ranslation Solutions

## PART II: SECOND ORDER EQUATIONS Section 12 Translation Solutions for the Constant Coefficient Equations James V. Herod*

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

2.7.5 -->pde93.10Section 12

Translation Solutions for the Constant Coefficient Equations

We identify solutions for a class of equations of the form

a F([[partialdiff]]2u,[[partialdiff]]x2) + b F([[partialdiff]]2u,[[partialdiff]]x[[partialdiff]]y) + c F([[partialdiff]]2u,[[partialdiff]]y2) + d F([[partialdiff]]u,[[partialdiff]]x) + e F([[partialdiff]]u,[[partialdiff]]y) + f u= 0. (12.1)

where b2 - 4ac > 0.

We suppose here that the coefficients are constants. Note that if

[[alpha]] = F(be-2cd,b2-4ac) and [[beta]] = F(bd-2ae,b2-4ac) (12.2)

and u(x,y) = exp(-[[alpha]]x-[[beta]]y) v(x,y) (12.3)

then u satisfies the above equation if and only if v satisfies an equation which has the form

a F([[partialdiff]]2v,[[partialdiff]]x2) + b F([[partialdiff]]2v,[[partialdiff]]x[[partialdiff]]y) + c F([[partialdiff]]2v,[[partialdiff]]y2) + Fv = 0. (12.4)

In effect, we removed the first order terms. To develop this initial understanding of solutions for constant coefficient, second order, partial differential equations, we first suppose that F in equation (12.4) is zero. We make solutions for an equation which has this form:

a F([[partialdiff]]2v,[[partialdiff]]x2) + b F([[partialdiff]]2v,[[partialdiff]]x[[partialdiff]]y) + c F([[partialdiff]]2v,[[partialdiff]]y2) = 0. (12.5)

To examine the types of solutions that are obtained for different choices of a, b, and d, we begin with a trial solution of the form

v(x,y) = f( y - mx )

for a twice-differentiable function f, and for a constant m to be determined by the requirement that u defined in this form is a solution to the partial differential equation. Using a prime to signify the derivative of f with respect to its argument, the required second partial derivatives are

m2f'', - m f'', and f''.

The implication is that

(am2 - bm + c) f'' = 0

from which we conclude that either f'' = 0 or the quantity enclosed in parentheses vanishes. The case where f'' = 0 yields the solution that must be a linear function. This is not the most general solution. Thus, we disregard this possibility and instead require

am2 - bm + c = 0. (12.6)

In other words, by choosing a particular linear combination of the independent variables x and y, we obtain a solution from any function that is twice differentiable. Solving the quadratic equation (12.6) for m, we obtain the two solutions if b2 - 4 a c != 0. The importance of the coefficients is now evident since the sign of the discriminate is crucial to determine the number and type of solutions for (12.5).

We suppose that b2 - 4ac > 0, so that we work with a hyperbolic equation. In this case, there are two distinct solutions m1 and m2 of the equation (12.6). Any function of either of the argument y-m1x or y-m2x solves the equation (12.5). Therefore, the general solution of the equation (12.5) is

v(x,y) = f(y-m1x)+g(y-m2x) (12.7)

where f and g are required to be twice-differentiable functions; otherwise, they may be chosen arbitrarily.

There is another way to see that this form is the most general solution. In Exercise 12.3 below it is verified that equation (12.5) can be transformed into the equation

F([[partialdiff]]w,[[partialdiff]][[xi]][[partialdiff]][[eta]]) = 0. (12.8)

By integrating this equation first with respect to [[xi]] and then with respect to [[eta]], we obtain the solution as w([[xi]],[[eta]]) = f([[xi]]) + g([[eta]]), where f and g are any twice-differentiable functions of integration. In terms of the original variables x and y, this solution is the same as that above.

Example. We solve the partial differential equation

2 F([[partialdiff]]2u,[[partialdiff]]x2) + 3 F([[partialdiff]]2u,[[partialdiff]]x[[partialdiff]]y) - 2 F([[partialdiff]]2u,[[partialdiff]]y2) + 1 F([[partialdiff]]u,[[partialdiff]]x) + 1 F([[partialdiff]]u,[[partialdiff]]y) + 3 u/25= 0 (12.9)

subject to the initial condition that

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

Let [[alpha]] and [[beta]] be as in equation (12.2). Then

m = (3 +/- 5 )/4.

The general solution of (12.10) is

u(x,y) = exp(- 7x/25 + y/25) [ f( y-m1x) + g( y -m2 x) ].

The initial conditions imply that

u(x,y) = [ exp(-3 x/10) sin( y + x/2 ) + exp(-x/5) sin( y -2 x)] / 5

Details of the arithmetic for the example were done in the MAPLE syntax.

The graphs that follows are more than graphs of the solution. There is a graph of exp([[alpha]]x + [[beta]]y) f( y + m1 x), of exp([[alpha]]x + [[beta]]y) g( y + m2 x) and of u(x,y). These three graphs, drawn together emphasize the fact that the solution is the sum of two waves, one moving along each characteristic curve.

Figure 12.a

Figure 12.b

Figure 12.c

Exercises:

12.1. Solve one of these. Explain why you can not solve the other with the methods of this section.

1 F([[partialdiff]]2u,[[partialdiff]]x2) + 3 F([[partialdiff]]2u,[[partialdiff]]x[[partialdiff]]y) + 1 F([[partialdiff]]2u,[[partialdiff]]y2) + 3 F([[partialdiff]]u,[[partialdiff]]x) + 1 F([[partialdiff]]u,[[partialdiff]]y) + u/5 = 0

and

1 F([[partialdiff]]2u,[[partialdiff]]x2) + 3 F([[partialdiff]]2u,[[partialdiff]]x[[partialdiff]]y) + 1 F([[partialdiff]]2u,[[partialdiff]]y2) + 3 F([[partialdiff]]u,[[partialdiff]]x) + 1 F([[partialdiff]]u,[[partialdiff]]y) - u/5= 0.

Each is subject to the conditions that

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

12.2. Verify that if u satisfies (12.1) and v is related to u by (12.3) then v satisfies (12.4).

12.3. Verify that equation (12.7) is a solution of equation (12.5).

12.4. Give the necessary transformations so that if one can solve (12.5) or (12.8), then that solution can be transformed into a solution for the other.

12.5. If the equation is hyperbolic, give the general solution in terms of the original variables.

(1) F([[partialdiff]]2u,[[partialdiff]]x2) + 3 F([[partialdiff]]2u,[[partialdiff]]x[[partialdiff]]y) + 2 F([[partialdiff]]2u,[[partialdiff]]y2) - F([[partialdiff]]u,[[partialdiff]]x) - F([[partialdiff]]u,[[partialdiff]]y) = 0.

(2) F([[partialdiff]]2u,[[partialdiff]]x2) + 3 F([[partialdiff]]2u,[[partialdiff]]x[[partialdiff]]y) + 2 F([[partialdiff]]2u,[[partialdiff]]y2) - 2 F([[partialdiff]]u,[[partialdiff]]x) - 4 F([[partialdiff]]u,[[partialdiff]]y) = 0.

(3) F([[partialdiff]]2u,[[partialdiff]]x2) + 2 F([[partialdiff]]2u,[[partialdiff]]x[[partialdiff]]y) + 2 F([[partialdiff]]2u,[[partialdiff]]y2) = 0.

(4) F([[partialdiff]]2u,[[partialdiff]]x2) + 2 F([[partialdiff]]2u,[[partialdiff]]x[[partialdiff]]y) + F([[partialdiff]]2u,[[partialdiff]]y2) + F([[partialdiff]]u,[[partialdiff]]x) + F([[partialdiff]]u,[[partialdiff]]y) = 0.

12.6. Solve the equation

2 F([[partialdiff]]2u,[[partialdiff]]x2) + 3 F([[partialdiff]]2u,[[partialdiff]]x[[partialdiff]]y) + 1 F([[partialdiff]]2u,[[partialdiff]]y2) = 0

subject to the condition that

u(0,y) = exp(-y2), with F([[partialdiff]] ,[[partialdiff]]x)u(0,y) = 0.

What is sup(u(x,y) for x > 0 and all y?