2l, we arrive at formula (8.1.8) forc.
The proof is complete.
In the next result it is quite helpful to make use of smooth Neumann data, which are also natural in several applications. We will consider
(8.1.9) g1∈Ck[0,T], g1(k−1)(0)=0,k=1,2, . . . .
First, we observe that due to time independence of our equations, a solutionu(;g1) to equation (8.1.6) inQwith the zero initial condition (8.0.2) and the Neumann condition (8.0.3) admits the following representation:
(8.1.10) u(x,t;g1)= t
0
g1(s)u(x,t−s;δ)ds.
Exercise 8.1.4. Prove relation (8.1.10) by using definition (8.0.51).
According to (8.1.10), the datag0(;g1) andg1(;δ) of the inverse problems with the Neumann datag1andδare related via
(8.1.11) g0(t;g1)= t
0
g1(s)g0(t−s;δ)ds,
which can be considered a Volterra-type integral equation with respect tog0(;δ). In addition to (8.1.9), one can assume thatg(m)1 (0)=0 whenm<k−1. Writing the convolution as the integral ofg1(t−s)g0(s) and differentiating equation (8.1.11) ktimes, we arrive at the Volterra equation
g(k)0 (t;g1)=g1(k−1)(0)g0(t;δ)+ t
0
g1(k)(s)g0(t−s;δ)ds,
which has a unique solution g0(t;δ)∈C[0,T] given any g0∈Ck[0,T]. This implies that the Dirichlet data g0 given for some Neumann data g1 satisfying
8.1. The one-dimensional case 227 condition (8.1.9) uniquely determine the Dirichlet datag0(;δ). On the other hand, the Dirichlet data for the delta function uniquely determine the general Dirichlet data via formula (8.1.11). Hence, only one set of lateral Cauchy data (withg1 =δ or satisfying condition (8.1.9)) uniquely determines the complete lateral Neumann- to-Dirichlet mapg1→g0.
Lemma 8.1.5. The data of the inverse problem uniquely determine a0(0), ∂xa0(0).
PROOF. First, we observe that ifa01,a02 are two coefficients producing the same data of the inverse problem, then
(8.1.12)
Q
(a202−a201)v1v2∗d Q =0
for any solutionv1∈ H(2)(Q) to the first equation inQsatisfying the initial condi- tionsv1=∂tv1=0 on (0,∞)× {0}and for any solutionv2∗to the second equation inQsatisfying the final conditionsv2 =∂tv2=0 on (0,∞)× {T}.
To prove (8.1.12), we subtract equations (8.1.6) with the coefficientsa02 and a01to obtain for the differenceuof their solutionsu2,u1the equation
a022∂t2u−∂x2u=(a201−a202)∂t2u1inQ.
We will take asu01,u02solutions to the hyperbolic problems (8.1.1), (8.0.2), (8.0.3) with zero initial data and with the same Neumann data g1∈C1[0,T],g1(0)= 0. According to our assumptions and to the remark before Lemma 8.1.5, two coefficients produce the same lateral Neumann-to-Dirichlet map, so u01=u02 on{0} ×(0,T). Therefore, the functionu has zero Cauchy data on this interval.
Multiplying the equation for u by any solutionu∗2 of the second equation with u∗2 =∂tu∗2 =0 on (0,∞)× {T}and integrating by parts two times on the left side of the equation and once (with respect tot) on the right side, we obtain the relation (8.1.12) with∂tu1, ∂tu∗2instead ofv1,v∗2. Observe that for anyv1,v2∗satisfying the conditions in (8.1.12) the functions
u1(x,t)= t
0
v1(x,s)ds, u∗2(x,t)= − T
t
v2∗(x,s)ds
solve the same hyperbolic equations and have zero Cauchy data on (0,∞)× {0}, (0,∞)× {T}. Using these functions in the orthogonality relations obtained, we arrive at (8.1.12).
To complete the proof, assume the opposite. Then there are two coefficients a01,a02witha01(0)<a02(0) or witha01(0)=a02(0) and∂xa01(0)< ∂xa02(0). In cases 0<(a02−a01) on the interval (0, ε) for some positive ε. Let v1 be the solution to the first equation inQ∗ =R×(0,T) with right side f ∈C1(Q∗) that is nonnegative and with supp f = {x≤0} ×[0,T0], and with zero Cauchy data on R× {0}. Letv∗2be the solution of the second equation with the same right side and with zero Cauchy data onR× {T}. By using Riemann’s function [CouH, p. 453], it is not difficult to show thatv1>0 near the origin above the characteristict = γ1(x) passing through (0,0)(γ1(x)=a−101(x), γ1(0)=0) and thatv2∗>0 near the
origin below the characteristict =γ2(x) passing through the point (0,T0)(γ2(x)=
−a02−1(x),γ2(0)=T0) whenT0is sufficiently small. For suchv1,v2∗ the integral (8.1.12) is negative. The contradiction shows that our assumption was wrong.
The proof is complete.
To formulate uniqueness results for the coefficient, we assume thata0≤a•. Corollary 8.1.6. Let=(0,1). The coefficient a0(x)∈C2([0,+∞))of the hy- perbolic equation(8.1.6), with zero initial conditions and the lateral Neumann condition−∂xu=δor−∂xu =g1on{0} ×[0,T], where g1satisfies conditions (8.1.9), is uniquely determined on[0,T/(2a•)]by the additional Dirichlet data u=g0on{0} ×[0,T].
PROOF. First, we consider the Dirac function as the Neumann data. From Lemmas 8.1.3, 8.1.5 and Corollary 8.1.2 we conclude that the coefficientc(y) given in (8.1.8) is uniquely determined on (0,T/2) by the data of the inverse problem.
By Lemma 8.1.5 the Cauchy datal(0), ∂yl(0) are uniquely determined as well.
Solving the Cauchy problem for the ordinary differential equation (8.1.8) with respect tol, we obtain uniqueness ofl(y) when 0<y<T/2. So we are given a0(x(y))=el(y). From the definition (8.1.7) of the substitutiony=y(x), we have d x/d y=a−01(x(y))=el(y), and therefore
x(y)= y
0
e−l(s)ds,0<y<T/2.
Sincea0<a•, the inverse functiony(x) is uniquely determined, at least on the interval 0<x<T/(2a•). Finally, we havea0(x)=el(y(x))on this interval.
The claim about the Neumann datag1satisfying condition (8.1.9) follows from the observation that the Dirichlet datag0forg1uniquely determine the Dirichlet data for the delta function.
The proof is complete.
Corollary 8.1.7. (i) The coefficient c∈C[0,1]of equation(8.1.1)with zero initial conditions(8.0.2)and Neumann condition(8.0.3)on{0} ×(0,T)with g1=δor satisfying condition(8.1.9) and zero on{1} ×(0,T)is uniquely determined on [0,T/2],T <2, by the additional Dirichlet data u=g0on{0} ×[0,T].
(ii) The coefficient a0∈C2[0,2]of equation(8.1.6)with the same initial and lateral boundary conditions as in (i) is uniquely determined on[0,T/(2a•)],T <
2a•, by the additional Dirichlet data on{0} ×(0,T).
PROOF. (i) First, we extenduandcas even functions with respect to (x−1) and use the same notation for the extended functions. From the zero Neumann boundary condition on {1} ×(0,T), we will obtain the same hyperbolic equation in the extended domain (0,2)×(0,T). Since in our problem the speed of propagation is 1 (Theorem 8.2) and the initial data are zero, its solution coincides on con(0,T) with the solution of the problem in the strip (0,∞)×(0,T). Now Corollary 8.1.7(i) follows from Corollary 8.1.2.