Laplace’s Equation1. Equilibrium PhenomenaConsider a general conservation statement for a region U in Rn containing a material whichis being transported through U by a flux field, F = Fx, t. Let u = ux, t denote the scalarconcentration field for the material (u equals the concentration at x, t). Note that u is ascalar valued function while Fx, t is a vector valued function whose value at each x, t is avector whose direction is the direction of the material flow at x, t and whose magnitude isproportional to the speed of the flow at x, t. In addition, suppose there is a scalar sourcedensity field denoted by sx, t. This value of this scalar at x, t indicates the rate at whichmaterial is being created or destroyed at x, t. If B denotes an arbitrary ball inside U, thenfor any time interval t1, t2 conservation of material requires that

∫B

ux, t2dx = ∫B

ux, t1dx − ∫t1

t2 ∫∂B

Fx, t ⋅ nxdSxdt + ∫t1

t2 ∫B

sx, tdxdt

Now

∫B

ux, t2dx − ∫B

ux, t1dx = ∫t1

t2 ∫B∂tux, tdxdt

and∫

t1

t2 ∫∂B

Fx, t ⋅ nxdSxdt = ∫t1

t2 ∫B

div Fx, t dxdt

hence

∫t1

t2 ∫B∂tux, t + div Fx, t − sx, tdxdt = 0 for all B ⊂ U, and all t1, t2. 1.1

Since the integrand here is assumed to be continuous, it follows that

∂tux, t + div Fx, t − sx, t = 0, for all x ∈ U, and all t. 1.2

Equation (1.1) is the integral form of the conservation statement, while (1.2) is thedifferential form of the same statement. This conservation statement describes a largenumber of physical processes. We consider now a few special cases,

a) Transport u = ux, t

Fx, t = ux, t V where V = constant,sx, t = 0.

In this case, the equation becomes ∂tux, t + V ⋅ gradux, t = 0

b) Steady Diffusion u = uxFx, t = −K∇ux where K = constant > 0,s = sx.

In this case, the equation becomes

− K div gradux = sx or − K ∇2ux = sx.

This is the equation that governs steady state diffusion of the contaminantthrough the region U. The equation is called Poisson’s equation if sx ≠ 0,

1

and Laplace’s equation when sx = 0. These are the equations we willstudy in this section.

Another situation which leads to Laplace’s equation involves a steady state vector fieldV = Vx having the property that div Vx = 0. When V denotes the velocity field for anincompressible fluid, the vanishing divergence expresses that V conserves mass. When Vdenotes the magnetic force field in a magnetostatic field, the vanishing divergence assertsthat there are no magnetic sources. In the case that V represents the vector field of electricforce, the equation is the statement that U contains no electric charges. In addition to theequation div Vx = 0, it may happen that V satisfies the equation, curl Vx = 0. Thiscondition asserts that the field V is conservative (energy conserving). Moreover, it is astandard result in vector calculus that curl Vx = 0 implies that V = −grad ux, for somescalar field, u = ux. Then the pair of equations,

div Vx = 0 and curl Vx = 0,

taken together, imply that

∇2ux = 0 and V = −grad ux.

We say that the conservative field V is ”derivable from the potential, u = ux”. To say that uis a potential is to say that it satisfies Laplace’s equation.

The unifying feature of all of these physical models that lead to Laplace’s equation is thefact that they are all in a state of equilibrium. Whatever forces are acting in each model,they have come to a state of equilibrium so that the state of the system remains constant intime. If the balance of the system is disturbed then it will have to go through anothertransient process until the forces once again all balance each other and the system is in anew equilibrium state.

2. Harmonic FunctionsA function u = ux is said to be harmonic in U ⊂ Rn if:

i) u ∈ C2U; i.e., u, together with all its derivatives of order ≤ 2, is continuous in U

ii) ∇2ux = 0 at each point in U

Note that in Cartesian coordinates,

div∇ux = ∂/∂x1, ... ,∂/∂xn ⋅

∂u/∂x1

⋮

∂u/∂xn

= ∂2u/∂x12 + ... + ∂2u/∂xn

2

= ∂∂ux = ∇2ux

It is clear from this that all linear functions are harmonic.A function depending on x only through the radial variable, r = x1

2 + ... + xn2 is said to

be a radial function. If u is a radial function then

2

∂u/∂xi = u′r∂r/∂xi and ∂r/∂xi =12 x1

2 + ... + xn2−

12 2xi = xi/r

∂2u/∂xi2 = u” r∂r/∂xi2 + u′r

r − xi xi/rr2 = u” r∂r/∂xi2 + u′r 1

r − xi2

r3

and

∇2ux = ∑i=1n ∂2u/∂xi

2 = u” r∑i=1n xi/r2 + u′r∑i=1

n 1r − xi

2

r3

= u” r + u′r nr − 1

r = u” r + n − 1r u′r

We see from this computation that the radial function u = unr is harmonic for various n if:

n = 1 u1”r = 0; i.e., u1r = Ar + B

n = 2 : u2”r + 1r u′r = 1

rddr ru2

′ r = 0; i.e., u2r = C ln r

n > 2 : un” r + n−1

r un′ r = r1−n d

dr rn−1 un

′ r = 0; unr = Cr2−n

Note also that since ∇2∂u/∂xi = ∂/∂xi∇2u, for any i, it follows that every derivative of aharmonic function is itself, harmonic. Of course this presupposes that the derivative existsbut it will be shown that every harmonic function is automatically infinitely differentiable soevery derivative exists and is therefore harmonic.

It is interesting to note that if u and u2 are both harmonic, then u must be constant. Tosee this, write

∇2u2 = divgradu2 = div2u∇u = 2∇u ⋅ ∇u + 2u∇2u = 2|∇u|2

Then ∇2u2 = 0 implies |∇u|2 = 0 which is to say, u is constant. Evidently, then, theproduct of harmonic functions need not be harmonic.

It is easy to see that any linear combination of harmonic functions is harmonic so theharmonic functions form a linear space. It is also easy to see that if u = ux is harmonic onRn then for any z ∈ Rn, the translate, vx = ux − z is harmonic as is the scaled function,w = wλx for all scalars λ. Finally, ∇2 is invariant under orthogonal transformations. To seethis suppose coordinates x and y are related by

Qx = y =

Q11x1 + ... + Q1nxn

⋮

Qn1x1 + ... + Qnnxn

Then∇x = ∂/∂x1, ... ,∂/∂xn

and∂xi = ∂y1/∂xi∂y1 + ... + ∂yn/∂xi∂yn = Q1i ∂y1 + ... + Qni ∂yn

= (i-th row of Q)⋅∇y

i.e., ∇x = Q∇y and ∇x = Q∇y

= ∇yQ

3

Then ∇x2 = ∇x

⋅ ∇x = ∇yQQ∇y = ∇y

∇y, for QQ = I. A transformation Q with thisproperty, QQ = I, is said to be an orthogonal transformation. Such transformationsinclude rotations and reflections.

Problem 6 Suppose u and v are both harmonic on R3. Show that, in general, the productof u times v is not harmonic. Give one or more examples of a special case where theproduct does turn out to be harmonic.

3. Integral IdentitiesLet U denote a bounded, open, connected set in Rn having a smooth boundary, ∂U. This issufficient in order for the divergence theorem to be valid on U. That is, if Fx denotes asmooth vector field over U, (i.e., F ∈ CŪ ∩ C1U and if nx denotes the outward unitnormal to ∂U at x ∈ ∂U, then the divergence theorem asserts that

∫U

div F dx = ∫∂U

F ⋅ n dSx 3.1

Consider the integral identity (3.1) in the special case that Fx = ∇ux foru ∈ C1Ū ∩ C2U. Then

div Fx = div∇ux = ∇2ux

and F ⋅ n = ∇u ⋅ n = ∂N ux the normal derivative of u)

Then (3.1) becomes

∫U

∇2uxdx = ∫∂U

∂N uxdSx 3.2

The identity (3.2) is known as Green’s first identity. If functions u and v both belong toC1Ū ∩ C2U and if Fx = vx∇ux, then

div Fx = div vx∇ux = vx∇2ux + ∇u ⋅ ∇v

and F ⋅ n = vx∇u ⋅ n = v∂N ux

and, with this choice for F, (3.1) becomes Green’s second identity,

∫Uvx∇2ux + ∇u ⋅ ∇v dx = ∫

∂Uvx∂N uxdSx 3.3

Finally, writing (3.3) with u and v reversed, and subtracting the result from (3.3), weobtain Green’s symmetric identity,

∫Uvx∇2ux − ux∇2vx dx = ∫

∂Uvx∂N ux − ux∂N vx dSx 3.4

Problem 7 Let u = ux,y,z be a smooth function on R3 and let A denote a 3 by 3 matrixwhose entries are all smooth functions on R3 Let F = A∇u. If U denotes a bounded openset in R3 having smooth boundary ∂U, then find a surface integral over the boundary whosevalue equals the integral of the divergence of F over U. If v = vx,y,z is also a smoothfunction on R3 then write the integral of v div F over U as the sum of 2 integrals, one of which

4

is a surface integral over ∂U.

4. The Mean Value Theorem for Harmonic FunctionsWe begin by introducing some notation:

Bra = x ∈ Rn : |x − a| < r the open ball of radius r with center at x=aBra = x ∈ Rn : |x − a| ≤ r the closed ball of radius r with center at x=a

Sra = x ∈ Rn : |x − a| = r the surface of the ball of radius r with center at x=a

Let An denote the n-dimensional volume of B10. Then A2 = π,A3 = 4π/3, and, ingeneral An = πn/2/Γn/2 + 1. Then the volume of the n-ball of radius r is rnAn. Also let Sn

denote the area of the (n-1)-dimensional surface of B10 in Rn, (i.e, Sn is the area of∂B10).Then Sn = nAn and the area of ∂Br0 is equal to nAnrn−1. In particular,S2r = 2πr, S3r2 = 4πr2, etc.

We will also find it convenient to introduce the notation

∫Bra

fxdx = 1Anrn ∫

Brafxdx = average value of fx over Bra

and

∫∂Bra

fxdŜx = 1Snrn−1

∫∂Bra

fxdSx = average value of fx over ∂Bra

Recall that it follows from Green’s first identity that if ux is harmonic in U, then for any ball,Bra contained in U, we have

∫∂Bra

∂N uxdSx = ∫Bra

∇2uxdx = 0.

This simple observation is the key to the proof of the following theorem.

Theorem 4.1 (Mean Value Theorem for Harmonic Functions)Suppose u ∈ C2U and ∇2ux = 0 for every x in the bounded, open set U in Rn. Then forevery Brx ⊂ U,

ux = ∫∂Brx

uydŜy = ∫Brx

uydŷ 4.1

i.e., 4.1 asserts that for every x in U, and r > 0, sufficiently small that Brx is contained inU, ux is equal to the average value of u over the surface, ∂Brx, and ux is also equal tothe average value of u over the entire ball, Brx. A function with the property asserted by(4.1) is said to have the mean value property.

Proof- Fix a point x in U and an r > 0 such that Brx is contained in the open set U. Let

gr = ∫∂Brx

uydŜy = ∫∂B10

ux + rzdŜz.

Here we used the change of variable, y = x + rz, or z = y − x/r so as y ranges over ,∂Brx, z ranges over ∂B10. Then

g′r = ∫∂B10

∇ux + rz ⋅ z dŜz. = ∫∂Brx

∇uy ⋅ y − xr dŜy

It is evident that as y ranges over ∂Brx, |y − x| = r, hence y − x/r is just the outwardunit normal to the surface ∂Brx which means that

5

∇uy ⋅ y − xr = ∂N uy.

Theng′r = ∫

∂Brx∂N uydŜy = ∫

Brx∇2uydŷ = 0 (since u is harmonic in U)

Now g′r = 0 implies that gr =constant which leads to,

gr = lim t→0 gt = lim t→0 ∫∂B10ux + tzdŜz = ux;

i.e., ux = ∫∂Brx

uydŜy for all r > 0 such that Brx ⊂ U.

Notice that this result also implies,

∫Brx

uydy = ∫0

r ∫∂Btx

uydSydt = ∫0

ruxSntn−1 dt = uxAnrn

or,

ux = 1Anrn ∫

Brxuydy = ∫

Brxuydŷ

which completes the proof of the theorem.■

The converse of theorem 4.1 is also true.

Theorem 4.2 Suppose U is a bounded open, connected set in Rn and u ∈ C2U has themean value property; i.e., for every x in U and for each r > 0 such that Brx ⊂ U,

ux = ∫∂Brx

uydŜy.

Then ∇2ux = 0 in U.

Proof- If it is not the case that ∇2ux = 0 throughout U, then there is some Brx ⊂ U suchthat ∇2ux is (say) positive on Brx. Then for gr as in the proof of theorem 4.1,

0 = g′r = ∫∂Brx

∂N uydŜy = rn ∫

Brx∇2 uydy > 0

This contradiction shows there can be no Brx ⊂ U on which ∇2ux > 0, and hence nopoint in U where ∇2ux is different from zero.■

For u = ux,y a smooth function of two variables, we have

∂xxux,y ux + h,y − 2ux,y + ux − h,y/h2

∂yyux,y ux,y + h − 2ux,y + ux,y − h/h2

hence h2∇2ux,y −4ux,y + ux + h,y + ux − h,y + ux,y + h + ux,y − h

Then the equation, ∇2ux,y = 0 in U, is approximated by the equation,

ux,y = ux + h,y + ux − h,y + ux,y + h + ux,y − h/4.

The expression on the right side of this equation is recognizable as an approximation for

∫∂Brx

uydŜy.

Thus, in the discrete setting, the connection between the property of being harmonic and

6

the mean value property is more immediate.

5. Maximum-minimum PrinciplesThe following theorem, known as the strong maximum-minimum principle, is animmediate consequence of the mean value property.

Theorem 5.1 strong maximum − minimum principle Suppose U is a bounded open,connected set in Rn and u is harmonic in U and continuous on, Ū, the closure of U. Let Mand m denote, respectively, the maximum and minimum values of u on ∂U. Then eitherux is constant on Ū (so then ux = m = M), or else for every x in U we havem < ux < M.

Proof Let M denote the maximum value of ux on Ū and suppose ux0 = M. If x0 is insideU then there exists an r > 0 such that Brx0 ⊂ U and ux ≤ ux0 for all x ∈ Brx0.Suppose there is some y0 in Brx0 such that uy0 < ux0. But this contradicts the meanvalue property since it implies

M = ux0 = ∫Brx0

uydŷ < M.

It follows that ux = ux0 for all x in Brx0. Similarly, for any other point y0 ∈ U, theassumption that uy0 < ux0 leads to a contradiction of the mean value property. Then ifx0 is an interior point of U we a force to conclude that ux is identically equal to M on Uand, by continuity, on the closure, Ū. On the other hand, if u is not constant on U, then x0

must lie on the boundary of U.■

Note that if u = ux,y satisfies the discrete Laplace equation,

ux,y = ux + h,y + ux − h,y + ux,y + h + ux,y − h/4,

on a square grid, then u can have neither a max nor a min at an interior point of the gridsince at such a point, the left side of the equation could not equal the right side. At aninterior maximum, the left side would be greater than all four of the values on the right side,preventing equality. A similar situation would apply at an interior minimum. Unless u isconstant on the grid, the only possible location for an extreme value is at a boundary pointof the grid.

There is a weaker version of theorem 5.1 that is based on simple calculus arguments.

Theorem 5.2 (Weak Maximum-minimum principle) Suppose U is a bounded open,connected set in Rn and u ∈ CŪ ∩ C2U. Let M and m denote, respectively, the maximumand minimum values of u on ∂U. Then

a − ∇2ux ≤ 0 in U implies ux ≤ M for all x ∈ Ū

b − ∇2ux ≥ 0 in U implies ux ≥ m for all x ∈ Ū

c − ∇2ux = 0 in U implies m ≤ ux ≤ M for all x ∈ Ū

Proof of (a): The argument we plan to use can not be applied directly to ux. Instead, letvx = ux + |x|2 for x ∈ U and note that

−∇2vx = −∇2ux − 2n < 0 for all x in U.

It follows that vx can have no interior maximum, since at such a point, x0 , we would have

7

∂v/∂xi = 0 and ∂2v/∂xi2 ≤ 0, for 1 ≤ i ≤ n, x = x0.

This is in contradiction to the previous inequality since it implies −∇2vx ≥ 0. This allowsus to conclude that vx has no interior max and vx must therefore assume its maximumvalue at a point on the boundary of U.

Now U is bounded so for some R sufficiently large, we have U ⊂ BR0 and thisimplies the following bound on maxx∈U vx,

maxx∈U

vx ≤ maxx∈∂U

vx ≤ M + |x|2 ≤ M + R2.

Finally, we have, ux ≤ vx ≤ M + R2 for all x in U and all > 0. Since this holds forall > 0, it follows that ux ≤ M for all x in Ū.Statement (b) can be proved by a similar argument, or, by applying (a) to -u. Then (c)follows from (a) and (b).■

In the special case, n = 1 , it is easy to see why theorem 5.2 holds. In that caseU = a,b and ∇2u = u”x and the figure illustrates (a), (b) and (c).

a ux ≤ M b ux ≥ m c m ≤ ux ≤ M

The following figure illustrates why it is necessary to have both of the hypotheses, u∈ CŪ, and u ∈ C2U.

u ∈ CŪ, u ∉ CŪ,

but u ∉ C2U but u ∈ C2U

If U is not bounded, then the max-min principle fails in general. For example, if U denotesthe unbounded wedge, x,y : y > |x| in R2 then ux,y = y2 − x2 is harmonic in U, equalszero on the boundary of U, but is not the zero function inside U. An extended version of themax-min principle, due to E Hopf, is frequently useful.

Theorem 5.3 Suppose U is a bounded open, connected set in Rn and u ∈ CŪ ∩ C2U.Suppose also that ∇2ux = 0 in U and that u is not constant. Finally, suppose U is suchthat for each point y on the boundary of U, there is a ball, contained in U with y lying onthe boundary of the ball. If uy = M, then ∂N uy > 0 and if uy = m, then ∂N uy < 0.

8

(i.e., at a point on the boundary of U where ux assumes an extreme value, the normalderivative does not vanish).

Problem 8 Let ux be harmonic on U and let vx = |∇ux|2. Show thatvx ≤ maxx∈∂U vx for x ∈ Ū.(Hint: compute ∇2v and show that it is non-negative on U)

6. Consequences of the Mean Value Theorem and M-mPrinciplesThroughout this section, U is assumed to be a bounded open, connected set in Rn. We listnow several consequences of the results of the previous two sections.

It is a standard result in elementary real analysis that if a sequence of continuous functionsum converges uniformly to a limit u on a compact set K, then u is also continuous.Moreover, for any open subset W in K, we have

limm→∞ ∫W

um dx = ∫W

udx.

Lemma 6.1 Suppose umx is a sequence of functions which are harmonic in U andwhich converge uniformly on Ū. Then u = limm→∞ um is harmonic in U.

Proof Since each um is harmonic in U, theorem 4.1 implies that for every ball, Brx ⊂ U ,we have

umx = ∫∂Brx

umydŜy = ∫Brx

umydŷ.

The uniform convergence of the sequence on U implies that

umx → ux, ∫∂Brx

umydŜy → ∫∂Brx

uydŜy, ∫Brx

umydŷ → ∫Brx

uydŷ

henceux = ∫

∂BrxuydŜy = ∫

Brxuydŷ.

But this says u has the mean value property and so, by theorem 4.2, u is harmonic.■

Lemma 6.2 Suppose u ∈ CŪ ∩ C2U satisfies the conditions

∇2ux = 0, in U, and ux = 0, on ∂U.

Then ux = 0 for all x in U.

Proof- The hypotheses, u ∈ CŪ ∩ C2U and ∇2ux = 0, in U, imply that m ≤ ux ≤ M,in Ū. Then ux = 0, on ∂U implies m = M = 0.■

Lemma 6.2 asserts that the so called Dirichlet boundary value problem

9

∇2ux = Fx, x ∈ U, and ux = gx, x ∈ ∂U,

has at most one solution in the class CŪ ∩ C2U. Solutions having this degree ofsmoothness are called classical solutions of the Dirichlet boundary value problem. Thepartial differential equation is satisfied at each point of U and the boundary condition issatisfied at each point of the boundary. Later we are going to consider solutions in a widersense.

Lemma 6.3 For any F ∈ CU and g ∈ C∂U, there exists at most one u ∈ CŪ ∩ C2Usatisfying

−∇2ux = F, in U, and ux = g, on ∂U.

Proof Suppose u1, u2 ∈ CŪ ∩ C2U both satisfy the conditions of the boundary valueproblem. Then w = u1 − u2 satisfies the hypotheses of lemma 6.2 and is therefore zero onthe closure of U. Then u1 = u2 on the closure of U.■

Lemma 6.4 Suppose u ∈ CŪ ∩ C2U satisfies

∇2ux = 0, in U, and ux = g, on ∂U,

where gx ≥ 0. If gx0 > 0 at some point x0 ∈ ∂U then ux > 0 at every x ∈ U.

Proof First, gx ≥ 0 implies that m = 0. Then gx0 > 0 at some point x0 ∈ ∂U impliesM > 0. It follows now from the strong M-m principle that 0 < ux < M at every x ∈ U.■

Note that lemma 6.4 asserts that if a harmonic function that is non-negative on theboundary of its domain is positive at some point of the boundary, then it must be positive atevery point inside the domain; i.e., a local stimulus applied to the ”skin” of the bodyproduces a global response felt everywhere inside the body. This could be referred to asthe organic behavior of harmonic functions. This mathematical behavior is related to thefact that Laplace’s equation models physical systems that are in a state of equilibrium. If theboundary state of a system in equilibrium is disturbed, even if the disturbance is very local,then the system must readjust itself at each point inside the boundary to achieve a newstate of equilibrium. This is the physical interpretation of ”organic behavior”.

Lemma 6.5 For F ∈ CŪ and g ∈ C∂U, suppose u ∈ CŪ ∩ C2U satisfies

−∇2ux = Fx, x ∈ U, and ux = gx, x ∈ ∂U.

Then maxx∈U |ux| ≤ Cg + MCF

where Cg = maxx∈∂U |gx|, CF = maxx∈U |Fx|, M = a constant depending on U.

Proof The estimate asserts that −Cg + MCF ≤ ux ≤ Cg + MCF for x ∈ Ū. First, let

vx = ux + |x|2 CF

2n

Then −∇2vx = −∇2ux − CF = Fx − CF ≤ 0 in U

and vx ≤ maxx∈∂U ux + |x|2 CF

2n for x ∈ Ū.

Since U is bounded, there exists some R > 0 such that |x|2 ≤ R2 for x ∈ U. Then

10

vx ≤ Cg + R2 CF

2n and ux ≤ vx ≤ Cg + MCF forx ∈ Ū.

Similarly, letwx = ux − |x|2 CF

2n

and show that ux ≥ wx ≥ −Cg + MCF for x ∈ Ū.■

If we define a mapping, S : CŪ × C∂U → CŪ ∩ C2U that associates the data pair,(F,g), for the boundary value problem of lemma 6.5 to the solution ux, then we wouldwrite u = SF,g. Evidently, lemma 6.5 asserts that the mapping S is continuous. To makethis statement precise, we must explain how to measure distance between data pairsF1,g1,F2,g2 in the data space CŪ × C∂U and between solutions u1,u2 in thesolution space CŪ. Although we know that the solutions belong to the spaceCŪ ∩ C2U, this is a subspace of the larger space, CŪ, so we are entitled to view thesolutions as belonging to this larger space. We are using the term ”space” to mean a linearspace of functions; that is, a set that is closed under the operation of forming linearcombinations.

Define the distance between u1,u2 in the solution space CŪ as follows

||u1 − u2||CŪ = maxx∈Ū |u1x − u2x|.

Similarly, define the distance from F1,g1 to F2,g2 in the data space CŪ × C∂U by

||F1,g1 − F2,g2||CŪ×C∂U = maxx∈Ū |F1x − F2x| + maxx∈∂U |g1x − g2x|..

Each of these ”distance functions” defines what is called a norm on the linear space whereit has been defined. In order to be called a norm, the functions have to satisfy the followingconditions,

i ||αu|| = |α| ||u|| for all scalars α and for all functions u

ii ||u + v|| ≤ ||u||+ ||v||, for all functions u,v

iii ||u|| ≥ 0, for all u and ||u|| = 0 if and only if u = 0.

One can check that the distance functions defined above both satisfy all three of theseconditions and they therefore qualify as norms on the spaces where they have beendefined. Now the estimate of lemma 6.5 asserts that if uj solves the boundary valueproblem with data Fj,gj, j = 1,2 then

maxx∈U |u1x − u2x| ≤ maxx∈∂U |g1x − g2x| + M maxx∈U |F1x − F2x|

i.e., ||u1 − u2||CŪ ≤ max1,M ||F1,g1 − F2,g2||CŪ×C∂U.

Evidently, if the data pairs are close in the data space, then the solutions arecorrespondingly close in the solution space. This is what is meant by continuousdependence of the solution on the data. Note that if we were to change the definition of thenorm in one or the other (or both) of the spaces, the solution might no longer dependcontinuously on the data.

Consider the solution for the following boundary value problem

∇2ux,y = 0 for 0 < x < π, y > 0,

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ux,0 = 0, ∂yux,0 = gx = 1n sin nx, 0 < x < π,

u0,y = uπ,y = 0, y > 0.

For any integer, n, the solution is given by ux,y = 1n2 sin nx sinhny.

Evidently, the distance between g and zero in the data space is

||gx − 0||CR = maxx∈R| 1n sin nx| ≤ 1

n ,

while the distance between ux,y and zero in the solution space is

||ux,y − 0||C 0<x<π, y>0, = max0<x<π, y>0,1n2 sin nx sinhny ≈ eny

n2

This means that the data can be made arbitrarily close to zero by choosing n large, whilethe solution can simultaneously be made as far from zero as we like by choosing y > 0,large. Then the solution to this problem does not depend continuously on the data sincearbitrarily small data errors could lead to arbitrarily large solution errors. This problem issaid to be ”not well posed”.

7. Uniqueness from Integral IdentitiesIntegral identities can be used to prove that various boundary value problems cannot havemore than one solution. For example, consider the following boundary value problem

∇2ux = Fx, x ∈ U, ∂N ux = gx, x ∈ ∂U.

This is known as the Neumann boundary value problem for Poisson’s equation. Green’sfirst identity leads to

∫U

Fxdx = ∫U∇2uxdx = ∫

∂U∂N uxdSx = ∫

∂UgxdSx.

Then a necessary condition for the existence of a solution to this problem is that the data,F,g satisfies

∫U

Fxdx = ∫∂U

gxdSx.

If this condition is satisfied, and if u1,u2 denote two solutions to the problem, thenw = u1 − u2 satisfies the problem with F = g = 0. Then we have

0 = ∫U

w∇2wdx = ∫∂U

w∂N wdSx − ∫U∇w ⋅ ∇wdx = − ∫

U|∇w |2 dx

But this implies that |∇w | = 0 which is to say, w is constant in U. Then the solutions to thisboundary value problem may differ by a constant, they are not unique. We should point outthat in order for the equation and the boundary condition to have meaning in the classicalsense, we must assume that the solutions to this problem belong to the class,C1Ū ∩ C2U.

On the other hand, consider the problem,

∇2ux = Fx, x ∈ U, ux = g1x, x ∈ ∂U1, ∂N ux = g2x, x ∈ ∂U2,

where ∂U is composed of two distinct pieces, ∂U1, and ∂U2. Now if u1,u2 denote twosolutions to the problem, and w = u1 − u2, then we have, as before

12

0 = ∫U

w∇2wdx = ∫∂U

w∂N wdSx + ∫U∇w ⋅ ∇wdx = ∫

∂U1w∂N wdSx + ∫

∂U2w∂N wdSx + ∫

U|∇w|2

In this case, w = 0 on ∂U1 and ∂N w = 0 on ∂U2, so we again reach the conclusion that w isconstant in U. Since w ∈ C1Ū ∩ C2U, it follows that if w = 0 on ∂U1, then w = 0 on Ū.Then the solution to this problem is unique.

Finally, consider the Neumann problem for the so called Helmholtz equation,

−∇2ux + cxux = Fx, x ∈ U, ux = gx, x ∈ ∂U,

where we suppose that cx ≥ C0 > 0 for x ∈ U. We can use integral identities to show thatthis problem has at most one smooth solution. As usual, we begin by supposing theproblem has two solutions and we let wx denote their difference. Then

−∇2wx + cxwx = 0, x ∈ U, wx = 0, x ∈ ∂U,and,

0 = ∫U

wx−∇2wx + cxwxdx = − ∫∂U

w∂N wdSx + ∫U∇w ⋅ ∇wdx + ∫

Ucxwx2dx

Since w = 0 on ∂U, it follows that

∫U|∇w|2 + cxwx2dx ≥ C0 ∫

Uwx2dx = 0,

and this implies that wx vanishes at every point of Ū. Notice that this proof of uniquenessdoesn’t work if we don’t know that the coefficient cx is non-negative. (How would the proofhave to be modified if we knew only that cx ≥ 0?.

Problem 9 Prove that the following problem has at most one smooth solution

−∇2ux = Fx, x ∈ U, and ux = gx, x ∈ ∂U.

Use first the Green’s identity approach and then use the result in lemma 6.5. Note that thisresult was already established by means of the M-m principle.

Problem 10 Prove that the following problem has at most one smooth solution

−∇2ux = Fx, in U, and ux + ∂N ux = gx, on ∂U.

Eigenvalues for the LaplacianThe eigenvalues for the Dirichlet problem for the Laplace operator are any scalars, λ, forwhich there exist nontrivial solutions to the Dirichlet boundary value problem,

−∇2ux = λux, x ∈ U, ux = 0, x ∈ ∂U.

Note that if ux = 0 then any choice of λ will satisfy the conditions of the problem.Therefore we allow only nontrivial solutions and we refer to these as eigenfunctions. If uxis an eigenfunction for this problem corresponding to an eigenvalue λ then

λ ∫U

ux2dx = − ∫U

ux∇2uxdx = − ∫∂U

u∂NudSx + ∫U|∇u|2dx.

13

Then λ satisfies

λ =∫

U|∇u|2dx.

∫U

ux2dx> 0.

Note that |∇u| ≠ 0 since this would lead to u = 0 which is not allowed if u is aneigenfunction. We have shown that all eigenvalues of the Dirichlet problem for the Laplaceoperator are strictly positive.

Problem 11 Show that the Neumann problem,

−∇2ux = λux, x ∈ U, ∂N ux = 0, x ∈ ∂U.

has a zero eigenvalue which has the corresponding eigenfunction, ux =constant.

Problem 12 Under what conditions on the function αx, does the boundary value problem,

−∇2ux = λux, x ∈ U, αxux + ∂N ux = 0, x ∈ ∂U.

have only positive eigenvalues?

Problem 13 Show that for each of the eigenvalue problems considered here, if ux is aneigenfunction corresponding to an eigenvalue, λ, then for any nonzero constant k,vx = kux, is also an eigenfunction corresponding to the eigenvalue, λ.

8. Fundamental Solutions for the LaplacianLet δx denote the ”function” with the property that for any continuous function, fx,

∫Rnδx fxdx = f0, or, equivalently, ∫

Rnδx − y fydy = fx

Of course this is a purely formal definition since there is no function δx which could havethis property. Later, we will see that δx can be given a rigorous, consistent meaning in thecontext of generalized functions. However, using the delta in this formal way, we can give aformal definition of a fundamental solution for the negative Laplacian as the solution of,

−∇x2Ex − y = δx − y, x,y ∈ Rn. 8.1

Formally, this definition implies

−∇x2 ∫

RnEx − yfydy = ∫

Rnδx − y fydy = fx

Then the solution of the equation

−∇2ux = fx, x ∈ Rn,is given by

ux = ∫Rn

Ex − yfydy. 8.2

Although these steps are only formal, they can be made rigorous. Note that since there areno side conditions imposed on Ex or on ux neither of these functions is unique. Forexample, any harmonic function could be added to either of them and the resulting functionwould still satisfy the same equation.

14

Since δx and ∇2 are both radially symmetric, it seems reasonable to assume that Ex isradially symmetric as well; i.e., Ex = Er, for r = x1

2 + ... + xn2 . Then a definition for Ex

which does not make use of δx can be stated as follows:

Enx is a fundamental solution for −∇2 on Rn if,

i Enr ∈ C2Rn\0

ii ∇2Enr = 0, for r > 0 8.3

iii lim →0 ∫∂B0

∂NEnxdSx = −1

The properties i) and ii) in the definition imply that

∇2Enr = En”r + n − 1r En

′ r = 0, for r > 0

i.e., En”r/En′ r = −n − 1/r

log En′ r = −n − 1 log r + C,

En′ r = Cr1−n,

Enr =C2 log r if n = 2

Cn r2−n if n > 2.

The constant Cn can be determined from part iii) of the definition. It is this part of thedefinition that causes −∇2Enx to behave like δx.For n = 2 we have

∫∂B0

∂NEnxdSx = ∫0

2π∂rC2 log r dθ = C2 ∫

0

2π 1 dθ = 2πC2.

Then lim →0 ∫∂B0

∂NE2xdSx = 2πC2. = −1

so C2 = −1//2π and E2r = − 12π log r.

When n = 3 we have

∫∂B0

∂NEnxdSx = ∫∂B0

∂rC3/ r 2 dω = −C3 ∫∂B012 2 dω = −4πC3.

Then lim →0 ∫∂B0

∂NE3xdSx = −4πC3. = −1

so C3 = 1//4π and E3r = 1/4πr.

We will now show that condition 8.3 iii) really does produce the δ behavior for −∇2En. Ofcourse we can’t try to show that −∇2En = δx since are not allowed to refer to δx.Instead, we will show equivalently that −∇2ux = fx, for u given by 8.2. Here, wesuppose that fx is continuous, together with all its derivatives of order less than or equal to2, and we suppose further that fx has compact support; i.e., for some positive K, fxvanishes for |x| > K. The notation for this class of functions is Cc

2Rn.

15

Theorem 8.1 Let Enr denote a fundamental solution for −∇2 on Rn. Then, for anyf ∈ Cc

2Rn,

ux = ∫Rn

Enx − yfydy,

satisfies

u ∈ C2Rn, − ∇2ux = fx for any x ∈ Rn.

Proof The smoothness of f implies the smoothness of u; i.e., for i = 1,2, ...,n

∂u/∂xi = lim h→0ux + hei − ux

h= lim h→0 ∫

RnEz

fx + hei − z − fx − zh

dz,

Nowfx + hei − z − fx − z

hconverges uniformly to ∂f/∂xi and it follows that for each i,

∂u/∂xi = ∫Rn

Ez∂xi fx − zdy,

Similarly, ∂2ux/∂xi∂xj exists for each i and j since the corresponding derivatives of f allexist.

To show the second assertion, write

−∇x2ux = ∫

Rn−Enz∇x

2fx − zdy = ∫Rn−Enz∇z

2fx − z dz.

Since Enz tends to infinity as |z| tends to zero, we treat this as an improper integral;

∫Rn

Enz∇z2fx − z dz = ∫

B0Enz∇z

2fx − z dz + ∫Rn\B0

Enz∇z2fx − z dz.

First, note that

∫B0

Enz∇z2fx − z dz ≤ maxB0|∇z

2fx − z| ∫B0

|Enz| dz.

But

∫B0

|Enz| dz. =1/2π ∫

0

∫0

2π|log r|rdrdθ = C2|log | if n = 2

Cn ∫0

∫ω

r2−nrn−1drdω = C2 if n > 2

hence

lim→0 ∫B0

Enz∇z2fx − z dz = 0.

Next,

∫Rn\B0

Enz∇z2fx − z dz = ∫

∂Rn\B0Enz∂Nfx − z dSz − ∫

Rn\B0∇Enz ⋅ ∇zfx − z dz,

and|∫

∂Rn\B0Enz∂Nfx − z dSz| ≤ maxz∈∂B0|∂Nfx − z|∫

−∂B0|Enz | dSz

16

≤ C1

1/2π ∫2π

0|log|dθ = C2|log| if n = 2

Cn ∫ 2−nn−1dω = C3 if n > 2

We used the fact that ∂Rn\B0 = −∂B0. Finally, since Enz is harmonic in Rn\B0,

∫Rn\B0

∇Enz ⋅ ∇zfx − z dz = ∫−∂B0

∂NEnz fx − z dSz − ∫Rn\B0

∇2Enz fx − z dz

= ∫−∂B0

∂NEnz fx − z dSz.

Now we can write

∫−∂B0

∂NEnz fx − z dSz = ∫−∂B0

∂NEnz fx − z − fx dSz + ∫−∂B0

∂NEnz fx dSz,

and note that because fx is continuous,

∫−∂B0

∂NEnz fx − z − fx dSz ≤ C maxz∈∂B0 |fx − z − fx| → 0 as → 0.

In addition,∫−∂B0

∂NEnzdSz = − ∫∂B0

∂NEnzdSz → 1 as → 0

because of 8.3iii, and then it follows that

−∇2ux = lim →0 ∫−∂B0∂NEnz fx − z dSz = fx ∀x ∈ Rn■

We remark again that since no side conditions have been imposed on ux, this solution isnot unique. Any harmonic function could be added to ux and the sum would also satisfy−∇2ux = fx.

9. Green’s Functions for the LaplacianThroughout this section, U is assumed to be a bounded open, connected set in Rn, whoseboundary ∂U is sufficiently smooth that the divergence theorem holds. Consider theDirichlet boundary value problem for Poisson’s equation,

−∇2ux = Fx, for x ∈ U, and ux = gx for x ∈ ∂U 9.1

We know that

ux = ∫Rn

Enx − yFydy,

satisfies the partial differential equation but this function, does not, in general, satisfy theDirichlet boundary condition. In order to find a function which satisfies both the equation andthe boundary condition, recall that for smooth functions ux and vx

∫Uvy∇y

2uy − uy∇y2vy dy = ∫

∂Uvy∂N uy − uy∂N vy dSy 9.2

For x in U fixed but arbitrary, let vy = Enx − y − φy in 9.2 where denotes a yet to bespecified function that is harmonic in U. Then since Enx − y is a fundamental solution and

17

φ is harmonic in U,

− ∫U

uy∇y2vy = ∫

Uuy−∇y

2Enx − y − 0 dy = ux

Since ux solves the Dirichlet problem, (9.2) becomes now,

ux = − ∫Uvy∇y

2uy dy + ∫∂Uvy∂N uy − uy∂N vy dSy

= ∫U

vyFydy − ∫∂U

gy ∂N vydSy + ∫∂U

vy∂N uydSy

If the values of ∂N uy were known on ∂U then this would be an expression for the solutionux in terms of the data in the problem. Since ∂N uy on the boundary is not given, weinstead choose the harmonic function φ in such a way as to make the integral containingthis term disappear. Let φ be the solution of the following Dirichlet problem,

∇y2φy = 0 for y ∈ U, φy = Enx − y, for y ∈ ∂U

where we recall that x denotes some fixed but arbitrary point in U. Thenvy = Enx − y − φy = 0 on the boundary and the previous expression for ux reduces to

ux = ∫U

Gx,y fydy − ∫∂U

∂NGx,ygydSy 9.3

where Gx,y = Enx − y − φy. Formally, Gx,y solves

−∇2Gx,y = −∇2Enx − y − 0 = δx − y for x,y ∈ U, 9.4

Gx − y = 0, for x ∈ U, y ∈ ∂U

and G(x,y) is known as the Green’s function for the Dirichlet problem for the Laplacian, or,alternatively, as the Green’s function of the first kind. Note that if there are two Green’sfunctions then their difference satisfies a completely homogeneous Dirichlet problem. Thiswould seem to imply uniqueness for the Green’s function except for the fact that theuniqueness proofs were for the class of functions C2U ∩ CU and it is not known thatGx,y is in this class. This point will be cleared up later.

It can be shown rigorously that Gx,y = Gy,x for all x,y ∈ U. However, a formaldemonstration based on (9.4) proceeds as follows. For x,z ∈ U, (be careful to note that xand y are points in Rn apply (9.2) with uy = Gy,z and vy = Gy,x,

∫Uuy∇y

2vy − vy∇y2uy dy = − ∫

UGy, zδy − x − Gy, xδy − zdz

∫∂Uuy∂N vy − vy∂N uy dSy = ∫

∂UGy, z∂Nvy − Gy, x∂NvydSy = 0

The last integral vanishes because Gy,z = Gy,z = 0, for y ∈ ∂U. Then (9.2) implies

0 = ∫UGy,zδy − x − Gy,xδy − zdz = Gx,z − Gz,x for all x,z ∈ U.

This proof will become rigorous when we have developed the generalized functionframework in which this argument has meaning.

Example 9.1 Let U = x1,x2 ∈ R2 : x2 > 0. The half space is the simplest example of aset having a boundary (i.e., the boundary of the half space is the x1 − axis, x2 = 0) and wewill be able to construct the Green’s function of the first kind for this simple set. Note that

18

the half space is not a bounded set (having a boundary is not the same as being bounded!).Since n = 2, we write

Ex − y = −1/2π log |x − y| = −1/2π log x1 − y12 + x2 − y22

For x = x1,x2 ∈ U, let x∗ = x1,−x2 and let

vy = −1/2π log |x∗ − y| = −1/2π log x1 − y12 + x2 + y22 .

Then v = vy is a harmonic function of y for y ∈ U. Moreover, v reduces to vy = Ex − yfor y ∈ ∂U; i.e., vy1,0 = Ex1,x2 − y1,0. Then

Gx, y = −1/2π log |x − y| − log |x∗ − y| = −1/2π log |x − y|/|x∗ − y|

= −1/2π log x1 − y12 + x2 − y22 / x1 − y12 + x2 + y22 . 9.5

Note that Gx, y = 0 for y ∈ ∂U; i.e.,Gx1,x2,y1,0 = 0. It is clear from the constructionthat for each fixed x = x1,x2 ∈ U, Gx, y is a harmonic function of y for y ∈ U.

Problem 14 Show that for the half-space U = x1,x2 ∈ R2 : x2 > 0,

∂NGx, y|y∈∂U = −1π

x2

x1 − y12 + x22

so that

ux1,x2 = 1π ∫

−∞

∞ x2

x1 − y12 + x22

gy1dy1

solves ∇2ux1,x2 = 0 in U, and ux1,0 = gx1, x1 ∈ R.

Example 9.2 Let U = r,θ : 0 < r < R, |θ| < π = DR0. Suppose u = ur,θ satisfies

−∇2ur,θ = 0, in U, and uR,θ = gθ on ∂U = r,θ : r = R, |θ| < π .

In an elementary course on PDE’s we would show that for all choices of the constants,an,bn,

ur,θ = 12 a0 +∑

n=1

∞rnan cosnθ + bn sinnθ

solves Laplace’s equation in the disc, U. Moreover, the boundary condition is satisfied if

uR,θ = 12 a0 +∑

n=1

∞Rnan cosnθ + bn sinnθ = gθ. 9.6

Then we would appeal to the theory of Fourier series which asserts that any continuous gcan be expressed as

gθ = 12 A0 +∑

n=1

∞An cosnθ + Bn sinnθ 9.7

whereAn = 1/π ∫

−π

πgscosnsds, Bn = 1/π ∫

−π

πgssinnsds.

19

Then, comparing (9.6) with (9.7), it follows that Rnan = An, Rnbn = Bn, and so

ur,θ = 12 A0 +∑

n=1

∞

r/RnAn cosnθ + Bn sinnθ

satisfies both the PDE and the boundary condition. By uniqueness, this must be thesolution of the boundary value problem. If we write

An cosnθ + Bn sinnθ = 1/π ∫−π

πgscosnsds cosnθ + 1/π ∫

−π

πgssinnsds sinnθ.

= 1/π ∫−π

πgscosns cosnθ + sinns sinnθ ds

= 1/π ∫−π

πgs cosnθ − s ds,

then ur,θ can be written as

ur,θ = 1/π ∫−π

π 1

2 +∑n=1

∞

r/Rn cosnθ − sgs ds,

= 12π ∫−π

π R2 − r2

R2 − 2Rrcosθ − s + r2 gsds

Here the series in n was summed by writing cosnθ − s in terms of exp±inθ − s andrecognizing that the series is a geometric series. Then

ur,θ = 12π ∫−π

π R2 − r2

R2 − 2Rrcosθ − s + r2 gsds = ∫−π

π∂NGr,θ,R,sgsds

where Gr,θ,R,s denotes the Green’s function for this problem. This representation isoften called the Poisson integral formula.

10. The Inverse Laplace OperatorWe are all familiar with problems of the form Ax = f where A denotes an n by n matrix andx, f denote vectors in the linear space Rn. In this situation, A can be viewed as a linearoperator from the linear space Rn into Rn. If the only solution of Ax = 0, is x = 0, thenAx = f has a unique solution x for every data vector f . This solution can be expressed asx = A−1f, where A−1 denotes the inverse of the matrix A. There are strong analogiesbetween the problem Ax = f on Rn and the problem (9.1).

Consider problem (9.1) in the special case g = 0; i.e.,

−∇2ux = fx, x ∈ U, ux = 0, x ∈ ∂U. 10.1

Recall that we showed that the only solution of (9.1) when g = f = 0, is u = 0, so thesolution to 10.1 is unique.In fact, the unique solution u = ux, can be expressed in terms of the Green’s function by

ux = ∫U

Gx, y fydy 10.2

20

If we define Kfx = ∫U

Gx, y fydy for any f ∈ CŪ,

then it is clear thatKC1f1 + C2f2 = C1Kf1 + C2Kf2 for all f1, f2 ∈ CU, C1,C2 ∈ R.

We say that K is a linear operator on the linear space CŪ. We recall that to say thatCŪ is a linear space is to say that for all f1, f2 ∈ CŪ and for all C1,C2 ∈ R., the linearcombination C1f1 + C2f2 is also in CŪ.

The problem (10.1) can be expressed in operator notation. Define an operator L by

Lux = −∇2ux for any u ∈ D = u ∈ C2Ū : ux = 0 for x ∈ ∂U .

Then for any u ∈ D it follows that Lux ∈ CŪ so L can be viewed as a function definedon D with values in CŪ. Since D is a subspace of CŪ we can even say that L is afunction from CŪ into CŪ but we should note that L is not defined on all of CŪ.It is also easy to check that L is a linear operator from D into CŪ, and 10.1 can beexpressed in terms of this linear operator as follows,

find u ∈ D such that Lu = f ∈ CŪ.

The uniqueness for 10.1, stated in the operator terminology, becomesLu = 0 if and only if u = 0.

Evidently, the operators K and L are related by,

a Kfx ∈ D for any f ∈ CŪ, and LKfx = fx

b for any u ∈ D, Lux ∈ CŪ, and KLux = ux.

These two statements together assert that K = L−1, K is the operator inverse to L.

If we use the notation ⟨x, z⟩ to denote the usual inner product between two vectors x, z,then

⟨Ax, z⟩ = ⟨x,Az⟩ for all x, z ∈ Rn.Here A denotes the matrix transpose of A. It is a fact from linear algebra that thedimension of the null space of the matrix A is equal to the dimension of the null space of thetranspose matrix, A. If the null space of A has positive dimension then the solution ofAx = f is not unique. What is more, if z denotes any vector in the null space of A then

f, z = ⟨Ax, z⟩ = ⟨x,Az⟩ = 0

and it is then evident that a necessary condition for the existence of a solution for Ax = f isthat f, z = 0 for all z in the null space of A. The matrix A is said to be symmetric if eitherof the following equivalent conditions applies, A = A or ⟨Ax, z⟩ = ⟨x,Az⟩ for all x, z. WhenA is symmetric, the null space of A not only has the same dimension as that of A, the twonull spaces are actually the same. In this case, Ax = f has no solution unless f, z = 0 for

all z in the null space of A. If this condition is satisfied, then any two solutions of Ax = f

21

differ by an element from the null space of A.We will now consider the analogue of these last results for the case of a boundary value

problem for Laplace’s equation. First, we have to have an inner product on the functionspace CŪ. The essential properties of the inner product are

i ⟨x, z⟩ = ⟨z, x⟩ for all x, z.ii ⟨Cx, z⟩ = C⟨x, z⟩ for all x, z, and all C ∈ R.iii ⟨x + y, z⟩ = ⟨x, z⟩ + ⟨y, z⟩ for all x, y, z.

iv ⟨x, x⟩ ≥ 0 for all x, and ⟨x, x⟩ = 0, if and only if x = 0.

and any mapping from Rn × Rn to R having these four properties is called an inner producton the linear space Rn.We can define an inner product on the function space CŪ, by letting

f1, f2 = ∫U

f1x f2xdx for all f1, f2 ∈ CŪ.

This is just a generalization of the vector inner product for vectors on Rn and it is easy tocheck that the four properties given above are all satisfied for this product.

We observe now, that

Kf1, f2 = ∫U

Kf1x f2xdx = ∫U

∫U

Gx, y f1ydy f2xdx for all f1, f2 ∈ CŪ.

Note further that

∫U

∫U

Gx, y f1ydy f2xdx = ∫U

∫U

Gx, y f2xdx f1ydy = ⟨f1,Kf2⟩

where Kf2 is defined by

Kf = ∫U

Gx, y fxdx for any f ∈ CŪ.

Clearly, Kf defines another linear operator on CU. WhenKf1, f2 = ⟨f1,Kf2⟩ for all f1, f2 ∈ CU,

we say that K is the adjoint of the operator K. Since we know thatGx, y = Gy, x for all x, y ∈ U,

it follows that Kf =Kf for any f ∈ CŪ.

We say that the operator K is symmetric. Since

Kf1, f2 = ⟨f1,Kf2⟩ for all f1, f2 ∈ CŪ,

and K = L−1, it seems reasonable to expect that ⟨Lu,v⟩ = ⟨u,Lv⟩ for all u,v ∈ D. That

22

this is, in fact, the case follows from (3.4).That is,

⟨Lu,v⟩ = ∫U−v∇2udx = ∫

U−u∇2v dx − ∫

∂Uv∂Nu − u∂NvdS

= ∫U−u∇2v dx = ⟨u,Lv⟩ for all u,v ∈ D.

Now consider the Neumann problem

−∇2ux = fx, x ∈ U, ∂Nux = 0, x ∈ ∂U. 10.3

Problem 10.3 can be expressed in terms of the following operator,

LNux = −∇2ux for any u ∈ DN = u ∈ C2Ū : ∂Nux = 0 for x ∈ ∂U .

asfind u ∈ DN such that LNu = f ∈ CŪ.

Although the action of this operator, LNu, is the same as that of the previously definedoperator, L, it is not the same operator since they have different domains. In particular, DN

contains all constant functions and these functions belong to the null space of LN. Then LN

is not invertible. However, the same argument used above shows that LN. is symmetric.Then LNu = f has no solution unless f satisfies ⟨f,v⟩ = 0 for all constant functions v. Ifthis condition is satisfied, then any two solutions differ by a constant. This fact was alreadymentioned in the beginning of section 7 but now we see it in a new setting. It is just theanalogue of the linear algebra result for singular matrices A.

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