Partial Differential Equations PowerTool

by Dr. Jim Herod

Section 1.3: Orthogonal Families

The notion that vectors, or functions, can be orthogonal is one of the important ideas in this set of notes. We recalled what orthogonality means in the previous Section 1.2. There, we connected the idea with perpendicularity through the dot product. This was the connection: two vectors are perpendicular if and only if the cosine of the angle between them is zero. Relating this to the dot product, we have that vectors are orthogonal if and only if their dot product is zero. We make this precise.

Definition:  The collection of vectors { , , , ... } is called an orthogonal family  if <   ,   > = 0 whenever . The family is o rthonormal  if each |  | = 1.

A glimmer into why this is such an important idea is suggested by the following example. Consider the collection of vectors

                         B  = { [1, 2, 1], [-2, 1, 0], [-1, -2, 5] }.

We verify that B  is an orthogonal family. To do this, three dot products need to be computed. Do them in your head, or let Maple calculate for you.

>	u:=<1,2,1>; v:=<-2,1,0>; w:=<-1,-2,5>;

>	DotProduct(u,v); DotProduct(v,w); DotProduct(w,u);

>

Suppose we want to find numbers a, b , and c  such that

                 [1, 2, 3] = a  [1, 2, 1] + b  [-2, 1, 0] + c  [-1, -2, 5].

There are many ways to do this. Linear algebra techniques come to mind. On the other hand, here is a way that takes advantage of the fact that B   is in orthogonal family. Take the dot product of both sides of the equation with the first vector [1, 2, 1]. This produces the equation

                        8 = a  6

from which a   is easily found. We use Maple to repeat this calculation, as well as the other two needed to compute b  and c.

>	xvec:=<1,2,3>; ort_bas:=(a*u+b*v+c*w);

>	DotProduct(xvec,u)=DotProduct(ort_bas,u); expand(%);

>	DotProduct(xvec,v)=DotProduct(ort_bas,v); expand(%);

>	DotProduct(xvec,w)=DotProduct(ort_bas,w); expand(%);

The result seems to be that a  = 4/3, b  = 0, and c  = 1/3. We check that this is correct by computing a u + b v + c w .

>	4/3*u+0*v+1/3*w;

>

Now imagine a large collection of orthogonal vectors, say 10,000 or an infinity of them. Solving for the correct coefficients in a problem such as the one above, where we had only three vectors, would be nearly impossible with the old linear algebra matrix methods. The method above used the fact that we had an orthogonal family. Having an orthogonal family reduced the work load.

We present some examples of orthogonal families containing an infinity of vectors.

Examples of orthogonal families

(1) The infinite collection of functions on C([0, ]) containing { sin(x), sin(2 x), sin(3 x), ...},
(2) The infinite collection of functions on C([0, ]) containing { 1, cos(x), cos(2 x), cos(3 x), ...},

(3) The infinite collection of functions on C([ , ]) containing the union of the above two sets.

We verify that the first collection is an orthogonal family and leave it to the student to verify that the second collection is an orthogonal family. We recall that the usual dot product in this space is an integral. What remains is to compute the following integral and determine if its value is zero. The assumption is that m  and n  are integers is needed:

                                        .

Here are the calculations.

>	assume(n,integer); assume(m,integer);

>	Int(sin(n*x)*sin(m*x),x=0..Pi)=int(sin(n*x)*sin(m*x),x=0..Pi);

You might be interested to know what happens if n = m.

>	Int(sin(n*x)*sin(n*x),x=0..Pi) = int(sin(n*x)^2,x=0..Pi);

>

The third set brings up two ideas. First, the third set is an orthogonal family on C([- , ]). And second, that third set is not an orthogonal family on C([0, ]). To see the first of these we need to evaluate three integrals.

>	Int(sin(n*x)*sin(m*x),x=-Pi..Pi) = int(sin(n*x)*sin(m*x),x=-Pi..Pi);

>	Int(cos(n*x)*cos(m*x),x=-Pi..Pi) = int(cos(n*x)*cos(m*x),x=-Pi..Pi);

>	Int(sin(n*x)*cos(m*x),x=-Pi..Pi) = int(sin(n*x)*cos(m*x),x=-Pi..Pi);

>

To show that the union of the two sets is not orthogonal on [0, ] we evaluate one more integral to see that the norm is not zero.

>	n:='n': m:='m': Int(sin(n*x)*cos(m*x),x=0..Pi)=int(sin(n*x)*cos(m*x),x=0..Pi); Int(sin(n*x)*cos(n*x),x=0..Pi)=int(sin(n*x)*cos(n*x),x=0..Pi);

We can say more about the norm than its being non-zero.  In particular, we know , because we knew from above that

>	Int(sin(n*x)*cos(m*x),x=-Pi..0)=int(sin(n*x)*cos(m*x),x=-Pi..0); Int(sin(n*x)*cos(n*x),x=-Pi..0)=int(sin(n*x)*cos(n*x),x=-Pi..0);

>

Why are we interested in orthogonal functions? We ask again: Why are they important?

We will find that they play a crucial role in some of the techniques for solving some of the classical partial differential equations. Indeed, they are the basis for the notion of Fourier Series. Consider these three ideas.

Idea 1.  If the set { , , , ... } is an orthogonal family, then any finite subset of the family is a linearly independent set.

Idea 2.  If you tell me that you have a function f that is a linear combination of {  }then I can tell you what the coefficients are. If fact, if

                   f =    +  +  ,

then               =   /  ||  || , n = 1, 2, or 3.

The student can make an argument for this by simply taking the dot product with  of both sides of the equation for f .

Idea 3.  If there are an infinity of 's and only a finite number of them are used, then it may be that we only approximate f.

Example.   If we take a function f on the interval [0,  ] and try to use the collection of sine terms, but only a finite number of them, we may get only an approximation for f. We illustrate this by drawing a graph of an f in black, two terms of sines in red, and six terms of sines in blue. The graphs show that this is only an approximation.

>	f:=t->Pi/2-abs(t-Pi/2);

>

a[1]:=int(f(t)*sin(t),t=0..Pi) / int(sin(t)^2,t=0..Pi); a[2]:=int(f(t)*sin(2*t),t=0..Pi) / int(sin(2*t)^2,t=0..Pi); a[3]:=int(f(t)*sin(3*t),t=0..Pi) / int(sin(3*t)^2,t=0..Pi); a[4]:=int(f(t)*sin(4*t),t=0..Pi) / int(sin(4*t)^2,t=0..Pi); a[5]:=int(f(t)*sin(5*t),t=0..Pi) / int(sin(5*t)^2,t=0..Pi); a[6]:=int(f(t)*sin(6*t),t=0..Pi) / int(sin(6*t)^2,t=0..Pi);

>	plot([f(t),       a[1] * sin(t),       add( a[n] * sin(n*t), n=1..3),       add( a[n] * sin(n*t), n=1..5)],   t=0..Pi,color=[black,red,green,blue]);

>

We end this section with four inequalities. I present no proof for the first two. The first two commonly appear in many texts -- both linear algebra texts and partial differential equations texts. The third is less common and I indicate a proof. The fourth follows from the third inequality. If it is true that all good analysis rests on an inequality, then the third inequality is the basis for Fourier Analysis.

Inequalities.

The triangle inequality:  || x + y || || x || + || y ||.

The Cauchy Swartz inequality:    | < x, y > | || x || || y ||.

The Fourier Inequality:   if { , , , ... } is an orthonormal  family and f is in the space, then

      0     =  +   - .

The Bessel's inequality:  if { , , , ... } is an orthonormal  family and f is in the space, then

                                               .

Two important consequences of The Fourier Inequality are given.

Important Consequence 1:  Suppose the 's are orthonormal. The way to choose the coefficients of the 's to make the approximation for f by

                       approximation for f   =  

as close as possible is to choose the 's to be < f ,  >. To see this, note that no choice you make on the 's changes the right most side of the Fourier Inequality except with the middle term, and that term can be made zero by choosing

                                                = < f ,  >.

Important Consequence 2:  If  = < f ,  > then the right most side of the Fourier Inequality gives that

             0    -  

which is Bessel's inequality.

The question that should arise now is what kind of functions can we approximate with an orthogonal family and how close can we get with the approximations. These are the issues that we examine before we start applying the ideas to partial differential equations.

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EMAIL: herod@math.gatech.edu   or    jherod@tds.net

URL: http://www.math.gatech.edu/~herod

Copyright ©  2003  by James V. Herod

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