Control of the wave equation at a rational point

Systems & Control Letters 19 (1992) 83-86 North-Holland

83

Control of the wave equation at a rational point T.P. Svobodny

Department of Mathematicsand Statistics, WrightState University, Dayton, USA Received 6 August 1991 Revised 18 January 1992

Abstract: The wave equation is controlled by a force at one point, which is a rational point of the interval [ - 1, 1]: geometric interpretations are given of the controllable subclasses.

steered to zero when the control point x~ is rational. We are not concerned with the topology of the steerable spaces and will consider f to be continuously differentiable and we will construct g as a function that is defined everywhere. It is well-known [1] that one can use the d'Alembert solution of the Cauchy problem to represent the solution of the boundary value problem in the following way. Let the data of the problem be extended in the odd way past each of the lines {x = odd integer}, then

Keywords: Distributed systems; controllability; wave equation; characteristics; Hadamard matrix.

+f

1. P r o b l e m s t a t e m e n t

We consider the forced wave equation

u,-u**=g(t)8(X-Xc)

: = F ( x , t),

xc~(-1, 1), with homogeneous boundary conditions, u(-a,

2 u ( x , t) = f ( x + t) + f ( x - t )

ff(x, t) dx dt

(x, t)

where f, if, ff are the odd extensions (in the sense we have described) of the data and A is the triangle with vertex set { ( x - t , 0), (x + t, 0), (x, t)}. The sides of A(x, t) are the x-axis and the two characteristics passing through the point (x, t). We can write

t) = u(1, t) = 0,

2ut(x, t) = f ' ( x + t) - f ' ( x -

t)

and the initial conditions,

u(x,

0) =f(x),

+ -~ ~ f f x , , F ( x '

u,(x, t) = o.

The control problem discussed here is that of choosing the function g in order to satisfy the conditions u(x, T ) = 0 = ut(x, T) for some time T. It is known [2] that this problem is soluble for general initial data, for time T = 4, in the case where xc is irrational. The purpose of this note is to characterize the initial functions that can be

2u~(x, t) = f ' ( x + t) + f ' ( x - t )

o

+ ~XX

F(x,t) dxdt, (x, t)

which is equivalent to the previous expression, because of the boundary condition and the assumed smoothness of f. In order to solve the control problem we set

Correspondenceto."T.P. Svobodny,Department of Mathematics and Statistics, Wright State University, Dayton, OH 45435, USA.

t) dx dt,

u,( x, 7") = O = ux( x, 7")

0167-6911/92/$05.00 © 1992 - Elsevier Science Publishers B.V. All rights reserved

84

T.P. Svobodny / Control of the wave equation at a rational point

and then solve the resulting equations for the control g. That is, the equations to solve are

O=f'(x+T)-~'(x-T) + -~t f f a(x, T)ff (x, t ) d x dt,

(1)

O=/'(x+r)+f'(x-V) + O_ff

OX a - ' A(x, T)

=x c

Associated to x c := xc(0) are the points Xc(1 ) = 2 - - ,

P q

xc(-1) =-2---

xc(2)=4+--, P , q

P q

xc(-2) = -4+--

P , q

P +x(m), q

m<~p,

P 4+---x(m),

m>p,

q

(2)

We will show that the equations (1), (2) will reduce to solving a family of finite dimensional linear systems parametrized by the interval [0, 1/q). We will only consider the case T = 4; other cases, including those for which T < 4 can be considered. For example, if p = 0 (we have control at the midpoint) then T = 2 works as well as T = 4 . We note that the problem could be solved by more sophisticated analytic techniques, but it is not clear that the resulting conditions can be related to the simple geometric conditions resulting from the elementary analysis given here.

p/q

4--t(0, m ) =

l~(x,t) d x d t .

2. Control at the point

Associated to each [x] is a set of t's, which are the times that the characteristics through the points (x(m), 4) intersect the lines x = x c ( P ) , P = 0 , +1, +2:

t(1, m ) = 2 + - t(-1,

P q

m) =2---

+x(m) P q

fx(m)

-x(m) -

t(2, m ) =

P q

--,

N.D.,

t(-2,

P-x(m), q N.D.,

m) =

Vm, Vm, m >p,

m ~p.

Note that each pair of characteristics corresponds to exactly 4 times and that each time in the set appears twice in the above description; these 4q times can be indexed from first to last: 0 < t I < t 2 < " ' " < t4q < 4. W e shall suppose that this has been done. The equations to determine the control (1,2) can be written 4

which are obtained from x c by reflections in the (x, t) plane about the lines x = +_ 1. To each x ~ (0, 1/q), we can associate a set of 2q points, denoted [x], which we will call the orbit of x. This set is obtained as the images of x, under reflection in the lines x = m/q, - q <~m <~ q, m 4: 0; the points x(j) of [x] are defined as

/m {m

x(m)=

m-1

In this way,

x(-m)

4

0 =f'(x

+ 4)

+f'(x -

4) + ~ ( - 1 ) l k g , ( ' r k ( X ) ) , l

(4)

g= +x,

m=2M+l,

lk= x,

= -x(m).

m = 2 M + 1.

g, -g,

P=0, +2, P= +1,

and

m = 2M,

m+l q

(3)

1

m = 2M,

q +x,

m<0,

4) + E ~ ( ~ ' k ( x ) ) ,

where

q-X,

m>O, x(m)=

0 = f ' ( x + 4) - f ' ( x -

1,

P= -1, -2,0(m>p),

2,

P=l,2,0(m<~p).

The symbol ~'k represents the mapping referred to previously, i.e. r :[x] ~ {t(P, m)} is the assignment, to each point in the orbit of x, of the set of

85

T.P. Svobodny / Control of the wave equation at a rational point

four times that the pair point ( x ( m ) , 4) intersect 0, +1, +2. In order to elucidate consider the example of = 5,2 we have

of characteristics at the the lines x = x c ( P ) , P =

We only have to consider x • (0, 1 / q ) and for each such x, we will have a system of linear equations to solve for the unknowns {~(%(x)): x [x]} as described below. For each orbit [x], we cycle through the times tl < " ' " < / 4 q (the columns of the matrix) twice. To make the construction of the matrix easier to see we describe this double cycling in some detail. The first cycle is

the above construction, 2 xc = 3. Here, with x~(0)

Xc(1 ) = 45, Xc(2 ) ----3-, 14 Xc(-1) =-7, 8 x~(-2) =

10 3 •

Now pick x ~ (0, ½). The orbit, [x], is found as the images of x under reflections in lines x = 3m, - 3 ~ < m ~ 3 , m=~0. Thus,

t I = t(2, p + 1) . . . . . tq_p = t(2, q),

[ x ] = {x(1) = x , x ( 2 ) = 32 - x ,

t2q+l = t ( - 1 ,

-p-

t3q-o+,

-q)

x ( 3 ) = ~ +2 x , x ( - 1 ) = - x , = - 7z + x , x ( - 3 )

x(-2)

= - 72 - x } .

2

4

.....

t4q_o = t ( 0 , ~-

-q),

- 1),

t(O, p ) .

The second cycle is

4

{x, 3 - x , 7 + x , ~ - x , 3 + x , 2 - x , 8

= t(0,

1) . . . . . t3q_ o = t ( - a ,

t4q_o+ , = t(O, 1) . . . . . t4q

The set of times is easily computed to be 2

q),...,t2q=t(-1, -p),

tq_o+, =t(-1,

8

t 1 = t(-2,

p),...,tp

= t(-2,

-1),

10

2 + x , 7 - x , 3 + x , ~ - x , 3- + x , 4 - x } . to+ , = t ( - 2 , In Figure 1, we illustrate this case: one can see the orbit of x on the line t = 4 ; the vertical dashed lines are the lines xc(P); x(1) has both characteristics drawn; one characteristic for x(3) is drawn to illustrate that these two points share a time, i.e., there exist k, l so that % ( x ( 1 ) ) = rl(x(3)).

~3X-2 ~

-'i" ..... I I

I I

xl

to+q+ , : t(1, - q ) . . . . . t2q = t(1, - p -

t2q+l

=

1),

t(1, - p ) . . . . . t3q+p = t(1, q ) ,

t3q+p+l = t(0, q) . . . . , t4q = t(0, p + 1).

Let yT = [g(tl) ' g(tz) . . . . . g(t4q)].

x 2 x~ -y "--"

Then the

t=4

i

i',

t

.'"

"'' /

""

[

.

i

.

1,' ,

..

I

"1. "-.

I i

i i I

i i'"

I

."

I • I

i

-q),

'

t

,

1) . . . . . to+ q-= t ( - 2 ,

" . " [ .~r -

o

I

i

I

i

I

I[

i

i

] -

, T

Fig. 1.

.. " '

" 1" 1_~

t=O

86

T.P. Svobodny / Control of the wave equation at a rational point

problem splits into the solution of the two systems of linear equations:

A~y=O,

is satisfied by the data if

vTz=O,

Vv~ker

AT;

(5)

A2y=z, but ker AT = ker[Az, .z~2]T= ker ,42T = ker X 1, which we know is given by

where zT = [-2f'(x(-q)),

-2f'(x(-q

+ 1)) . . . . ,

{P E R 2 q : L', = v 2 . . . . .

L'2q }.

-2f'(x(q))], Thus the orthogonality condition is

and

E Z k = O, k~O

A, =

where

which for our data means 11

alj!) = 1,

i+j=p+q+l Vi+j=p+5q+l Vj- i = 3q-p V i-j =p, i+j=3q-p+lVj-i=p+q, otherwise,

1

f ' ( y ) = 0, yc[x]

0
(6) q

and

3. Conclusion

A 2 = aij(2) ,

The points x = n / q can be dealt with similarly, and when the required conditions hold for f ( x ) , it is easily seen that the resulting control g(t) is continuous. It would be nice to have a complete characterization of the condition (6). Because of the geometric way in which it was derived, it should not be surprising that there is a geometric interpretation. The following result, which is easily established, goes part way to such a geometric characterization. Let ~b denote the periodic extension of a function ~b defined on [ - 1, 1], i.e., ~b(x + 2) = qS(x). If f ' is odd-invariant under any of the lines {m/q}, then (6) is satisfied. One can also investigate this problem using Fourier series, obtaining the condition on f that the multiples of 2q in its transform must vanish. It is not clear whether there is a direct connection between this condition and (6).

where '1, --1,

(2) aij =

L0,

i + j = 3q - p + 1 v i - j = p v j - i = 3q - p , i+j=p+q+ l vi+j=5q+p+ l vj-i=p+q, otherwise.

These are examples of Hadamard matrices, which appear frequently in combinatorics. If A (j) denotes the j-th column of A, then we have A(1k) =A~ 4q-k+l), thus if we split the matrix A 1 = [Al, A1], and setting w~ =Yk +Y4q-k+l, we have

which, from the form of A, gives W1 = W2 =

...

= W2q ,

which, for the control function g, means

g( tl) + g(t4q) . . . . . The equation

A2Y = z

References

g(t2q ) + g(t2q+L). [1] R. Courant and D. Hilbert, Methods of Mathematical Physics, Vol. II (Interscience, New York, 1962). [2] J.L. Lions, Exact controllability, stabilization and perturbations for distributed systems, SlAM Rev. 30 (1988) 26.