Advanced topics.

Linear Methods of Applied Mathematics

Evans M. Harrell II and James V. Herod*

*(c) Copyright 1994,1995,1996 by Evans M. Harrell II and James V. Herod. All rights reserved.

version of 1 June 1996


Some advanced topics.

A maximum principle

At the beginning of calculus, one obtains a first understanding of the importance of the second derivative. The second derivative predicts where the graph of a function is concave up or down. In particular, if u is a function defined on the interval [0,1], and u'' is continuous and not negative, then the curve is concave up. As a consequence of this, the values of u on [0,1] do not exceed the maximum of its values at the end points 0 and 1.

A similar result holds for a function of more variables. The role of the second derivative is played by the Laplacian operator. We assume --2u >= 0 in a bounded region in Rn and show that it must take on its maximum value on the boundary.

THEOREM (MAXIMUM PRINCIPLE). Let D be a bounded region and u be continuous on the closed set D[[union]][[partialdiff]]D with

--2u = f on D,

u = g on [[partialdiff]]D,

If f(x,y) >= 0 then u assumes its maximum on [[partialdiff]]D.

Here's a way to think of why this is so. First suppose that f > 0. Any function which is continuous on a closed and bounded set in Rn has a maximum value on that set. Since u is continuous on the closed and bounded set D[[union]][[partialdiff]]D, it has a maximum on D[[union]][[partialdiff]]D. Suppose {[[alpha]],[[beta]]} is in the interior of D and and the maximum of u occurs at that place : max u = u([[alpha]],[[beta]]). Then

F([[partialdiff]]u,[[partialdiff]]x)([[alpha]],[[beta]]) = 0 = F([[partialdiff]]u,[[partialdiff]]y)([[alpha]],[[beta]]) and F([[partialdiff]]2u,[[partialdiff]]x2)([[alpha]],[[beta]])< 0 , F([[partialdiff]]2u,[[partialdiff]]y2)([[alpha]],[[beta]])< 0.

This contradicts f > 0 since

F([[partialdiff]]2u,[[partialdiff]]x2)([[alpha]],[[beta]]) + F([[partialdiff]]2u,[[partialdiff]]y2)([[alpha]],[[beta]]) = f([[alpha]],[[beta]]).

Therefore, {[[alpha]],[[beta]]} must be on the boundary of D.

Now assume f > 0. We will show that it remains true that u must have its maximum on [[partialdiff]]D. For, suppose u satisfies

--2u = f on D, with u = g on [[partialdiff]]D.


V[[epsilon]](x,y) = u(x,y) + [[epsilon]] (x2+y2).


--2V[[epsilon]]= --2(u+[[epsilon]] (x2+y2)) = f + 4[[epsilon]] > 0 on D.

By the previous paragraph, V[[epsilon]] has a maximum on [[partialdiff]]D. Let this maximum occur at {c[[epsilon]],d[[epsilon]]} and u([[alpha]],[[beta]]) = max u. We have

u([[alpha]],[[beta]]) < u([[alpha]],[[beta]]) + [[epsilon]] ([[alpha]]2+[[beta]]2) = v[[epsilon]]([[alpha]],[[beta]]) <

v[[epsilon]](c[[epsilon]],d[[epsilon]]) = u(c[[epsilon]],d[[epsilon]]) + [[epsilon]] (c[[epsilon]]2+d[[epsilon]]2).

Also, because D[[union]][[partialdiff]]D is bounded, lim[[epsilon]]->0 [[epsilon]] (c[[epsilon]]+d[[epsilon]]) =0, so that

u([[alpha]],[[beta]]) = max u < u(lim(c[[epsilon]], d[[epsilon]])) < max u.

Since [[partialdiff]]D is closed, lim{c[[epsilon]], d[[epsilon]]} will be in [[partialdiff]]D and u(lim{c[[epsilon]], d[[epsilon]]}) = max u, we have the result.

EXAMPLE. Consider

--2u = 0 on the unit disk in the plane,

u(1,[[Theta]]) = sin([[Theta]]) on the closed unit circle.

Since the maximum value of u occurs on the boundary and u = sin([[Theta]]) on the boundary of D then -1 < u(x,y) < +1. (Actually, u(r,[[Theta]]) = r sin([[Theta]]).)

EXAMPLE. This proof for the maximum principle just given used that D is bounded. The result may fail to be true if D is not bounded as this example will show. Let D be the strip D = { {x,y}: 0 < x < [[pi]] and y > 0}. Let g(x,y) = sin(x) on the boundary of D. A solution of --2u = 0 on D and u = g on [[partialdiff]]D is u(x,y) = sin(x) ey, and this function goes unbounded in D...certainly not taking on its maximum value on the boundary.

Several questions now come to mind. Is there a minimum principle in case f <= 0 ? Yes, see exercise 1 below. What can be done in case f is positive at some places in D and negative at others? The next theorem addresses this question and finds the maximum value of |u| in terms of f, g, and D. This is a useful result for, in addition to saying something about the maximum value of |u|, we will see that it provides a chance to see how u changes with small changes in f, g, or D. Also, this theorem is used in establishing that solutions for this type problem are unique.

THEOREM (Continuous Dependence on the Data). Let D be a bounded region and u be continuous on the closed set D[[union]][[partialdiff]]D. There is a number K such that if f and g are continuous on D and

--2u = f on D,

u = g on [[partialdiff]]D.

then, |u| <= max |g| + K max |f|.

Here's a way to see this:

--2( + u + (x2+y2) max |f|/4) = + --2u + max |f| = + f + max |f| > 0.

Therefore, by the maximum princple, + u + (x2+y2) max |f|/4 assumes its maximum on the boundary of D. Since D is bounded, let K >= x2+y2 for x and y in [[partialdiff]]D;

|u(x,y)| < max( + u + (x2+y2) max |f| /4) < max |g| + K max |f|.

It follows that the solution depends continuously on the data f and g. [[florin]]

The ideas in this section have been laying out what we would always hope for in a PDE problem: we would hope that the problem has a solution, that the solution changes little if we make small errors in the data, and that the problem only has one solution. (Being able to find that solution is important, of course.) To this point we have not addressed the question of uniqueness of solutions.

THEOREM (UNIQUENESS OF SOLUTIONS) If each of u and v is a continuous function on the bounded region with a piecewise smooth boundary and

--2u = f on D and --2v = f on D

with u(x,y) = g(x,y) on [[partialdiff]]D with v(x,y) = g(x,y) on [[partialdiff]]D

then u = v.

Here's why. Let W = u - v. Note the equations that W must satisfy:

0 = --2W = [[partialdiff]]2W/[[partialdiff]]x2 + [[partialdiff]]2W/[[partialdiff]]y2 with W = 0 on [[partialdiff]]D.

By Green's first identity, 0 = òòD W.0 dA

= òòD W.--2W dA = ò[[partialdiff]]D < W--W , [[eta]] > ds - òòD < --W , --W > dA.

Thus, 0 = < --W, --W > = ([[partialdiff]]W/[[partialdiff]]x)2 + ([[partialdiff]]W/[[partialdiff]]y)2 and [[partialdiff]]W/[[partialdiff]]x = 0 = [[partialdiff]]W/[[partialdiff]]y. It follows that W is constant. Since it is zero on the boundary of D, it must be zero everywhere. Hence, u - v = W = 0. [[florin]]

These three ideas of existence, uniqueness, and continuous dependence on the data are desirable properties for a differential equation to have. One says that a partial differential equation is WELL- POSED if it has these three properties.

A problem is well-posed if

(a) it has a solution,

(b) the solution is unique, and

(c) the solution depends continuously on the initial data.

The PDE in this section is called a Dirichlet problem We have shown that it is well posed. That is, if D is a bounded region with a piecewise smooth boundary then the Dirichlet problem, --2u = f on D, with u = g on [[partialdiff]]D, is well posed.

We have seen examples of differential equations which did not have a solution and for which the solution was not unique. Here is an example of a partial differential equation where the solution does not depend continuously on the data.

EXAMPLE. The problem

--2u(x,y) = 0 for y > 0 and x in R,

u(x,0) = 0 for x in R

[[partialdiff]]u/[[partialdiff]]y|x=0 = sin(nx)/n for x in R.

A solution is un(x,y) = sin(nx) sinh(ny) /n2.

Note that limn->*sin(nx) sinh(ny)/n2 = *.

but, --2u(x,y) = 0 for y > 0 and x in R,

u(x,0) = 0 for sin R

[[partialdiff]]u/[[partialdiff]]y(x,0) = 0 for x in R.

has u(x,y) = 0 for a solution. [[florin]]

An example: The shape of a drum

We are now in a position to determine the equation which describes the shape of a drum. Let D be a region in the plane and U(x,y) be the height of a membrane with a prescribed boundary. That is, we assume that the values of U are known on the boundary of D. If we take g to be that function which describes the values of U on the boundary of D, then we have assumed that U(x,y) = g(x,y) for {x,y} on [[partialdiff]]D. For the interior of D, we assume that the potential energy of the membrane is proportioned to the surface area

E(U) = K òòD R(1 + ([[partialdiff]]U/[[partialdiff]]x)2 + ([[partialdiff]]U/[[partialdiff]]y)2) dA.

The question is, how can U be chosen to minimize E? Let U be a surface that minimizes energy. Let [[Phi]] be any smooth function with [[Phi]](x,y) = 0 on [[partialdiff]]D. Then, a possible shape is U + [[epsilon]][[Phi]], [[epsilon]] > 0. The potential energy of this surface changes with [[epsilon]] and is given as a function of [[epsilon]] by the equation

E(U+[[epsilon]][[Phi]]) = KòòD R(1+[[[partialdiff]](U+[[epsilon]][[Phi]])/[[partialdiff]]x}2 + [[[partialdiff]](U+[[epsilon]][[Phi]])/[[partialdiff]]y]2 ) dA.

By hypothesis, E(U) < E(U+ [[epsilon]][[Phi]]). That is,

F([[partialdiff]]E,[[partialdiff]][[epsilon]]) |[[epsilon]] = 0 = 0.

Use the approximation R(1+z) ~ 1 + z/2. Then

E(U) ~ K òòD [ 1 + (([[partialdiff]]U/[[partialdiff]]x)2 + ([[partialdiff]]U/[[partialdiff]]y)2)/2 ] dA


E(U+ [[epsilon]][[Phi]]) ~ K òòD [ 1 + (([[partialdiff]](U+[[epsilon]][[Phi]])/[[partialdiff]]x)2 + ([[partialdiff]](U+[[epsilon]][[Phi]])/[[partialdiff]]y)2)/2 ] dA.

Now, compute dE/d[[epsilon]] and evaluate at [[epsilon]]=0. Since E(U) is minimum, this derivative should be zero.

0 = dE/d[[epsilon]]|[[epsilon]]=0 = K/2 òòD[ 2 [[partialdiff]]U/[[partialdiff]]x [[partialdiff]][[Phi]]/[[partialdiff]]x + 2 [[partialdiff]]U/[[partialdiff]]y [[partialdiff]][[Phi]]/[[partialdiff]]y ] dA

= K òòD <--U, --[[Phi]] > dA.

Use Green's first identity to get

dE/d[[epsilon]]|[[epsilon]]=0 = K ò[[partialdiff]]D < [[Phi]] --U , [[eta]] > - K òòD [[Phi]] --2U dA

Since [[Phi]] = 0 on [[partialdiff]]D then 0 = -K òòD [[Phi]] --2U dA for all [[Phi]] with [[Phi]] = 0 on [[partialdiff]]D. Therefore, --2U = 0.

This result on the shape of a drum shows that a drum at rest, not changing in time, will be situated so that it satisfies a Dirichlet problem. It is resting in the steady state. A drum in the transition state - moving from some initial conditions to the steady state - will satisfy

F([[partialdiff]]2u,[[partialdiff]]t2) = --2u in D

u = g on [[partialdiff]]D

u(0,x,y) = initial distribution on D,

F([[partialdiff]]u,[[partialdiff]]t) (0,x,y) = initial velocity on D.

Other physical situations which arrange themselves inorder to minimize energy will often lead to elliptic problems, too. For example, consider the following heat problem: Suppose the temperature at each point in the upper- right quarter plane has assumed a value u(x,y) such that u has a continuous second partial derivative and the temperature is constant 0 along the positive x axis and constant 1 along the positive y axis. What must be the values of u for other {x,y}? Or, what is the shape of the graph of u?

The mathematical formulation of this problem is as follows:

Find u such that

--2u = 0 for x > 0 and y > 0,

with u(x,0) = 0 for x > 0,

u(0,y) = 1 for y > 0.

The answer is u(x,y) = F(2,[[pi]]) arctan( F(y,x) ). (Exercise 1 below asks you to check this.)

Because the steady state - time independent - heat equation satisfies the same Dirichlet equations, all the results of this and the previous section apply to the heat equation, as well as to the equation for a drum.

We now investigate the uniqueness of solutions for the problem

[[partialdiff]]u/[[partialdiff]]t = --2u on D

with u(x,y,0) = f(x,y) on D,

and u(t,x,y) = g(t,x,y) on [[partialdiff]]D.

This equation represents the transition of temperature from an initial distribution of g to the steady state with prescribed boundary conditions. The claim is that this problem has no more than one solution.


There is only one solution to the equation

[[partialdiff]]u/[[partialdiff]]t = --2u on D

with u(x,y,0) = f(x,y) on D,

and u(x,y,t) = g(x,y,t) on [[partialdiff]]D.

Here's a way to see that solutions are unique. Let u and v be solutions and w = u - v. Then

[[partialdiff]]w/[[partialdiff]]t - --2w = 0 on D

with w(x,y,0) = 0 on [[partialdiff]]D,

and u(x,y,t) = 0 on [[partialdiff]]D.

Consider E(t) = òòDw2(x,y,t) dx dy/2.

We have E'(t) = òòD w(x,y,t) [[partialdiff]]w/[[partialdiff]]t(x,y,t) dx dy

= òòDw(x,y,t) --2w(x,y,t) dx dy

= ò[[partialdiff]]D< w--w , [[eta]] > ds - òòD ||--w|| 2 dx dy.

This last comes from Green's first identity. Recall w = 0 on [[partialdiff]]D. Then, E'(t) < 0 and E is not increasing. Also E(t) > 0 and E(0) = 0. This means E(t) = 0 for all t. [[florin]]

Exercises XIX

XVIII.1.    Suppose u is as in the Maximum Principle identity and that f <= 0.

Show that the minimum value of u occurs on [[partialdiff]]D. ( Hint: consider v = -u.)

XIX.2.    Suppose that | g1([[theta]]) - g2([[theta]]) | < .01 for 0 <= [[theta]] <= 2[[pi]] and that u1 and u2 are continuous on [[partialdiff]]D and satisfy

--2u1=0 on D and --2u2=0 on D

with u1 = g1 on [[partialdiff]]D with u2 = g2 on [[partialdiff]]D.

Show that

max |u1(r,[[theta]]) - u2(r,[[theta]])| < .01 for 0 < r < 1 and 0 <= [[theta]] <= 2[[pi]].

XIX.3.    Suppose that u(x,y) = e-y sin(x) for y >= 0 and all real x.

(a) Show that --2u = 0 and u(x,0) = sin(x).

(b) If possible give { [[alpha]], [[beta]] } such that u([[alpha]],[[beta]]) is minimum.

(c) If possible give { [[alpha]], [[beta]] } such that u([[alpha]],[[beta]]) is maximum.

(d) Show that u(x,y) = ey sin(x) solves the same equation.

(e) Explain why this does not contradict the uniqueness identity.

XIX.4.    Let u(r,[[theta]]) = r sin([[theta]]) for 0 <= r <= 1 and 0 <= [[theta]] <= 2[[pi]].

(a) Show that --2u = 0 and u(1,[[theta]]) = sin([[theta]]).

(b) If possible give { [[alpha]], [[beta]] } such that u([[alpha]],[[beta]]) is minimum.

(c) If possible give { [[alpha]], [[beta]] } such that u([[alpha]],[[beta]]) is maximum.

(d) Show that if u(r,[[theta]]) = F(1,r) sin([[theta]]) then --2u = 0.

(e) Explain why this does not contradict the uniqueness identity.

XIX.5.    Show that if u(x,y) = F(2,[[pi]]) arctan( F(y,x) ) for x > 0, y > 0, then

--2u = 0 for x > 0 and y > 0,

with u(x,0) = 0 for x > 0,

u(0,y) = 1 for y > 0.

XIX.6.    Show that if u(r,[[theta]]) = ln(r)/ln(2) then

--2u = 0 for 1 < r < 2, 0 <= [[theta]] <= 2[[pi]].

BRC}(A(u(1,[[theta]]) = 0,u(2,[[theta]]) = 1)) for 0 <= [[theta]] <= 2[[pi]].

XIX.7.    Let K(t,x) = exp(-x2/4t) / R(4[[pi]]t).

(a) Show that F([[partialdiff]]K,[[partialdiff]]t) = F([[partialdiff]]2K,[[partialdiff]]x2) for t > 0 and -* < x < *.

(b) Sketch the graph of K(t,x) for t = 1, F(1,2), F(1,4).

(c) Suppose that f is continuous and bounded for all real numbers and that u(t,x) =I(-*,*, )K( t , x-y ) f(y) dy for t > 0 and all real x.

Show that F([[partialdiff]]u,[[partialdiff]]t) = F([[partialdiff]]2u,[[partialdiff]]x2) [ Using the methods of Laplace transforms, one can show that u(0,x) = f(x). Take MATH 4581 or see page 230 of AN INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS AND HILBERT SPACE METHODS by Karl E. Gustafson.]

XIX.8.    Let K(x,y) = F(1,[[pi]]) F(y,x2 + y2) .

(a) Show the --2K = 0 for x > 0, y > 0 and K(x,0) = 0 for x different from 0.

(b) Sketch the graph of K(x,y) for y = 1, F(1,2), F(1,4).

(c) Suppose that f is continuous and bounded for all real numbers and that u(x,y) = I(-*,*, ) K( x-s , y ) f(s) ds for x > 0, y > 0.

Show that --2u = 0 for x > 0, y > 0. [We will show that u(x,0) = f(x). See also page 128 of the book cited above or Chapter 11 of Churchill & Brown.]

Back to Chapter XVIII

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