Second Order Equations as First Order Systems

PART II: SECOND ORDER EQUATIONS
Section 18
Second Order Equations as First Order Systems
James V. Herod*

*(c) Copyright 1993,1994,1995 by James V. Herod, herod@math.gatech.edu. All rights reserved.

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


We recall Section 3 where we were concerned with systems of partial differential equations. In this section, we will again investigate systems of equations using the methods of characteristics.

We consider a partial differential equation expressed by

F([[partialdiff]]u,[[partialdiff]]t) + A F([[partialdiff]]u,[[partialdiff]]x) = F(t,x,u) (18.1)

defined for - * < x < *, t > 0,

with the initial condition u(0,x) specified. (18.2)

In the equation (18.1), A denotes an nxn matrix whose entries depend on t, x, and u. The function F is vector valued and depends on t, x, and  u also -- but not [[partialdiff]]u/[[partialdiff]]t or [[partialdiff]]u/[[partialdiff]]x. The solution u is a vector function of t and x. In Section 4 we wrote systems of first order partial differential equations in this form. Here, we will re-write second order equations as these matrix systems and use techniques that are suggested by the Jordon form representations of A and by the methods of characteristics to analyze these equations.

Re-writing second order equation as first order systems is not a new idea. Recall that this notion was used in sophomore ordinary differential equations. Here's an example to recall the form that the method took for ordinary differential equations.

We re-write the second order, ordinary differential equation

y'' + 3 y' + y = 0, y(0) = 5, y'(0) = 7

as a first order system. The procedure is standard: let u = y and v = y'. Then

u' = y' = v

and v' = y'' = - y - 3 y' = - u - 3 v.

Hence B(A( u, v))' = B(ACO2( 0, 1, -1, -3)) B(A( u, v))

with

B(A( u(0), v(0))) = B(A( 5, 7)).

The methods for solving systems are richer and encompass more varieties than just the second order equation converted to this system. Also, the system's method is the structure that generalizes to more complicated situations.

We now illustrate these methods by re-writing the wave equation in this form.

Example 18.1

Consider

utt - [[gamma]]2 uxx = 0. (18.3)

We express (18.3) as a first order system as follows: let v1 = ut and v2 = [[gamma]] ux. Then equation (18.3) implies that

F([[partialdiff]] v1,[[partialdiff]]t) = [[gamma]] F([[partialdiff]] v2,[[partialdiff]]x).

But, assuming the equality of mixed partials,

F([[partialdiff]] v2,[[partialdiff]]t) = [[gamma]] F([[partialdiff]] ux,[[partialdiff]]t) = [[gamma]] F([[partialdiff]] ut,[[partialdiff]]x) = [[gamma]] F([[partialdiff]]v1,[[partialdiff]]x).

Writing this as a system,

F([[partialdiff]] ,[[partialdiff]]t) B(A(v1,v2)) = B(ACO2( 0, [[gamma]], [[gamma]], 0)) F([[partialdiff]] ,[[partialdiff]]x) B(A(v1,v2))

or F([[partialdiff]] ,[[partialdiff]]t) B(A(v1,v2)) + B(ACO2( 0, -[[gamma]], -[[gamma]], 0)) F([[partialdiff]] ,[[partialdiff]]x) B(A(v1,v2)) = B(A(0,0)).

Thus,

F([[partialdiff]]v,[[partialdiff]]t) + A F([[partialdiff]]v,[[partialdiff]]x) = 0

where v = B(A(v1,v2)) and A is the matrix indicated.

Example 18.2

Another second order partial differential equation often encountered is Laplace's equation:

F([[partialdiff]]2u,[[partialdiff]]x2) + F([[partialdiff]]2u,[[partialdiff]]y2) = 0

This second order equation can be written as the system

F([[partialdiff]] ,[[partialdiff]]x) B(A(v1,v2)) + B(ACO2( 0, 1, -1, 0)) F([[partialdiff]] ,[[partialdiff]]y) B(A(v1,v2)) = B(A(0,0)).

In their book

Introduction to Partial Differential Equations with Applications, Dover Publications, Inc. New York (1976)

Zachmanoglou and Thoe define hyperbolic and elliptic systems. See page 362 of that book.

Definition: A system such as (18.1) with A an nxn matrix is

(1) hyperbolic if the eigenvalues of A are real and the Jordon form of A is diagonal; that is, if the eigenvalues of A are real and A has n linearly independent vectors.

(2) totally hyperbolic if the eigenvalues of A are real and distinct.

(3) elliptic if A has no real eigenvalues.

Remark. The wave equation leads to a hyperbolic system and Laplace's equation leads to an elliptic system, unremarkably.

In Section 11, we defined characteristics for a system of the form

autt + butx + cuxx +d ut + e ud + f u = g. (18.4)

The characteristics were given by equations (11.3) and (11.4). We now establish what might have been expected.

Theorem: When (18.4) is re-written as the system as (18.1), then the eigenvalues of A determine the characteristics of (18.4).

To verify this result, we identify v1 = ut and v2 = ux. Equality of mixed partial derivatives gives that

F([[partialdiff]] v2,[[partialdiff]]t) - F([[partialdiff]] v1,[[partialdiff]]x) = 0.

and, from (18.4),

F([[partialdiff]] v1,[[partialdiff]]t) + F(b,a) F([[partialdiff]] v1,[[partialdiff]]x) + F(c,a) F([[partialdiff]] v2,[[partialdiff]]x) = F(1,a) (g - f u - e v1 - d v2).

The resulting matrix form is

F([[partialdiff]] ,[[partialdiff]]t) B(A(v1,v2)) + B(ACO2( b/a, c/a, -1, 0)) F([[partialdiff]] ,[[partialdiff]]x) B(A(v1,v2)) = B(A(f1,0)).

The eigenvalues of this matrix A are solutions for the equation

([[lambda]] - b/a)[[lambda]] + c/a = 0.

That is,

a [[lambda]]2 - b [[lambda]] + c = 0.

This agrees with (11.4) and (11.5) and completes the verification of the Theorem.

In what follows we suppose that the system is totally hyperbolic. This implies that the characteristic curves will be real and distinct. The key feature of the method of characteristics is that along the characteristic curves system (18.1) can be reduced to a system of ordinary differential equations.

Example 18.3

Suppose that the coefficients of A in (18.1) depend on t, x, and u, that F = 0, and that the system is totally hyperbolic. Let [[lambda]] and u be the two eigenvalues of A. Construct the matrix K so that

KAK-1 = B(ACO2( [[lambda]], 0, 0, u)) = D,

or, what is the same,

A = K-1 D K.

Note that all of [[lambda]], u, K, and D are not constant in t or x since A is not. We recall that K is constructed as the matrix with columns the eigenvectors of A. If u = Kv then since

0 = F([[partialdiff]]u,[[partialdiff]]t) + A F([[partialdiff]]u,[[partialdiff]]x)

we have 0 = BBC[( F([[partialdiff]]K,[[partialdiff]]t) v + KF([[partialdiff]]v,[[partialdiff]]t)) + ABBC[(F([[partialdiff]]K,[[partialdiff]]x) v + K F([[partialdiff]]v,[[partialdiff]]x) )

= KF([[partialdiff]]v,[[partialdiff]]t) + A K F([[partialdiff]]v,[[partialdiff]]x) + F([[partialdiff]]K,[[partialdiff]]t) v + AF([[partialdiff]]K,[[partialdiff]]x) v .

Multiplying by K-1

0 = F([[partialdiff]]v,[[partialdiff]]t) + D F([[partialdiff]]v,[[partialdiff]]x) + K-1 F([[partialdiff]]K,[[partialdiff]]t) v + K-1AF([[partialdiff]]K,[[partialdiff]]x) v (18.5)

= F([[partialdiff]]v,[[partialdiff]]t) + D F([[partialdiff]]v,[[partialdiff]]x) + F(t,x,v)

Then

0 = F([[partialdiff]]v1,[[partialdiff]]t) + [[lambda]] F([[partialdiff]]v1,[[partialdiff]]x) + F(t,x,v)1

and

0 = F([[partialdiff]]v2,[[partialdiff]]t) + u F([[partialdiff]]v2,[[partialdiff]]x) + F(t,x,v)2.

This is a system of differential equations such as we solved with first order systems and can be solved with similar techniques.

Example 18.4: Consider the partial differential equation

uxx + 4 uxy + 3 uyy = 0.

We find the general solution by the techniques of this section.

The problem is rewritten as a system

F([[partialdiff]]U,[[partialdiff]]t) + A F([[partialdiff]]U,[[partialdiff]]x)

where A is the matrix

A = B(ACO2( 4, 3, -1, 0)).

This has Jordan form

D = B(ACO2( 1, 0, 0, 3)),

where the matrix K such that

D = K A K-1

is

K = B(ACO2( 1, 3, 1, 1)).

Choose

v(x,y) = f(y + x) and w(x,y) = g( y + 3 x).

Take u(x,y) to be the first component of

K-1 B(A(v(x,y),w(x,y))).

This provides a solution of the partial differential equation.

Exercises

18.1 Find the general solution for these two partial differential equations:

(a) uxx + 2 uxy - 3 uyy = 0.

(b) uxx - 4 uxy + 4 uyy = 0.

18.2 Solve the partial differential equation

uxx + 4 uxy + 3 uyy = 0

subject to these conditions:

(a) u(0,y) = sin(y) and ux(0,y) = 0.

(b) u(x,-x) = j(x) and [[partialdiff]]u/[[partialdiff]][[eta]] = h(x) where [[eta]] = {1,1}/R(2). (Recall that

[[partialdiff]]u/[[partialdiff]][[eta]] = < {ux, uy}, [[eta]] >.

(c) u((x,x2) = j(x) and [[partialdiff]]u/[[partialdiff]][[eta]] = h(x). In this case [[eta]](x) = {-2x,1}/R(1+4x2).

18.3 For each of the following matrices A solve the system

Zt = AZx

(a) A = B(ACO2( 3/2, -1/2, -1/2, 3/2))

(b) A = B(ACO2( -3/2, 1/2, 1/2, -3/2))

(c) A = B(ACO2(41/25, -12/25, -12/25, 34/25))

(d) A =B(ACO2(-41/25, 12/25, 12/25, -34/25))

18.4 For each of the following matrices A, solve the system

Zt = A Zx.

(a) A = B(ACO2( (1-x)/2, (1+x)/2, (1+x)/2, (1-x)/2))

(b) A = B(ACO2( (9-16x)/25, 12(1+x)/25, 12(1+x)/25, (16-9x)/25))

(c) A = B(ACO2( (t-x)/2, (t+x)/2, (t+x)/2, (t-x)/2))

(18.5) (a) Write the parabolic system

F([[partialdiff]]u,[[partialdiff]]t) = F([[partialdiff]]2u,[[partialdiff]]x2) (18.5)

as the system

B(ACO2( 0, 0, 1, 0)) F([[partialdiff]] ,[[partialdiff]]t) B(A(v1,v2)) + B(ACO2( 0, 1, 1, 0)) F([[partialdiff]] ,[[partialdiff]]x) B(A(v1,v2)) = B(ACO2( 1, 0, 0, 0)) B(A(v1,v2)).

(b) In his book

Partial differential Equations, Oxford Applied Mathematics and Computing Series, Oxford University Press, New York (1980)

W. E. Williams classifies n dimensional systems of the form

Aux + B uy = c. (18.6)

See page 297. Consider the equation

det(A[[lambda]] - B) = 0 (18.7)

and the vector equation

(A[[lambda]] - B)v = 0 . (18.8)

The system (18.6) is

hyperbolic if the roots of equation (18.7) are all real (but not necessarily distinct) and such that there exists n independent solutions of equation (18.8).

totally hyperbolic when the roots of equation (18.6) are distinct and real.

parabolic when the roots are real but not distinct and there exist less than n independent solutions for (18.8).

elliptic if none of the roots of (18.7) are real.

With these classifications, is (18.5) parabolic?

(c) Write F([[partialdiff]]2u,[[partialdiff]]x2) + F([[partialdiff]]2u,[[partialdiff]]y2) = 0 and F([[partialdiff]]2u,[[partialdiff]]x2) - F([[partialdiff]]2u,[[partialdiff]]y2) = 0

in the form (18.6) and classify them with the scheme of Williams.


Return to James Herod's Table of Contents

Back to preceding lecture

Onward to the next lecture

Return to Evans Harrell's