Solving Nonhomogeneous PDEs (Eigenfunction Expansions)

Transcription

1 Chapter 1 Solving Nonhomogeneous PDEs (Eigenfunction Expansions) 1.1 Goal We know how to solve diffusion problems for which both the PDE and the s are homogeneous using the separation of variables method. Unfortunately, this method requires that both the PDE and the s be homogeneous. We also learned how to apply certain transformations so that nonhomogeneous s are transformed into homogeneous ones. Unfortunately, these transformations may in some cases, transform the PDE into a nonhomogeneous one. To complete the set of tools we have to solve diffusion problems, we must learn how to handle nonhomogeneous PDEs. We will begin by reviewing some concepts and giving new definitions. Then, we will show how to solve the IBVP PDE u t = α 2 u xx + f (x, t) <x<1 <t< α 1 u x (,t)+β 1 u (,t)= <t< (1.1) α 2 u x (1,t)+β 2 u (1,t)= IC u (x, ) = φ (x) x 1 by finding a series solution of the form u (x, y) = T n (t) X n (t) where X n (t) are the eigenfunctions we find when solving the associated homo- 73

2 74CHAPTER 1. SOLVING NONHOMOGENEOUS PDES (EIGENFUNCTION EXPANSIONS) geneous problem PDE u t = α 2 u xx <x<1 <t< α 1 u x (,t)+β 1 u (,t)= <t< α 2 u x (1,t)+β 2 u (1,t)= IC u (x, ) = φ (x) x 1 and T n (t) are functions which can be found by solving a sequence of ODEs. 1.2 Outline Recall that the solution of 1.2 is of the form u (x, t) = A n e ( λnα)2t X n (x) Where λ n and X n (x) are the eigenvalues and eigenfunctions of the problem X + λ 2 X = α 1 X () + β 1 X () = α 2 X (1) + β 2 X (1) = For the problem in 1.1, we will look for a solution of the form u (x, t) = T n (t) X n (x) (1.2) The physical reason for this is that without f (x, t), there is no heat source, so it is normal to expect the temperature to decrease with time, hence the damping term e ( λnα)2t. With a heat source, temperature will no longer decrease, hence we might expect the part which depends on t to be different. 1.3 General Idea We illustrate this method with the following nonhomogeneous IBVP: PDE u t = α 2 u xx + f (x, t) <x<1 <t< u (,t)= <t< u (1,t)= IC u (x, ) = φ (x) x 1 The general idea is to decompose f (x, t) into simple components f (x, t) =f 1 (t) X 1 (x)+f 2 (t) X 2 (x) f n (t) X n (x)+... (1.3) and find the response u n (x, t) =T n (t) X n (x) to each of these individual components. The solution to our problem will then be u (x, t) = u n (x, t)

3 1.3. GENERAL IDEA 75 We will break the procedure of solving this problem into several steps. Step 1 Find the functions X n (x). It turns out that the functions X n (x) are the eigenfunctions of the associated homogeneous problem when we solve it by separation of variables. We derive this problem one more time. The associated homogeneous PDE is u t = α 2 u xx. Ifwelookfor asolutionoftheformu (x, t) =T (t) X (x). Replacing in the PDE gives T (t) X (x) =α 2 T (t) X (x). Dividing each side by α 2 T (t) X (x) gives T (t) α 2 T (t) = X (x) X (x). We concluded these had to be equal to a negative constant we called λ 2. Thus, to find X, wesolvethesecondorderode X + λ 2 X =. For X, the boundary condition meant that X () = and X (1) =. Thus, we see that finding X amounts to solving the initial value problem X + λ 2 X = X () = X (1) = When we do so, we say that we are finding the eigenfunctions of this problem. You will recall that the solutions are X (x) =A sin λx+b cos λx. Using the boundary conditions gives = X () = B, sothatx (x) = A sin λx. Also, we have =X (1) = A sin λ. It follows that we must have sin λ =which means that λ = nπ for n =1, 2, 3, 4,... If we call λ n = nπ for each n =1, 2, 3, 4,... then we have X n (x) =sinnπx. Note that we omitted the constant, it will be part of the other components f n (t). Step 2 Find the functions f n (t). So far, we have f (x, t) =f 1 (t)sinπx + f 2 (t)sin2πx f n (t)sinnπx +... To find f n (t) we simply multiply each side by sin mπx and integrate from to 1 with respect to x. We have already used this method. We will have f (x, t)sinmπxdx = f n (t) = 1 2 f m (t) sin mπxdx sin nπxdx Thus f n (t) =2 f (x, t)sinnπxdx (1.4) Step 3 Find the response u n (x, t) = T n (t) X n (x). nonhomogeneous term f (x, t) by its decomposition Wecanreplacethe f (x, t) = f n (t)sinnπx

4 76CHAPTER 1. SOLVING NONHOMOGENEOUS PDES (EIGENFUNCTION EXPANSIONS) and we try to find the individual responses u (x, t) = T n (t)sinnπx So,wehavetofindthefunctionsT n (t) which solve the IBVP 1.3. If we replace u in that problem with the expression we have, we obtain PDE T n (t)sinnπx = α 2 (nπ) 2 T n (t)sinnπx + f n (t)sinnπx <x<1 <t< IC T n (t)sin= T n (t)sinnπ = T n () sin nπx = φ (x) x 1 The s do not give us any information, they simply say =. We are left with [ ] T n (t)+(nπα) 2 T n (t) f n (t) sin nπx = T n () sin nπx = φ (x) Thus, T n must satisfy the initial value problem { T n (t)+(nπα) 2 T n (t) f n (t) = T n () = 2 φ (x)sinnπxdx Let a n =2 φ (x)sinnπxdx. ThisisafirstorderlinearODEwhichcan be solved using the integration factor technique. Recall, if we multiply each side of the ODE by e (nπα)2t,weobtain T n (t) e (nπα)2t +(nπα) 2 T n (t) e (nπα)2t = f n (t) e (nπα)2 t which is ( T n (t) e (nπα)2 t ) = fn (t) e (nπα)2 t Integrating from to t on each side, we get (T ) t n (τ) e (nπα)2 τ dτ = f n (t) e (nπα)2t dτ T n (t) e (nπα)2t T n () = f n (τ) e (nπα)2τ dτ <t<

5 1.4. A SPECIFIC PROBLEM 77 Recall we set a n = T n (), sowehave T n (t) = a n e (nπα)2t + e (nπα)2 t = a n e (nπα)2t + Thus, the solution to the IBVP 1.3 is u (x, t) = T n (t)sinnπx = f n (τ) e (nπα)2τ dτ f n (τ) e (nπα)2 (τ t) dτ a n e (nπα)2t sin nπx + sin nπx f n (τ) e (nπα)2 (τ t) dτ This shows in particular that the temperature in the rod is due to two parts. One comes from the initial condition. The other one from the heat source. 1.4 A specific Problem We now apply the above procedure to a specific example. Consider the following IBVP: PDE u t = α 2 u xx +sin3πx <x<1 <t< u (,t)= <t< u (1,t)= IC u (x, ) = sin πx x 1 The eigenfunctions X n (x) depend on the corresponding homogeneous PDE and the s. Since they are the same in this problem as in the previous one, the X n (x) will be the same. Thus, we have to compute the coefficients T n (t) in the expansion u (x, t) = T n (t)sinnπx. Using our work from the previous example, we see that T n (t) must satisfy And T n +(nπα) 2 T n = f n = 2 sin 3πx sin nπxdx { if n 3 = 1 if n =3 T n () = 2 sin πx sin nπxdx { if n 1 = 1 if n =1

6 78CHAPTER 1. SOLVING NONHOMOGENEOUS PDES (EIGENFUNCTION EXPANSIONS) If we write these equations for each n, wehave T (n =1) 1 +(πα) 2 T 1 = T 1 () = 1 T (n =2) 2 +(2πα) 2 T 2 = T 2 () = T (n =3) 3 +(3πα) 2 T 3 =1 T 3 () = T (n 4) n +(nπα) 2 T n = T n () = Solution for n =1 T 1 (t) =Ae (πα)2t.sincet 1 () = 1, it follows that A =1. Thus, T 1 (t) =e (πα)2t. Solution for n =2 T 2 (t) =Ae (2πα)2t. Since T 2 () =, it follows that A =. Thus,T 2 (t) =. Solution for n =3 We use the integrating factor technique. We get T 3 (t) = 1 ( ) (3πα) 2 1 e (3πα)2 t. Solution for n 4 T 2 (t) =Ae (nπα)2t.sincet n () =, it follows that A =. Thus,T n (t) =. Thus, we see that the solution is u (x, t) =e (πα)2t sin πx Problems 1 ( ) (3πα) 2 1 e (3πα)2 t sin 3πx 1. In the last example, find the solution in the case n =3. 2. Solve the problem PDE u t = u xx +sinπx +sin2πx <x<1 <t< u (,t)= <t< u (1,t)= IC u (x, ) = x 1 3. Solve the problem PDE u t = u xx +sinπx <x<1 <t< u (,t)= <t< u (1,t)= IC u (x, ) = 1 x 1

7 1.5. PROBLEMS Solve the problem PDE u t = u xx +sinλ 1 x <x<1 <t< u (,t)= <t< u x (1,t)+u(1,t)= IC u (x, ) = x 1 where λ 1 is the first root of the equation tan λ = λ. eigenfunctions X n in this problem? What are the 5. Solve the problem PDE u t = u xx <x<1 <t< u (,t)= <t< u (1,t)=cost IC u (x, ) = x x 1 by: (a) Transforming it to one with homogeneous s. (b) Solving the resulting problems using the techniques of this chapter.