Transforming Nonhomogeneous BCs Into Homogeneous Ones

Transcription

1 Chapter 8 Transforming Nonhomogeneous s Into Homogeneous Ones 8.1 Goal In this chapter we look at more challenging example of problems which can be solved by separation of variables. A restriction of the method described in the previous chapter is that both the boundary conditions and the PDE must be homogeneous. In this chapter we will see what can be done if the boundary conditions are not homogeneous. We will handle the case when the PDE is not homogeneous in a separate chapter. 8.2 Introduction In this section, we will address problems of the form PDE u t = α 2 u xx 0 <x< αu x 0,t)+βu0,t)=g 1 t) γu x, t)+δu, t) =g 2 t) IC u x, 0) = φ x) 0 x 8.1) The general idea is that we will perform a change of variable of the form u x, t) = S x, t) +U x, t) where S will also satisfy the boundary conditions. This will ensure that the new boundary conditions are homogeneous. We will prove it below. The general form of S x, t) will be S x, t) =A t) 1 x ) + B t) x 53

2 54CHAPTER 8. TRANSFORMING NONHOMOGENEOUS S INTO HOMOGENEOUS ONES where A t) and B t) will be determined using the fact S must satisfy the boundary conditions. We will rewrite the IBVP 8.1 in terms of U and then solve it. Proposition 51 If u x, t) =S x, t) +U x, t) and S x, t) also satisfies the boundary conditions, then the boundary conditions of the new IBVP written in terms of U will be homogeneous. Proof. We verify it only for the first boundary condition. We have Thus g 1 t) = αu x 0,t)+βu0,t) = α S x 0,t)+U x 0,t)) + β S 0,t)+U 0,t)) = αs x 0,t)+αU x 0,t)+βS 0,t)+βU 0,t) = αs x 0,t)+βS 0,t)+αU x 0,t)+βU 0,t) = g1t)+αu x 0,t)+βU 0,t) since S satisfies the s and this is a homogeneous condition. αu x 0,t)+βU 0,t)=0 The examples below will show that two elements play an important role. They are: 1. Whether the nonhomogeneous part of the s g 1 t) and g 2 t)) arefunctions of t or constants. That will determine in part the exact form S has. It will also determine if the PDE of the new problem is homogeneous or not. If they are constant, then we will see that S is only a function of x and the PDE of the new problem will be homogeneous. 2. Whether the s have a derivative term or not. That will determine in part the exact form S has. Webeginwithasimplecasewhichshouldhelpusunderstandwhyweuse this form for S x, t). 8.3 The Nonhomogeneous Terms Are Constants, no Mixed Boundary Conditions Consider an insulated rod for which the ends boundaries) are kept at constant temperatures T 1 and T 2. Then, the IBVP becomes PDE u t = α 2 u xx 0 <x< u 0,t)=T 1 8.2) u, t) =T 2 IC u x, 0) = φ x) 0 x

3 8.3. THE NONHOMOGENEOUS TERMS ARE CONSTANTS, NO MIXED BOUNDARY CONDITIONS55 Because the s are not homogeneous, we cannot apply the technique used in the previous chapter. You will recall that in many diffusion problems we have discussed so far, we noticed that there was a steady state solution, one which only depended on x for large time. So, it makes sense to think of our temperature u x, t) as the sum of two parts: u x, t) =S x)+u x, t) where S x) is the steady state part of the solution, the one that will remain for large time. U x, t) is the part of the solution which depends on the IC and will eventually go to 0. Experience and many trial and error attempts show that the following expression for S x) will accomplish what we want, that is change our s into homogeneous ones. S x) = T 1 + x T 2 T 1 ) = T 1 1 x ) x + T 2 You will note that this is just the line through the points 0,T 1 ) and, T 2 ). You will also note that S x) satisfies the s. With this value of S x), u x, t) becomes: u x, t) = T 1 + x ) T 2 T 1 ) + U x, t) 8.3) By substituting into equations 8.2, we will get a new problem in U x, t) with homogeneous boundary conditions. We show how this is done in details. For this, we need to compute u t and u xx. u t = U t u x = T 2 T 1 + U x u xx = U xx Also, u 0,t)=T 1 + U 0,t) But from the original, we know that u 0,t)=T 1 Combining the two gives U 0,t)=0 Similarly, u, t) =T 2 + U, t) But from the original, we know that u, t) =T 2

4 56CHAPTER 8. TRANSFORMING NONHOMOGENEOUS S INTO HOMOGENEOUS ONES Combining the two gives Finally, u x, 0) = U, t) =0 But from the original IC, we know that Combining the two gives T 1 + x ) T 2 T 1 ) + U x, 0) u x, 0) = φ x) U x, 0) = φ x) = ψ x) T 1 + x ) T 2 T 1 ) Thus, the IBVP becomes PDE U t = α 2 U xx 0 <x< U 0,t)=0 U, t) =0 IC U x, 0) = ψ x) 0 x This is now a problem we can solve using separation of variables. In fact, we solved it with a different IC) in the previous chapter. The IC only plays a role in the computation of the coefficients A n. Thus, we can use the solution we found. It was nπx ) U x, t) = A n e nπα)2t sin where Therefore where u x, t) = A n = 2 n=0 0 ψ x)sin T 1 + x ) T 2 T 1 ) + A n = 2 0 nπx ) dx A n e nπα)2t sin n=0 ψ x)sin nπx ) dx nπx ) 8.4 The Nonhomogeneous Terms Are Constants, Mixed Boundary Conditions Here, we consider the IBVP PDE u t = α 2 u xx 0 <x< u 0,t)=T1 u x, t)+hu, t) =T 2 IC u x, 0) = φ x) 0 x 8.4)

5 8.4. THE NONHOMOGENEOUS TERMS ARE CONSTANTS, MIXED BOUNDARY CONDITIONS57 We use a similar idea as before, except that it is more complicated this time. The solution we want will be of the form u x, t) =S x, t)+u x, t) where the steady state part, S x, t) is given by S x, t) =A t) 1 x ) + B t) x A t) and B t) are found by requiring that S satisfies the boundary conditions. Requiring that S satisfies the s will ensure that the new problem has homogeneous boundary conditions. This is easy to see. From u x, t) =S x, t)+u x, t), we get that u 0,t)=S 0,t)+U 0,t) We know that u 0,t)=g 1 t). IfS also satisfies the boundary conditions, then S 0,t)=g 1 t) and therefore, the above equation becomes U 0,t)=0 The same is true for the other boundary condition. You will also note that this form of S x, t) is similar to the one in the previous example. The first step is to use the s to find A t) and B t). S must satisfy Also S 0,t)=T 1 S 0,t)=A t) Combining both gives A t) =T 1. To find B t), we use the second boundary condition for S. S x, t)+hs, t) =T 2 We need to compute S x, t). Therefore So, S x x, t) = S x, t) = A t) A t) + B t) + B t) T 2 = S x, t)+hs, t) = A t) + B t) + hb t) = T 1 +1+h) B t)

6 58CHAPTER 8. TRANSFORMING NONHOMOGENEOUS S INTO HOMOGENEOUS ONES Solving for B t) gives T 2 = T 1 +1+h) B t) B t) = T 1 + T 2 1+h hence S x, t) =T 1 1 x ) + T 1 + T 2 x 1+h Note that is is only a function of x. Therefore u x, t) =T 1 1 x ) + T 1 + T 2 x + U x, t) 1+h The next step is to rewrite the problem shown in equation 8.4 as a problem where the unknown function is U. For this, we rewrite the PDE, the s and the IC. Rewriting the PDE. For this, we need to compute d T 1 1 x ) + T ) 1 + T 2 x + U x, t) 1+h u t = dt = U t and so d T 1 1 x ) + T ) 1 + T 2 x + U x, t) 1+h u x = dx = T 1 + T 1 + T h) + U x u xx = = U xx T1 d + T ) 1 + T h) + U x dx So, the new PDE is U t = α 2 U xx Rewriting the s. SinceS satisfies the s, as noted in the introduction,wewillhave U 0,t) = 0 U x, t)+hu, t) = 0

7 8.5. THE NONHOMOGENEOUS TERMS ARE FUNCTIONS OF T 59 Rewriting the IC. φ x) = u x, 0) = S x, 0) + U x, 0) So U x, 0) = φ x) S x, 0) It follows that the new problem is PDE U t = α 2 U xx 0 <x< U 0,t)=0 U x, t)+hu, t) =0 IC U x, 0) = φ x) S x, 0) 0 x This is a problem which can be solved with the separation of variable method. See the next chapter for an example. 8.5 The Nonhomogeneous Terms Are Functions of t We now look at another example in which the s are functions of t. Consider the IBVP PDE u t = α 2 u xx 0 <x< u 0,t)=g 1 t) 8.5) u x, t)+hu, t) =g 2 t) IC u x, 0) = φ x) 0 x We use a similar idea as before, except that it is more complicated this time. The solution we want will be of the form u x, t) =S x, t)+u x, t) where the steady state part, S x, t) is given by S x, t) =A t) 1 x ) + B t) x The unknown functions A t) and B t) will be chosen such that S satisfies the s. As before, requiring that S satisfies the s will ensure that the new problem has homogeneous boundary conditions. You will also note that this form of S x, t) is similar to the one in the previous example. The first step is to use the s to find A t) and B t). S must satisfy S 0,t) = g 1 t) S x, t)+hs, t) = g 2 t)

8 60CHAPTER 8. TRANSFORMING NONHOMOGENEOUS S INTO HOMOGENEOUS ONES Computing S x gives Thus, we get S x x, t) = A t) + B t) A t) = g 1 t) A t) + B t) + hb t) = g 2 t) SincewealreadyhaveA t), wecansolveforb t). ) 1+h B t) = g 2 t)+ A t) = g 2 t)+g 1 t) It follows that B t) = g 1 t)+g 2 t) 1+h And therefore u x, t) =g 1 t) 1 x ) + g 1 t)+g 2 t) x + U x, t) 1+h The next step is to substitute in the original problem and obtain a new problem in terms of U x, t). For this, we first compute u t and u xx. u t = g 1 t) 1 x ) u x = g 1 t) u xx = U xx Thus, the new PDE is or Or, in terms of S g 1 t) 1 x ) + g 1 t)+g 2 t) 1+h + g 1 t)+g 2 t) 1 + h) + U x + g 1 t)+g 2 t) 1+h U t α 2 U xx = g 1 t) 1 x ) U t = α 2 U xx S t x + U t x, t) x + U t α 2 U xx =0 g 1 t)+g 2 t) 1+h Unfortunately, this is no longer homogeneous. We will have to wait a few chapters before we know how to solve this. et us find the new s and IC for this problem, even though we cannot yet solve it. Recall that u x, t) =S x, t)+u x, t) where S was derived so it would solve the s. x

9 8.6. CONCUSION 61 So, we have u 0,t) = S 0,t)+U 0,t) = g 1 t)+u 0,t) We also know that Combining both gives We also know that u 0,t)=g 1 t) U 0,t)=0 u x, t)+hu, t) = S x, t)+hs, t)+u x, t)+hu, t) = g 2 t)+u x, t)+hu, t) We also know that Combining both gives Finally, for the IC. We also know that Combining both gives u x, t)+hu, t) =g 2 t) U x, t)+hu, t) =0 u x, 0) = S x, 0) + U x, 0) u x, 0) = φ x) U x, 0) = φ x) S x, 0) Thus, the new IBVP is PDE U t = α 2 U xx S t 0 <x< U 0,t)=0 U x, t)+hu, t) =0 IC U x, 0) = φ x) S x, 0) 0 x 8.6 Conclusion 1. The techniques of this chapter allow us to transform nonhomogeneous boundary conditions into homogeneous ones. Unfortunately in the process, we might obtain a PDE which is no longer homogeneous. Such a PDE will be solved in a few chapters, using eigenfunction expansions. 2. Obtaining homogeneous s is only relevant for methods which require it. If you solve the IBVP using a technique not requiring it, then it is not necessary to try to obtain homogeneous boundary conditions. We will study such techniques in a few chapters.

10 62CHAPTER 8. TRANSFORMING NONHOMOGENEOUS S INTO HOMOGENEOUS ONES 3. The technique used in the second and third examples also works for the most general nonhomogeneous s such as αu x 0,t)+βu0,t)=g 1 t) γu x, t)+δu, t) =g 2 t) Unfortunately, the resulting PDE will more than likely be nonhomogeneous. 4. For simpler s such as u 0,t)=g 1 t) u, t) =g 2 t) the method of the second example will give us the transformation u x, t) =S x, t)+u x, t) where S x, t) =g 1 t)+ x g 2 t) g 1 t)) You will note that this is very similar to the transformation used in the first example. 5. In summary, if the boundary conditions only involve u, then use u x, t) =S x, t)+u x, t) where S x, t) =g 1 t) 1 x ) + g 2 t) x Where g 1 t) and g 2 t) are given in the boundary conditions. However, if the boundary conditions involve u x,then u x, t) =S x, t)+u x, t) where S x, t) =A t) 1 x ) + x B t) and A t) and B t) are found by requiring that S satisfy the boundary conditions as in the second and third examples). 6. If the nonhomogeneous part of the s is a constant, then S x, t) will only be a function of x which we can write as S x). Thus,S t =0. Furthermore, since the highest poser of x in S x, t) is 1, it follows that S xx =0.Thus, the new PDE will also be homogeneous.

11 8.7. PROBEMS Problems 1. Completely solve the IBVP PDE u t = α 2 u xx 0 <x<1 u 0,t)=1 u x 1,t)+hu 1,t)=1 IC u x, 0) = sin πx + x 0 x 1 Sketch the solution for various values of t and α =1. 2. Completely solve the IBVP PDE u t = u xx 0 <x<1 u 0,t)=0 u 1,t)=1 IC u x, 0) = x 2 0 x 1 What is the steady state solution? 3. Transform the IBVP below into one with homogeneous s. Is the new problem homogeneous? PDE u t = u xx 0 <x<1 u x 0,t)=0 u x 1,t)+hu 1,t)=1 IC u x, 0) = sin πx 0 x 1