5.1 Second-Order linear PDE

Consider a second-order linear PDE

L[u] = auxx + 2buxy + cuyy + dux + euy + fu = g, (x, y) ∈ U (5.1)

for an unknown function u of two variables x and y. The functions a, b andc are assumed to be of class C1 and satisfying a2 +b2 +c2 6= 0. The operator

L0[u] := auxx + 2buxy + cuyy

consisting of the second order terms of L is called the principal part of L.Many of the fundamental properties of the solutions of (5.1) are determinedby the sign of the discriminant of L. The discriminant ∆(L)(x, y) is definedby

∆(L)(x, y) = det

[b ac b

]= b2 − ac

where a, b and c are evaluated at the point (x, y).We are interested in how the PDE is transformed under changes of coor-

dinates. We consider a C1- map F (x, y) = (ξ(x, y), η(x, y)) whose Jacobiansatisfies

detJ(x, y) = det

[ξx ξyηx ηy

]6= 0

at each point of (x0, y0) ∈ U . The inverse function theorem implies that nearthe point (x0, y0) the map F has an inverse F−1(ξ, η) = (x(ξ, η), y(ξ, η)).The inverse is of class C1. Now, assuming that u is a solution of (5.1),define w(ξ, η) = u(x(ξ, η), y(ξ, η)). Then u(x, y) = w(ξ(x, y), η(x, y)) and,by the chain rule,

ux = wξξx + wηηx

uy = wξξy + wηηy

uxx = wξξ(ξx)2 + 2wξηξxηx + wηη(ηx)2 + wξξxx +wηηxx

uyy = wξξ(ξy)2 + 2wξηξyηy + wηη(ηy)

2 + wξξyy + wηηyy

uxy = uyx = wξξξxξy + wξη(ξxηy + ηxξy) + wηηξxηy + wξξxy + wηηxy

Substituting into (5.1), we find that

L̃[w] = Awξξ + 2Bwξη + Cwηη +Dwξ + Ewη + Fw = G (5.2)

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with the coefficients of the principal part L̃0[w] = Awξξ + 2Bwξη + Cwηη

given byA(ξ, η) = aξ2x + 2bξxξy + cξ2y

B(ξ, η) = aξxηx + b(ξxηy + ξyηx) + cξyηy

C(ξ, η) = aη2

x + 2bηxηy + cη2

y

(5.3)

Observe that[A BB C

]=

[ξx ξyηy ηy

]·[a bb c

]·[ξx ηx

ξy ηy

]t

,

where t denotes the transpose of the matrix. Recalling that the determinantof the product of matrices is equal to the product of the determinants ofmatrices and that the determinant of a transpose of a matrix is equal to thedeterminant of a matrix, we get

det

[A BB A

]= det

[a bb c

]· (J(x, y))2. (5.4)

This shows that the discriminant of L has the same sign as the discriminantof the transformed equation and so it is an invariant of the change of coor-dinates. Consequently, we can classify equations (5.1) according to the signof the discriminant.

Definition 5.1. The equation (5.1) is called

• hyperbolic at (x, y) if ∆(L)(x, y) > 0.

• parabolic at (x, y) if ∆(L)(x, y) = 0.

• elliptic at (x, y) if ∆(L)(x, y) < 0.

Example 5.2. (i) The wave equation utt −uxx = 0 is a hyperbolic equa-tion.

(ii) The heat equation ut − uxx = 0 is a parabolic equation.

(iii) The Laplace equation uxx = uyy = 0 is an elliptic equation.

Example 5.3. Consider the Tricomi equation

yuxx + uyy = 0. (5.5)

Here a = y, b = 0, c = 1 and d = e = f = g = 0. Its discriminant is equal to

det

[b ac b

]= det

[0 y1 0

]= −y.

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Hence the equation (5.5) is hyperbolic for y < 0, parabolic when y = 0, andelliptic for y > 0.

Next we shall show that we can find changes of coordinates in which the(5.1) takes a simple form.

• Hyperbolic equations.

Suppose that equation (5.1) is hyperbolic on the domain U . This meansthat b2 − ac > 0 at each point of U . We shall show that in this we canchoose (x, y) 7→ (ξ(x, y), η(x, y)) so that

A(ξ, η) = aξ2x + 2bξxξy + cξ2y = 0 (5.6)

C(ξ, η) = aη2

x + 2bηxηy + cη2

y = 0. (5.7)

Under such a change of coordinates and dividing by 2B the hyperbolic equa-tion (5.1) takes its canonical form

L̃[w] = wξη + ℓ[w] = G, (5.8)

where ℓ is a first-order linear operator and G is function.Note that if a and c are equal to 0, then the equation (5.1) is already in itscanonical form (just divide by 2b). Hence without loss of generality we mayassume that a 6= 0. Note also that the equation (5.7) for η is the same as(5.6) for ξ. Hence it suffices to consider only one of the equations, say (5.6).It can be written as a product

a

[ξx − −b−

√b2 − ac

aξy

]·[ξx − −b+

√b2 − ac

aξy

]= 0

and so, we need to solve the following two linear equations

ξx − µ1ξy = 0 (5.9)

andξx − µ2ξy = 0, (5.10)

where we have abbreviated

µ1 =−b−

√b2 − ac

aand µ2 =

−b−√b2 − ac

a.

Note that µ1 and µ2 are the real solutions of the equation

aµ2 + 2bµ+ c = 0. (5.11)

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In order to obtain a nonsingular map (x, y) 7→ (ξ(x, y), η(x, y)), wechoose ξ to be the solution of (5.9) and η to be the solution of (5.10).To solve (5.9), we use the method of characteristics (except that we don’tspecify the initial condition). The characteristic equations are

dx

dt= 1,

dy

dt= −µ1,

dz

dt= 0.

The last equation says that the solution ξ is constant along each of the char-acteristics (x(t), y(t)). In view of the first two equations, the characteristicscan be obtain as curves y = y(x) solving

dy

dx=

dydtdxdt

= −µ1, (5.12)

and then the solution ξ is constant at points (x, y(x)). Similarly, one solves(5.10) to obtain η.

In summary, to choose ξ and η one solves the (5.11) to obtain two realroots µ1 and µ2. Then, denoting by

f(x, y) = C1 and g(x, y) = C2

the solutions of characteristics equations

dy

dx= −µ1(x, y and

dy

dx= −µ2(x, y), (5.13)

the variables ξ and η are defined by

ξ(x, y) = f(x, y), η(x, y) = g(x, y).

The solutions of both equations in (5.13) are called the two families of char-

acteristics of (5.1).

Example 5.4. Consideryuxx + uyy = 0

In the region where y < 0, the equation is hyperbolic. Solving yµ2 + 1 = 0,one finds two real solutions

µ1 = − 1

(−y)1/2and µ2 =

1

(−y)1/2

We look for two real families of characteristics, dydx +µ1 = 0 and dy

dx +µ2 = 0,

dy

dx− 1

(−y)1/2= 0 and

dy

dx+

1

(−y)1/2= 0.

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The solutions of the equations are

2

3(−y)3/2 + x = C1 and − 2

3(−y)3/2 + x = C2.

Therefore, we set

ξ =2

3(−y)3/2 + x and η = −2

3(−y)3/2 + x.

The derivatives of ξ and η are,

ξx = 1, ξy = −(−y)1/2, ηx = 1, ηy = (−y)1/2.

With u(x, y) = v(ξ(x, y), η(x, y), one gets

ux = vξ + vη

uy = −(−y)1/2vξ + (−y)1/2vη

uxx = vξξ + 2vξη + vηη

uyy = −yvξξ + 2yvξη − yvηη +1

2(−y)−1/2[vξ − vη].

Substituting into the equation, one obtains

0 = yuxx + uyy

= 4yvξη +1

2(−y)−1/2[vξ − vη] = 4y

[vξη −

1

8(−y)−3/2(vξ − vη)

].

Since ξ − η = 4

3(−y)3/2, one concludes that

vξη −1

6(ξ − η)(vξ − vη) = 0.

• Parabolic equations.

Suppose that (5.1) is parabolic on the domain U . Hence b2 − ac = 0 ateach point of U . As before assume that a 6= 0 on U . We find a map(x, y) 7→ (ξx, y), η(x, y)) so that B(ξ, η) = A(ξ, η) = 0. It suffices to makeA = 0 since 0 = B2 − AC = B2 implies that B(ξ, η) = 0. Under sucha change of coordinates the parabolic equation (5.1) can be brought to itscanonical form

L̃[w] = wξξ + ℓ[w] = G(ξ, η)

where ℓ is a first-order linear operator and G is function.

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To do this we look for (x, y) 7→ ξ(x, y) so that

A(ξ, η) = aξ2x + 2bξxξy + cξ2y = 0.

Since b2 = ac, we have

aξ2x + 2bξxξy + cξ2y = a

(ξ2x + 2

b

aξxξy +

b2

a2ξ2y

)= a (ξx − µξy)

2 ,

where µ = − ba is the double root of

aµ2 + 2bµ+ c = 0.

So, we look for the solution ξ of the first-order linear equation,

ξx − µξy = 0.

The solution ξ is constant along each characteristic which is determined bythe equation

dy

dx= −µ =

b

a. (5.14)

For the map (x, y) 7→ η(x, y) we can take any map so that ξxξ − ξyξx 6= 0.In summary, in the parabolic case to choose ξ and η one solves the (5.11)

to obtain double root µ = −b/a. Then, denoting by

f(x, y) = C

the solution of characteristics equation

dy

dx= −µ (5.15)

the variable ξ is defined by

ξ(x, y) = f(x, y)

and the variable η is chosen so that ξxξ − ξyξx 6= 0.

Example 5.5. Reduce the following equation to its canonical form and then

find the general solution,

x2uxx − 2xyuxy + y2uyy + xux + yuy = 0

for x > 0.

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The discriminant is equal to

det

[−xy x2

y2 −xy

]= 0

so that the equation is parabolic. The quadratic equation x2µ2−2xyµ+y2 =0 has exactly one solution, µ = y

x . Next we look for characteristics. Theseare solutions of the equation

dy

dx= µ, i.e.,

dy

dx= −y

x.

The family of solutions is given by xy = C. Therefore, we define ξ(x, y) = xyand we take as the second independent variable η(x, y) = x. Then

ξx = y, ξy = x, ηx = 1, ηy = 0.

The Jacobian of the map (x, y) 7→ (ξ(x, y), η(x, y)) is nonzero. Let v(ξ, η) =u(x(ξ, η), y(ξ, η)), that is, u(x, y) = v(ξ(x, y), η(x, y)). Using the chain rule,

ux = yvξ + vη

uy = xvξ

uxx = y2vξξ + 2yvξη + vηη

uxy = xyvξξ + xvξη + vξ

uyy = x2vξξ.

Substituting into the equation, one gets

x2vηη + xvη = 0,

so that, using x = η,

vηη +1

ηvη = 0.

To solve this equation introduce the function w = vη. Then

wη = −1

ηw

which has the general solution w = 1

ηA(ξ). So,

vη =1

ηA(ξ)

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which after integration gives

v(ξ, η) = A(ξ) ln η +B(ξ).

Since ξ(x, y) = xy and η(x, y) = x and u(x, y) = v(ξ(x, y), η(x, y)), thegeneral solution of the equation has the form

u(x, y) = A(xy) ln x+B(xy)

where A,B are two arbitrary functions of class C2.

• Elliptic equations.

Suppose that (5.1) is elliptic on the domain U . Then b2 − ac < 0 at eachpoint of U . This time we look for the map (x, y) 7→ (xξ(x, y), η(x, y)) sothat

A(ξ, η) = aξ2x + 2bξxξy + cξ2y = C(ξ, η) = aη2

x + 2bηxηy + cη2

y

B(ξ, η) = aξxηx + b(ξxηy + ξyηx) + cξyηy = 0.

Under such a change of coordinates and dividing by A, the elliptic equa-tion (5.1) can be brought to its canonical form

L̃[w] = wξξ + wηη + ℓ[w] = G(ξ, η)

where ℓ is a first-order linear operator.The above system consists of two nonlinear first-order equations. Sub-

tracting C(ξ, η) form A(ξ, η) and multiplying B(ξ, η) by 2i leads to thefollowing system,

a(ξ2x − η2

x) + 2b(ξxξy − ηxηy) + c(ξ2y − η2

y) = 0

aξx(2iηx) + b(ξx(2iηy) + ξy(2iηx)) + cξy(2iηy) = 0.

Consequently, setting φ = ξ + iη, one finds that the above system is equiv-alent to

aφ2

x + 2bφxφy + cφ2

y = 0.

This can be written as a product,

a [φx − µ1φy] · [φx − µ2φy] = 0,

where we have abbreviated

µ1 =−b− i

√ac− b2

aand µ2 =

−b− i√ac− b2

a.

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These are complex roots of

aµ2 + 2bµ+ c = 0.

Note that µ1 and µ2 are conjugated, i.e., µ1 = µ2. As in the hyperbolic casewe solve the characteristics equations

dy

dx= −µ1 and

dy

dx= −µ2. (5.16)

This time the solutions are complex. If f(x, y) = C1 and g(x, y) = C2 arecomplex solutions of (5.16), then

φ(x, y) = f(x, y) and ψ(x, y) = g(x, y).

Then set

ξ =1

2(φ+ ψ) and η =

1

2i(φ− ψ).

Example 5.6. Consideryuxx + uyy = 0

In the region where y > 0, the equation is elliptic. Solving yµ2 + 1 = 0, onefinds two complex solutions

µ1 =i

y1/2and µ2 = − i

y1/2

We look for two complex families of characteristics, dydx+µ1 = 0 and dy

dx+µ2 =0,

dy

dx+

i

y1/2= 0 and

dy

dx− i

y1/2= 0.

The solutions of the equations are

2

3y3/2 + ix = C1 and

2

3y3/2 − ix = C2.

Therefore, we set

φ =2

3y3/2 + x and ψ =

2

3y3/2 − x,

and then

ξ =1

2(φ+ ψ) =

2

3y3/2 and η =

1

2i(φ− ψ) = x.

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The derivatives of ξ and η are,

ξx = 0, ξy = y1/2, ηx = 1, ηy = 0.

With u(x, y) = v(ξ(x, y), η(x, y), one gets

ux = vη

uy = y1/2vξ

uxx = vηη

uyy = yvξξ +1

2y−1/2vξ.

Substituting into the equation, one obtains

vξξ + vηη +1

2y3/2vξ = 0.

Finally, since ξ = 2

3y3/2, the equation becomes

vξξ + vηη +3

ξvξ = 0.

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