9 maths 2 - legendre transforms€¦ · 09.09.2018 · hyperplasticity legendre transforms 1 9/1...

Hyperplasticity Legendre transforms 1

9/1

Before we go any further though we need to learn some more maths. The technique we shalllearn is the Legendre transform – an incredibly powerful method that underpins a lot of what weshall do in the remainder of this course. It is widely used in both thermodynamics and plasticitytheory.


9/2

I often find it convenient to see things both algebraically and geometrically, so I shall introducethe transform from both points of view. In fact there are two different geometricinterpretations.

We shall start with the simple transform, but I shall then deal with more complex cases –passive variables, chains of transformations, and finally the special case of transforms of firstorder functions, for which case there is a degenerate form of the fransform.


9/3

Suppose we have a function big X of little x_i

… and we have some other variables y_i which ar dX / dx_i

Then define a new function big Y therough X + Y = x_i y_i (where the term on the right hand sideis an inner product, i.e. the summation rule applies). We will write big Y as a function of the littley_i

Now consider a small increment, which from the product rule we can write as shown… or as in the second line.

We group the terms in dx_i and dy_i together, and note that the first term disappears because ofthe definition of y_i

… and if the equation is to be true for all dy_i then the contents of the bracket must be zero

… so x_i = dY / dy_i

In which effectively we have just interchanged the roles of x and y


9/4

The result looks cleaner if we display it in a symmetric way.

Very importantly, once one defines big X, one can derive big Y, or vice versa. Both essentiallycontain exactly the same mathematical information.


9/5

Let us look at this geometrically. On the horizontal axis are the x_i, and on the vertical axis y_iwhich are some function of x_i.

big X is the area below the graph and big Y the area above it. It is easy to see that X + Y = theproduct x_i y_i

If we consider a small increment, it is easy to see that dX = y_i dx_i and dY = x_i dy_i


9/6

Let us now take an example.

Suppose X = Ax^4

Then y = 4Ax^3

… which we should invert to get x in terms of y as we shall need this later.

We then take X + Y = xy to get an expression for Y, which we can simplify, and then substitute forx in terms of y, since big Y has to be a function of little y.

Now starting from big Y, we can differentiate to get x

… which is of course the same expression that we had before, confirming that the whole processhas worked.


9/7

There is also another geometric interpretation. This time we plot X as a function of x_i, andnothe that y_i is the slope of this curve.

If we subtract x_i y_i from X we get –Y.

Now we normally think of the definition of a curve “pointwise” – that is we specify X as afunction of the coordinate x_i. However, we could specify the same curve by specifying theintercept –Y as a function of the slope y_i – this is the “planewise” definition of the curve. Insome applications this way of thinking of the transform may prove useful.


9/8

Now let us move on to partial transforms.

Suppose that X is a function not just of x_i but also some other variables a_j, where y_i = dX /dy_i and b_j = dX / da_j

We now form a partial transform in which we interchange the x’s and y’s as before, but we leavethe a_’s untransformed – so Y is a function of y_i and a_j.

We consider a small increment as before, and again group the terms.

… and eliminate the brackets that are zero.

Then if the expression is to be generally true for all dy_i and da_j, we have the results at the end.

Note that the transform from x to y is exactly as before, but that b_j is minus dY / d a_j. Note theminus sign which is very important!


9/9

The partial Legendre transform allows us to establish a chain of four transformations.

If we start with X at the top, we can transform to Y by interchanging x and y. Then we cantransform again to W by interchanging and b

Or we can first interchange and b to get V, and then interchange x and y to get W

So we can go either way round the loop.

Note that because of the minus signs on the passive variables you need to be rather carefulabout all the signs in the transformations.


9/10

There is an important special case that we need to consider. If the original function ishomogeneous first order the transform is a degenerate case. Remember that the dissipation ishomogeneous first order, so this will be an important case for us.

From Euler’s theorem we can show that in this case X = x_i y_i, so it follows that Y is a funcyionthat is identically zero. It further follows that dY/dy_i times dy_i is zero.

Furthermore we can show, following the methods we had before, that x_i dy_i is also zero.

However, instead of being able to deduce that x_i = dy/dy_i, this time we can just deduce thatone must be a multiple of the other, in other words x_i = lambda dY/dy_i, where lambda is someunknown constant (which can always be made positive by appropriate choice of Y).

Note that we still have the same amount of information as before - on the one hand we haveintroduced one extra unknown, lambda, but on the other hand we know that Y is identically zerofor all values of y_i.


9/11

There is another way of dealing with the first order case. I have only come across this relativelyrecently and have not yet pursued all the possibilities, but it looks very promising. Giaquinta andHildebrandt call this a “generalised transform”, but that is not really a very good terminology asit is not a generalisation, it is a different treatment.

Suppose we have a first order function X, with as usual y = dx / dx_i.

We form a second order function P = ½ X^2, and define the quantities q_i = dP d x_i. It is easy tosee that q_i = X times y_i

We take the Legendre transform of P to obtain Q, and it is easy to show that as P was secondorder in X, Q is second order in q

And then effectively take the square root so that Q = ½ Z^2. Z is of course first order.

It is easy to show that, because they are second order transforms, P = Q, and so it also followsthat X(x_i) = Z(q_i).

Now with a little manipulation and change of variable we can show that Z(y_i) = 1. It isconvenient to define a new function Y(y_i) = Z -1, and Y(y_i) =0. It also follows that x_i = lambdadZ/dy_i = lambda dY / d(y_i)

Y(y_i) is therefore one particular form of the Y that we had in the degenerate case, but it is inthis case uniquely defined, and so we call it the CANONICAL case – and denote it by the overbar.Furthermore it is considerably easier to find Z (and hence Y_bar) than to use the degeneratetransform.


9/12

Let us just consider an example of the degenerate transform.

We take a simple first order function that is quite common – a square root of sum of squares. Wederive y from X, and it is easy to verify that x_i y_i = X, so Y is identically zero.

We can see that if we square and add y_1 and y_2 we just get k^2, so we can identify a suitable formof Y, but this is a somewhat ad hoc observation for this particular function.

… and from Y we can differentiate to get x, with the unknown factor lambda.

Alternatively we can use the G-Q transform

We form P by squaring X.

This is now a quadratic function, and the transform is well known as another quadratic function – theonly difference is that the k^2/2 turns into 1/2k^2.

Q is originally written as a function of q_i, but we can just change the variable to y_i. In fact we couldeasily omit the intermediate step. Then we calculate Z, which is first order, and from then Y_bar, thecanonical form of Y. Note that this is in the form of a homogeneous first order function minus one.

And we can easily determine x from Y_bar, and also verify for this case the interpretation thatlambda_bar = X.

Note that our original Y was perfectly acceptable, but just one of many forms of appropriate function.Y_bar selects one particular well defined form.

Finally note that later when we look at convex analysis we can identify X with what is called a supportfunction and Z with a gauge function. Thus we can embed our new result in standard convex analysis.


9/13

In summary:

We shall make a lot of use of the simple Legendre transform which allows us to go back andforth between different sets of independent and dependent variables.

In many cases we shall have passive variables that do not take part in the transformation, andwe need to be careful about the sign change as the transformation is made.

And finally there will be many cases – specifically involving the dissipation – where we need thetransform of a first order function, and this is a special case. Later on when we make use ofconvex analysis we shall gain a further insight into this problem.

9 maths 2 - legendre transforms€¦ · 09.09.2018 · hyperplasticity legendre transforms 1 9/1...

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