Definitions, Theorems, Examples, Topics for Asmar 2nd edition
Chapter 7: 7.4 only

 7.4 The Heat Equation and Gauss' Kernel
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  u_t = c^2 u_xx on the line, t>0,
  u(x,0)=f(x)

  PREVIOUS RESULT from Example 2, section 7.3:
  
    u(x,t) = inverse FT of F(w)exp(-c^2w^2t)

 DEF. GAUSS' Kernel is g_t(x) = (1/c/sqrt(2t))exp(-x^2/(4c^2t))

 THEOREM 1. [Solution of the Heat Equation as a Convolution]
  The heat equation solution on the line is the convolution of Gauss'
  kernel with the initial temperature f(x). That is,
     u(x,t) = (g_t * f)(x)
            = constant * integral of g_t(x-t)f(t)dt on the line
            
 PROOF. An equivalent formulation of the conclusion is the frequency 
   domain formula FT[u(x,t)]=FT[g_t*f], because we can cancel FT from
   each side to obtain u=f*g_t. Apply Theorem 5 in section 7.2, which
   says that
               FT[exp(-ax^2/2)]=(1/sqrt(a)) exp(-w^2/(2a)
   Then
    c sqrt(2t) FT[g_t(x)] = FT[exp(-ax^2/2)] where 1/(4c^2t)=a/2
                          = exp(-w^2/(2a))/sqrt(a)
                          = exp(-c^2w^2t)(c sqrt(2t))
   The answer from Example 2 section 7.3 is then re-written:
                    FT[u(x,t)] = FT[f]exp(-c^2w^2t)
                               = FT[f]FT[g_t]
                               = FT[f*g_t]

 EXAMPLE 1. Error function.

 DEF. erf(w)=(2/sqrt(Pi)) times the integral of exp(-z^2) on z=0 to z=w.

    u_t = u_xx, u(x,0)=f(x)=100*unit pulse on |x|<1.
    ANSWER. u(x,t)=50(erf(A)-erf(B)), A=(x+1)/2/sqrt(t), 
    B=(x-1)/2/sqrt(t).
    DETAILS. Write out the convolution formula for u(x,t) with Gauss' 
    kernel, then change variables in the integral.

 EXAMPLE 2. Heat problem with nonconstant coefficients
    u_t = t u_xx on the line, t>0, u(x,0)=f(x)
    ANSWER. u(x,t)=f convolution exp(-x^2/2/t^2)/t
    DETAILS. Let y(t)=FT[u(x,t)]. Get ODE y'+tw^2 y=0, y(0)=F(w). Solve
    the ODE for y(t)=F(w)exp(-t^2w^2/). This is a convolution product
    provided we can solve for g in the relation FT[g]=exp(-t^2w^2/2). 
    The answer is provided by Theorem 5, section 7.2, which says 
    g =(1/t)exp(-x^2/(2t^2)). Then FT[u]=FT[f]FT[g]=FT[f*g] implies
    u(x,t)=f*g.

 EXAMPLE 3. The two plots compare problems and answers as follows. The
            initial temperature is f(x)=100*pulse(x,-1,1).
    Problem 1. u_t = u_xx, u(x,0)=f(x), with answer
               u_1(x,t)=50(erf(v1)-erf(v2)), 
                   v1=(x+1)/(2sqrt(t)), v2=(x-1)/(2sqrt(t))
    Problem 2. u_t = t u_xx, u(x,0)=f(x), with answer
               u_2(x,t)=50(erf(w1)-erf(w2)), 
                   w1=(x+1)/(t sqrt(2)), w2=(x-1)/(t sqrt(2))
    Plots are made showing the different shapes of the two answers.

 EXAMPLE 4. A Periodic initial temperature f(x)
            u_t = u_xx on the line, u(x,0)=f(x)=1+cos(2x)
   ANSWER. u(x,t)=1+exp(-4t)cos(2x)
   DETAILS. The FT of f(x) is a problem. We avoid the trouble by using 
   the convolution formula of Theorem 1, then verify that the answer 
   works. We deal with the relation

                         u(x,t) = [g_t * f] (x)

   The deep theoretical troubles are met with work done in exercise 7.4-19 and 
   Example 3 in 7.3.