230 CHAPTER 4. LINEAR PARTIAL DIFFERENTIAL EQUATIONS Only if A(t) = 0 will they all satisfy the same boundary condition! Principle 24 Any sum of solutions to a homogeneous partial dierential equation and homogeneous boundary conditions is a solution. If the sum is innite, convergence of the sum must be checked. The importance of this Principle 24 is that it gives us the opportunity to compose a general solution, such as (4.1.33), that we can use to satisfy the remaining condition, the initial condition (4.1.34) or (4.1.36). The main purpose of the method of separation of variables, then, is to construct all the solutions to the homogeneous dierential equation and boundary conditions in order to form a general solution. Then use the general solution to satisfy any remaining condition that may contain some function, for example, the initial condition. 4.1.6 Exercises A fundamental property of the diusion equation is revealed in the next exercise; 1. In the solutions (4.1.35) and (4.1.39), the constant in the Fourier series doesn't change in time. This meanss that the average value of remains constant in time, a natural consequence of the conservation of the amount of salt. There is another way this result can be seen. Integrate (4.1.6) over the length of the pipe, and use the boundary conditions (4.1.8) to show that Z L d (x, t) dx = 0 . dt 0 Since the integral is just a0 L, you will have shown that a0 doesn't change in time and that it must agree with the value determined by the initial prole. Convergence of the Fourier series 2. Note the form of the Fourier series in the solution (4.1.39). What is the nature of the decay of the Fourier coe cients at t = 0? Is this the expected result? How does the convergence change when t > 0? When t = 0.001 sec, estimate the number of terms needed to obtain the graph in Fig. 4.4. The possibility of a point source for an initial condition is considered in the next exercise: 3. Suppose that the initial salt concentration in the pipe with sealed ends is given by ( Q/(AH) , for (L H)/2 < x < (L + H)/2 , (x, 0) = 0, elsewhere . 4.1. DIFFUSION: PART I 231 3 H/L=0.5 H/L=0.25 H/L=0.125 H/L=0. 2.5 AL/Q 2 1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1 x/L Figure 4.5: Concentration proles at t = 0.01 sec for dierent values of H/L: D/L2 = 1 per sec. The salt is deposited in the middle over a region of length H < L. The cross-sectional area is A. Show that the total amount of salt deposited is Q. Find the Fourier coe cients in the solution (4.1.33). Plot the results for the choices D/L2 = 1 per sec, and H/L = 1/2, 1/4, 1/8 at times t = 0.01, 0.1 sec: you need only consider the rst three terms in the Fourier series for the solution. What do you observe happening? Can you take the limit as H ! 0; this describes the deposit of a point source at L/2. What does the solution look like at t = 0.01, 0.1? Compare it to the sequence in H. Answer: Results at t = 0.01 sec are shown in Fig. 4.5. Observe that the prole for small H looks somewhat like a Gaussian. The next exercise explores the importance of the origin of the coordinate system.: 4. Let's reconsider the salt concentration problem (4.1.6) and (4.1.8) but on a shifted interval ( L/2 < x < L/2). In other words, the boundary conditions are @ ( L/2, t) = 0 , @x The initial condition is (x, 0) = m 2 1 and cos @ (L/2, t) = 0 . @x 2 L x+ L 2 . 232 CHAPTER 4. LINEAR PARTIAL DIFFERENTIAL EQUATIONS Construct the solution by the method of separation of variables. Plot the solution for the same circumstances as in Fig. 4.2. You should expect the same proles just shifted over. Is there a connection between the Fourier series for the shifted region and the Fourier series in (4.1.35)? Which choice of coordinates proves easier to treat during separation of variables? Dierent boundary conditions 5. Suppose the contamination has lled the sealed pipe so that the density is uniform m . Now study the consequence of removing the contamination at the ends. Imagine there are exchange pumps that have the ability to remove the contaminated water, a ltering process removes the contaminant, and the exchange pump returns clean water so that the volume of water is unchanged. Assume the process can reduce the density to zero at the ends and maintain zero density for the rest of time. How long will it take to reduce the maximum density in the pipe to below a tolerance level c ? Answer: For the choice D/L2 = 1 per sec and c = 0.01m , the time will be 0.5 sec. 4.2 Diusion: Part II There are other ways the diusion equation (4.1.16) arises but with the same main idea, namely, that the ux is proportional to a gradient. Another important example is the transport of energy. For solids, the internal energy is a measure of the agitation of the molecules making up the solid: The more agitated the molecules the higher the internal energy. Unfortunately, the internal energy is di cult to measure: Instead the temperature is used. For most solids, a slight increase in the temperature, produced by adding heat, increases the internal energy slightly. Mathematically, if T K is the temperature and e(T ) joules/kg is the internal energy per unit mass, then de (T ) = c(T ) , dT where the slope c(T ) joules/(kg K) is called the heat capacity per unit mass. Provided the changes in T are not too large, c(T ) is a constant, dependent only on the material properties of the solid. The temperature in a solid, for example a rod, may vary along the rod and change in time. The concentration of internal energy is then e T (x, t) 3 joules/m3 where kg/m is the density of the solid, assumed a constant. Consequently, the rate of change of the energy concentration will be @ de @T @T e(T (x, t)) = (T ) (x, t) = c (x, t) . @t dT @t @t (4.2.1) Fourier (1822) introduced the idea that the ux of internal energy q(x, t) (, ) = 0 + =1 (,) = 1 [1 2 (, ) = (0, ) = 42 2 2 (, 2 (, 2 )] cos( ) ) = 0 2 2 /2 cos( ) (4.1.35) (4.1.39) (4.1. 6) (4.1.8) (, ) = 0 + =1 (,) = 1 [1 2 (, ) = (0, ) = 42 2 2 (, 2 (, 2 2 /2 cos( ) 2 )] cos( (4.1.39) ) (4.1. 6) ) = 0 (, ) = 0 + =1 (4.1.35) (4.1.8) 2 2 /2 cos( ) (4.1.33)