Variable Coefficients

## PART II: SECOND ORDER EQUATIONS Section 13 Hyperbolic, Second Order PDE with Variable Coefficients James V. Herod*

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

In this section, we develop the method of characteristics for the general second order partial differential equation. We do not suppose the equation has constant coefficients. We do suppose that it is linear, however. The implication is that each of the coefficients depends on x and y, but not u.

We seek solutions for the equation

a uxx + buxy + c uyy + d ux + e uy + f u = 0.

We might suppose that s and t are functions of x and y and that there is a function v related to u by

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

It follows that

a uxx + b uxy + c uyy + d ux + e uy + f u =

A vss + Bvst + C vtt + D vs + E vt + F v = 0 (13.1)

where

A = a sx2 + b sxsy + c sy2

B = 2a sxtx + b (sxty + sytx) + 2c syty.

C = a tx2 + b txty + c ty2 (13.2)

D = a sxx + b sxy + c syy + d sx + e sy

E = a txx + b txy + c tyy + d tx + e ty, and

F = f.

(The verification that the coefficients are as stated are the tedious calculations left to MAPLE .)

We suppose that there are functions y1 and y2 changing with x so that s(x,y1(x)) and t(x,y2(x)) are constant. The coefficient A and C given above vary now with x. We ask that those functions y1 and y2 be defined so that exactly A and C are zero. Consider what requirement this request places on y1 and y2. Because s(x,y1(x)) and t(x,y2(x)) are constant, then

sx + sy y1'(x) = 0 and tx + ty y2'(x) = 0.

Hence, y1'(x) = - F(sx,sy) and y2'(x) = - F(tx,ty). (13.3a)

We now impose the requirement that A = 0. This implies that

a sx2 + b sxsy + c sy2 = 0.

Dividing by (sy)2 and using (13.3a)

a [F(dy1,dx)]2 - b F(dy1,dx) + c = 0. (13.3)

In a similar manner, to require that C = 0 is to ask that

a [F(dy2,dx)]2 - b F(dy2,dx) + c = 0. (13.4)

We have from these two quadratic functions that

F(dy1,dx) = F(b + R(b2 - 4ac),2a)

and

F(dy2,dx) = F(b - R(b2 - 4ac),2a) .

Take note of the fact that in order that we should have two real numbers in the equations above, the partial differential equation must be hyperbolic.

Example 13.1: Find the characteristic equations and find a general solution for the equation

4uxx - 16uyy + 4ux + u = 0.

Note this is a hyperbolic partial differential equation with constant coefficients. In fact, the techniques of the previous section work here, too. We remove the first order term by defining a function v with the equation

u(x,y) = exp([[alpha]]x) v(x,y).

(We did not have to include a y term in the exponential because there was no first order term in y.) Upon substituting this value of u into the equation in question, we see that if [[alpha]] = -1/2 then the first order term for v is zero, and so is the coefficient of v equal to zero. Hence, we wish to find v so that

4vxx - 16 vyy = 0.

The next job is to find w(s,t) so that if v satisfies the previous equation and v(x,y) = w(s(x,y),t(x,y)), then w is of the form ws,t = 0. In the discussion preceeding this example, we found that if we look for the locus of point (x,y) where s(x,y) and t(x,y) are constant, and so that y is a function of x then we can find the derivatives of y. The formula for the derivatives of y1 and y2 are

F(dy1,dx) = -2 and F(dy2,dx) = 2

so that y1(x) = - 2 x + c1 and y2(x) = 2 x + c2. This means that w satisfies ws,t = 0 and, thus, can be solved as w(s,t) = f(s) + g(t). Consequently, v can be written as the sum of solutions of the form

v(x,y) = [f(y + 2x) + g(y - 2x)].

and u can be written as

u(x,y) = exp(-x/2) [f(y + 2x) + g(y - 2x)].

Check this result.

Example 13.2: We find the characteristics for the equation

y2 uxx - x2 uyy + 3 F(y2,x) ux - 3 F(x2,y) uy = 0, x > 0, y > 0

and reduce this equation to a standard form.

This equation is hyperbolic for b2 - 4ac = 4 x2 y2 > 0. We simply compute the characteristics by seeking functions s and t so that s(x,y) and t(x,y) are constant and so that the equation has form with ws,s = wt,t = 0. Because s and t are to be constant along characteristics and because y is a function of x,

0 = F([[partialdiff]]s,[[partialdiff]]x) + F([[partialdiff]]s,[[partialdiff]]y) F(dy,dx) (13.5)

and

0 = F([[partialdiff]]t,[[partialdiff]]x) + F([[partialdiff]]t,[[partialdiff]]y) F(dy,dx). (13.6)

As in (13.3) and (13.4), we have that

a [F(dy,dx)]2 + b F(dy,dx) + c = 0

or y'(x) must be +/- x/y. We can either solve for y by separation of variables directly from this last - - to get that

y2 +/- x2 = c.

Or, we can solve for s and t in the previous two equations (13.5) and (13.6) by recognizing that these are two first order partial differential equations and use the methods of previous sections. In either case, s(x,y) = y2 - x2 and t(x,y) = y2 + x2.

Take u(x,y) = w(s(x,y),t(x,y)) and ask what partial differential equation w must satisfy. The answer will have the form

A wss + B wst + C wtt + D ws + E wt + F = 0,

where, as in (13.2),

A = C = 0,

B = - 2 y2 (-2 x)((2x) + 2 (-x2)(2y)(2y) = -16 x2y2

C = 0,

D = - 8 (y2+x2),

E = 8 (y2 - x2), and

F = 0.

We have then that

4 x2 y2 wst + 2 (y2 + x2) ws - 2 (y2 - x2) wt = 0

or, by substituting y2 = (s+t)/2 and x2 = (t-s)/2,

(t2 - s2) wst + 2 t ws - 2 wt = 0.

Since

[ (t2 - s2) w(s,t) ]st = (t2 - s2) wst + 2 t ws - 2 s wt

we have

(t2 - s2) w(s,t) = f(s) + g(t).

Example 13.3: We will produce a general solution to this equation:

y uxx + (x+y) uxy + x uyy = 0.

Note that equation is hyperbolic. The characteristics are defined by

y [F(dy,dx)]2 - (x+y) F(dy,dx) + x = 0.

Hence, F(dy,dx) = F(x,y) or 1.

We have s(x,y) = y2 - x2 and t(x,y) = y-x. We now compute the new equation for v(s,t) where u(x,y) = v(s(x,y1),t(x,y2)):

A = C = 0,

B = 2 y (-2x)(-1) + (x+y)(-2x - 2y) + 2x (2y) = -2(x-y)2 = -2t2.

C = 0,

D = y (-2) + x (2) = 2(x - y) = - 2t,

E = y (0) +x (0) = 0, and

F = 0.

The equation in this {s,t} coordinates system is

0 = -2t2 vst - 2t vs= -2t (t vst + vs).

Hence, 0 = t vst + vs

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

It follows that t vt + v is constant in s:

f(t) = t vt + v = F([[partialdiff]](tv),[[partialdiff]]t).

Hence, g(s) + F(t) = t v

or F(g(s),t) + H(t) = v(s,t).

Thus, there are functions g and H so that

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

This means that

u(x,y) = g(y2 - x2)/(y-x) + H(y-x).

Check this result.

The methods of the previous section start work with constant coefficients and are, perhaps, conceptually easier. The methods here are more general and can handle some problems that have variable coefficients. We have seen what should be no surprise: with variable coefficients, the methods of this section may even fail.

Exercises:

13.1 Find the general solution for each of these partial differential equations. Draw graphs to illustrate how the solutions are different.

(a) uxy = 0

(b) uxy + 2 uy = 0. (Hint: ux + 2u is independent of y.)

(c) uxy + 2 ux = 0. (Hint: uy + 2u is independent of x.)

(d) y uxy + ux = 0. (Hint: y ux is independent of y.)

(e) x uxy + uy = 0. (Hint: x uy is independent of x.)

13.2 Find the general solution for each of these partial differential equations:

(a) 3 uxx + 10 uxy + 3 uyy = 0.

(b) y uxx + (x-y) uxy - x uyy = 0.

(c) uxy + 2 x uyy = 0. (Hint:[[partialdiff]]/[[partialdiff]]y (F([[partialdiff]]u,[[partialdiff]]x) + 2x F([[partialdiff]]u,[[partialdiff]]y)) = 0 leads to a first order system.)

(d) (x-y) uxy - ux + uy = 0. (Hint: [[partialdiff]]2(x u)/[[partialdiff]]x[[partialdiff]]y = x uxy + uy,

so consider (x-y) u.)

(e) (x-y) uxy - 2 ux + 2 uy = 0. (Hint: [[partialdiff]]2( (x-y) u(x/2,y/2) )/[[partialdiff]]x[[partialdiff]]y .)