Factorization

## PART II: SECOND ORDER EQUATIONS Section 21 Toward Finding a General Solution for Second Order, Constant Coefficient PDE's: Factorization James V. Herod*

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

During the next sections, we organize methods to obtain solutions for all second order constant coefficient differential equations in two variables. We have seen that if the system can be transformed to

uxy = 0

then the solution comes easily. Factorization as developed in this section provides an alternate method for solution,

Consider the wave equation:

[[gamma]]2 uyy - uxx = 0.

This equation can be "factored" to the following expression:

([[gamma]] F([[partialdiff]] ,[[partialdiff]]y) - F([[partialdiff]] ,[[partialdiff]]x) ) ([[gamma]] F([[partialdiff]] ,[[partialdiff]]y) + F([[partialdiff]] ,[[partialdiff]]x)) u = 0.

Hence, we argue that solving the second order wave equation has been reduced to solving this system:

([[gamma]] F([[partialdiff]] ,[[partialdiff]]y) + F([[partialdiff]] ,[[partialdiff]]x)) u = v

([[gamma]] F([[partialdiff]] ,[[partialdiff]]y) - F([[partialdiff]] ,[[partialdiff]]x) ) v = 0.

This can be an easier problem. Think of the first half of these notes!

Example 21.1 In more detail, here's how to solve a standard problem in the context of this Chapter. We solve

0 = utt - c2 uxx

with u(0,x) = f(x) and ut(0,x) = 0.

in the context of factoring.

Change the problem to this pair of linear equations:

(1) v = (F([[partialdiff]] ,[[partialdiff]]t) + c F([[partialdiff]] ,[[partialdiff]]x) ) u, u(0,x) = f(x), and

(2) 0 = (F([[partialdiff]] ,[[partialdiff]]t) - c F([[partialdiff]] ,[[partialdiff]]x) ) v, v(0,x) = ut(0,x) + c ux(0,x) = 0 + c f'(x).

Solve the 2nd problem to get that v(t,x) = c f' (x + ct). Use this in the 1st problem:

z'(t) = c f'(ct + [[eta]] + ct).

Integrate both sides from 0 to t to get

z(t) - z(0) = z(t) - f([[eta]])

= I(0,t, ) c f'(2 c s + [[eta]]) ds = F(f(2 c t + [[eta]]) - f([[eta]]),2) ,

or

z(t) = F(f(c t+ x) - f(x - c t),2).

This is the solution we expected by methods already known.

One should ask what kind of problems can be written this way and what are their solutions. To answer that question, consider the kind of factorizations that are wanted and ask what can be written in that form.

We could hope for

(a F([[partialdiff]] ,[[partialdiff]]x) + b)(c F([[partialdiff]] ,[[partialdiff]]y) + d ) u = 0.

Expanding this, we have the partial differential equation should have the form

ac uxy + ad ux + (bc + acx) uy + (bd + adx) u = 0.

We can solve such an equation by solving the system

c uy + du = v

a vx + bv = 0.

Example 21.2: An example of where this technique has been used before in in Example 13.3. There

0 = [F([[partialdiff]] ,[[partialdiff]]s) + 0] [ t F([[partialdiff]] ,[[partialdiff]]t) + 1]v.

We solve this as follows

t vt + v = w

ws = 0.

Then w = h(t) and t vt + v = h(t). This is an ordinary differential equation:

F([[partialdiff]] [t v(s,t)],[[partialdiff]]t) = h(t).

Thus,

t v(s,t) - 0 = H(t) + G(s)

and

v(s,t) = H(t) + G(s)/t.

Example 21.3: Find the characteristics, and solve the system

3uxx + 3uxy - 6 uyy + 13 ux + 14 uy + 12u = 0.

We know that if we suppose that characteristics are given by

0 = a(y')2 - by' + c = 3(y')2 - 3y' - 6 = 3 (y' - 2) (y' + 1)

so that characteristics are y - 2x = constant or y + x is constant.

Let

s(x,y) = y - 2x and t(x,y) = y + x

and let v be defined by

u(x,y) = v(s,t).

If we insist that u satisfies the above differential equation then

- 9 vst - 4vs + 9vt + 4v = 0.

We ask if this equation can be factored. If it can, we write the coefficients as

ac = - 9

ad = - 4

bc + acs = 9

bd + adt = 4.

A solution is a = - 1, c = 9, d = 4, b = 1.

Hence, we factor the equation as

(- F([[partialdiff]] ,[[partialdiff]]s) + 1)(9 F([[partialdiff]] ,[[partialdiff]]t) + 4) v = 0.

We need to solve these system of equations:

(9 F([[partialdiff]] ,[[partialdiff]]t) + 4) v = w

(- F([[partialdiff]] ,[[partialdiff]]s) + 1) w = 0.

Solve the second one first:

w(s,t) = exp(- s) f(t).

Now solve

(9 F([[partialdiff]] ,[[partialdiff]]t) + 4) v = e-s f(t).

We see

F([[partialdiff]] ,[[partialdiff]]t)(exp(4t/9) v) = exp(s+4t/9)f(t)

Integrating, we get

v(s,t) = exp(s) F(t) + exp(-4t/9) G(s)

where F is related to f through an integral. This means

u(x,y) = exp(y - 2x) F(y+x) + exp(- 4(y+x)/9) G(y - 2x).

Work in Progress: Exercise: Exercises can be made up to use this technique by appropriate choices of the coefficients. Here is one choice: a = 1, c = s, b = 1, d = t2. This leads to the equation

svst + t2 vs + (s+1)vt + (t2 + 2t)v = 0.

Work in Progress: What about the factorization

(aF([[partialdiff]] ,[[partialdiff]]x) + b F([[partialdiff]] ,[[partialdiff]]y) + c) (dF([[partialdiff]] ,[[partialdiff]]x) + e F([[partialdiff]] ,[[partialdiff]]y) +f) u = 0?

Work in Progress: Solve some problems where the right side != 0 and factorization work for the left side.

Work in Progress: Find an example where the left side cannot be changed to uxy = 0 but it can be factored.