Solutions of the Problems

Let us now solve the problem stated above systematically. We shall focus our attention on the linear model (12.2.1). 

we set up the extended matrix from the matrices B, A, and vector c occurring in the model (12.2.1), see Fig. 12-2. We assume that the variables (columns of matrices B and A) have been arranged into groups forming submatrices according to whether the values are or are not required. Further, we have separately ordered the columns of both submatrices of B (unmeasured variables) according to their measurability the relatively easily measurable variables are on the respective right-hand sides. 

To the initial macromatrix arranged in this manner we shall now apply Gauss-Jordan elimination. The procedure is shown in Fig. 12-3. 

The elimination succeeds using the pivots from the submatrix of unmeasured nonrequired variables in an analogous way. One thus obtains unit submatrix Z2 3, zero submatrix Z, 3, and general submatiices Z, 4 and Zj 4. 

In addition, we try to find certain submatrices (when they exist) as is shown in the following Fig. 12-4. 

We perform the solution of the preceding problem introducing a number of simplifying assumptions whose validity is checked in the following text. 

It is obvious that the value of p under the main part of the spreading droplet is independent of the coordinate but changes only with time due to the interaction of the amphiphilic molecules with the aqueous phase. Let us denote this coordinate-independent value by p/.t), which, according to Equation 5.215, satisfies the following equation  

It follows from Equation 5.217 that the characteristic time scale of molecule overturning tr is equal to t, = ViSK + 

In a narrow region with width 8 near the droplet edge (from the inner side), the diffusion term in Equation 5.215 becomes of the same order of magnitude as the term describing the overturning of molecules due to their interaction with water. Let us introduce dimensionless values y = rQ t)r)l5, X = aja x = t/u % t) = r(t)/r., where r. = (6V/ji), and new unknown function g (t, y)=p it, r) -Pj it). The time scale t is selected below. 

Rewriting Equation 5.215 in dimensionless form using the introduced variables, we obtain 

---

Several manufacturers of transducers and equipment were contacted to give their best solution of the problems. Two instruments and five types of transducers were selected for further experiments. The equipments were very different concerning the possibility for controlling the measuring conditions and the transducer frequencies where from 10 MHz to 25 MHz with different crystal diameters and focussing lenses. 

Theorem 1.12. The solution of the problem (1.81) exists if and only if there exists a solution of the variational inequality... 

Theorem 2.3. Let the above hypotheses be fulfilled. Then a solution of the problem (2.23) exists. 

Theorem 2.14. From the sequence x = of solutions of the problem (2.116) one can choose a subsequence, still denoted byto, such that as 5 0 the convergence (2.119) takes place and, moreover, the limiting function satisfies (2.123). 

This result enables us to investigate the extreme crack shape problem. The formulation of the last one is as follows. Let C Hq 0, 1) be a convex, closed and bounded set. Assume that for every -0 G the graph y = %j) x) describes the crack shape. Consequently, for a given -0 G there exists a unique solution of the problem... 

This means that the function is a solution of the problem (2.134), (2.131), which completes the proof. 

Here the solution of the problem of minimizing the functional H over the set Ko is equivalent to the following variational inequality ... 

Let x,0 be the solution of the problem (3.48). Additional regularity of the solution in the vicinity of T, clT, is proved provided that T, is a segment of a straight line. The following statement holds. 

Theorem 3.11. There exists a solution of the problem (3.144), (3.147), (3.148) provided that 5 is small enough. 

To construct the solution of the problem (3.224), we shall apply the arguments of the previous subsection. Let us define the Lagrange function... 

We first have to prove that the right-hand side of (4.84) does not depend on 9. As we know (see Yakunina, 1981) the solution of the problem (4.62)-(4.63) has an additional regularity up to the crack faces. For any x G Ei there exists a neighbourhood V of the point x such that... 

---