James V. Herod*

Page maintained by Evans M. Harrell, II, harrell@math.gatech.edu.

Section 3.2. **Standard Forms**

In this section, we review the procedures to get second-order equations into standard forms, and compare these with the techniques to get second-order partial differential equations with constant coefficients into standard form.

Performing these algebraic procedures corresponds to the geometric notions of translations and rotations.

For example, the equation

x^{2} - 3y^{2} - 8x + 30y = 50

represents a hyperbola. To draw the graph of the hyperbola, one algebraically factors the equation or, geometrically, translates the axes:

(x - 4)^{2} - 3(y - 5)^{2} = 1.

Now, the response that this is a hyperbola with center {4,5} is expected. More detailed information about the direction of the major and minor axes could be made, but these are not notions that we will wish to carry over to the process of getting second order partial differential equations into standard forms.

There is another idea more appropriate. Rather than keeping the hyperbola in the Euclidean plane where it now has the equation

x^{2} - 3y^{2} = 1

in the translated form, think of this hyperbola in the Cartesian plane, and do not insist that the x axis and the y axis have the same scale. In this particular case, keep the x axis the same size and expand the y axis so that every unit is the old unit multiplying by R(3). Algebraically, one commonly writes that there are new coordinates {x',y'} related to the old coordinates by

x = x' , R(3) y = y'.

The algebraic effect is that the equation is transformed into an equation in {x',y'} coordinates:

x' ^{2 } - y' ^{2 }= 1.

Pay attention to the fact that it is a mistake to carry over too much of the geometric language for the form. For example if the original quadratic equation had been

x^{2} + 3y^{2} - 8x + 30y = 50

and we had translated axes to produce

x^{2} + 3y^{2} = 50,

and then rescaled the axes to get

x^{2} + y^{2 }= 50

we have not changed an ellipse into a circle for a circle is a geometric object whose very definition involves the notion of distance. The process of changing the scale on the x axis and the y axis certainly destroys the entire notion of distance being the same in all directions.

Rather, the rescaling is an idea that is* algebraically *simplifying.

Before we pursue the idea of rescaling and translating in second order partial differential equations in order to come up with standard forms, we need to recall that there is also the troublesome need to rotate the axis in order to get some quadratic forms into the standard one. For example, if the equation is

xy = 2,

we quickly recognize this as a quadratic equation. Even more, we could draw the graph. If pushed, we would identify the resulting geometric figure as a hyperbola. We ask for more here since these geometric ideas are more readily transformed into ideas about partial differential equations if they are converted into algebraic ideas. The question, then, is how do we achieve the algebraic representation of the hyperbola in standard form?

One recalls from analytic geometry, or recognizes from simply looking at the picture of the graph of the equation, that this hyperbola has been rotated out of standard form. To see it in standard form, we must rotate the axes. One forgets the details of how this rotation is performed, but should know a reference to find the scheme. From here on, the following may serve as your reference!

Here is the rotation needed to remove the xy term in the equation

a x^{2} + b xy + c y^{2} + d x + e y + f = 0.

(3.3)

The new coordinates {x', y'} are given by

(3.4)

where

(3.5)

And equivalent way to write (11.2) is to multiply both sides by the inverse of the matrix:

(3.6)

Thus, substitute x = x' cos(a) - y' sin(a) and y = x' sin(a) + y' cos(a) into the equation (11.1), where a is as indicated above and the cross term, bxy, will disappear.

Given a general quadratic in two variables, there are three things that need
to be done to get it into standard form: get rid of the xy terms, factor all
the x terms and the y terms separately, and rescale the axes so that the
coefficients of the x^{2} term and the y^{2} terms are the
same. Geometrically this corresponds, as we have recalled, to a rotation,
translation, and expansion, respectively. From the geometric point of view, it
does not matter which is done first: the rotation and then the translation, or
vice versa. Algebraically, it is better to remove the xy terms first, for then
the factoring is easier.

Finally, it should also be remembered that the sign of b^{2} - 4ac can
be used to predict whether the curve is a hyperbola, parabola, or ellipse.

What follows is a Maple program for removing the cross term. Using the program assures fast accurate computation for otherwise tedious calculations to determine the rotation for equations of the form

a x^{2} + b xy + c y^{2} +d x +ey + f = 0.

The coefficients are read in first.

* a:=?: b:=?:c:=?:d:=?:e:=?:f:=?:

The angle is determined and the rotation is performed.

*if a=c then alpha := Pi/4 else alpha:=arctan(b/(a-c))/2 fi; * si:=sin(alpha); * co:=cos(alpha); * x:=co*u-si*v; y:=si*u+co*v; * q:=a*x^2+b*x*y+c*y^2+d*x+e*y+f; * simplify("); * evalf(");

The purpose of the previous paragraphs recalling how to change algebraic equations representing two dimensional conic sections into standard form was to suggest that the same ideas carry over almost unchanged for the second degree partial differential equations. The techniques will change these equations into the standard forms for elliptic, hyperbolic, or parabolic partial differential equations.

Here are the forms we want:

Elliptic Equations:

Hyperbolic Equations:

Parabolic Equation:

The choice for which form to use is determined by the techniques used to solve the equations.

We turn now to the techniques of arriving at standard forms for partial differential equations.

If one has a second order partial differential equation with constant coefficients that is not in the standard form, there is a method to change it into this form. The techniques are similar to those used in the analytic geometry. Having the standard form, one might then solve the equation. Finally, the solution should be transformed back into the original coordinate system.

We will illustrate the procedure for transformation of a second order equation into standard form. Consider the equation

In the original equation, if we think of the equation as

Then, a = 11, b = 4 R(3), c = 7, so that b^{2} - 4 a c = -260 and we
identify this as an elliptic equation.

We would like to transform the equation into the form

Introduce new coordinates (x,h) by rotation of axes so that in the transformed equation the mixed second partial derivative does not appear. Let

or,

Using the chain rule,

and

It follows that

so that

In a similar manner,

and

The original equation described u as a function of x and y. We now define v as a function of x and h by v(x,h) = u(x,y). The variables x and h are related to x and y as described by the rotation above:

v(x,h) = u(x(x,h), y(x,h))

= u(cos(a) x - sin(a) h, sin(a) x + cos(a) h)

Of course, we have not specified a yet. This comes next.

The equation satisfied by v is

where we have used the abbreviations s = sin(a) and c = cos(a). The coefficient of the mixed partials will vanish if a is chosen so that

that is,

tan(2a) = 3^{1/2}/2

This means

After substitution of these values, the equation satisfied by v becomes

This special example, together with the foregoing discussions of analytic geometry makes the following statement believable: Every second order partial differential equation with constant coefficients can be transformed into one in which mixed partials are absent. The angle of rotation is, again, given by (3.3).

It is left as an exercise to see that in the general case, b^{2} - 4 a
c is left unchanged by this rotation. Surely,
Maple is to be used to do this
calculation. (See Exercise 4.)

We are now ready for the second step: to remove the first order term. For economy of notation, let us assume that the given equation is already in the form

Define v by

where b will be chosen so that the transformed equation will have the first order derivative removed. Differentiating u and substituting into the equation we get that

If we choose b = - 3/2, we have

Notice that this transformation to achieve an equation lacking the first derivative with respect to x is generally possible when the coefficient on the second derivative with respect to x is not zero, and is otherwise impossible. The same statements hold for derivatives with respect to y.

The final step is rescaling. We choose variables \xi and \eta by

\xi = \mu x and \eta = \nu y, where m and n are chosen so that in the
transformed equation the coefficients of v \xi \xi , v \eta \eta ^{,
}and v are equal in absolute value. We have

Our equation becomes

The condition that

\mu^{2} = 4 \nu^{2} = 5/2

will be satisfied if \mu = (5/2)^{1/2} and \nu = (5/8)^{1/2} .
Then, we obtain the standard form

v \xi \xi - v \eta \eta - v = 0.

**Generalized Rotations to Remove Cross Terms**

The removal of cross terms as typically presented is complicated by the necessity of finding the angle a of rotation and of working with the ensuing trigonometric functions. These ideas seem inappropriate for higher dimensions.

An alternative approach is to address the problem from the perspective of linear algebra, where the ideas even generalize for large dimensions. We will need the linear algebra package to make this work.

`
> with(linalg);
`

We choose the constants as the coefficients for the partial differential equations

a u_{xx} + b _{xy} + c_{yy} + d u_{x} + e u_{y} + f = 0.

`
> a:=2: b:=3: c:=-2: d:=1: e:=1: f:=3/25:
`

Make a symmetric matrix with a, b/2, and c as the entries.

`
> A:=matrix([[a,b/2],[b/2,c]]):
`

It is a theorem in linear algebra that symmetric matrices have a diagonal Jordan form.

`
> jordan(A,P);
`

We want K to be a matrix of eigenvectors.

`
> K:=inverse(P);
`

In order to make this process work, we ask that the eigenvectors -- which form the columns of K -- should have norm 1. This will cause the eigenvectors to be orthonormal.

`
> N1:=norm([K[1,1],K[2,1]],2): N2:=norm([K[1,2],K[2,2]],2):`

`
> L:=matrix([[K[1,1]/N1,K[1,2]/N2],[K[2,1]/N1,K[2,2]/N2]]);
`

It will now be the case that (transpose(L) * A * L) is the Jordan form.

> evalm(transpose(L) &* A &* L);

To remove the cross-product term, we now perform a change of variables defining s and t.

> s:=(x,y)->evalm(transpose(L) &* vector([x,y]))[1]:

t:=(x,y)->evalm(transpose(L) &* vector([x,y]))[2]:

If we define v as indicated below, then it will satisfy a PDE with the cross-term missing and with u satisfying the original partial differential equation.

`
> u:=(x,y)->v(s(x,y),t(x,y)):`

`
> a*diff(u(x,y),x,x)+b*diff(u(x,y),x,y)+c*diff(u(x,y),y,y)+`

`
d*diff(u(x,y),x) + e*diff(u(x,y),y) + u(x,y):`

`
> simplify(");`

`
> collect(",D[1,2](v)(s(x,y),t(x,y)));`

`
> pde:=collect(`

`
collect(`

`
collect(`

`
collect(",D[1](v)(s(x,y),t(x,y)))`

`
,D[2](v)(s(x,y),t(x,y)))`

`
,D[2,2](v)(s(x,y),t(x,y)))`

`
,D[1,1](v)(s(x,y),t(x,y)));`

`
`
Thus, we solve this simpler system and create u that satisfies the original
equation.

**EXERCISES**:

1. Transform the following equations into standard form: