The lecture of 1/23/2013: WHY STURM-LIOUVILLE?

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1 The lecture of 1/23/2013: WHY STURM-LIOUVILLE? 1 Separation of variables There are several equations of importance in mathematical physics that have lots of simple solutions. To be a bit more specific, many equations in which the unknown function u is a function of the variables x and t (x being space, t time) have lots of solutions in which the variables separate; that is solutions of the form u(x, t) = X(x)T (t), where X is just a function of x, T of t. The idea is then (very roughly) to find all such solutions and see if we can combine them to solve BIVP s for the equation. One such equation is the heat equation u tt = ku xx, (1) and I will illustrate the method with it. This equation is linear, homogeneous. An important consequence of being linear is that from a few solutions one gets more because: 1. If u = u(x, t) is a solution of (1), so is cu(x, t) for all constants c. 2. If u 1 = u 1 (x, t) and u 2 = u 2 (x, t) are solutions of (1), so is u 1 (x, t) + u 2 (x, t). 3. Combining these two facts, we get: If u 1 (x, t), u 2 (x, t),..., u n (x, t) are solutions of (1), so is c 1 u 1 (x, t) + c 2 u 2 (x, t) + + c n u(x, t) for all choices of constants c 1, c 2,..., c n. 4. We can usually go even further: If u 1 (x, t), u 2 (x, t), u 3 (x, t),... is a sequence of solutions of (1), then if the constants c 1, c 2,... are such that the series c n u(x, t) converges for all x, t, then u(x, t) = c n u(x, t) has a good chance of being also a solution of (1). 5. u(x, t) 0 (the function that is identically 0) is a solution of (1); mostly a totally useless one, but one that has to be occasionally kept in mind.

2 1 SEPARATION OF VARIABLES 2 A problem: If we do not impose further conditions, we simply end with too many solutions. Too much of a good thing can sometimes be as bad as too little. The ideal situation is have also homogeneous boundary conditions. A simple example of a linear equation with homogeneous boundary conditions is Because the boundary conditions are homogeneous, solutions to this problem, satisfy all the properties of the previous page; a solution to the problem being a function defined for 0 x L, t 0 such that it satisfies the equation and the boundary conditions. Here are the properties repeated for the BVP. 1. If u = u(x, t) is a solution of (2), so is cu(x, t) for all constants c. 2. If u 1 = u 1 (x, t) and u 2 = u 2 (x, t) are solutions of (2), so is u 1 (x, t) + u 2 (x, t). 3. Combining these two facts, we get: If u 1 (x, t), u 2 (x, t),..., u n (x, t) are solutions of (2), so is c 1 u 1 (x, t) + c 2 u 2 (x, t) + + c n u(x, t) for all choices of constants c 1, c 2,..., c n. 4. We can usually go even further: If u 1 (x, t), u 2 (x, t), u 3 (x, t),... is a sequence of solutions of (2), then if the constants c 1, c 2,... are such that the series c n u(x, t) converges for all x, t, then u(x, t) = c n u(x, t) has a good chance of being also a solution of (2). 5. u(x, t) 0 (the function that is identically 0) is a solution of (2); mostly a totally useless one, but one that has to be occasionally kept in mind. (2)

3 1 SEPARATION OF VARIABLES 3 Solving (2); that is, the problem: Main idea: Try to find ALL solutions of (2) that are products of a function of x by a function of t; that is, all solutions of the form u(x, t) = X(x)T (t), where X(x) is defined for 0 x L and T (t) for t 0. Really all? Well, we ignore the useless 0 solution; we try to find ALL non-zero solutions of (2) that are products of a function of x by a function of t. When is a function u(x, t) = X(x)T (t) a solution of (2)? We have: for u to solve the equation we must have u t (x, t) = X(x)T (t), u xx (x, t) = X (x)t (t); X(x)T (t) = kx (x)t (t). And now comes if not the stroke of genius, a clever thing to do: Divide both sides by kx(x)t (t): end we get: X(x)T (t) kx(x)t (t) = kx (x)t (t) kx(x)t (t), cancelling what can be canceled, kx(x) T (t) kx(x) T (t) = X (x)t (t) kx(x)t (t) T (t) kt (t) = X (x) X(x). (3) Lo and behold! (voilà as a French person might say), the variables have been separated. Here s the thing. In equation (3), the variables x, t are independent. We could move t while keeping x fixed. If x is fixed, the right hand side of (3)is fixed, which means that the left side can t move either, how ever much we may move t. In other words, the left hand side must be a constant. But so must the right hand side. In other words, we should have T (t) kt (t) = X (x) = λ, (4) X(x) where λ is a constant. The reason for writing λ rather then λ is technical. Not all values of λ will work once the boundary conditions get into play; writing λ makes the working λ s positive (as we are about to see). Nobody likes negative numbers. It is tradition to use Greek letters for these constants.

4 1 SEPARATION OF VARIABLES 4 We haven t touched the boundary conditions yet. We found that a solution u = XT of u t = ku xx must satisfy T (t) kt (t) = X (x) X(x) = λ where λ is a constant. This breaks up into two equations: X + λx = 0, (5) T + λkt = 0. (6) It is time to invite the boundary conditions to join the party. On the left we have 0 = u(0, t) = X(0)T (t) for all t > 0. Either T (t) = 0 for all t > 0, or X(0) = 0. If T (t) 0, then u(x, t) = X(x)T (t) is always zero, and we didn t do all this work to get what we could get for nothing. So we must demand X(0) = 0. Similarly, from 0 = u(l, t) = X(L)T (t) for all t > 0 we conclude that we must have X(L) = 0. At this point we see: u(x, t) = X(x)T (t) solves the BVP (2) if and only if X, T are non-zero solutions of X + λx = 0, X(0) = 0, X(L) = 0, (7) T + λkt = 0. (8) for some constant λ. Let s concentrate first on (7). It is an example of what is called a Sturm Liouville problem; to be more precise, as we are about to see there are only so many constants λ for which (7) has a non-zero solution; determining all these constants and the corresponding solutions is what is known as a Sturm-Liouville problem. We must determine all values of λ for which (7) has non-zero solutions, and find these non-zero solutions.

5 1 SEPARATION OF VARIABLES 5 We saw: u(x, t) = X(x)T (t) solves the BVP (2) if and only if X, T are non-zero solutions of X + λx = 0, X(0) = 0, X(L) = 0, T + λkt = 0. for some constant λ. We were going to concentrate first on the X part, known as a Sturm-Liouville problem: Determine all λ s for which the problem X + λx = 0, X(0) = 0, X(L) = 0 (7) has non-zero solutions; find those solutions. Can λ be negative? If λ < 0, we can write λ = µ 2, where µ > 0 and the equation is X µ 2 X = 0. From our knowledge of ODE, we know that the general solution of this equation is X(x) = C 1 e µx + C 2 e µx, where C 1, C 2 are constants. For x = 0, we have X(0) = C 1 + C 2, so to satisfy X(0) = 0 we need to have 0 = C 1 + C 2 = 0, or C 2 = C 1. The solution now looks like X(x) = C 1 (e µx e µx ) = 2C 1 sinh(µx). To satisfy now X(L) = 0 we must have 2C 1 sinh(µl) = 0. We assume L > 0 (Heat distribution in a bar of zero length is not of great interest). The only zero of sinh is at zero, so sinh(µl) 0. 2, of course, is not 0. So we need to have C 1 = 0. But then X(x) = 0 for all x! We have the useless zero solution once more. The conclusion is: If λ < 0 there is no non-zero solution to problem (7).

6 1 SEPARATION OF VARIABLES 6 We saw: u(x, t) = X(x)T (t) solves the BVP (2) if and only if X, T are non-zero solutions of for some constant λ. We saw has no non-zero solutions if λ < 0. X + λx = 0, X(0) = 0, X(L) = 0, T + λkt = 0. X + λx = 0, X(0) = 0, X(L) = 0 (7) Can λ be zero? If λ = 0, the equation for X is X = 0; with general solution X(x) = C 1 x + C 2. The condition X(0) = 0 forces C 2 = 0; with C 2 = 0, the condition X(L) = 0 becomes C 1 L = 0. This can only be satisfied if C 1 = 0, and we are again left with the zero solution. If λ = 0 there is no non-zero solution to problem (7).

7 1 SEPARATION OF VARIABLES 7 We saw: u(x, t) = X(x)T (t) solves the BVP (2) if and only if X, T are non-zero solutions of for some constant λ. We saw has no non-zero solutions if λ 0. X + λx = 0, X(0) = 0, X(L) = 0, T + λkt = 0. X + λx = 0, X(0) = 0, X(L) = 0 (7) Can λ be positive? (Our last chance!) If λ > 0, we can write λ = µ 2, where µ > 0 and the equation is X + µ 2 X = 0. Invoking the ODE genie, we know that the general solution of this equation is X(x) = C 1 cos µx + C 2 sin µx, where C 1, C 2 are constants. For x = 0, we have X(0) = C 1, so to satisfy X(0) = 0 we need to have C 1 = 0. The solution now looks like X(x) = C 2 sin µx. To satisfy X(L) = 0 we need to have C 2 sin µl = 0 and this can be achieved without having X(x) become the zero solution! What is necessary, and sufficient, is that µl be a zero of the sine function. sin x = 0 if and only if x = 0, ±π, ±2π, ±3π,.... Since µ > 0 and L > 0, we will have sin µl = 0 if µl = nπ, so µ = nπ L hence λ = n2 π 2, n = 1, 2, 3,.... L2 If λ > 0 there are non-zero solutions of (7) (precisely) for the corresponding solutions are for constants C n. λ = n2 π 2, n = 1, 2, 3,... ; L2 X n (x) = C n sin nπx, n = 1, 2, 3,..., L

8 1 SEPARATION OF VARIABLES 8 (2) We saw: u(x, t) = X(x)T (t) solves the BVP (2) if and only if X, T are non-zero solutions of for some constant λ. We saw X + λx = 0, X(0) = 0, X(L) = 0, T + λkt = 0. X + λx = 0, X(0) = 0, X(L) = 0 (7) has non-zero solutions if and only if λ = n 2 π 2 /L 2, the solutions being C n sin(nπx/l), n = 1, 2, 3,.... The numbers n 2 π 2 /L 2 are called the eigenvalues of the Sturm-Liouville problem (7), any function C sin(nπx/l) being an eigenfunction corresponding to the eigenvalue n 2 π 2 /L 2. In some ways the main one is the one where the constant C is 1; all others are a constant times this one. Having solved (7) we now solve T + λkt = 0 (6) for λ = n 2 π 2 /L 2, n = 1, 2, 3,.... From our journey through ODEland we know these solutions are: T n (t) = A n e n2 π 2 k L 2 t, n = 1, 2,... Putting everything together we found that the solutions we are looking for are of the form A n C n sin nπx n 2 π 2 k L e L 2 t, n = 1, 2, 3,... Constant times constant is just constant; it is silly to have A n C n in front. We replace the product of the two with a single constant and state our almost final result. All the non-zero solutions of (2) of the form u(x, t) = X(x)T (t) are given by u n (x, t) = b n sin nπx n 2 π 2 k L e L 2 t for n = 1, 2, 3, 4,..., arbitrary constants b 1, b 2, b 3,....

9 1 SEPARATION OF VARIABLES 9 The story concluded We found that all the non-zero solutions of (2) of the form u(x, t) = X(x)T (t) are given by u n (x, t) = b n sin nπx L n 2 π 2 k e L 2 t for n = 1, 2, 3, 4,..., arbitrary constants b 1, b 2, b 3,.... Now we can try to get a more general type of solution if we remember that all is linear, homogeneous. We can combine all these solutions into u(x, t) = b n sin nπx L n 2 π 2 k e L 2 t, (9) which, if it converges, is also a solution. We have found, it turns out, the most general solution of (2). The b n s are free, any choice will do as long as the series converges, giving us a massive amount of solutions. All this is very nice (depending on your aesthetic criteria, of course) but what one REALLY wants to solve is a IBVP: u(l, t) = 0, t > 0, u(x, 0) = f(x) 0 < x < L, where f(x) is a prescribed function. The function in (9) satisfies all but the initial condition, so the big question is: can we select the b n s so the initial condition holds? Yes, we can! Setting t = 0 and equating we see that we must select the b n s so that f(x) = b n sin nπx L. (11) In other words, we must expand f(x) into a sine series! Having paid a lot of attention to the course so far, we see at once that the choice is b n = 2 L f(x) sin nπx dx, n = 1, 2, 3, 4, 5,.... (12) L 0 L Example. Solve the BVP Solution. and the solution is In this case, u t (x, t) = 3u xx (x, t), 0 < x < 1, t > 0, u(1, t) = 0, t > 0, u(x, 0) = x 0 < x < L, b n = u(x, t) = 2 π x sin nπx dx = 2 ( 1)n πn = 2( 1)n 1 nπ ( 1) n 2 n e 3n π 2t sin nπx This series converges like the Fourier series it is for t = 0, but for t > 0 the negative exponentials kick in and convergence is super fast. (10)