23bdry.html

Partial Differential Equations PowerTool

by Dr. Jim Herod

Section 5.5: Different Boundary Conditions

Maple Packages for Section 5.5

> restart;

> with(plots):

Warning, the name changecoords has been redefined

>

We examine a variety of boundary conditions that can be associated with the wave equation. We will provide illustrations for how these boundary conditions would look in a vibrating string.

Fixed endpoints. Fixed endpoints is the situation we have considered to this point in these notes. Here is a graph of a function that has this boundary condition. Mathematically, the conditions have been

u ( t , 0) = 0 and u ( t , L ) = 0.

> L:=Pi;
u:=(t,x)->2*sin(x)*cos(t);

We check that this function satisfies the wave equation and is zero at the boundaries: 0 and L .

> diff(u(t,x),t,t)-diff(u(t,x),x,x);
u(t,0);
u(t,L);

We graph the initial condition.

> plot(u(0,x),x=0..L);

Here is an animation of the solution. Remember that it is what happens at the boundaries that is our interest in this Section.

> animate(u(t,x),x=0..L,t=0..8, frames=40);

Finally, here is a graph of the solution surface.

> plot3d(u(t,x),x=0..L,t=0..2*Pi,axes=NORMAL,orientation=[-160,60]);

>

Elastic attachment. This situation assumes we have a spring, or other elastic device attached to the string. Such an arrangement would tend to bring the string back from any displacement. How hard the elastic device pulls the string back depends on how far it is displaced. Such conditions could be written

(t, 0) = k u ( t , 0) and (t, L) = - k u (t, L).

Such conditions are reminiscent of radiation cooling for heat diffusion problems. Here is an illustration for how such a system would behave.

> L:=Pi/2;
u:=(t,x)->cos(t)*sin(x+Pi/4);

We check that this u satisfies the wave equation and the boundary conditions.

> diff(u(t,x),t,t)-diff(u(t,x),x,x);
D[2](u)(t,0)-u(t,0);
D[2](u)(t,Pi/2)+u(t,L);

Here is the initial displacement.

> plot(u(0,x),x=0..L,y=0..1);

Watch the boundaries.

> animate(u(t,x),x=0..Pi/2,t=0..L, frames=40);

Here is the solutions surface.

> plot3d(u(t,x),x=0..L,t=0..2*Pi,axes=NORMAL,orientation=[-30,70]);

>

Frictionless sleeve. This models having the string attached to carts which hold the string with a horizontal tangent at the ends, but allows the string to move up and down as the cart runs along a frictionless track. This situation can be described with

(t, 0) = 0 and (t, L) = 0.

In this illustration, the right end is fixed, the left end is on a frictionless sleeve.

> L:=Pi/2;
u:=(t,x)->sin(t+Pi/2)*cos(x);

We check the PDE and the boundary conditions.

> diff(u(t,x),t,t)-diff(u(t,x),x,x);
D[2](u)(t,0);
u(t,L);

Here is the initial distribution.

> plot(u(0,x),x=0..L);

The animation is a good way to get an understanding for the boundary conditions.

> animate(u(t,x),x=0..L,t=0..2*Pi, frames=40);

> plot3d(u(t,x),x=0..L,t=0..2*Pi,axes=NORMAL,orientation=[-40,55]);

>

Changing Boundary Condition . We suppose we have a half infinite string in this model and we change the boundary condition with time. Take the initial conditions to be zero, and take U(t, 0) = b(t). We expect to see a signal move down the string. Here is an analysis of the problem. With no appeal to initial conditions or boundary conditions, we found in Section 5.2 that solutions should have the form (x + c t) + (x - c t). Using that the initial conditions are zero for x > 0, recall that we can conclude that (x) = 0 and (x) = 0 for x > 0. The term (x + c t) will be zero since c > 0 and t > 0. Thus, whatever happens,

u(t, x) = (x - c t),

and this is zero as long as x > c t. We ask how to extend to the negative numbers. The answer must lie in the boundary condition. Recall that it did earlier, also.

We know that u(t, 0) = b(t), so

(0 - c t) = u(t, 0) = b(t).

So, for negative numbers n, we have (n) = b(-n/c). Here is an example. Think of standing at the end of a long rope and moving the end up and down.

> b:=x->sin(x);

From what came above, here would be the solution for the wave equation with this boundary.

> u:=(t,x)->piecewise(x<t,sin((t-x)/c),0);

> c;

Watch the wave move down the rope.

> animate(u(t,x),x=0..20, t=0..20, frames=40);

> plot3d(u(t,x),x=0..20,t=0..20,axes=NORMAL,orientation=[-115,55]);

>

The intent of this Section was to recall that there can be a variety of boundary conditions that change the character of the solutions for the wave equation. Some of these were illustrated above. Before the reader grows weary with the endless variations possible, we move to a new chapter and step up the dimension. We next consider partial differential equations associated with the steady state equation in multidimensions.

Unassisted Maple

This moving boundary problem suggests ideas about sending signals. Suppose we know the speed of the wave along the string is c = 3. We ask, how long will it take for the peak of the wave to first reach the point x = 10? This is a problem that Maple can solve unassisted. We ask Maple to solve an equation involving the unknown as part of the argument of a trigonometric function.

First we define u.

> u:=(t,x)->piecewise(x<t,sin((t-x)/c),0);

Take x to be 10 and find t such that the u ( t , 10) is first 1.

> solve(sin((t-10)/c)=1,t);

To get a visualization of this answer, we choose a particular c.

> c:=3;

Having this c , watch to see that at the value of t found above, the wave first peaks at x = 10.

> animate(u(t,x),x=0..10,t=0..10+Pi*c/2, frames=40);

>

EMAIL: herod@math.gatech.edu or jherod@tds.net

URL: http://www.math.gatech.edu/~herod