Computer Algebra Recipes

216 CHAPTER 5. LINEAR PDE MODELS. PART 1
The plane wave is assumed to be traveling in the positive x-direction, its time
part being e ¡I!t ,whereI = p ¡1. In the ¯rst region, x psi[1]:=exp(I*k[1]*x)+a*exp(-I*k[1]*x);
Ã1 := e (KxI) (¡I Kx)
+ ae
The ¯rst term is the incident plane wave, while the second term is the re°ected
wave with amplitude a. Since the transmission and re°ection coe±cients involve
ratios of squared amplitudes, the amplitude of the incident wave has been set
equal to one without loss of generality. The energy re°ection coe±cient then is
given by R = jaj2 =1=jaj2 .
In region 2 (0 · x · L), the wave form must be made up of waves traveling
in the positive and negative x-directions, viz., Ã2 = beIk2x + ce ¡Ik2x ,with
undetermined amplitudes b and c.
> psi[2]:=b*exp(I*k[2]*x)+c*exp(-I*k[2]*x);
Ã2 := be (3 IKx) (¡3 IKx)
+ ce
In the third region, x>L, there will be only a transmitted plane wave, with
spatial part Ã3 = deIk1x ,thewavenumberbeingk1 = K since the string
density is the same as in the ¯rst region. The fraction of the energy incident in
region 1 that is transmitted into region 3 is given by the transmission coe±cient
T = jdj2 =1=jdj2 .
> psi[3]:=d*exp(I*k[1]*x);
Ã3 := de (KxI)
To evaluate the four unknown coe±cients a, b, c, and d, four independent
equations are needed. The ¯rst two equations, eq1 and eq2 ,followfromthe
physical continuity of the string. The string segment in region 2 is joined to
the segment in region 1 at x =0andtothesegmentinregion3atx = L.
> eq1:=eval(psi[1]=psi[2],x=0);
eq1 := 1 + a = b + c
> eq2:=eval(psi[2]=psi[3],x=L);
eq2 := be (3 IKL) + ce (¡3 IKL) = de (KLI)
Since the wave equation, and therefore the second spatial derivative, remains
¯nite everywhere along the string, the ¯rst derivative of à with respect to x
must be continuous. So, continuity of the slope at x =0andx = L yields the
third and fourth equations.
> eq3:=eval(diff(psi[1],x)=diff(psi[2],x),x=0);
eq3 := KI¡ aKI =3IbK¡ 3 IcK
> eq4:=eval(diff(psi[2],x)=diff(psi[3],x),x=L);
eq4 := 3 IbKe (3 IKL) ¡ 3 IcKe (¡3 IKL) = dKe (KLI) I
The system of four equations is now solved for the four unknown amplitudes,
and the solution is assigned.