Diffusion - The Heat Equation

Transcription

1 Chapter 6 Diffusion - The Heat Equation 6.1 Goal Understand how to model a simple diffusion process and apply it to derive the heat equation in one dimension. We begin with the fundamental conservation law given in equation 5.4. The diffusion model will result in a different expression for the flux, φ (x, t), hence we will have a different PDE for the diffusion model. In the process, we will explain what diffusion is. We will not solve the heat equation yet. 6.2 Diffusion In the previous model, advection, the flow of the quantity being studied was caused by the motion of the supporting fluid. In the example of a pollutant in a river, the flow of the pollutant is caused by the current of the water in the river. Flow may be the result of other physical phenomena. In this section, we consider another model, diffusion. Diffusion is the transport of a material or chemical by molecular motion. Molecules of a substance exhibit microscopic erratic motions due to being randomly struck by other molecules of that substance, or other substances also present. As a result, individual particles or molecules will follow paths sometimes known as "random walks". This is possible because whether a substance is in the gas state, the fluid state or the solid state, its molecules can have some motion with respect to one another. In the case of a gas, individual molecules can move freely with respect to one another. In the case of a liquid, the bonds between molecules are weak and molecules can have a wide range of motion with respect to one another. For solids, the bonds between molecules are stronger, but they can still vibrate with respect to one another. The important point is that molecules can exhibit some motion with respect to one another. If 37

2 38 CHAPTER 6. DIFFUSION - THE HEAT EQUATION the substance is in a steady state, all the molecules move or vibrate at the same rate, there is equilibrium. If for some reason, some molecules of a substance move at a higher speed, then they will strike other molecules and cause them to move faster. These molecules will in turn strike other molecules and cause them to move faster. This will create a flow. The flow is always from the molecules which move at higher speed to the ones which move at lower speed. This phenomenon can be observed in the following experiment. Consider a box separated in two halves with a membrane. Both halves contain the same gas. In one half, the molecules are much more agitated than in the other half. If we remove the membrane, the more agitated molecules will strike the less agitated ones and cause them to move faster. After a while, an equilibrium will be reached. To model this random motion, we make two observations. The first one was already mentioned above: flow is always from more agitated molecules (higher kinetic energy) to less agitated molecules. Since the agitation of molecules is related to the temperature of the substance as well as the density of the substance, the greater the temperature or density, the greater the agitation. This leads to the second observation. The steeper the density gradient of the substance being studied, the greater the flow. In other words, the flux, which measures the speed of the flow, will be proportional to the gradient of the concentration of the substance being studied. In one dimension, where there is only one spatial variable, the gradient is simply the first derivative with respect to the spatial variable. Thus we can write φ (x, t) = Du x Where D>0 is a constant of proportionality. It is called the diffusion constant. u is the density of the quantity being studied. We see that if u x > 0, that is the density increases from left to right, the the flow should be from right to left, that is will be negative. If u x < 0, that is the density decreases from right to left, then the flux will be from left to right, thus will be positive. If we use this expression for φ in equation 5.4, then we get u t Du xx = f (x, t) For now, we will focus on the diffusion process only and assume there is no sources, that is f (x, t) =0. Thus, we obtain the general diffusion equation Equation 6.1 is known as Fick s law. 6.3 Diffusion and Other Models Again, we start with the conservation law u t Du xx =0 (6.1) u t + φ x = f (x, t)

3 6.4. HEAT EQUATION WITH ONLY DIFFUSION PRESENT If both diffusion and advection are present and there are no sources, then the flux is given by φ = cu Du x Thus, the conservation law becomes u t + cu x Du xx =0 (6.2) This is the advection-diffusion equation. This equation could govern the density of a pollutant in a river. Advection would be caused by the flow of the river with speed c. As the pollutant is flowing, it would also diffuse according to Fick s law. 2. In the above example, if the pollutant also decays at rate λ>0, then f (x, t) = λu and the equation governing this model becomes u t + cu x Du xx = λu (6.3) This is the advection-diffusion-decay equation. 6.4 Heat Equation With Only Diffusion Present We now apply equation 6.1 to the study of heat flow. Heat can flow according to the diffusion model. The warmer a substance is, the more agitated its molecules are. If we consider a rod and assume that one part of the rod is warmer, then the molecules of that warmer part will be more agitated. As they strike surrounding molecules, they will cause these molecules to be more agitated thus the heat will spread. This is diffusion. Let us introduce some notation. Consider a thin rod having a constant density ρ and specific heat C. The specific heat of a substance is the amount of energy needed to raise the temperature of a unit mass of the substance by one degree. Both ρ and C are known for known substances, they can be found in engineering and physics handbooks. If u (x, t) is the energy density, then u (x, t) =ρcθ (x, t) If we apply the conservation law with no sources (equation 5.4), we obtain ρcθ t + φ x =0 Because of heat flow follows a diffusion model, φ = Kθ x where K is the thermal conductivity, another constant which can be found for each known substance in engineering and physics handbooks. This is known as Fourier s heat law. This, combining these two equations gives ρcθ t Kθ xx =0 (6.4)

4 40 CHAPTER 6. DIFFUSION - THE HEAT EQUATION or θ t kθ xx =0 (6.5) where k = K,itiscalledthediffusivity or thermal diffusivity. This is ρc known as the heat equation. It is the same equation you were given earlier. Now, we have derived it using the conservation law. Remark 44 In deriving this equation, we made some assumptions about the rod.theseassumptionsleadustoassumethatρ and C and K were constants. If the rod is not homogeneous, K will also depend on x. In addition, if we consider wide ranges of temperatures, K mayalsodependonθ. So, equation 6.4 would become ρcθ t (K (x, θ) θ x ) x =0 (6.6) This is a non-linear model. 6.5 Auxiliary Conditions As we have already discussed, there are two kinds of auxiliary conditions. They are the boundary conditions and the initial conditions Initial Conditions These conditions specify the value of the unknown function for some value of t, usually when t =0. If one were to study the temperature of a thin rod of length L, an initial condition would have the form u (x, 0) = u 0 (x) for 0 x L Where u 0 (x) is the given initial density distribution Boundary Conditions There are three kind of boundary conditions. 1. Dirichlet Condition. In this case, the value of the unknown function is specified at the boundary. Again, in the case of a thin rod of length L, the Dirichlet condition might be u (0,t)=g 1 (t), u (L, t) =g 2 (t) for t>0 2. Neumann Condition. In this case, the flux at the boundary is specified. Again, in the case of a thin rod of length L, the Neumann condition might be Du x (0,t)=h 1 (t), Du x (L, t) =h 2 (t) for t>0

5 6.6. STEADY STATE SOLUTION Mixed Condition. This condition expresses Newton s law of cooling which states that the heat flux is proportional to the temperature difference between the end of the bar and the given temperature which would represent the temperature of the surrounding medium. Again, in the case of a thin rod of length L, the mixed condition might be Du x (0,t)= β 1 (u (0,t) ψ 1 (t)), Du x (L, t) = β 2 (u (L, t) ψ 2 (t)) for t>0 6.6 Steady State Solution Many PDE models, in particular diffusion problems, have the property that after a long time, they approach a steady state, that is a solution which is no longer evolving with time. In other words, for large t, u (x, t) becomes a function of x only and u t =0. Even if we do not yet know how to solve PDE s, we can, in most cases, find the steady state solution. To do so, we set u t =0 in the diffusion equation. The resulting equation is a second-order linear ODE, which we know how to solve. Remember that the solution should be a function of x only. The example below illustrate how. Example 45 Find the steady state solution of equation 6.1. The steady state solution is reached when u t =0. Thus, we solve u xx =0 The steady state solution is u (x) =ax + b forsomeconstanta and b which can be determined from the boundary conditions. For example, in the case of a rod of length L, ifweknowthatu (0,t)=T 1 and u (L, t) =T 2 for all t>0, thenit follows for the steady state solution that u (0,t)=a (0) + b = T 1 So, And b = T 1 u (L, t) = al + b So = al + T 1 = T 2 al + T 1 = T 2 Thus a = T 2 T 1 L Therefore, the steady state solution is u (x) = T 2 T 1 x + T 1 L

6 42 CHAPTER 6. DIFFUSION - THE HEAT EQUATION 6.7 Problems 1. Show that the diffusion-advection-decay equation u t Du xx + cu x = λu can be transformed into the diffusion equation by the transformation u (x, t) =w (x, t) e αx βt where α = c c2 and β = λ + 2D 4D. 2. Let u = u (x, t) satisfy the PDE model u t = ku xx 0 <x<l, t>0 u (0,t)=u(L, t) =0 t>0 u (x, 0) = u 0 (x) 0 x L (a) Find the steady-state solution. (b) Show that L 0 u (x, t)2 dx L 0 u 0 (x) 2 dx. (hint: LetE (t) = L 0 u (x, t)2 dx and show that E (t) 0). What can be said of u (x, t) if u 0 (x) =0? 3. Consider an initial value problem for the diffusion-decay equation u t Du xx + ru =0 0<x<L, t>0 u (0,t)=0, Du x (L, t) = 1 t>0 u (x, 0) = g (x) 0 <x<l Find the steady-state solution. 4. Heat flow in a metal rod with an internal heat source is given by { u t ku xx =1 0<x<1, t>0 u (0,t)=0, u (1,t)=1 t>0 Find the steady state. Does it matter that no initial condition is given?