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we can see that the characteristics y=

B+√

B2−4AC 2A

x+c1, y=

B−√

B2−4AC 2A

x+c2, (4.3.2) are two families of straight lines. Consequently, the characteristic coordi- nates take the form

ξ=y−λ1x, η=y−λ2x, (4.3.3) where

λ1,2= B+√

B2−4AC

2A . (4.3.4)

The linear second-order partial differential equation with constant coeffi- cients may be written in the general form as

Auxx+Buxy+Cuyy+Dux+Euy+F u=G(x, y). (4.3.5) In particular, the equation

Auxx+Buyy+Cuyy = 0, (4.3.6) is called theEuler equation.

(A) Hyperbolic Type

IfB2−4AC >0, the equation is of hyperbolic type, in which case the characteristics form two distinct families.

Using (4.3.3), equation (4.3.5) becomes

uξη =D1uξ+E1uη+F1u+G1(ξ, η), (4.3.7) where D1, E1, and F1 are constants. Here, since the coefficients are con- stants, the lower order terms are expressed explicitly.

WhenA = 0, equation (4.3.1) does not hold. In this case, the charac- teristic equation may be put in the form

−B(dx/dy) +C(dx/dy)2= 0, which may again be rewritten as

dx/dy= 0, and −B+C(dx/dy) = 0.

Integration gives

x=c1, x= (B/C)y+c2,

wherec1and c2 are integration constants. Thus, the characteristic coordi- nates are

4.3 Equations with Constant Coefficients 101

ξ=x, η=x−(B/C)y. (4.3.8)

Under this transformation, equation (4.3.5) reduces to the canonical form uξη =D1uξ+E1uη+F1u+G1(ξ, η), (4.3.9) whereD1,E1, andF1 are constants.

The canonical form of the Euler equation (4.3.6) is

uξη = 0. (4.3.10)

Integrating this equation gives the general solution

u=φ(ξ) +ψ(η) =φ(y−λ1, x) +ψ(y−λ2, x), (4.3.11) whereφandψare arbitrary functions, andλ1andλ2 are given by (4.3.3).

(B) Parabolic Type

WhenB2−4AC = 0, the equation is of parabolic type, in which case only one real family of characteristics exists. From equation (4.3.4), we find that

λ12= (B/2A), so that the single family of characteristics is given by

y = (B/2A)x+c1, wherec1 is an integration constant. Thus, we have

ξ=y−(B/2A)x, η=hy+kx, (4.3.12) whereη is chosen arbitrarily such that the Jacobian of the transformation is not zero, andhandkare constants.

With the proper choice of the constantshandkin the transformation (4.3.12), equation (4.3.5) reduces to

uηη=D2uξ+E2uη+F2u+G2(ξ, η), (4.3.13) whereD2,E2, andF2are constants.

IfB= 0, we can see at once from the relation B2−4AC = 0,

thatC or Avanishes. The given equation is then already in the canonical form. Similarly, in the other cases whenA orC vanishes,B vanishes. The given equation is is then also in canonical form.

The canonical form of the Euler equation (4.3.6) is

uηη= 0. (4.3.14)

Integrating twice gives the general solution

u=φ(ξ) +η ψ(ξ), (4.3.15) whereξ andη are given by (4.3.12). Choosingh= 1,k= 0 andλ=B

2A

for simplicity, the general solution of the Euler equation in the parabolic case is

u=φ(y−λx) +y ψ(y−λx). (4.3.16) (C) Elliptic Type

WhenB2−4AC <0, the equation is of elliptic type. In this case, the characteristics are complex conjugates.

The characteristic equations yield

y=λ1x+c1, y=λ2x+c2, (4.3.17) whereλ1and λ2 are complex numbers. Accordingly,c1andc2 are allowed to take on complex values. Thus,

ξ=y−(a+ib)x, η=y−(a−ib)x, (4.3.18) whereλ1,2=a+ibin whichaandbare real constants, and

a= B

2A, and b= 1 2A

4AC−B2. Introduce the new variables

α= 1

2(ξ+η) =y−ax, β= 1

2i(ξ−η) =−bx. (4.3.19) Application of this transformation readily reduces equation (4.3.5) to the canonical form

uαα+uββ =D3uα+E3uβ+F3u+G3(α, β), (4.3.20) whereD3,E3,F3are constants.

We note thatB2−AC <0, so neitherAnorC is zero.

In this elliptic case, the Euler equation (4.3.6) gives the complex char- acteristics (4.3.18) which are

ξ= (y−ax)−ibx, η= (y−ax) +ibx=ξ. (4.3.21) Consequently, the Euler equation becomes

uξξ = 0, (4.3.22)

with the general solution

u=φ(ξ) +ψ

ξ . (4.3.23)

The appearance of complex arguments in the general solution (4.3.23) is a general feature of elliptic equations.

4.3 Equations with Constant Coefficients 103 Example 4.3.1.Consider the equation

4uxx+ 5uxy+uyy+ux+uy= 2.

Since A = 4, B = 5, C = 1, and B2−4AC = 9 > 0, the equation is hyperbolic. Thus, the characteristic equations take the form

dy

dx = 1, dy dx =1

4, and hence, the characteristics are

y=x+c1, y= (x/4) +c2. The linear transformation

ξ=y−x, η=y−(x/4), therefore reduces the given equation to the canonical form

uξη= 1 3uη−8

9. This is the first canonical form.

The second canonical form may be obtained by the transformation α=ξ+η, β =ξ−η,

in the form

uαα−uββ = 1 3uα−1

3uβ−8 9. Example 4.3.2.The equation

uxx−4uxy+ 4uyy =ey,

is parabolic sinceA = 1, B = −4, C = 4, and B2−4AC = 0. Thus, we have from equation (4.3.12)

ξ=y+ 2x, η=y,

in which η is chosen arbitrarily. By means of this mapping, the equation transforms into

uηη= 1 4eη. Example 4.3.3.Consider the equation

uxx+uxy+uyy+ux= 0.

Since A = 1,B = 1, C = 1, and B2−4AC = −3 <0, the equation is elliptic.

We have

λ1,2= B+√

B2−4AC

2A =1

2+i

√3 2 , and hence,

ξ=y− 1

2 +i

√3 2

x, η=y− 1

2−i

√3 2

x.

Introducing the new variables α= 1

2(ξ+η) =y−1

2x, β= 1

2i(ξ−η) =−

√3 2 x, the given equation is then transformed into canonical form

uαα+uββ = 2

3uα+ 2

√3uβ. Example 4.3.4.Consider the wave equation

utt−c2uxx= 0, c is constant.

Since A = −c2, B = 0, C = 1, and B2 −4AC = 4c2 > 0, the wave equation is hyperbolic everywhere. According to (4.2.4), the equation of characteristics is

−c2 dt

dx 2

+ 1 = 0, or

dx2−c2dt2= 0.

Therefore,

x+ct=ξ= constant, x−ct=η= constant.

Thus, the characteristics are straight lines, which are shown in Figure 4.3.1.

The characteristics form a natural set of coordinates for the hyperbolic equation.

In terms of new coordinates ξandη defined above, we obtain uxx=uξξ+ 2uξη+uηη,

utt=c2(uξξ−2uξη+uηη), so that the wave equation becomes

4.3 Equations with Constant Coefficients 105

Figure 4.3.1Characteristics for the wave equation.

−4c2uξη= 0.

Sincec= 0, we have

uξη = 0.

Integrating with respect toξ, we obtain uη1(η).

where ψ1 is the arbitrary function ofη. Integrating with respect to η, we obtain

u(ξ, η) =

ψ1(η)dη+φ(ξ). If we setψ(η) =*

ψ1(η)dη, the general solution becomes u(ξ, η) =φ(ξ) +ψ(η),

which is, in terms of the original variablesxandt, u(x, t) =φ(x+ct) +ψ(x−ct),

providedφandψare arbitrary but twice differentiable functions.

Note thatφis constant on “wavefronts”x=−ct+ξthat travel toward decreasingxastincreases, whereasψis constant on wavefrontsx=ct+η that travel toward increasingx as t increases. Thus, any general solution can be expressed as the sum of two waves, one traveling to the right with constant velocitycand the other traveling to the left with the same velocity c.

Example 4.3.5.Find the characteristic equations and characteristics, and then reduce the equations

uxx+

sech4x uyy = 0, (4.3.24ab) to the canonical forms.

In equation (4.3.24a),A= 1,B= 0 and C=−sech4x. Hence, B2−4AC = 4 sech4x >0.

Hence, the equation is hyperbolic. The characteristic equations are dy

dx = B+√

B2−4AC

2A = + sech2x.

Integration gives

y+ tanhx= constant.

Hence,

ξ=y+ tanhx, η=y−tanhx.

Using these characteristic coordinates, the given equation can be trans- formed into the canonical form

uξη= (η−ξ)

"

4−(ξ−η)2#(uξ−uη). (4.3.25) In equation (4.3.24b),A= 1,B = 0 andC= sech4x. Hence,

B2−4AC= +isech2x.

Integrating gives

y+itanhx= constant.

Thus,

ξ=y+itanhx, η=y−itanhx.

The new real variablesαandβ are