Definitions, Theorems, Examples, Topics for Asmar 2nd edition
Chapter 3: 3.3 and 3.4 only          

 3.3  Wave Equation, Separation of Variables
====
 The BVP 

    u_tt = c^2 u_xx
    u(0,t)=0, u(L,t)=0,
    u(x,0)=f(x), u_t(x,0)=g(x)

 PRODUCT SOLUTIONS
   Assume u(x,t)=X(x)T(t) satisfies the PDE and u(0,t)=0, u(L,t)=0,
   then
          X'' -k X =0, X(0)=X(L)=0
          T'' -kc^2 T=0, T(0) nonzero
 THEOREM. 
   The boundary value problem
                       X'' + p X =0, X(0)=X(L)=0
   has only the zero solution for p<0 or p=0. For p=a^2>0, the solution
   is X(x)=c sin(ax) subject to sin(aL)=0. All other p=a^2 for which
   sin(aL) is nonzero have only the zero solution.

 THEOREM. 
   A nonzero product solution u(x,t)=X(x)T(t) is found only in case 
   -k = n Pi/L for n=1,2,3, ..., in which case
           X(x) = sin(n Pi x/L)
           T(t)=b[n] cos(n c Pi t/L) + b*[n] sin(n c Pi t/L)
 THEOREM. Solution of the 1-dimensional wave equation.
   The BVP [see above] has solution u(x,t) given by the series
    sum(sin(Pi nx/L)[b[n] cos(Pi nct/L)+b*[n] sin(Pi nct/L),n=1..inf)
   where
      b[n] = (2/L)int(f(x) sin(Pi nx/L),x=0..L)
      b*[n] = (2/(Pi nc))int(g(x) sin(Pi nx/L),x=0..L)

 DEF. A NORMAL MODE is the product solution
       u[n](x,t)=sin(Pi nx/L)[b[n] cos(Pi nct/L)+b*[n] sin(Pi nct/L)
     that appears in the preceding theorem.

 DEF. The FUNDAMENTAL MODE is u[1](x,t)    
 DEF. An OVERTONE is u[n](x,t) for n>1.
 DEF. The NATURAL FREQUENCY of normal mode u[n](x,t) is the coefficient 
      of t in the mode: Pi nc/L. The value of c^2 = tau/rho, hence the 
      pitch of the sound from a guitar string can be changed by changing
      the tension tau or the mass density rho.

 EXAMPLE. Normal modes of vibration.
   String shape f(x)=sin(Pi mx/L), velocity g(x)=0. We find the mth mode
                 u[m](x,t) = sin(Pi mx/L)cos(Pi cmt/L)

 EXAMPLE. A nonzero initial velocity.
   Let L=Pi/2, c=1, f(x)=0, g(x)=x cos(x). Then
     u(x,t) = sum(v[n],n=1..infinity)
     v[n]=(8/Pi)((1)^(n+1)/(4n^2-1))sin(2nx)sin(2nt)

 EXERCISES. The result for the damped finite string in 3.3-12 is of 
   physical interest. The answer can be viewed as a generalization of
   the damped oscillator my''+cy'+ky=0 in ODE. See also 3.3-13, 3.3-14,
   3.3-15.

 3.4 d'Alembert's Method
====
 THEOREM [d'Alembert's formula] 
   The infinite string problem 
                            u_tt = c^2 u_xx
   has general solution
                        u(x,t) = F(x+ct)+G(x-ct)
   where F, G are sufficiently smooth functions of one variable.

 THEOREM [d'Alembert's Solution]
   Let f(x) and g(x) be odd 2L-periodic functions on the line. Then
   the finite string problem
                       u_tt = c^2 u_xx
                       u(0,t)=u(L,t)=0,
                       u(x,0)=f(x), u_t(x,0)=g(x)
   has solution u(x,t) given by the formula
     u = [f(x-ct)+f(x+ct)]/2 + int(g(s),s=x-ct .. x+ct)/(2c)

 QUESTION. The 2L-periodic extension f(x) is chosen to be odd. How does
   oddness arise from the initial conditions and boundary conditions?

   ANSWER: Consider the case g(x)=0. Then u(x,t)=[f(x+ct)+f(x-ct)]/2
   implies 0=u(0,t)=[f(ct)+f(-ct)]/2, which says f(w)=-f(-w) where w=ct.    
   Therefore, f(w) has to be odd.

 EXAMPLE. Product solutions
   Let f(x)=sin(Pi mx/L), g(x)=0. They are odd and 2L-periodic. Then
                  u(x,t) = sin(Pi mx/L) cos(Pi mct/L)
   which is the product solution obtained by the Fourier series method.

 TRAVELING WAVES, SPECIAL CASE
   Consider the wave velocity zero case: g(x)=0. Then
               u = (1/2)f(x-ct) + (1/2)f(x+ct) = u1 + u2
   Then for fixed t, 
     u1 is a translate of f(x) by ct units to the right, followed by
       scaling by 1/2.
     u2 is a translate of f(x) by ct units to the left, followed by
       scaling by 1/2.
   Both u1, u2 are traveling waves moving at velocity c, as t increases.   
 TRAVELING WAVES, GENERAL CASE
   Consider the case g(x) nonzero. Then
               u =  u1 + u2
               u1 = (1/2)f(x-ct)-(1/2c)G(x-ct)
               u2 = (1/2)f(x+ct)+(1/2c)G(x+ct)
               G(x)=int(g(s),s=a..x)
   Then the waves u1, u2 move right and left, resp, with shapes
   u1(x,0), u2(x,0). The geometry is unchanged when g(x)=0, but
   then the two shapes are identical.

 EXAMPLE. Let f0(x)=0.3x on [0,1/3] and f0(x)=0.3(1-x)/2 on [1/3,1].
          Let L=1, c=1/Pi. Let g(x)=0. Let f(x) be the odd 2L-periodic
          of f0(x) to the whole real line.
            (a) Find the string shape at t=Pi/3 and t=2Pi/3.
            (b) Find the first time t=T when the string returns to its 
                initial shape. Answer: t=2Pi.

 EXAMPLE. Let L=c=1. Let f(x)=0, g0(x)=x on [0,1]. Let g(x) be the odd 
          2L-periodic extension of g0(x). Solve the string problem with
          d'Alembert's method.
          ANSWER: u(x,t)=(1/2)G(x-ct)-(1/2)G(x+ct)
                  G0(x)=(x^2-1)/2 for |x| < 1 
                  G(x) is the 2L-periodic extension of G0(x)

 CHARACTERISTIC LINES.
   This is a study of the dependence of u(x,t) on the periodic 
   extensions in its d'Alembert formula. We don't study it, because of
   the complexity. Much of the use for the idea is to simplify the
   formula for u(x,t).
 CHARACTERISTIC PARALLELOGRAM.
   This is a deeper study of the dependence of u(x,t) on the periodic 
   extensions which appear in its d'Alembert formula. We don't study
   it, because of the complexity. Simplifications to the formula for 
   u(x,t) can be dramatic.

====