Euro. Jnl of Applied Mathematics (2004), vol. 15, pp. 129-[46. © 2004 Cambridge University Press 129001: 10.1017/S0956792503005345 Printed in the United Kingdom

Inverse problems of mixed type in linearplate theory

I Department ()f Biological Sciences. University of Warwick. Coventry. CV4 7AL. UKemail: salazar~aths.warwick.ac .uk

2 Dept. of Marhematics and Sralislics. 2500 University Drive NW, Calgary, Alberta. T2N IN4. Canadaemail: westbroo@math.ucalgary.ca

The characterisation of those shapes that can be made by the gravity sag-bending manufac-turing process used to produce car windscreens and lenses is modelled as an inverse problemin linear plate theory. The corresponding second-order partial differential equation for theYoung's modulus is shown to change type (possibly several times) for certain target shapes.We consider the implications of this behaviour for the existence and uniqueness of solutionsof the inverse problem for some frame geometries. In particular, we show that no generalboundary conditiohs for the inverse problem can be prescribed if it is desired to achievecertain kinds of target shapes.

It is advantageous for glass manufactures to shape lenses and car windscreens by thegravity sag-bending process: a rectangular sheet of glass is simply-supported on a hori-zontal frame and is non-uniformly radiatively heated until it sags under its own weighttoward the desired shape. This process, when successful, introduces much less stress in theglass than the more common mould press-bending process and is consequently cheaperand more desirable.

It is the experience of the manufacturers that not every shape can be made by thegravity sag-bending process. Unwanted consequences of trying to manufacture certainshapes are undesirable changes of curvature (which introduce optical defects in the glass)and possibly the lift-off of the glass sheet from the frame, especially near the corners.

Our main goal is to determine what shapes can be made by the gravity sag-bendingprocess. As such, we would like to know whether there is a solution to the inverse problem:"given a certain target shape, compute the temperature field for which that target shape isat steady state." We also wish to know whether or not this inverse problem is well-posedand to establish a strategy to compute its solution.

The glass sheet will be modelled as a linear plate with variable Young's modulus,the Young's modulus being a prescribed function of temperature. Although the as-sumption of linearity may be questionable for certain target shapes due to the largedisplacements required to achieve them, it is the obvious starting point for the devel-opment of a theory for these problems. As described below, the inverse problem in

linear plate theory is surprisingly rich and complex and it deserves study in its own

right . 3 . h h .. h d d· IWe take a coordinate system (x, y, z) 10 R Wit t e z aXIs 10 t e ownwar vertIca

direction. Our problem can now be formulated as follows: given a certain target surfacedefined by z = w(x,y), a positive body force (typically gravity) q = q(x,y), with bothwand q defined in a domain Q U R2, and a value for the (constant) Poisson's ratio v,determine whether the differential equation (02W 102 x2 == Wxx, etc.)

[D(wxx + VWyy)lxx + 2(1 - v)[Dwxylxy + [D(wyy + VWxx)]yy = q, (1.1)

has a solution D = D(x, y) > 0, also defined in Q. D is the flexural rigidity of the plateand is a multiple of the Young's modulus E = E(x,y):

Eh3

D = 12(1 _ v2)'

where h is the plate thickness. Physically, the Young's modulus must be posItive ando ~ v ~ 1/2. Hence we require that D must be greater than zero everywhere in Q.

Equation (1.1) is a linear second-order partial differential equation for D and we wouldlike to formulate a well posed problem for the determination of D and to obtain someidea of the set of functions W for which a solution exists.!

If we non-dimensionalise x and y in (1.1) with a typical length-scale I, q with thegravity load ahg (where a is the surface density and g is the acceleration due to gravity)and W with a typical vertical displacement w·, then the non-dimensionalisation constantfor D would be D' = ahgl4 Iw·. Poisson's ratio is already non-dimensional. Proceedingthis way we find that all the terms of (1.1) are of the same order, so no term can beasymptotically neglected. From now on we will assume that all the quantities involvedhave been non-dimensionalised in this way.

One of the reasons why sag-bending introduces less stress in the glass than press-bendingis that the glass sheet is simply-supported on the frame. Consequently, we concentratein this paper on simply-supported target shapes. We note here that equation (1.1) isunaltered if a linear function of (x,y) is added to w. Since the optical quality of theresulting target shape is of paramount importance in this problem, we will always assumethat w is at least in C4.

The implications of the simple support condition are revealed when we write equation(1.1) in the more explicit form

A = Wxx + VWyy,

B = (1 - v)wxy,

C = Wyy + VWxx,

a = 2(wxxx + Wxyy),

b = 2(wxxy + Wyyy),

c = Wxxxx + 2wxxyy + Wyyyy = V4w;

1 Of course, the forward problem is a linear fourth-order partial differential equatidn for thedisplacement w. For examples of solutions of the forward problem with variable Young's modulus,see chapter V of Mansfield [8].

FIGURE1. Characteristics curves and interior elliptic region of an inverse problem on a flat squareframe (see § 5). For this frame geometry, the frame itself is a characteristic curve.

On any simply-supported boundary with outward normal (1,m), the simple supportcondition becomes Ai2 + 2Bim + Cm2 = O. Since this equation must then have a realsolution (1,m), we must have L1 = B2 - AC ~ 0 near any of the edges where the plateis simply-supported and that implies that (1.2) is hyperbolic or parabolic there. But,inside the domain we usually wish to have target shapes with positive Gaussian curvaturebecause changes in the sign of the Gaussian curvature may induce undesirable opticaleffects. Hence, WxxWyy - W~y > 0 and, as we can see from (1.3), away from the boundariesof the domain, (1.2) is elliptic. We must therefore deal with an equation of mixed type(see Figure 1).

Moreover, (1.2) is degenerate in the hyperbolic regions since for any frame where thenormal is (1,m) and the tangent is (m, -i), simple support implies Ai2 + 2Bim + Cm2 = 0and then dyldx = -ilm is always a characteristic direction. So the frame is an envelopeof characteristic curves or a characteristic curve itself.

As can be seen from this introductory discussion, to understand the analytical propertiesof (1.2), it is necessary to consider simultaneously the elastic properties of the target shapeas well as its geometry.

There have already been some numerical explorations of the gravity sag-bendingprocess. In Manservisi [6] and Manservisi & Gunzburger [7], finite-element codes weredeveloped to compute the Young's modulus for certain target shapes; unfortunately,the changes of type of the inverse problems that they considered and their possible ill-posedness are not addressed. In Engl & Kugler [3], for a simplified model inspired in (1.2),they explore the different numerical performance of their algorithm when their equation

is elliptic versus when it is hyperbolic, and error propagation is also considered in amixed case. In particular, they showed there that changes of type in this kind of inverseproblem are relevant for least-square approaches. Neither of these papers allows for theglass sheet to lift off the frame; for constant Young's modulus, such contact problemswere considered in Westbrook [11, 12]. A preliminary analysis of the changes of type ofthis inverse problem was done in Temple [9].

In this paper we are not able to arrive at a standard formulation of the problem forD, and the relatively simple examples which we examine, display enough complexity tosuggest that there may not be one. The bulk of this paper concerns the analysis of thepossible changes of type for (1.2) and the consequences of these changes of type for thewell-posedness of the inverse problems for certain specific domains. In § 2, we study indetail the one-dimensional inverse problem (beam theory), and in § 3 and §4, we presentand discuss some results for circular plates when w is axisymmetric. In the axisymmetriccase we only have to deal with an ordinary differential equation and we can obtain resultsanalytically. Finally, in § 5 we summarise our previous discussions and describe somefeatures of the inverse problems in square domains.

2 Inverse problems in beam theory

The inverse problem for a beam equation is surprisingly rich In behaviour The beamequation is written as

d2

( d2w(x))dx2 B(x)~ = q(x),

where B = B(x) > 0 is the beam constant (B = EI, where E is the Young's modulus andI is the moment of inertia of the cross-section of the beam). The inverse problem is thesecond-order linear ordinary differential equation

It is clear from (2.1) that whenever w" is zero, we may have a singularity of B. Changesof the sign of w" for the beam equation correspond to changes of type for equation (1.2).It is also clear that the equation is unaltered if a linear function is added to w so that theboundary values of ware irrelevant.

The exact solution of (2.1) is

B(x) = P(x) + ax + f3w"(x) ,

where f3 = B(O)w"(O), a = B'(O)w"(O)+B(O)wll/(O) and P(x) = J;(x-s)q(s)ds. Since q > 0,P(x) > 0 and P'(x) > 0 for x E (0, L], P(O) = P'(O) = 0 and since PII(x) = q(x) > 0 forx E [0, L], P is a COnvex function.

Given a target shape w = w(x), the determination of the constants a and f3 wouldgive us B, but how many degrees of freedom do we really have when we insist in havingB > O? It turns out that this depends on the target shape that we want to achieve through

wI! (because the beam bending moment is M = -Bwl! and B must be positive, the zerosof wI! correspond to the zeros of M). We can distinguish five cases:

(1) wl!(x) =1=0 for all x E [O,L).

In this case, we cannot give precise values to the constants ex and 13 without someextra information. There are two sub-cases that we must consider, each leading toconstraints in ex and 13:

(a) wl!(x) > 0 for all x E [0, L).

Since B(O) = 13/ wI!(0), 13 must be positive and exhas to be chosen so thenumerator of (2.2) is always positive.

(b) wl!(x) < 0 for all x E [O,L).

Then 13 must be negative and we require ex < -(P(L) + 13)/ L.

In either case, we have many choices of ex and 13 for which the numerator in (2.2)will never vanish. To determine B, we are free to complement formula (2.2) withtwo boundary conditions.

(2) wl!(xo) = 0 for a unique Xo E [0, L) and wll/(xo) =1=O.

In this case, the numerator of (2.2) has a simple zero at x = Xo and Xo is aninflection point of w. We can allow this by setting 13 = -P(xo) - exxo. Notice thatnow B(xo)wl!'(xo) = P'(xo) + ex.There are again two possibilities:

(a) wl!(x) < 0 for all x E [O,xo) and wl!(x) > 0 for all x E (xo,L).

Then 13 must be negative so ex > -P(xo)/xo. Since wll/(xo) > 0, theJ:lex > -P'(xo). But we must also avoid any more zeros in B(x), so ex >-[P(L) - P(xo))/(L - xo). Since P is convex, it is enough to require

P(xo)ex> ---.

Xo

(b) wl!(x) > 0 for all x E [O,xo) and wl!(x) < 0 for all x E (xo,L).

Then 13 must be positive and

P(L) - P(xo)ex<------.L-xo

Consequently, all these target shapes ware achievable. In both sub-cases, westill have a one-dimensional family of solutions and we are free to prescribe oneboundary condition for B.

(3) wl!(xo) = wll/(xo) = 0 and wl!(x) =l=0 for x E [O,L) - {xo}.

Now we must have ex= -P'(xo), 13 = -P(xo) - P'(xo)xo. We no longer have anydegree of freedom for B. Indeed, smoothness completely determines B.

Notice that

Consequently, we must have wiV(xo) > 0 which implies that we can only achievetarget shapes for which w"(x) ;::0 for x E [0, L] in this case.

(4) w"(xo) = w"(xd = 0, wlll(xo) * 0 * wlll(xd for xo, Xl E [0, L], and w"(x) * 0 for allX E [O,L] - {xo,xI}.

Then the numerator of (2.2) must have simple zeros at Xo and Xl. This implies

thatIX = - P(xd - P(xo) fJ = -P(xo) + P(xd - P(xo) xo.

Xl - Xo XI -xoAgain, we do not have any degree of freedom left and consequently B is completelydetermined by (2.1).

Notice that now, if x· = Xo or x· = Xl,

B(x') = _ P(xd - P(xo) - P'(X')(XI - xo).(Xl - XQ)w'''(x')

Using the convexity of P(x), we can show that we must have w"'(xo) < 0 andw"'(xd > 0, in order to have B positive everywhere. So w" must be positive in[O,xo), negative in (xo,xd and positive again in (xl,L]. We could not achieve anytarget shape other than one of this kind. We see also that the values of B closeto the points (xo, xd can be unstable with respect to small changes in w". A smallrapidly changing variation in w" could lead substantial change in wlll(x') and hencein B. In fact it is possible to have an order e change in w", which gives an order1fe change in B.

(5) w" has more than two zeros or a zero of more than double multiplicity.

We cannot achieve any of these shapes with a positive B.

As the previous analysis has revealed, the number of degrees of freedom present in theinverse problem depends on the number of zeros of the target shape bending moment M.But the boundary conditions that the target shape satisfies may already introduce zerosof M.

To understand the effects of the boundary conditions on the well-posedness of theinverse beam problem, let us assume that the beam is simply supported at X = 0 sow"(O) = O. We then have several possibilities:

(i) If wlll(O) * 0 and w"(x) * 0 for X E (0, L], then we are in case 2 of the previous list.We now have fJ = 0 and we must have

{IX < -P(L)fL, if w"(x) < 0

IX> 0, if w"(x) > 0X E (O,L];X E (O,L].

We can achieve any of these shapes and we are free to prescribe one boundarycondition on B.

(ii) If wlll(O) = 0 and w"(x) * 0 for X E (0, L], then we are in case 3 of the previous list.We now have IX = fJ = O. In order for the solution of the inverse problem to exist,w"(x) must be positive for X E (0, L], but for those target shapes we cannot prescribeany boundary conditions on B.

(iii) If w" has two simple zeros, x = 0 and x = Xl, we are in case 4. Now ex = -P(Xd/XIand fJ = O.We can only achieve target shapes for which w is concave between 0 andXl and convex between Xl and L. Then

B(x) = P(X)Xl - P(xdx.XIW"(X)

As a sub-case, we have the possibility of having Xl = L, that is, we have a simplysupported beam at both endpoints. Then w must be concave everywhere in [0,L].

Either way, we cannot impose boundary conditions on B and again B will beunstable to small rapid changes near the endpoints.

In case (ii), we have existence of solutions which are completely determined by theinverse equation; however, a slight change in w could put us in cases (i) or (iii), which aredifferent inverse problems. So case (ii) is ill-posed.

Cases (i) and (iii) have solutions for target shapes that satisfy the correspondingboundary conditions and are at least twice differentiable, but they are unstable near thezeros of w". Cases (i) and (iii) are different inverse problems; for certain shapes we mayprescribe one boundary condition for B and for others we may not.

From a numerical point of view, all cases may be treated by first solving the equationM" = -q with the two boundary conditions that M(O) and M(L) are given. In all cases(i)-(iii), we must set M(O) = 0; also in case (i), M(L) would only be constrained by aninequality, but in case (ii) we must have M(L) = P(L) and in case (iii), M(L) = 0 fora simply-supported plate at X = L. Finally, B can be obtained by solving the algebraicequation M = -Bw".

3 Inverse problems for circular plates

We use polar coordinates (r, e) with r = R as the outer boundary of the circular plate.For a target shape w which is a function of r only, (1.2) for D reduces to

2 2'a D 1 a D ~(d 2 me) aD 4mr or2 + me r2 ae2 + 2 dr (V w) + -;:- a;: + DV w = q

where the bending moments in the rand e directions are Mr = -Dmr and Me = -Dme,respectively, with

wrmr(r) = Wrr + V-,r

wrme(r) = VWrr + -

r2 wrand V w(r) = wrr + -.

r

In this section, q is taken to be constant for simplicity.The discriminant for this equation is L1 = -mrme/r2. The equation is therefore elliptic

when mrme > 0 and hyperbolic when mrme < O. Since both mr and me are functions ofr only, the parabolic lines for these inverse problems are always circles. Additionally, acircular frame where the plate is simply-supported is always a parabolic line.

We can also see from (3.1), that on a parabolic line where me = 0 but mr '*' 0, thecharacteristic curves will form cusps (oriented in the direction of the normal to theparabolic line). These are similar to the characteristic curves of the classical Tricomi's

-1.5-1.5 -1 -0.5 0

-1.51.5 -1.5 -1 --{).5 0

FIGURE 2. Two chains of characteristics depicted for Example 1 and one for Example 2 of §4.None of these chains perfectly close on itself even though it looks that way.

equation that arises from the study of transonic flows in gas dynamics [2, 4, 5]; it has theform

Uxx + XUyy = O. (3.2)

This equation is hyperbolic for x < 0, elliptic for x > 0 and parabolic along the linex = O. Its characteristic curves are of the form

2 3/2Y - c = ±3( -x) ,

where c is a constant.On the other hand, as we said in § I, at least one characteristic curve must be tangential

to a simply-supported frame. Here this implies that the characteristic curves near a simply-supported frame where me * 0 are going to be similar to the characteristic curves of thecounterpart equation to Tricomi's, that is

where c is a constant (see Figure 2 for examples). Consequently, the inverse problem forcircular plates may have several parabolic lines simultaneously where the characteristics ofthe equation (as wel1 as the solution of the inverse problem itself) behave very differently.

First let us deduce some consequences of assuming that the target shape w(r) issufficiently smooth (say C4([0, R])) with wr(O) = wrrr(O) = 0 due to symmetry. Then:

lim Wr = wrr(O) so that hm mr = hm mo = (1 + y)w,,(O),

r.....•O r , .....•0 , ....•0

Inverse problems of mixed type in linear plate theory

and also limr->o(w,.r - wr/r)/r = limr->o wrrr/2 = 0 so that

I· (dmr mr - me) -I' ( Wrr - wr/r) - 01m -d + - 1m Wrrr + - .r->O r r r->O r

Provided mr(O) =l=0, the first result implies that, in the neighbourhood of the on gill,mrme > 0 and hence that there is always an elliptic region at the centre of the plate. Aswe will see in § 4, if D is a function of r only, then the second result says that if there is asolution D which is C' in a neighbourhood of the origin then D'(O) = 0 by (4.1). We willmake use of one further result which is given in the following lemma.

Proof We deal only with the case mr < 0; the proof for mr > 0 is similar.Assuming then that mr = Wrr + vwr/r < 0 we have that

:r(rVWr) = rV(Wrr + vwr/r) <0 ~ wr<O on (O,ro);

This lemma tell us that the most interior parabolic line in any circular plate is alwaysdue to m,. becoming zero and me will be different from zero there (see again Figure 2).

We can provide now a uniqueness result for the interior elliptic part of any circularplate. This theorem allows us to restrict our search for solutions to the axisymmetric case.It also tells us that there will be no freedom to impose a boundary condition on D. Wewill assume then that mr < 0 on (0, ro) with mr(ro) = O. From Lemma 3.1, we thereforehave me < 0 on (0, ro).

Theorem 3.2 Let D(r,8) be the solution of the homogeneous version oj' (3.1) for a targetshape w such that mr < 0 on (0, ro) with mr(ro) = O. If V4w ;::: 0 on (0, ro), then D == 0 insidethe circle r = roo

Then we integrate the first two terms by parts - noting that mr(ro) = 0, assuming D and~~ to be continuous and using the fact that r~ + m,. - me = rfr(V2w) - to obtain

12][ lro [ (OD)2 mo (OD)2 d 2 oD 2 4 ]-rmr - -- - +r-(V w)D-+rD V w drd8=0.o 0 or r 08 dr or

We now write D ~~ as i Og,2 and integrate this term by parts. We note that rV4w

12"[1 d ] 121ro[ (OD)2 me (OD)2 1 ]-r-(V2w)D2 de + -rmr - - - -- + -rD2V4w dr de = o.

o 2 dr r=ro 0 0 or r oe 2

From our hypothesis that V4w > 0 and the fact that rfr(V2w) = 0 at r = 0 we see thatrfr(V2w) > 0 at r = ro; since mr < 0 and me < 0, all the terms on the left-hand side ofthe previous equality are non-negative. Thus we must have D == O. D

This theorem implies that if the most interior elliptic region for a homogeneous inverseproblem in a circular plate is surrounded by a parabolic line, then D and all its derivativesare identically zero on it. This fact would allow us to establish an exterior Cauchy problemthere for the following region of that inverse problem and perhaps, at least in some cases,prove uniqueness there too.2 But as example 2 in the following section shows, the ideathat uniqueness in the most interior elliptical region implies uniqueness in the remainingregions that may be present in a particular inverse problem is not a forgone conclusion.

4 Circular symmetric inverse problems

If we restrict ourselves to solutions where D is only a function of r, (3.1) can be writtenas

~~ (r~(DV2w)) _ (1- v) ~ (dD dW) = q.r dr dr r dr dr dr

This equation can be integrated once to give

d 2 dD dw r2

rdr(DV w)-(I-v)~~=q2'

The smoothness condition wr(O) = 0 has been used to account for the constant ofintegration. This first-order equation for D is finally written in the form3

dD (dmr mr -me) _ qrmr dr + dr + r D - 2

with mr and me as above. Singularities in this equation will occur when mr = 0; they areparabolic curves for (3.1). Note that parabolic curves for (3.1) also occur if me = 0 butthese are not singularities of (4.1).

The nature of the solutions D of (4.1) will depend on the singularities of this equationwhich occur in [0, R] and we follow the procedure used in § 2 to discuss different cases(here mr is going to playa similar role than w" there). At the end of the section we givesome simple examples. We note that boundary values of w are irrelevant since a constant

2 Notice that a curve where mr = 0 is both a parabolic line and an envelope of characteristicscurves. So Holmgren's uniqueness theorem does not apply there (see Courant & Hilbert [2]). If it isused on some curve inside the elliptic region, its statement would be valid only up to the parabolicline. See also Bitsadze [1, p. 40].

3 See sections 5.4 and 5.5 of Mansfield [8] for a derivation of (4.1), which appears there as (5.56)'-::'but with different notation.

(1) mr(r) '* 0 for all r E [0,R).

In this case (3.1) is elliptic throughout the domain. The differential equation (4.1)will have a solution with one arbitrary parameter. An inequality constraint on thisparameter will be needed to ensure that D > O.

(2) mr(ro) = 0 for a unique ro E [0,R) and m~(ro) '* O.We write mr = -(r - rO)J.l(r), where J.l(r) has one sign throughout [0,R). We see that(3.1) is elliptic in the region 0 < r < ro and has a parabolic curve at r = roo Ingeneral, when ro < R there will be a hyperbolic region outside r = roo

Equation (4.1) may now be written in the form

d [r-ro me(r)] qr(r - ro)-(DJ.l) + 1+ -- + -- DJ.l = --.dr r rJ.l(r) 2

Defining f> = 1+ mo(ro) and f(r) = 1.+ [mo(r) - mo(ro») _1_ we write (4.2) as:rOI'(ro) r rl'(r) rOI'(ro) r-ro'

d qr(r - rO)-d (DJ.l) + [f> + (r - ro)f(r))DJ.l = --.

r 2

We note here that by using the earlier results on limits as r -4 0, f(r) is a continuousfunction on [0,R).

We note also that me(O) = mr(O) = roJ.l(O) and hence, using Lemma 3.1, mo(ro)and J.l(ro) have the same sign and therefore f> > 1. If (4.3) is to have a solution Dbounded at ro with a left and right derivative at ro then:

qroD(ro) = -~( ).2uJ.l ro

For D(ro) > 0 we must have J.l(r) < 0 so that mr < 0 on [0,ro) and mr > 0 on (ro, R).Because f> > 1, the constant of integration in the solution of (4.3) is determined bythe requirement that D(r) is bounded and then

q 1r (s-ro)8-1D(r) = --2-(-)(--) s -- exp[F(s) - F(r)) ds,J.l r r - ro ro r - ro

where F' = f. One can make the substitution s = ro + t(r - ro) to obtain thealternative form

D(r) = - 2J.l~r) 11

t8-1(ro + t(r - ro)) exp[F(ro + t(r - ro)) - F(r)) dt.

As expected, D(r) > 0 only when J.l(r) < O.Summing up, we see that in this case a positive solution D exists only when

J.l(r) < 0 and that no extra condition on D may be specified. If ro = R, the interioris everywhere elliptic. It can be shown that small rapidly changing perturbations ofmr may be made which result in J.l(ro) becoming small, and hence resulting in largechanges in D(ro) so that again we have instability near the singularities.

(3) mr(ro) = mr(rd = 0, m~(ro) 9= 0 9= m~(rd for ro,r, E [0, R] and fnr(r) 9= 0 for allI' E [0, R] - fro, rd·Here we write mr = (I' - 1'0)(1'- rd/l(r) and again /l(r) has one sign on [0, R] ando < 1'0 < 1'1 ~ R. Equation (4.1) for D can be written in the form

i.(D/l) + [_Y- + _fJ_ + f(r)] D/l = __ q_r__ ,dr I' - rl I' - 1'0 2(1' - 1'0)(1'- 1'1)

1 [ mo(rl ) mo(r)] 1 [ mo(ro) mo(r)] 1f = -;:+ rl/l(rd - r/l(r) (I' - rd(rt - 1'0) - ro/l(ro) - r/l(r) (I' - 1'0)(1'1- 1'0)'

We see that f(r) has well defined limits as I' -4 0, I' -4 1'0 and I' -4 1'1. Also byLemma 3.1, mo(ro) and /l(ro) haV'p the same sign, so that fJ > 1. We also seethat if D is bounded with left and right derivatives at 1'0 and 1'1 then D(ro) =-qro/2fJ /l(ro)(rl - 1'0) and D(rl) = qrJ!2y /l(rd(rl - 1'0)'

For a positive solution D, we will now require /l(r) < 0 and y < 0 whichmeans that mo(rd < O. This is consistent with (3.1) having an elliptic region wheno < I' < 1'0 and a hyperbolic region when 1'0 < I' < 1'1. The solution of (4.4) is

1r ( )J-l ( ),-1q S - 1'0 S - 1'1D(r) = 2 ( )( )( ) s -- -- exp[F(s) - F(r)] ds,

J.L I' I' - 1'0 I' - 1'1 ro I' - 1'0 I' - 1'1

for 0 < I' < 1'1; where again F' = f.The lower limit of integration 1'0 is required for convergence of D at 1'0, and

convergence at 1'1 is obtained only when y < O. The values of D(ro), D(rd are thenseen to be the same as those given above. For 1'1 < I' < R,

lr ( )J-I( ),-1q S - 1'0 S - 1'1D(r) = ()( )( ) s -- -- exp[F(s) - F(r)] ds

2/l I' I' - 1'0 I' - 1'1 R I' - 1'0 I' - 1'1

K exp[-F(r)]+ (I' - ro)J(r - I'd"

where K is an arbitrary constant which could be determined by specifying D(R).Although D(r) is continuous at I' = 1'1, its smoothness at I' = 1'1 is the same as that of1(1'-1'1)1-', and its derivative could be unbounded (see Bitsadze [1, p. 63]). The valuesof D will again be unstable with respect to small changes in mr near its zeros so thatany attempt to solve the differential equation (4.1) numerically would require care.

Consequently, the sign of y offers a simple criterion for the existence of abounded positive solution for D: it exists only when y < O. On the other hand, fJis always positive in this case.

According to this analysis, it is possible to have a solution of(4.1) for a simply-supportedplate even if tnr has an interior zero. This would correspond to a 'hard' parabolic linefor (3.1): a parabolic line that produces a singularity in (4.1). There may also be 'soft'

parabolic lines, which do not produce singularities in (4.1) and dQ not seem to playany role in relation to the existence and uniqueness or otherwise of solutions'to (4.1).Consequently, perhaps against expectations, the symmetric circular inverse problem isvery different from the inverse beam problem.

Finally, we may also now prove a uniqueness theorem for certain hyperbolic regions ofthe circular symmetric inverse problem.

Theorem 4.1 Let D = D(r) be the solution of the homogeneous version of (4.1) in a hyper-bolic region 1'0 < I' ~ 1'1 where mr > 0 and mr(ro) = O. then D(r) == 0 on (1'0, rd.

lr

l

[Dmrdd (Dmr) + mr_m_r_-_m_oD2] dr = O.ro I' I'

The first term above may be integrated to give [!m;D2]~~, which is non-negative becauseof the condition mr(ro) = O. The remaining term is positive because mrmo < 0; henceD==Q 0

4.1 Some examples

We will discuss two examples. All the quantities involved In the examples have beennon-dimensionalised as explained in the introduction.

where Wo is a constant; see Figure 3(a). For this target we find mr = -(1+v)(1-r2)

and, [1 + 3v 2]mo = -(1+ v) 1- --I' .

3+vIf R < I, mr has no zeroes so this example would correspond to case 1 in theprevious list. For R = 1, the plate is simply-supported and then this example wouldcorrespond to case 2. If R > 1, the plate is not simply-supported and there is aninterior 'hard' parabolic line.

The general solution of (4.1) for this example is

q3+v 2-bD(r) = --- +K(I-r)161 + v

4b=3+v'

Here K is an arbitrary constant when R < 1 but K must be zero if R ? 1 and thenthere is a unique D(r) which is constant (see Figure 3(b)). This agrees with Theorem3.2, since here V4w(r) = 32(1 + v)/(6 + 2v) > O.

With the notation of our second case, J.l(r) = -( 1 + v)(1 + 1'), b = 4/(3 + v) andf(r) = (15 -1)/(1 + 1').

IIJ)/

FIGURE 3. Displacements and flexural rigidities for example 1 with v = 0.22 and q = 1. If R = 0.8,all three D obtained from K = 0,0.1,0.2 are valid. If R = 1.5, only D with K = 0 is valid.

For this example we can verify that m(l has the same sign as mr on [0,1) anda hyperbolic region exists for r > 1. In fact, m(l will remain negative on [0, S),

where S = V ?,:;~.If we also have R > S, (3.1) will have an elliptic region [0,1), ahyperbolic region (1, S) and an additional elliptic region (S, R], where the solutionabove will still hold. Notice how the existence, uniqueness and regularity of thesolution for R > 1 is not affected at all by the 'soft' parabolic line at r = S.

(2) The second example is for the target shape:

where Wo is a constant. This gives mr = -(1 + v)(r - l)(r - ro) with zeros at r = roand r = 1. We note here that wrrr(O) =l=0 for this function but we are mainlyinterested in the solution for r E (ro, 1).

To enable us to write an exact solution we take the case where v = 1/3, andro = 10/11. Then:

{

4 ( 10) .mr = -3 r - IT (r - 1),

m = -~ + ~r - ~r2.(I 33 Il S'

see Figure 4(a). In the formalism of our third case, J1(r) = -4/3, y = -3/5, b = 3,and f(r) = O. Hence

D(r) = 3q {625 _ 125 + 125 2 _ 2.r3 _ 1875 [11(1- )]3/S} (4.5)8(r _ ¥l)3 10164 3388r 308r 12 37268 r ,

for 0< r ~ R; see figure 4(b).

FIGURE4. Displacements and flexural rigidities in Example 2. If 10/11 ~ R ~ 1, we must use Dwith K = O.For R > 1, any D with K big enough is valid.

Then, for R> 1 we may add K(r - 1)3/5/(r - 10/11)3 to (4.5) in the region 1 <r < R, where K is any constant big enough to make D > 0 in (I,R); see figure 4(c).But now D(r) is not smooth around r = 1 because it has a vertical tangent there.

For this example, me is always negative and for R > 1, we have an elliptic region[0,10/11), a hyperbolic region (10/11,1) and an elliptic region again (1,R). SinceV4w(r) is unbounded at r = 0, Theorem 3.2 does not apply to this example; but aswe have seen, we do have uniqueness of solutions in the interior elliptic region (ifthe plate is simply-supported at R = 10/11), and even in the next hyperbolic region(if the frame is anywhere between 10/11 and 1); but not if R > 1.

Observe that if ro < 13/15 < 10/11, then 'Y > 0 and bounded solutions are notpossible. This is an indication that the hyperbolic region must be narrow for asolution to exist.

It is interesting to plot the characteristics in the hyperbolic regions for Examples 1 and2, see Figure 2. As expected, the characteristics of example 1 form cusps on the 'soft'parabolic line r = S and are tangential to the 'hard' parabolic line r = 1. In Example 2,also as expected, the parabolic lines at r = 10/11 and r = 1 are both 'hard,' and all thecharacteristics are tangential to both of them.

We have modelled the gravity sag-bending process as the inverse problem for a linear plateequation with variable Young's modulus. As such, we have shown that the second-orderpartial differential equation for the Young's modulus changes type, it mayor may nothave a solution and, under certain circumstances, it may have a unique solution withoutthe imposition of any general boundary conditions.

Indeed, we can deduce from the expression for D(r) in cases 2 and 3 of § 3 that the valuesof both D and dD/ dr on simply-supported frames depends on the target shapes throughthe functions J1(r). No single boundary condition for D would be valid for all cases. Thefact that the specification of boundary values for D may be replaced at the boundary and

perhaps at other singularities of the differential equation by smoothness conditions whichare also dependent on the target shape and that the solutions may also be unstable tosmall changes of the target function near the singularities suggests that from a numericalpoint of view, this inverse problem presents considerable difficulties.

We have also seen that these mixed type problems are of a very different nature thanthe ones that have been analysed before, and in particular, than the classical mixed typeproblems that arise in the context of gas dynamics (see chapter 12 of Garabedian [4]).

We have studied in detail the inverse beam problem and the inverse problem for circularplates. For the beam inverse problem, we can only define one bending moment M andits zeros correspond bijectively with the zeros of wI! and with the changes of type ofthe inverse equation; their number determine what shapes can be made and how manyboundaries conditions are needed for the inverse problem to have a unique solution.On the other hand, for the circular symmetric inverse problem, we have two bendingmoments, Mr and Mo, but only Mr determine the existence, uniqueness and regularityproperties of D. This seem to allow the existence of solutions of inverse problems withmultiple type changes due to zeros of Mo. The questions of uniqueness and regularity ofsolutions are more tricky, as shown by the examples in §4.

Let us now compare the previous situations with the case of a square plate with a flatframe. All the general remarks that we made in relation to (1.2) apply here. To illustratethis problem, let us consider an example taken from Temple [9]: w(x, y) = p(x)p(y) wherep(z) = z - 2z3 + z4. This target shape is simply-supported in a flat square frame thatbounds the region [0, I] x [0,1] since f(O,y) = f(l,y) = f(x,O) = f(x, I) = 0 for f = w,

f = Wxx and f = Wyy.

For this target shape

Ll(x,y) = (I - v)2(1_ 6x2 + 4x3)2(1 - 61 + 4y3)2

- [(-12x + 12x2)(y - 2y3 + i) + v(x - 2x3 + x4)( -12y + 12y2)]

* [(x - 2x3 + x4)( -12y + 121) + v( -12x + 12x2)(y - 2y3 + i)].

As expected, Ll < 0 in the interior of the domain. We have that Ll > 0 in the cornerregion, which is also expected for this kind of domain since in the interior of a rightcorner region, the product of the bending moments tends to zero (AC ~ 0) while thesquare of the twisting moment is bounded away from zero (B2 > 0, see pp. 85 and 107 ofTimoshenko & Woinowsky-Krieger [10]).

Naturally, we have two real characteristic curves going through every point in thehyperbolic regions and none in the elliptic one (see Figure 1). Also, the frame edges arecharacteristic curves themselves, and at any point on an edge, the second characteristiccurve is orthogonal to the edge.

Perhaps more surprisingly, the parabolic line meets the boundary, dividing the domainin one elliptic part and four hyperbolic parts. This is a general fact for this kind of frameas can be seen analysing the sign of wxy along an edge.

The characteristic curves (other than the edge itself) that start at a point on the edgewhich lies up to a quarter of the length of the edge from the nearest corner accumulates·at the middle point of the adjacent edge. They do so being tangent to that adjacent edge(see Figure 5(a)). The characteristic curves of the next quarter edge intercept the parabolic

line forming cusps with other characteristic curves that accumulate at the middle point ofthe edge under consideration (see Fig. 5b).

If we move along a section of the parabolic line in between two adjacent edges (seeFigure 5(b)), we will see the characteristic curves approaching the parabolic line andforming cusps in almost every point of the parabolic line; that is, locally the characteristiccurves of this problem behave similarly to the characteristic curves of the Tricomi's (3.2).However, at the middle point of every section of the parabolic line, the characteristiccurves of this problem are tangential to the parabolic line like those of (3.3). So thetwo possible model behaviours of the characteristic curves near a parabolic line are nowpresent along the same parabolic line.

The structure of this example leads us to believe that for the inverse problems insquare domains, as for the inverse problems in some of the circular symmetric shapes thatwe have analysed, if there is a solution then it is unique without imposing any generalboundary conditions on D.

This kind of uniqueness property is obviously very different from the classical one sinceis related only to the equation, not to a full problem which would include boundaryconditions. Consequently, any attempt to imposed boundary conditions to equation (1.1)could very easily end up producing an empty solution set. It is not the case, as withmost other inverse problems, of being able to select a solution between the many thatexist, but of being able to approximate the one and only solution of the equation bysome procedure that would not require the a priori knowledge of boundary valuesfor D.

It remains an open problem to establish what shapes can be made by the gravity sag-bending process on flat square frames. It is the manufacturer experience that frames thatcontains corners that are very flat and near ninety degrees produce the greatest difficulties.That indicates that the flat square frames should be the object of further research.

Acknowledgements

The authors wish to thanks David Allwright, Peter Howell and John Ockendon (fromOCIAM, Mathematical Institute, Oxford University) for some very interesting discussionsand suggestions from the very beginning of the modelling of the gravity sag-bendingprocess. More recently, the authors have benefited from suggestions from Gregory Kozyteff-also from OCIAM and from the interaction with Heinz Engl and his group fromJohannes Kepler University in Linz (Austria) which has been considering the applicationof regularisation methods to this inverse problem (see Engl & Kugler [3]).

Domingo Salazar's research was carried out when he held a post-doctoral position atOCIAM and was funded by the glass company Pilkington and by the European UnionTMR network on Differential Equations in Industry and Commerce, grant ERBFMRXCT97-0117.

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