Maple Packages for Section 4.3

>

restart;

>	with(PDEtools):

>

Solution for the First Problem

     We have discussed a physical interpretation for the PDE   and for the initial condition . Here is an idea to give intuition for the boundary conditions in this problem. Having the derivative with respect to x  to be zero at these two ends suggests that there is no movement of heat across the boundary x  = 0 or x  = 1.

     From this, what do we expect for solutions of this problem? Since the problem acts as though there is no radial transfer of heat and no end point transfer of heat, we expect the entire system to retain the total heat, and to move to an equal distribution of heat throughout the rod. We will solve the equation with a particular f  and sketch the graph of the solution. We expect to see that the solution has limit a solution that is constant in t  and in x . Also, because the total heat in the system should remain constant, it should be that

                 =  , for all t > 0.

Here goes making a solution.

First, we find two ODE's, with boundary conditions, which arise from separation of variables.

The PDE leads to

          X T ' = X '' T,

or

          T ' / T = X '' / X.

Since the left side is independent of x , the right side must be constant. In a similar manner, the left side is constant and these constants are the same. Thus,

          X '' =  X, with X'(0) = 0 and X '(1) = 0,

and

          T ' =  T.

Using methods we have seen earlier, this constant must be negative and have the form

             with X ( x ) =  , for each non-negative integer n .

Also, T  ( t ) = exp( ).

Thus, the general solution for the problem is

          u(t ,x) =  +  .

To find a particular solution, we need to specify f . We do that now. Using this f , we write a solution and check that the answer is right, and seems right!

>	f:=x->x*(x-1)+1;

Because this f   is u (0, x ), then

           f ( x ) =    +  .

The computation of the c  's is a job for Fourier Series.

>	for n from 0 to 10 do      c[n]:=int(f(x)*cos(n*Pi*x),x=0..1)/              int(cos(n*Pi*x)^2,x=0..1); od; n:='n':

To see how good our fit is, we plot f  and the Fourier Approximation.

>	plot([f(x),c[0]+sum(c[n]*cos(n*Pi*x),n=1..10)],x=0..1);

>

Now we make the u ( t , x ).

>	u:=(t,x)-> c[0]+sum(c[n]*exp(-n^2*Pi^2*t)*cos(n*Pi*x),n=1..10);

This completes a solution of this first problem.

Solution for the Second Problem

An interpretation for this problem is that there is no heat loss laterally, that the left end is held constant at temperature A, and that the right end is insulated.

We should recognize that this problem does not have zero boundary conditions and so we should break the problem into two: one problem gives a particular solution and the other provides a general solution for the homogeneous equation.

Take v ( x ) = A. This solutions satisfies the PDE, and the boundary conditions, but not the initial conditions, of course. The function v ( x ) = A is the steady state solution.

Now look for w to satisfy

The same PDE:            ,

The Homogeneous Boundary Conditions:     (t, 0) = 0 and  (t,1) = 0.

Finding this w, we will add w and v and then choose coefficients to satisfy the initial conditions.

By now, we see that this homogeneous problem leads to two ODE 's, one having boundary conditions:

          X '' =  X, with X(0) = 0 and X '(1) = 0,

and

          T ' =  T.

Be reminded that in Section 3.3, we listed this differential equation and boundary conditions among the mixed boundary conditions . We derive the results listed there.

It is no surprise that we should take .  The solution for the X '' =  X is

            cos(  x) +  sin(  x).

The requirement that X(0) = 0 implies that  = 0. The requirement that X '(1) = 0 implies that

            cos( ) = 0.

To have a solution that is not zero for all x , it must be true that cos( ) = 0. This happens if  is an odd multiple of . That is,

                 , n = 1, 2, 3, ... .

Thus, these are the eigenvalues for this homogeneous problem corresponding to eigenfunctions sin(  x). It is clear that the corresponding solution for the T equation is exp(-  t) .

We can now write the general solution for the original problem:

          w (t, x) + v (x) =   +  A.

To determine the c 's, let t = 0, and use the Fourier idea.

To be specific, we choose a particular f and A.

>	A:=1; f:=x->cos(Pi/2*x);

>	n:='n': for n from 1 to 5 do      c[n]:=int((f(x)-A)*sin((2*n-1)*Pi/2*x),x=0..1)/               int(sin((2*n-1)*Pi/2*x)^2,x=0..1); od; n:='n';

>	plot([f(x)-A,sum(c[n]*sin((2*n-1)*Pi/2*x),n=1..5)],x=0..1);

>

In passing, note the nature of convergence.

     We now define u .

>	u:=(t,x)->A+sum(c[n]*exp(-(2*n-1)^2*Pi^2*t/4)*sin((2*n-1)*Pi/2*x),n=1..5);

This completes the solution for the second problem.

Unassisted Maple

We check to see what Maple 8 can do with this equation unassisted. As usual, we have called in the PDE(tools) at the start of this Section. The partial differential equation has not changed from that of Section 4.1.

>

u:='u';

>	PDE:=diff(u(t,x),t)-diff(u(t,x),x,x)=0;

By now, we are familiar with the suggestion that Maples makes for the form for a solutions: products of two functions just as we used above.

>	ans := pdsolve(PDE);

>

It may be that the reader becomes impatient to see the other techniques that Maple uses. Those techniques will use information about the boundary conditions. See Section 8.2.