1.2.1.
(a) Explain the minus sign in conservation law
    (1.2.3) or (1.2.5) if Q=0.

Solution.
  The equation to be explained is de/dt = - (d/dx)phi 
  To make sense of the equation we replace derivatives
  by Newton difference quotients, as follows:
     e_t approx (e(x,t+h)-e(x,t))/h
     phi_x approx (phi(x+k,t)-phi(x,t))/k
  Suppose the thermal energy density e(x,t) increases in the time
  interval t to t+h. Then the difference quotient for e(x,t) is
  positive. On the other side of the equation, the amount of heat 
  energy per unit time flowing to the right had to decrease, that is
  more heat had to be trapped in the rod section in order to increase
  the heat density e(x,t). So the right side of the equation is 
     - (d/dx)phi = (minus)(phi(x+k,t)-phi(x,t)) = (minus)(minus) = plus 
  which accounts for the minus sign.
  
(b) Explain the minus sign in Fourier's law
    (1.2.8), phi = - K0 (du/dx) 

  Replace du/dx by the Newton quotient (u(x+h,t)-u(x,t))/h for h>0. If
  the temperature u(x+h,t) is higher than u(x,t), then the quotient is
  positive and heat energy flows from hot to cold, which is to the left. 
  Symbol phi is the heat flux, the amount of thermal energy per unit
  time flowing to the right, which is the opposite direction, hence the
  minus sign in Fourier's law.

(c) Explain the minus sign in conservation law(1.2.12), du/dt = - (d/dx)phi 

  A positive derivative du/dt means the expected change in temperature
  from u(x,t) to u(x,t+h) is approximately h*du/dt > 0. Then u(x,t+h) is
  a higher temperature than u(x,t). This happens when more heat energy
  is locally stored in the rod near location x over time interval t to
  t+h. The amount of thermal energy escaping to the right has to decrease
  in order to store energy in the rod, so (d/dx)phi < 0. This explains the
  sign in the conservation law.

(d) Explain the minus sign in Fick's law (1.2.13), phi = - k (du/dx)

  This is a chemical re-formulation of Fourier's Law, already explained
  in part (b) for temperature u(x,t). The details remain the same for a
  chemical concentration u(x,t). New in the explanation is the use of 
  terminology like chemical flow and a physical explanation in terms of 
  migration of atoms.

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1.2.2. Derive the heat equation for a rod with constant thermal
       properties, constant cross-section A and no heat sources. 
       Suggestions follow, one method suffices.

 (a) Consider the total thermal energy e(x+dx,t)-e(x,t) in rod section
     [x,x+dx]. 

 (b) Consider the total thermal energy between x=a and x=b, which
     is int(A e(x,t),x=a..b) for fixed t.

1.2.3. Derive the heat equation for a rod assuming constant thermal 
       properties with variable cross-sectional area A(x) assuming no
       sources by considering the total thermal energy between x = a and
       x = b.

1.2.4. Derive the diffusion equation for a chemical pollutant.

  (a) Consider the total amount of the chemical in a thin region between
      x and x + dx.

  (b) Consider the total amount of the chemical between x = a and x = b.
  
1.2.5. Derive an equation for the concentration u(x,t) of a chemical
       pollutant if the chemical is produced due to chemical reaction at
       the rate of  alpha u(beta - u) per unit volume.
       
1.2.6. Suppose that the specific heat is a function of position and
       temperature, c(x,u).

  (a) Show that the heat energy per unit mass necessary to raise the
      temperature of a thin slice of thickness dx from 0 Celsius to
      u(x,t) is not c(x,u(x,t)), but instead int(c(x,v),v=0..u(x,t)).

  (b) Rederive the heat equation in this case. Show that (1.2.3) remains
      unchanged, that is, de/dt = - (d/dx) phi + Q.

1.2.7. Consider conservation of thermal energy (1.2.4) for any segment
       of a one-dimensional rod a < x < b. This is the equation

(d/dt) int(e,x=a..b) = phi(a,t)-phi(b,t) + int(Q,x=a..b);

By using the fundamental theorem of calculus, (d/db) int(f(x),x=a..b) = f (b), derive the
       heat equation (1.2.9). This is equation
 
c rho (du/dt) = (d/dx)(K0 (du/dx)) + Q.

1.2.8. If u(x,t) is known, give an expression for the total thermal
       energy int(A(x)e(x,t),x=0..L) contained in a rod x=0 to x=L, which is time-dependent,
because the energy can escape at the ends of the rod. 

Validate the answer using a uniform rod of length L and cross-sectional area A, held at steady-state temperature u=u0. Reference: Serway and Vuille, College Physics, Chapter 11.

1.2.9. Consider a thin one-dimensional rod without sources of thermal
       energy whose lateral surface area is not insulated.

 (a) Assume that the heat energy flowing out of the lateral sides per
     unit surface area per unit time is w(x,t). Derive the partial
     differential equation for the temperature u(x,t).
      
 (b) Assume that w(x, t) is proportional to the temperature difference
     between the rod u(x,t) and a known outside temperature gamma(x,t).
     Derive that 
     
     c rho (du/dt) = (d/dx)(K0 (du/dx)) 
                     - (P/A) [u(x,t) - gamma(x,t)] h(x),      (1.2.15) 

     where h(x) is a positive x-dependent proportionality, P is the
     lateral perimeter, and A is the cross-sectional area.
     
 (c) Compare (1.2.15) to the equation for a one-dimensional rod whose
     lateral surfaces are insulated, but with heat sources. 
     
 (d) Specialize (1.2.15) to a rod of circular cross section with
     constant thermal properties and 0 Celsius outside temperature.

 (e) Consider the assumptions in part (d). Suppose that the temperature
     in the rod is uniform [i.e., u(x,t) = u(t)]. Determine u(t) if
     initially u(0) = u0.