Section 3.5 ADJOINTS OF DIFFERENTIAL OPERATORS IN TWO DIMENSIONS

In the chapters that came before, an understanding of the adjoints of linear functions was critical in determining when certain linear equations would have a solution....and even in computing the solutions for some cases. It is, then, no surprise that we shall be interested in the computation of adjoints in this setting, too.

In the general inner product space, the adjoint of the linear operator L is defined in terms of the dot product:

< L(u) , v > = < u , L*(v) >.

For ordinary differential equations boundary value problems, the dot product
came with the problem in a sense: it was an integral over an appropriate
interval on which the functions were defined. For partial differential
equations with boundary conditions, the dot product will be over the
*region* of interest:

< f , g > = òòD f(x,y) g(x,y) dx dy.

For ordinary differential equations, integration-by-parts played a key role in
deciding the appropriate boundary conditions to impose so that the formal
adjoint would be the __real__ adjoint. Now, Green's identities provide the
appropriate calculus.

In fact, Green's second identity can be used to compute adjoints of the Laplacian. We will see that the divergence theorem is useful for the more general second-order, differential operators.

EXAMPLE: Consider the problem

--^{2}u = f,

u(x,0) = u(x,b) = 0

[[partialdiff]]u/[[partialdiff]]x(0,y) = [[partialdiff]]u/[[partialdiff]]x(a,y) = 0

on an interval [0,a] x [0,b] in the plane. This problem invites consideration of the operator L defined on a manifold as given below:

L(u) = --^{2}u

and M = {u: u(x,0) = u(x,b) = 0, [[partialdiff]]u/[[partialdiff]]y(0,y) = [[partialdiff]]u/[[partialdiff]]y(a,y) = 0}.

The second identity presents a natural setting in which the operator L is self-adjoint in the sense that L = L*. Let u and v be in M:

òòD [v L(u) - L(v) u] dA = ò[[partialdiff]]D [v [[partialdiff]]u/[[partialdiff]][[eta]] - [[partialdiff]]v/[[partialdiff]][[eta]] u] ds = 0.

^{ }

^{}This last equality follows because, on the boundary, all of u,
[[partialdiff]]u/[[partialdiff]][[eta]], v, and
[[partialdiff]]v/[[partialdiff]][[eta]] = 0.
[[florin]]

In order to discuss adjoints of more general second order partial
differential equations, let A, B, C, and c be scalar valued functions. Let
**b** be a vector valued function. Let L(u) be given by

L(u) = A[[partialdiff]]^{2}u/[[partialdiff]]x^{2} +
2B[[partialdiff]]^{2}u/[[partialdiff]]x[[partialdiff]]y +
C[[partialdiff]]^{2}u/[[partialdiff]]y^{2} + < **b** ,
--u > + cu.

DEFINITION: The __FORMAL__ ADJOINT is given by

L*(v) = [[partialdiff]]^{2}(Av)/[[partialdiff]]x^{2} +
2[[partialdiff]]^{2}(Bv)/[[partialdiff]]x[[partialdiff]]y +
[[partialdiff]]^{2}(Cv)/[[partialdiff]]y^{2} - --*(**b**v) +
cv.

Take A, B, and C to be constant. What would it mean to say that L is formally
self-adjoint? That L = L* (formally)? Then <**b**, --u> must be -
--***b**u = < -**b** , --u > - u --***b**. Thus, 2 <
**b** , --u > = -u (--***b**) for all u. Since this must hold for all
u, it must hold in the special case that u [[equivalence]] 1, which implies
that --***b** = 0. Taking u(x,y) to be x, or to be y gets that each of
**b**1 and **b**2 = 0. Hence, if L is formally self adjoint, then
**b** = **0**.

EXAMPLES:

1. Let L[u] = 3
F([[partialdiff]]^{2}u,[[partialdiff]]x^{2})^{ + 5 }
F([[partialdiff]]^{2}u,[[partialdiff]]y^{2})^{ . }The
formal adjoint of L is L. Note that

L[u] v - u L[v] = F([[partialdiff]] ,[[partialdiff]]x) ( 3[F([[partialdiff]]u,[[partialdiff]]x) v - u F([[partialdiff]]v,[[partialdiff]]x) ] ) + F([[partialdiff]] ,[[partialdiff]]y) ( 5[F([[partialdiff]]u,[[partialdiff]]y) v - u F([[partialdiff]]v,[[partialdiff]]x) ] )

= --.( 3[F([[partialdiff]]u,[[partialdiff]]x) v - u F([[partialdiff]]v,[[partialdiff]]x) ] , 5[F([[partialdiff]]u,[[partialdiff]]y) v - u F([[partialdiff]]v,[[partialdiff]]x) ] )

2. Let L[u] = 3
F([[partialdiff]]^{2}u,[[partialdiff]]x^{2})^{ + 5 }
F([[partialdiff]]^{2}u,[[partialdiff]]y^{2}) ^{ +} 7
F([[partialdiff]]u,[[partialdiff]]x) + 11 F([[partialdiff]]u,[[partialdiff]]y)
+ 13 u. The formal adjoint of L is

L*[v] = 3 F([[partialdiff]]^{2}v,[[partialdiff]]x^{2) + 5 }
F([[partialdiff]]^{2}v,[[partialdiff]]y^{2) -} 7
F([[partialdiff]]v,[[partialdiff]]x) - 11 F([[partialdiff]]v,[[partialdiff]]y)
+ 13 v. Note that

L[u] v - u L*[v] = F([[partialdiff]] ,[[partialdiff]]x) ( 3[F([[partialdiff]]u,[[partialdiff]]x) v - u F([[partialdiff]]v,[[partialdiff]]x) ] + 7 uv) + F([[partialdiff]] ,[[partialdiff]]y) ( 5[F([[partialdiff]]u,[[partialdiff]]y) v - u F([[partialdiff]]v,[[partialdiff]]x) ] + 11 uv)

= --.( 3[F([[partialdiff]]u,[[partialdiff]]x) v - u F([[partialdiff]]v,[[partialdiff]]x) ] +7 uv, 5[F([[partialdiff]]u,[[partialdiff]]y) v - u F([[partialdiff]]v,[[partialdiff]]x) ] + 11 uv).

3. Let L[u] = e^{x}
F([[partialdiff]]^{2}u,[[partialdiff]]x^{2}) + 5
F([[partialdiff]]u,[[partialdiff]]y) + 3u. The formal adjoint of L is L* given
by

L*[v] = e^{x} F([[partialdiff]]^{2}v,[[partialdiff]]x^{2)
}+ 2e^{x} F([[partialdiff]]v,[[partialdiff]]x)
-5F([[partialdiff]]v,[[partialdiff]]y) + (e^{x}+3) u. Note that

L[u] v - u L*[v] = F([[partialdiff]] ,[[partialdiff]]x) (e^{x} v
F([[partialdiff]]u,[[partialdiff]]x) - u F([[partialdiff]]
e^{x}v,[[partialdiff]]x) ) + F([[partialdiff]] ,[[partialdiff]]y)
(5uv)

= --.( e^{x} v F([[partialdiff]]u,[[partialdiff]]x) - u
F([[partialdiff]] e^{x}v,[[partialdiff]]x) , 5uv).

THE CONSTRUCTION OF M*

We now come to the important part of the construction of the __real__
adjoint: how to construct the appropriate adjoint boundary conditions. We are
given L and M; we have discussed how to construct L*. We now construct M*.
To see what is M* in the general case, the divergence theorem is recalled:

òòD --*F dx dy = ò[[partialdiff]]D < F, [[eta]] > ds.

The hope, then, is to write v L(u) - L*(v) u as --*F for some suitable chosen F.

**Theorem**. If L is a second order differential operator and L* is the
formal adjoint, then there is F such that vL(u) - L*(v)u = --.F.

Here's how to see that. Note that

(v A[[partialdiff]]^{2}u/[[partialdiff]]x^{2} + v
C[[partialdiff]]^{2}u/[[partialdiff]]y^{2}) - (u
[[partialdiff]]^{2}(Av)/[[partialdiff]]x^{2} + u
[[partialdiff]]^{2}(Cv)/[[partialdiff]]y^{2})

= [[partialdiff]]/[[partialdiff]]x(vA[[partialdiff]]u/[[partialdiff]]x - u[[partialdiff]](Av)/[[partialdiff]]x) + [[partialdiff]]/[[partialdiff]]y (vC[[partialdiff]]u/[[partialdiff]]y - u[[partialdiff]](Cv)/[[partialdiff]]y)

= --*{ v(A[[partialdiff]]u/[[partialdiff]]x, C[[partialdiff]]u/[[partialdiff]]y) - u([[partialdiff]]Av/[[partialdiff]]x, [[partialdiff]]Cv/[[partialdiff]]y)}.

Also, v < **b** , --u > + u --*(**b**v)

= v **b**1 [[partialdiff]]u/[[partialdiff]]x + **b**2
[[partialdiff]]u/[[partialdiff]]y v + u
[[partialdiff]](**b**1v)/[[partialdiff]]x +
u [[partialdiff]](**b**2v)/[[partialdiff]]y

= --*(v**b**u).

**COROLLARY**: òòD [vL(u) - uL*(v)]

= òòD
--*{v(A[[partialdiff]]u/[[partialdiff]]x,C[[partialdiff]]u/[[partialdiff]]y) -
u([[partialdiff]](Av)/[[partialdiff]]x,[[partialdiff]](Cv)/[[partialdiff]]y) +
v**b**u)

=ò[[partialdiff]]D
[v{A[[partialdiff]]u/[[partialdiff]]x,C[[partialdiff]]u/[[partialdiff]]y}-u{[[patialdiff]](Av)[[partialdiff]]x,[[partialdiff]](Cv)/[[partialdiff]]y}
+v**b**u}]*[[eta]] ds.

**EXAMPLES**.

1(cont). Let L[u] be as in example 1 above for {x,y} in the rectangle D = [0,1]x[0,1] and M = {u: u=0 on [[partialdiff]]D}. Then, according to Example 1, L = L* and

òòD [vL(u) - uL*(v)] dA=

ò[[partialdiff]]D <{3 [ F([[partialdiff]]u,[[partialdiff]]x) v - u F([[partialdiff]]v,[[partialdiff]]x) ], 5 [ F([[partialdiff]]u,[[partialdiff]]y) v - u F([[partialdiff]]v,[[partialdiff]]y)] } , [[eta]] > ds.

Recalling that the unit normal to the faces of the rectangle D will be {0.-1}, {1,0}, {0,1}, or {-1,0} and that u = 0 on [[partialdiff]]D, we have that

òòD [vL(u) - uL*(v)] dA=

= - I(0,1, ) 5F([[partialdiff]]u,[[partialdiff]]y) v dx + 3 I(0,1, )F([[partialdiff]]u,[[partialdiff]]x) v dy

+ 5 I(1,0, ) F([[partialdiff]]u,[[partialdiff]]y) v |dx| - 3 I(1,0, ) F([[partialdiff]]u,[[partialdiff]]y) v |dy|.

In order for this integral to be zero for all u in M, it must be that v = 0 on [[partialdiff]]D. And M = M*. Hence, {L, M} is (really) self adjoint.

2.(cont) Let L[u] be as in Example 2 above and M = {u: u(x,0) = u(0,y) = 0, and F([[partialdiff]]u,[[partialdiff]]x)(1,y) = F([[partialdiff]]u,[[partialdiff]]y)(x,1) = 0, 0 < x < 1, 0 < y < 1}. Using the results from above,

òòD [L(u)v - uL*(v)] dA=

-I(0,1, ) 5 F([[partialdiff]]u,[[partialdiff]]y) v dx + I(0,1, ) [-3 u F([[partialdiff]]v,[[partialdiff]]x) + 7uv] dy

+ I(1,0, ) [5u F([[partialdiff]]v,[[partialdiff]]x) + 11 uv] |dx| - I(1,0, )[3 F([[partialdiff]]u,[[partialdiff]]x) v ] |dy|.

It follows that M* = {v: v(x,0) = 0, v(1,y) = F(3,7) F([[partialdiff]]v,[[partialdiff]]x) (1,y), v(x,1) =

F(5,11) F([[partialdiff]]v,[[partialdiff]]x) (x,1), and v(0,y) = 0}.

3(cont). Let L[u] be as in Example 3 above for [x,y} in the first qaudrant. Let M = {u: u = 0 on [[partialdiff]]D}. Then

òòD [vL(u) - uL*(v)] dA=

=ò[[partialdiff]]D <{e^{x} v
F([[partialdiff]]u,[[partialdiff]]x) - u F([[partialdiff]] e^{x}v
,[[partialdiff]]x) , 5uv }, [[eta]] > ds = - I(0,*, )v
F([[partialdiff]]u,[[partialdiff]]x) dy.

Thus, M* = {v: v(0,y) = 0 for y > 0}.

**EXERCISES**

1. Suppose that L(u) =
[[partialdiff]]^{2}u/[[partialdiff]]x^{2} -
[[partialdiff]]^{2}u/[[partialdiff]]t^{2} restricted to M = {u:
u(0,t) =u(a,t) = 0 and u(x,0) = [[partialdiff]]u/[[partialdiff]]t(x,0) = 0}.
Classify L as parabolic, hyperbolic, or elliptic. Find L* . Find F such that
v L[u] - L*[v} u = --F. What is M*?

2. Suppose that L[u] =
[[partialdiff]]^{2}u/[[partialdiff]]x^{2} +
[[partialdiff]]^{2}u/[[partialdiff]]y^{2} -
[[partialdiff]]^{2}u/[[partialdiff]]z^{2} restricted to

M = {u: u(0,y,z) = 0, u(1,y,z) = 0, u(x,y,0) = u(x,y,1), [[partialdiff]]u/[[partialdiff]]z (x,y,0) = [[partialdiff]]u/[[partialdiff]]z(x,y,1),

[[partialdiff]]u/[[partialdiff]]y(x,0,z) = 3 u(x,0,z), [[partialdiff]]u/[[partialdiff]]y(x,1,z) = 5 u(x,1,z)}

Classify L as parabolic, hyperbolic, or elliptic. Give L*. Find F such that v L[u} - L*[v] u = --.F. What is M*?