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]]


1. 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.

2. 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]].

3. 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.]

4. 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 != 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.]