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
and
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.
THEOREM (UNIQUENESS OF SOLUTION FOR THE TIME DEPENDENT HEAT EQUATION)
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]]
EXERCISE.
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.]