Definitions, Theorems, Examples, Topics for Asmar 2nd edition
Chapter 7: 7.3 only

 7.3 The Fourier Transform Method
====
 FOURIER TRANSFORM in the x-VARIABLE
   Hold t fixed in the function u(x,t) and take the Fourier transform in 
   the x-variable. 

     FT[u(x,t)](w,t) = (1/sqrt(2Pi))(integral of u(x,t)exp(-iwx))
                       with integration x=-inf to x=inf.

 FOURIER TRANSFORM and PARTIAL DERIVATIVES
   A Fourier transform equation can be formally differentiated on the 
   t-variable, differentiation passing through the integral sign, with
   expected results:
   
          FT[(d/dt)u(x,t)](w) = (d/dt) FT[u(x,t)](w)

   The Fourier transform of (d/dx)u(x,t) is related to the Fourier 
   transform of u(x,t) the same way as functions of x alone:
         
          FT[(d/dx)u(x,t)](w) = (iw) FT[u(x,t)](w)

 EXAMPLE 1. Solve the wave equation on the line by Fourier transform
            methods.
     u_tt = c^2 u_xx  on -inf < x < inf, t>0
     u(x,0) = f(x)   [given shape]
     u_t(x,0) = g(x) [given velocity]

  QUESTION. What happened to the endpoint conditions?
    ANSWER. This is not a finite string. The ends are at -inf and +inf.
    DETAILS. Fix t>0. Take the Fourier transform across all three 
    equations to get the w-domain equations (for fixed t>0)

              y''(t) + c^2 w^2 y(t) = 0, y(0)=F0, y'(0)=F1

    where the symbols are defined by

            y(t) = FT[u(x,t)](w)  Depends on t and w, but not on x.
            w = constant, held fixed
            t = independent variable, 0<t<inf
            F0=FT[f(x)](w)  Depends on w, but not on x or t
            F1=FT[g(x)](w)  Depends on w, but not on x or t

    This is an initial value problem for a linear second order 
    homogeneous ODE. The ODE answer is a linear combination of cos(cwt), 
    sin(cwt).Coefficients in this linear combination are A(w), B(w), 
    because w is fixed while t=variable.

    The initial values F0, F1 give rise to explicit answers for A,B.

    Then u(x,t) equals the inverse Fourier transform of y(t) [which 
    depends on w].

    EXAMPLE. g(x)=0
    Then solving for A, B gives B=0 and A=F0. The solution u(x,t) equals 
    (1/sqrt(2Pi)) times the integral of A(w)cos(cwt)e^{iwx}dw

    REMARKS. The d'Alembert solution gives a different formula. Fourier 
    transforms give a solution that applies to different situations.

 EXAMPLE 2. Heat Equation for an Infinite Rod.
    
    Assume u_t = c^2 u_xx on |x| < inf, t>0 with u(x,0)=f(x), which is 
    the initial temperature along the infinite od.

    Let y(t)= FT[u(x,t)](w) for fixed w.

    Then y' + c^2 w^2 y = 0, y(0)=FT[f(x)](w) [independent of f], with 
    solution y = constant/(integrating factor) = A(w)exp(-c^2w^2t).

    The factor A(w) = FT[f] = F(w), by setting t=0 in the answer.
    The answer is presented as a Fourier transform integral

         u(x,t) = inverse Fourier transform of y(t)
                = c1*integral of F(w)*exp(-c^2w^2t+iwx)
                   over w=-inf to w=inf, where
                   c1 = 1/sqrt(2Pi)

    Section 7.4 rewrites the integral as a convolution of f(x) and the
    Gauss heat kernel.

 EXAMPLE 3. Solve u_tx = u_xx on the line, t>0, u(x,0)=f(x) where
            f(x)=sqrt(Pi/2)*exp(-|x|).

   ANSWER. u(x,t) = f(x+t).
   DETAILS. Let y(t)=FT[u(x,t)], get iw y'(t) = -w^2 y(t) from Transform 
   rules. Solve the DE for y(t)=A(w)exp(iwt). Then A(w)=FT[f]=1/(1+w^2).
   So u(x,t)=inverse FT of exp(iwt)/(1+w^2) = f(x+t) [some work needed 
   to see this fact].

 EXAMPLE 4. u_tt = -c^2 u_xxxx on the line, u(x,0)=f(x), u_t(x,0)=g(x)

   ANSWER. u(x,t)=inverse FT of 
                  FT[f]*cos(cw^2t)+FT[g]*sin(cw^2t)/(cw^2).
   DETAILS.
     Let y(t)=FT[u(x,t)]. Take transforms to get y''(t) + c^2w^4 y(t)=0.    
     And y(0)=FT[f], y'(0)=FT[g]. Solve the initial value problem for 
     y(t).

 EXAMPLE 5. tu_x + u_t = 0 on the line, u(x,0)=f(x)
            The is the advection equation studied in week 1.  
   ANSWER. u(x,t)=f(x-t^2/2)
   DETAILS.
     Let y(t)=FT[u(x,t)]. Then FT of the PDE gives the ODE 
     y'(t) + iwt y(t) = 0, with initial condition y(0)=F(w).
     Solve it for y(t)=F(w)exp(-iwt^2/2). Now apply the $w$-shifting 
     property of the FT to get the answer.