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Gâteaux differentials and EulerLagrange equations using Maple
This article describes how to solve simple problems in the calculus of
variations using Maple. Maple actually has a function module for this,
but it's more illustrative — and perhaps more reliable — to
perform the algebraic manipulations stepbystep.
In what follows I assume that the reader is basically familiar with the
terminology and notation of calculus of variations. If that isn't the
case, you might prefer to look first at my related article
Calculus of variations: a lunchbreak guide.
I also assume knowledge of calculus (particular integration by
parts) and some familiarity with the computer algebra system
Maple. I've tested the examples with version 16 on Linux, but I have no
reason to think that other versions would behave differently.
The problem
In this example problem, we want to find a value of y —
that is, a function
y(x) — which is a stationary path
of the functional S[y]:
$$ S[y] = \int^1_0 y'^2 + y^2 + 2te^x \, dx $$
The path must pass through the points (0,0) and (1,0); that is,
the boundary conditions are y(0)=0 and y(1)=0.
I've picked this example because it's relatively straightforward to
solve — with or without Maple — and produces a compact, comprehensible
answer. Many variational problems are not so obliging. In particular,
there may be multiple stationary paths; in this example, there is
only one.
Any stationary path may be a maximum or minimum, but need not necessarily
be either — determining this requires a consideration of second variations,
which I will touch on only briefly in this article.
The basic procedure will be:
1. Find the Gâteaux differential of the functional;
2. determine the EulerLagrange equation from the Gâteaux differential;
3. solve the EL equation to determine the stationary path y;
4. consider second variations to get an idea whether the stationary
path is a maximum or minimum, if either.
The Gâteaux differential
If y=y(x) is a stationary path of the functional S[y],
and h(x) is any alternative path that satisfies the same
boundary conditions, then we write the
Gâteaux differential as follows:
$$ \Delta S[y,h] = \frac{\partial}{\partial \epsilon} S[y+\epsilon h] \,  _{\epsilon=0} $$
This is the analogue in the calculus of variations of finding a turning point
(critical point) of a function — in that case we consider changes to
the value of the function caused
by a small change ε in its argument, looking for a place where
a small displacement in the argument causes no change in the function's
value. At this point, the function must be at a turning point. In the
calculus of variations, we're looking for a function, rather than an
argument to a function, for which small changes in the function cause no
change in the functional.
When the functional is of the form
$$ S[y] = \int^1_0 F(x,y,y') \, dx $$
then we derive the
Gâteaux differential as follows:
$$ \Delta S[y,h] = \frac{\partial}{\partial \epsilon} \int^1_0 F(x,y+\epsilon h, y'+\epsilon h') \, dx \,  _{\epsilon=0} $$
Note that we have to consider variation by εh of both
y and y'. Provided all the derivaties are wellbehaved, we
can change the order of integration in this expression, and write it thus:
$$ \Delta S[y,h] = \int^1_0 \frac{\partial}{\partial \epsilon} F(x,y+\epsilon h, y'+\epsilon h') \,  _{\epsilon=0} $$
This formulation is much easier to handle, and with luck and a bit or work
we should end up with an expression like this:
$$ \Delta S[y,h] = \int^1_0 \frac{\partial}{\partial \epsilon} G(x,y,y') h(x) \,  _{\epsilon=0} $$
From here, we can get the EulerLagrange equation immediately, as
$$ G(x,y,y')=0 $$
For reasons that will be outlined below.
However, we've got quite a bit of work to do first. Happily, Maple can
handle most of it.
Start by defining in Maple the expression which we will differentiate
with respect to ε, that is, the integrand of the functional:
(This is a side issue, but remember that Maple does not care to have
exponentials entered as e^{x}, even though it displays them
that way).
Now define a new variable, F2, to hold the value of F when
y is displaced by εh :
I'm using the variable eps to represent ε in Maple; the
program does support Greek letters, but I find English easier to
type. Now differentiate F2 with respect to ε, and set
ε to zero:
Subtituting back into the integral gives us the Gâteaux differential:
$$ \Delta S[y,h] = \int^1_0 2y'h' + 2yh + 2e^x h \, dx $$
Deriving the EL equation
Annoyingly, we're not quite ready to get the EL equation from this
expression for the Gâteaux differential, because the integrand has
terms in both h and h',
so we can't just factor h out of each term. To convert the
h' term into an h term, we need to split the integral
up into a sum of integrals, and apply integration by parts to the
integral that contains h'.
So, to begin:
$$ \Delta S[y,h] = \int^1_0 2y'h' \, dx + \int^1_0 2(y + e^x) h \, dx $$
Performing the integration by parts of the first integral is trivial in
this case, but the Maple intparts function might be helpful
in less straightforward cases; this function is in the 'student' package.
In the screenshot above, I've cutandpasted the relevant term from the
output of the differentiation expression, to avoid typing errors (which
is why it appears to be formatted already). With this result,
we can rewrite the
Gâteaux differential:
$$ \Delta S[y,h] = [2y' h]^1_0  \int^1_0 2y''h dx + \int^1_0 2(y + e^x) h dx $$
In this example, I'm assuming that the alternative path h must
meet the stationary path y at the boundaries. That is, h(a)=h(b)=0. In this case, the boundary term of the above equation vanishes, and
we can recombine the two integrals, both of which now have integrands that
are multiples of h.
$$ \Delta S[y,h] = 2 \int^1_0 (y + e^x y'') h \, dx $$
The EulerLagrange equation is obtained simply by equating the term
multiplied by h to zero, as we require the value of y
that will make the Gâteaux differential zero for any alternative
path h. This gives us:
$$ y + e^x  y'' = 0 $$
Or, more conventionally:
$$ y''  y = e^x $$
Solving the EL equation
In this example, we've ended up with a fairly benign differential
equation, which is rather unusual for a practical variational problem.
This DE is a secondorder, linear one with constant coefficients — just the kind of thing covered in undergraduate DE courses, and for
which a welldefined solution exists. The DE is nonhomogenous, which means
that the general solution will be comprised of a homogeneous solution
and some particular solution. The homoegeneous solution to this
kind of DE is nearly always of the form:
$$ y_h = Ae^ax + Be^bx $$
and the particular solution can be guessed from the righthand side
of the equation to be something like
$$ y_p = Cxe^x $$
Actually, our first guess might be
$$ y_p = Ce^x $$
This is, indeed, a solution to the DE, but that doesn't help us,
since this solution
is already part of y_{h}. Multiplying the solution by
x is a pretty common strategy such cases.
Anyway, we can solve this EL equation by a simple application of
Maple's dsolve function, as follows:
The constants _C1 and _C2 fall to be determined from the
boundary conditions, which we are told are
y(0)=0 and y(1)=0. Rather than solving for them explicitly,
we can just list the boundary conditions in the set of equations to be
solved by dsolve , and that gives the final answer:
The form of y will be shaped to some extent by the values of
the boundary conditions, but this class of differential equation only
has one nontrivial solution (and this one doesn't even have the
trivial solution y=0). So we can be reasonably sure that, whether
the y we have found is a maximum or a minumum or neither, there
isn't a 'better' solution that we haven't spotted. This might not be the
case in more complex problems.
Second variations
The functional
$$ S[y+\epsilon h] $$
is, for any particular choice of y and h, a function of
ε. We have already considered how to find a stationary value
of y by finding the first derivative of the functional with
respect to ε and setting ε to zero. However, to determine
the nature of the stationary path requires consideration of higherorder
derivatives.
There is a clear analogy here with finding turning points of functions;
the first derivative of the function will tell us the turning points,
but we need the second (or possibly higher) derivative to determine
whether the turning point is a maximum or minimum, if either.
The analog of the second derivative in the calculus of variations is
termed the second variation, and is typically derived using the following
procedure.
We have already seen that we can change the order of integration of
the function — putting the derivative with respect to ε
inside, rather than outside, the integration with respect to x.
Where it is possible to do that, we can express the integrand as
a Taylor expansion in ε. Where the functional is of this form:
$$ S[y] = \int^1_0 F(x,y,y') \, dx $$
the function obtained by displacing y by εh is:
$$ S[y + \epsilon h] = \int^1_0 F(x,y+\epsilon h, y'+\epsilon h') \, dx $$
and the Taylor expansion is:
$$ S[y + \epsilon h] = \int^1_0 F_{\epsilon=0} + \epsilon{\frac{\partial F}{\partial \epsilon}_{\epsilon=0}} + \frac{\epsilon^2}{2!}{\frac{\partial^2 F}{\partial \epsilon^2}_{\epsilon=0}} + ...\, dx $$
Because we can change the order of integration, this expression can
also be written:
$$ S[y + \epsilon h] = \int^1_0 F_{\epsilon=0} \, dx + \int^1_0 \epsilon{\frac{\partial F}{\partial \epsilon}_{\epsilon=0}} \, dx + \int^1_0 \frac{\epsilon^2}{2!}{\frac{\partial^2 F}{\partial \epsilon^2}_{\epsilon=0}} \, dx + ... $$
The first term in the expression is just S[y]. The second
term, usually called the first variation, must be zero for there to be a
stationary path — because it is based on the first derivative of
F with respect to ε, it is zero for the same values
of h and ε that make the Gâteaux differential zero.
It is the third term — the second derivative — we need to consider.
In Maple we can use the taylor function to get an Taylor
expansion of F. As before, we define F2 based on the
supplied functional:
Then we take a Taylor expansion to second order (the argument
'3' includes the zerothorder term):
The first four terms in this series are just the zerothorder term of the
expansion, corresponding to S[y].
The next term, (...) eps, is the firstorder
term, and we can see it is just Gâteaux differential derived earlier,
multiplied by ε It is the next term, (...) eps^2,
that is relevant here  this is the secondorder term in the Taylor
expansion. With a value of y that makes the variation
stationary, the nature of the stationary path is categorized by
the behaviour of this term with respect to ε and h.
As luck (and a bit of careful problem selection) would have it,
it's easy to analyse this second variation. Because the
values of h, h' and eps are all
squared, the second variation will be positive, whatever
these values. Therefore, any displacement from the stationary path
can only increase the value of the functional. Consequently,
this is a minimum. Moreover, because we know that the differential
equation has only one solution, it is a global minimum.
Summary
This article has explained in outline a method for solving simple
variational problems using Maple and in a way that I hope is reasonably
illustrative. One of the problems with working with calculus of variations
in Maple is that the techniques we use involve treating a the
derivatives of the arguments to a function as if they were separate
independent variables; this isn't a thing that Maple is particularly
happy with, and so care has to be taken.
In his book Physics with Maple, Frank Wang suggests a completely
different approach than the one I have used in this article. His method is
to use Maple's subst function to temporarily turn Maple
functions of the form diff(y,x) into ordinary variables which can
then be used as the independent variables in other differentiations.
With this
approach, he substitutes the integrand of a functional into the generic
EulerLagrange equation, and solves it. This is almost a onestep process,
but it has (in my view) the disadvantage that mistakes are less
easily spotted in the intermediate results.
