Introduction

We now investigate the mechanical prospection of tissue, an application extending techniques developed in the electrical analysis of a nerve cell. In this application, one applies traction to the edges of a square sample of planar tissue and seeks to identify, from measurement of the resulting deformation, regions of increased `hardness' or `stiffness.' For a sketch of the associated apparatus, visit the Biaxial Test site .

A Uniaxial Truss

Figure 1

A uniaxial truss

As a precursor to the biaxial problem let us first consider the uniaxial case. We connect 3 masses with four springs between two immobile walls, apply forces at the masses, and measure the associated displacement. More precisely, we suppose that a horizontal force, f j f j , is applied to each m j m j , and produces a displacement x j x j , with the sign convention that rightward means positive. The bars at the ends of the figure indicate rigid supports incapable of movement. The k j k j denote the respective spring stiffnesses. The analog of potential difference (see the electrical model) is here elongation. If e j e j denotes the elongation of the jjth spring then naturally, e 1 = x 1 e 1 x 1 e 2 = x 2 − x 1 e 2 x 2 x 1 e 3 = x 3 − x 2 e 3 x 3 x 2 e 4 =- x 3 e 4 x 3 or, in matrix terms, e=Ax e A x , where A=100-1100-1100-1 A 100 -110 0-11 00-1 We note that e j e j is positive when the spring is stretched and negative when compressed. This observation, Hooke's Law, is the analog of Ohm's Law in the electrical model.

Definition 1: Hooke's Law

1. The restoring force in a spring is proportional to its elongation. We call the constant of proportionality the stiffness, k j k j , of the spring, and denote the restoring force by y j y j .

2. The mathematical expression of this statement is: y j = k j e j y j k j e j , or,

3. in matrix terms: y=Ke y K e where K= k 1 0000 k 2 0000 k 3 0000 k 4 K k 1 000 0 k 2 00 00 k 3 0 000 k 4

Definition 2: force balance

1. Equilibrium is synonymous with the fact that the net force acting on each mass must vanish.

2. In symbols, y 1 − y 2 − f 1 =0 y 1 y 2 f 1 0 y 2 − y 3 − f 2 =0 y 2 y 3 f 2 0 y 3 − y 4 − f 3 =0 y 3 y 4 f 3 0

3. or, in matrix terms, By=f B y f where f= f 1 f 2 f 3 and B=1-10001-10001-1 f f 1 f 2 f 3 and B 1-100 01-10 001-1

As in the electrical example we recognize in BB the transpose of AA. Gathering our three important steps:

Gaussian Elimination and the Uniaxial Truss

Although Matlab solves systems like the one above with ease our aim here is to develop a deeper understanding of Gaussian Elimination and so we proceed by hand. This aim is motivated by a number of important considerations. First, not all linear systems have solutions and even those that do do not necessarily possess unique solutions. A careful look at Gaussian Elimination will provide the general framework for not only classifying those systems that possess unique solutions but also for providing detailed diagnoses of those defective systems that lack solutions or possess too many.

In Gaussian Elimination one first uses linear combinations of preceding rows to eliminate nonzeros below the main diagonal and then solves the resulting triangular system via back-substitution. To firm up our understanding let us take up the case where each k j =1 k j 1 and so Equation 4 takes the form

Alternate Paths to a Solution

Although Gaussian Elimination remains the most efficient means for solving systems of the form Sx=f S x f it pays, at times, to consider alternate means. At the algebraic level, suppose that there exists a matrix that `undoes' multiplication by SS in the sense that multiplication by 2-1 2 undoes multiplication by 2. The matrix analog of 2-12=1 2 2 1 is S-1S=I S S I where II denotes the identity matrix (all zeros except the ones on the diagonal). We refer to S-1 S as:

Definition 3: Inverse of S

Also dubbed "S inverse" for short, the value of this matrix stems from watching what happens when it is applied to each side of Sx=f S x f . Namely, Sx=f⇒S-1Sx=S-1f⇒Ix=S-1f⇒x=S-1f S x f S S x S f I x S f x S f

Gauss-Jordan Method: Computing the Inverse of a Matrix

Let us now consider how one goes about computing S-1 S . In general this takes a little more than twice the work of Gaussian Elimination, for we interpret S-1S=I S S I as nn (the size of SS) applications of Gaussian elimination, with ff running through nn columns of the identity matrix. The bundling of these nn applications into one is known as the Gauss-Jordan method. Let us demonstrate it on the SS appearing in Equation 5. We first augment SS with II. 2-10│100-12-1│0100-12│001 2-10 │ 100 -12-1 │ 010 0-12 │ 001 We then eliminate down, being careful to address each of the three ff vectors. This produces 2-10│100032-1│12100043│13231 2-10 │ 100 0 3 2 -1 │ 1 2 1 0 00 4 3 │ 1 3 2 3 1 Now, rather than simple back--substitution we instead eliminate up. Eliminating first the (2,3) element we find 2-10│1000320│3432340043│13231 2-10 │ 100 0 3 2 0 │ 3 4 3 2 3 4 00 4 3 │ 1 3 2 3 1 Now, eliminating the (1,2) element we achieve 200│321120320│3432340043│13231 200 │ 3 2 1 1 2 0 3 2 0 │ 3 4 3 2 3 4 00 4 3 │ 1 3 2 3 1 In the final step we scale each row in order that the matrix on the left takes on the form of the identity. This requires that we multiply row 1 by 12 1 2 , row 2 by 32 3 2 , and row 3 by 34 3 4 , with the result 100│341214010│12112001│141234 100 │ 3 4 1 2 1 4 0 1 0 │ 1 2 1 1 2 00 1 │ 1 4 1 2 3 4

Now in this transformation of SS into II we have, ipso facto, transformed II to S-1 S ; i.e., the matrix that appears on the right after applying the method of Gauss-Jordan is the inverse of the matrix that began on the left. In this case, S-1=34121412112141234 S 3 4 1 2 1 4 1 2 1 1 2 1 4 1 2 3 4 One should check that S-1f S f indeed coincides with the xx computed above.

Invertibility

Not all matrices possess inverses:

Definition 4: singular matrix

A matrix that does not have an inverse.

Example

A simple example is: 1111 11 11

Definition 5: Invertible, or Nonsingular Matrices

Matrices that do have an inverse.

Example

The matrix SS that we just studied is invertible. Another simple example is 0111 01 11