1.5 First Order PDEs and Method of Characteristics

Transcription

1 1.5. FIRST ORDER PDES AND METHOD OF CHARACTERISTICS First Order PDEs and Method of Characteristics We finish this introductory chapter by discussing the solutions of some first order PDEs, more specifically the equations we obtained from the advection model. The remaining chapters will be devoted to the solution of second order, linear PDEs. Before we look at the general first order equation, we look at a simplified version and show how its solution is easy to find Advection Equation With no Source This equation is extremely simple to solve as we will see. Recall, we derived this equation above. Its most general form is: u t + cu x = 0 where u = u (x, t). The term cu x corresponds to the flux of the quantity being studied (see chapter on conservation law). This means that the rate at which the quantity changes (u t ) only depends on advection. In other words, if we are studying the amount of a pollutant in a river, this equation would tell us that the pollutant only depends on the current carrying it. It turns out that solving this equation is not diffi cult. before deriving its solution analytically, let us see what it might be simply by reflecting upon this problem. Consider a river flowing and carrying a pollutant, where the amount of pollutant only comes from the pollutant being carried. There is no pollutant being added or removed. The only motion of the pollutant is that given by the current. The current of the river is uniform. This is exactly the situation this equation is describing. In this case then, the amount of pollutant at a certain place in the river will be the same as what it was upstream a little earlier. This means if we know the amount of pollutant at a certain time and a certain place, then we will know it everywhere in the river. This is captured by the next proposition. Proposition 49 If F is any differentiable function, then F (x ct) is a solution of u t + cu x = 0 for any value of t. Proof. Let u (x, t) = F (x ct) and assume F is differentiable. Then Similarly u t = F (x ct) (x ct) t = cf (x ct) u x = F (x ct) = F (x ct) (x ct) x

2 36CHAPTER 1. INTRODUCTION TO PDES: NOTATION, TERMINOLOGY AND KEY CONCEPTS Thus, we see that u t + cu x = cf (x ct) + cf (x ct) = 0 Such solutions are called right-traveling waves. The graph of F (x ct) is the graph of F (x) shifted ct spatial units to the right which is exactly the distance covered by the river in t seconds if the speed of the current is c. This shows that whatever the shape of the initial solution is, it will travel to the right as time increases. Hence, once we know the initial condition u (x, 0), we also know the solution u (x, t), as indicated below. The pure initial value problem for the advection equation is ut + cu x = 0 t > 0 (1.25) u (x, 0) = u 0 (x) Where u 0 (x) is a given initial density. From proposition 49, it follows that u (x, t) = u 0 (x ct) is the solution to the problem given by equations Example 50 Consider the problem ut + 2u x = 0 t > 0 Then u (x, t) = sin (x 2t) is a solution to this problem. We now look at more complex first order equations for which the method of characteristics we are about to learn will be needed Methods of Characteristics for Linear First Order Linear PDEs A linear first order PDE is of the form α (x, t) u t +β (x, t) u x +γ (x, t) u = g (x, t). Assuming that α (x, t) 0, we can divide each side by it to obtain u t (x, t) + c (x, t) u x (x, t) + a (x, t) u (x, t) = f (x, t) (1.26) We develop the method of characteristics for this equation. More specifically, we consider the initial value problem (IVP) ut (x, t) + c (x, t) u x (x, t) + a (x, t) u (x, t) = f (x, t) t > 0 (1.27) u (x, 0) = u 0 (x) We seek a classical solution that is a function u (x, t) such that the derivatives u t, u x exist and are continuous. The PDE in equation 1.27 is satisfied for all points (x, t) and the initial condition is satisfied for all x.

3 1.5. FIRST ORDER PDES AND METHOD OF CHARACTERISTICS 37 We begin by noting that u t (x, t)+c (x, t) u x (x, t) is the directional derivative of u (x, t) in the direction of the vector (c (x, t), 1) in the (x, t) plane. If we plot these direction vectors in the (x, t) plane, we obtain a direction field. The curves x = X (t) which fit this direction field are the solutions of the ODE X (t) = c (X (t), t) (1.28) The characteristic passing through the point (x 0, 0) satisfies the initial condition X (0) = x 0. Definition 51 The curves which solve equation 1.28 are called characteristic curves or simply characteristics. Given an arbitrary poin t (x, t), we now consider a solution u (x, t) of the IVP given by equation 1.27 and the characteristic curve x = X (t) with X (0) = x 0 which goes through the point (x, t). We restrict u (x, t) to the characteristic curve and define the function Using the chain rule, we see that v (t) = u (X (t), t) v (t) = u x (X (t), t) X (t) + u t (X (t), t) = u x (X (t), t) c (X (t), t) + u t (X (t), t) The last line being obtained from equation Combining equation 1.26 with the one above, we see that v (t) satisfies v (t) + a (X (t), t) v (t) = f (X (t), t) t > 0 (1.29) v (0) = u 0 (x 0 ) If we solve the ODE, we will have solved the PDE. The important point is that we can solve the PDE by solving a family of first order ODEs. The general idea is the following: To solve the PDE at an arbitrary point (x, t) we first find the characteristic which goes through (x, t), we then solve the PDE along the characteristic. We do this because along the characteristic, the PDE becomes a family of first order ODEs. We then write the solution in terms of our arbitrary point (x, t). Before looking at specific examples, we outline the technique to follow. ut (x, t) + c (x, t) u Algorithm To solve the IVP problem x (x, t) + a (x, t) u (x, t) = f (x, t) t > 0 u (x, 0) = u 0 (x) follow the steps below:, Step 1 Find the characteristics, for this solve X (t) = c (X (t), t) with initial condition X (0) = x 0. Since the solution depends on x 0, we will call it X x0 (t).

4 38CHAPTER 1. INTRODUCTION TO PDES: NOTATION, TERMINOLOGY AND KEY CONCEPTS Step 2 For a given point (x, t), find the starting point x 0 of the characteristics through (x, t) in other words, for a given point (x, t), solve the equation X x0 (t) = x for x 0. Call the solution x 0 = p (x, t). v Step 3 With the function X (t) = X x0 (t), solve the IVP (t) + a (X (t), t) v (t) = f (X (t), t) v (0) = u 0 (x 0 ) Since the solutions depend on x 0, we will denote them v x0 (t). Be careful here. Once you have written the ODE, if some of the coeffi cients involve x, x will have to be replaced by what you found in part a for X x0 (t) since we are solving the equation along a characteristic curve where x = X x0 (t). See example 55. Step 4 The solution to our original problem is u (x, t) = v x0 (t) x0=p(x,t). Let us first apply this technique to simple examples. ut + 2u Example 52 Solve the IVP x = 0 t > 0 1. Find the characteristics, for this we solve X (t) = 2 with X (0) = x 0. The solution is X (t) = 2t + C and x 0 = X (0) = C we have X x0 (t) = 2t + x To find x 0, we solve x = X x0 (t) = 2t + x 0 or x 0 = x 2t. 3. Next, we solve the IVP v (t) = 0 with v (0) = sin x 0. The solution to this ODE is v (t) = C hence v x0 (t) = v (0) = sin x 0 4. Replace x 0 by its value found in step 2 that is u (x, t) = v x0 (t) x0=x 5t = sin (x 2t) which is the solution we had found above. ut + tu Example 53 Solve the IVP x = 0 t > 0 u (x, 0) = e x 1. We find the characteristic curves by solving X (t) = t with X (0) = x 0. The solution is X (t) = 1 2 t2 + C and X (0) = x 0 = C. Thus, X x0 (t) = 1 2 t2 + x We find x 0 by solving x = X x0 (t) = 1 2 t2 + x 0. So, x 0 = x 1 2 t2. 3. Next, we solve the IVP v (t) = 0 with v (0) = e x0. The solution of this ODE is v (t) = C hence v x0 (t) = v (0) = e x0 4. Replace x 0 by its value found in step 2 that is x 0 = x 1 2 t2 so u (x, t) = v x0 (t) x0=p(x,t) = ex 1 2 t.

5 1.5. FIRST ORDER PDES AND METHOD OF CHARACTERISTICS 39 Now, we are ready to solve more challenging problems. ut + 5u Example 54 Solve the IVP x + 2u = 0 t > 0 u (x, 0) = e x 1. Find the characteristics, for this solve X (t) = 5 with X (0) = x 0. The solution is X (t) = 5t + C and x 0 = X (0) = C so X x0 (t) = 5t + x We find x 0 by solving for x 0 the equation x = X x0 (t) = 5t + x 0. So x 0 = x 5t. 3. Next, solve the IVP v (t) + 2v (t) = 0 with v (0) = e x0. The ODE can be written as v (t) = 2v (t) hence v (t) = Ce 2t. We find C from the initial condition. e x0 = v (0) = C. So v x0 (t) = e x0 e 2t. 4. Finally, replace x 0 by what we found in step 2 to obtain u (x, t) = v x0 (t) x0=x 5t = e x 5t e 2t = e x 7t ut + xu Example 55 Solve the IVP x + u = 3x t > 0 u (x, 0) = tan 1 x 1. To find the characteristics, we solve X (t) = X (t) with initial condition X (0) = x 0. The general solution of this ODE is X (t) = Ce t and X (0) = C = x 0 thus X x0 (t) = x 0 e t. 2. The next step is to find the starting point x 0 by solving X x0 (t) = x that is x 0 e t. The solution is x 0 = xe t. So, p (x, t) = xe t. 3. Next, we solve the IVP v (t) + v (t) = 3x 0 e t with v (0) = tan 1 (x 0 ). Note that on the right side of the ODE, we have used 3x 0 e t instead of 3x. This is because we are solving the ODE along the characteristic curve where x = X x0 (t) = x 0 e t from part a. We can solve this ODE using the integrating factor technique. For this, we multiply each side by e t to obtain v (t) e t + v (t) e t = 3x 0 e 2t which is the same as (v (t) e t ) = 3x 0 e 2t hence v (t) e t = 3x 0 e 2t dt = 3 2 x 0e 2t + C Thus v (t) = 3 2 x 0e t + Ce t

6 40CHAPTER 1. INTRODUCTION TO PDES: NOTATION, TERMINOLOGY AND KEY CONCEPTS We can get C from the initial condition. On one hand, v (0) = 3 2 x 0 + C but on the other hand, v (0) = tan 1 x 0 thus C = tan 1 x x 0 and v x0 (t) = 3 ( 2 x 0e t + tan 1 x 0 3 ) 2 x 0 e t 4. Now, replace x 0 by what we found in step 2, that is x 0 = xe t and we get u (x, t) = v x0 (t) x0=x 5t = 3 2 x 3 2 xe 2t + e t tan 1 ( xe t) ut + 2u Example 56 Solve the IVP x u = t t > 0 1. Find the characteristics, that is solve X (t) = 2 with X (0) = x 0. The solution is X (t) = 2t + C and x 0 = X (0) = C we have X x0 (t) = 2t + x Find x 0 by solving X x0 (t) = x that is x = 2t + x 0 or x 0 = x 2t. 3. Next, solve the IVP v (t) v (t) = t with v (0) = sin (x 0 ). We can solve this ODE using the integrating factor technique. For this, we multiply each side by e t to obtain v (t) e t v (t) e t = te t which is the same as (v (t) e t ) = te t hence v (t) e t = te t dt = e t (t + 1) + C So v (t) = (t + 1) + Ce t We can find C with the initial condition sin x 0 C = sin x hence = v (0) = 1 + C or v x0 (t) = (t + 1) + e t (sin x 0 + 1) 4. Replace x 0 by what we found in step 2 that is u (x, t) = v x0 (t) x0=x 2t = (t + 1) + e t (sin (x 2t) + 1) 5. The reader will verify that this solves the initial value problem given.

7 1.6. PROBLEMS Conclusion In the previous sections we derived the equation for the very important conservation law. We applied it to derive the equation corresponding to the advection model, a first order PDE. In this section, we looked how to solve some first order PDEs which come from the advection model and in the process introduced the method of characteristics. 1.6 Problems 1. Solve the initial value problem ut + cu x = 0 t > 0 u (x, 0) = e x2 Pick c = 2 and sketch the solution surface using your favorite software at several time snapshots. 2. Solve the initial value problem ut 2u x = 0 t > 0 3. From problems 1 and 2, can you write a formula for the solution of the initial value problem ut + cu x = 0 t > 0 u (x, 0) = u 0 (x) ut + 2tu 4. Solve the IVP x = 0 t > 0 5. Show that the decay term in equation 1.17 can be removed by making a change of the dependent variable to w = ue λt. ut + αtu 6. Solve the IVP x + βu = 0 u (x, 0) = f (x) t > 0 where α and β are constants and f (x) is a given function. 7. Solve the IVP u t + 3xu x = 2t t > 0 u (x, 0) = ln ( 1 + x 2) 8. Solve the IVP ut + u x 3u = t t > 0 u (x, 0) = x 2