Analysis III (BAUG) Assignment 3 Prof. Dr. Alessandro Sisto Due 13th October 2017

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1 Analysis III (BAUG Assignment 3 Prof. Dr. Alessandro Sisto Due 13th October 2017 Question 1 et a 0,..., a n be constants. Consider the function. Show that a 0 = 1 0 φ(xdx. φ(x = a 0 + Since the integral is additive, we have 0 a 0 + N ( nπx a n cos dx = It is easily seen that a 0 = 1 for each n > 0. 0 cos N ( nπx a n cos 0 a 0 dx + N ( nπx a n cos dx. a 0 0dx, thus we only need to show that cos ( nπx 0 dx = 0 ( nπx [ ( nπx ] dx = nπ sin = 0 0 = Question 2 Consider the following IBVPs: { PDE : ux (x, t + e (1 t u xt (x, t = 0 for 0 < x < 3 and t > 0 BC : u(0, t = 0, u(3, t = 0 for t 0 { PDE : ux (x, t + e (2 t u xt (x, t = 0 for 0 < x < 3 and t > 0 BC : u(0, t = t, u(3, t = t 2 for t 0 and suppose v(x, t satisfies (1 and w(x, t satisfies (2. (i Does v(x, t + w(x, t satisfy (1? No, because for example v(0, t + w(0, t = 0 + t 0 for t > 0. 1

2 (ii Does v(x, t + w(x, t satisfy (2? Yes. Since v(x, t and w(x, t satisfy the same homogeneous PDE as in (2, then so does v(x, t + w(x, t. Moreover, since v(0, t + w(0, t = 0 + t = t and v(3, t + w(3, t = 0 + t 2 = t 2 for all t 0, v(x, t + w(x, t also satisfies the boundary conditions of (2. (iii Does αv(x, t + βw(x, t satisfy (2 for any constants α and β? No, because for example αv(0, t + βw(0, t = 0 + βt t for t > 0 and β 1. (iv Does αv(x, t + w(x, t satisfy (2 for any constant α? Yes. Since v(x, t and w(x, t satisfy the same homogeneous PDE as in (2, then so does αv(x, t + w(x, t for any constant α. Moreover, since αv(0, t + w(0, t = 0 + t = t and αv(3, t + w(3, t = 0 + t 2 = t 2 for all t 0, αv(x, t + w(x, t also satisfies the boundary conditions of (2 for any constant α. Question 3 For each of the corresponding physical situations, write the corresponding IBVP and solve it for the given initial condition. A Consider a laterally insulated rod of length 3, parametrised using the coordinate x with x (0, 3, and with diffusion coefficient α 2 = 4. Suppose heat flows in this rod starting with an initial temperature distribution given by φ(x for x (0, 3 whilst both ends have their temperature fixed at 0 degrees at all times. Write the corresponding IBVP. PDE : u t (x, t = 4u xx (x, t for 0 < x < 3 and t > 0 BC : u(0, t = 0, u(3, t = 0 for t 0 IC : u(x, 0 = φ(x for 0 x 3 2

3 Solve the IBVP in the case where φ(x = 10 sin(2πx 4 sin(4πx. In this problem we have α 2 = 4, = 3 and φ(x = 10 sin(2πx 4 sin(4πx. Thus, from the formulas seen in the lectures we conclude that the function u(x, t = 10e 16π2t sin (2πx 4e 64π2t sin (4πx is the unique solution to the IBVP. B Consider a laterally insulated rod of length 4, parametrised using the coordinate x with x (0, 4, and with diffusion coefficient α 2 = 4. Suppose heat flows in this rod starting with an initial temperature distribution given by φ(x for x (0, 4 whilst both ends are insulated, meaning that they do not allow any heat to pass. Write the corresponding IBVP. PDE : u t (x, t = 4u xx (x, t for 0 < x < 4 and t > 0 BC : u x (0, t = 0, u x (4, t = 0 for t 0 IC : u(x, 0 = φ(x for 0 x 4 Solve the IBVP in the case where φ(x = 5 2 cos ( πx cos(πx. In this problem we have α 2 = 4, = 4 and φ(x = 5 2 cos ( πx 4 +3 cos(πx. Thus, from the formulas seen in the lectures we conclude that the function u(x, t = 5 2e ( ( π t πx cos + 3e (π2 2 2t cos (πx 4 ( = 5 2e π2 πx 4 t cos + 3e 4π2t cos (πx 4 is the unique solution to the IBVP. C Consider an insulated circular wire of length 10, parametrised using the coordinate x with x ( 5, 5, and with diffusion coefficient α 2 = 16. Suppose heat flows in this wire starting from an initial temperature distribution given by φ(x for x ( 5, 5. Write the corresponding IBVP. 3

4 PDE : u t (x, t = 16u xx (x, t for 5 < x < 5 and t > 0 BC : u( 5, t = u(5, t, u x ( 5, t = u x (5, t for t 0 IC : u(x, 0 = φ(x for 5 x 5 Solve the IBVP in the case where φ(x = 3 4 sin(3πx + 30 cos(2πx. In this problem we have α 2 = 16, = 5 and φ(x = 3 4 sin (3πx + 30 cos(2πx. Thus, from the formulas seen in the lectures we conclude that the function u(x, t = 3 4e (3π2 4 2t sin (3πx + 30e (2π2 4 2t cos (2πx = 3 4e 144π2t sin (3πx + 30e 64π2t cos (2πx is the unique solution to the IBVP. Question 4 The goal of this exercise is to derive the solution of the following IBVP: PDE : u t (x, t = u xx (x, t for 0 < x < and t > 0 BC : u x (0, t = 0, u(, t = 0 for t 0 IC : u(x, 0 = φ(x for 0 x for a suitable initial distribution φ(x. (i Use the Ansatz u(x, t = f(xg(t and proceed as in the lecture (for a more precise reference, ecture 2, part I to obtain a non-zero solutions of the above PDE that depends on 3 parameters c 1, c 2, c 3. If u(x, t = f(xg(t then u t (x, t = f(x dg (t = dt f(xg (t and u xx (x, t = d 2 f (xg(t = f (xg(t. Thus, since u dt 2 t (x, t = u xx (x, t it follows that f(xg (t = f (xg(t and hence g (t = f (x. g(t f(x Since the left-hand side only depends on t and the right-hand side only depends on x, it follows that both sides must equal a constant K. Hence, we conclude 4

5 that Kf(x = f (x and Kg(t = g (t for some constant K. Now, we have to differentiate cases depending on K. We know form previous courses that g(t = c 1 e Kt is a solution for Kg(t = g (t. Now consider Kf(x = f (x. As in the lecture (ecture notes 2, part 2, step I, we have different cases depending on the sign of K. If K = 0, then f (x = 0 and g(t = c 1. Proceeding as in the lecture notes, we get that this would give the zero-solution. If K > 0, then there is λ such that K = λ 2. We know from previous courses that f(x = c 2 e λx + c 3 e λx is a solution for λ 2 f(x = f (x, for some constants c 2 and c 3. Since, in this case, we can write g(t = c 1 e λ2t, we get that a solution in this case is u(x, t = c 1 e λ2 t ( c 2 e λx + c 3 e λx. If K < 0, then again there is λ such that K = λ 2. The general solution of this ODE is f(x = c 2 sin(λx + c 3 cos(λx for some constants c 1 and c 2. Again, we obtain that a solution of the PDE is u(x, t = c 1 e λ2x (c 2 sin(λx + c 3 cos(λx. (ii Plug in the expressions for f and g obtained into the boundary conditions and obtain conclusions on the parameters c 1, c 2, c 3. Remember that we are only interested in non-zero solutions. Since u x (x, t = f (xg(t and u x (0, t = 0 from the boundary conditions, we conclude that f (0g(t = 0 for all t 0. This implies that one of the following two cases holds: g(t = 0 for every t 0 f (0 = 0 If g(t = 0 for every t 0 then it follows that u(x, t = f(xg(t 0. Since we are interested in the non-zero solutions, let s assume that this is not the case. We recall that at the moment we have two possible solutions, depending on the value of K. If K = λ 2, for λ > 0, then f(x = c 2 e λx +c 3 e λx. Imposing the condition f (0 = 0 we get c 2 λ c 3 λ = 0. Since λ > 0, we conclude that c 2 = c 3. 5

6 From u(, t = 0 for each t, we get f(g(t = 0 for each t. As before, we can assume that f( = 0, otherwise we will find the zero solution. We get f( = c 2 (e λ + e λ = 0, thus c 2 = 0. In particular, this implies that every solution with K > 0 has to be the zero solution. Thus we can only consider the case K < 0. Imposing the condition f (0 = 0, we get Thus, since λ > 0, we have c 2 = 0. (c 2 λ cos(0 c 3 sin(λ0 = 0. Imposing the condition f( = 0, we get c 3 cos(λ = 0. Since cos(x = 0 if and only if x = ( n π for some integer n, we conclude that for some integer n. λ = ( n + 1 ( π λ = n + 1 π 2 2 Finally, note that if c 1 = 0 then g(t 0 and so u(x, t 0 as well. Thus, we have showed that either u(x, t 0, or u(x, t = c 1 c 3 e λ 2 t (cos (λx, where λ = ( n π, for some integer n. (iii Conclude that for each integer n the function u n (x, t = e ((n+ 1 2 π 2t cos (( n + 1 πx 2 satisfies the PDE and the boundary conditions of the IBVP above. By taking c 1 = 0, c 2 = c 3 = 1 and λ = ( n π c we get f(x = cos (( n πx g(t = e ((n+ 1 2 π 2 t in the expressions obtained in and so in that case u(x, t is exactly the function u n (x, t defined above. We conclude that u n (x, t satisfies the PDE and the boundary conditions of the IBVP. 6

7 (iv Using the superposition principle deduce that if φ(x = N a n cos (( n + 1 πx 2, for some constants a 1,..., a N and for some positive integer N, then u(x, t = satisfies the IBVP above. N (( a n e ((n+ 1 2 π 2t cos n + 1 πx 2 We know that the functions u n (x, t satisfy the PDE and the boundary conditions of the IBVP above. Note that we have: u n (x, 0 = cos (( n πx. Thus, the function u n (x, t satisfies the IBVP above if the initial condition is given by the function (( φ n (x = cos n + 1 πx. 2 Since the PDE is linear and homogeneous and the boundary conditions are homogeneous we conclude from the superposition principle that if the initial condition φ(x is of the form N φ(x = a n φ n (x then the function u(x, t u(x, t = N a n u n (x, t satisfies the corresponding IBVP. Given the defitions of u n (x, t and φ n (x, this is exactly what we were asked to deduce, hence we are done. (v Can you conjecture which heat problem this IBVP models? Use the interpretation of the BCs used in the examples you saw in lectures. This IBVP can model the heat problem where we have a laterally insulated rod of length with diffusion coefficient α 2 = 1 where one end is insulated, the other is kept fixed at temperature 0 and heat flows starting from an initial temperature distribution given by φ(x. 7

8 Question 5 1. Indicate which of the following PDEs is linear: (i 10xu xt + e x+t u x u = 0 (ii u xx + t 100 u tt x 3 u x = e xt u t (iii u xxy 20u xyy + 100u xx + e x2 u = x + 3y (iv u xx + 30u xy + 2u x u y = x 10 + u 2 The first PDE is not linear because of the non-linear term u x u. The fourth PDE is not linear because of the non-linear terms u x u y and u 2. The second and third PDEs are linear because all the terms consist of u, its partial derivatives or 1 multiplied by functions independent of u or its derivatives. 2. Consider the following PDE: u xx + 2u xy + 2u yy + u y = 0 This PDE is inear/nonlinear? Homogenous/not homogenous? Parabolic/elliptic/hyperbolic? [Exam question, ] The PDE given is linear because it can be written in the form: Au xx + Bu xy + Cu yy + Du x + Eu y + F u = G where A, B, C, D, E, F and G are functions independent of u and its partial derivatives. Moreover, since G = 0, the PDE is homogeneous. Finally, since B 2 4AC = = 4 < 0, the PDE is elliptic. 3. Which of the following PDEs are homogeneous? (Check all that apply. (i u xy + e 1+y2 u x + u = 2 (ii u xy u yz + 100u x 2u y = 30u (iii u xxx + e x2 u xyx y 2 u x = 0 (iv ye x u xxy + u xy 10u y = e x+y3 (v u xy + x 3 yu xx + log(1 + x 2 10yu yy + u = 0 8

9 The first PDE has the term 2 which makes it not homogeneous. The fourth PDE has the term e x+y3 which makes it not homogeneous. The fifth PDE has the term log(1 + x 2 (which is not u or one of its partial derivatives multiplied by some function which makes it not homogeneous. The remaining two PDEs are linear combinations of u and its partial derivatives and hence are homogeneous. Question 6 Note! This question is harder then the ones above, but it is more interesting. It is not compulsory to solve it, but we encourage you to try to do at least some parts of it. Consider the following IVBP: PDE u t u xx = 0 for (x, t Ω = [0, ] [0, ; BC u(0, t = u(, t = 0 for all t > 0; IC u(x, 0 = φ(x. The physical intuition tells us that, no matter which is the initial temperature φ(x of the rod, after enough time the temperature will stabilize to 0 degrees. Indeed, for some explicit values of φ(x, we are able to compute an explicit solution and hence compute the limit lim t u(x, t and observe that tends to zero, regardless of the point x. The goal of this exercise is to compute the, so called, solution at infinity for a general (! function φ(x, and show that is, indeed, the constant function 0. Part I et ψ, ψ : [0, ] R be two continuous functions such that, for each x [0, ] one has ψ(x ψ (x. et u be a solution of the IVBP with φ(x = ψ(x and v a solution φ(x = ψ (x. Show that for each (x, t Ω, one has u(x, t v(x, t. By the superposition principle, the function v u satisfy the IBVP with initial condition φ(x = ψ (x ψ(x. In particular, one has that min x [0,] {φ(x} 0 (in fact it is 0 since the temperature at the endpoints is the same. We want to show that for each (x 0, t 0 Ω, one has (v u(x 0, t 0 0, that is v(x 0, t 0 u(x 0, t 0. 9

10 Applying the maximum principle (Question 5 of the last exercise sheet with domain Ω = [0, ] [0, t 0 + 1], we get that (v u(x 0, t 0 min {φ(x} = 0. x [0,] Part II et ψ(x: [0, ] R be a function such that ψ (0 <, ψ ( <. This is a technical condition that can be assumed to be true in any physical system. Show that for u satisfying the above IVBP with φ(x = ψ(x, for each x 0 [0, ], we have lim u(x 0, t = 0. t Consider the function ( π ψ (x = C sin x. Given any continuous function ψ(x such that ψ(0 = ψ( = 0, we can always find a C big enough such that ψ (x ψ(x for all x [0, ]. Indeed, let g 1 (x be the function φ(x. Intuitively, g x 1 measures the slope of a segment in the plane connecting the origin to the point (x, φ(x. It is easily seen that g 1 is continuous and, by hypotheses we have that lim x 0 g 1 (x <. Moreover, g 1 ( = 0. Thus, it has a maximum C 1. This imply that for each x [0, ], we have ψ(x C 1 x. Similarly, define g 2 as the function g 2 (x = ψ(x x. As above, it has a maximum C 2. et C = max{c 1, C 2 }. Then the graph of the function ψ is below the lines emanating from (0, 0 and (0, with slopes C and C respectively. This implies that for each x [0, ], we have that ψ(x C sin ( π x. et v be a solution of the IVBP with φ(x = ψ (x. By part I, we know that for each (x 0, t 0 we have that u(x 0, t 0 v(x 0, t 0. However, we can explicitly compute v and we get that nπ v(x, t = Ce ( 2t ( nπ sin x. Since, for each x 0 we have that lim t v(x 0, t = 0, we get the result. Question 7 The goal of this exercise is to derive the solution of the following IBVP: PDE u t u xx = 0 for (x, t Ω = [0, ] [0, ; 10

11 BC u(0, t = c 1, u(, t = c 2 for all t > 0; IC u(x, 0 = φ(x. (i For a suitable φ, the above IBVP admits a solution u 0 that does not depend on t. Try to guess such an u 0 and verify that satisfy the PDE and BC. Hint: A more formal strategy may be to use the Ansatz u(x, t = f(xg(t and consider the case g (t g(t = K = 0. et u 0 (x, t = ( x c 1 + x c 2. It is easily seen that u 0t = u 0xx = 0, and that the BC are satisfied. Choosing φ(x = ( x c 1 + x c 2, we get that u 0 is a solution of: PDE u t u xx = 0 for (x, t Ω = [0, ] [0, ; BC u(0, t = c 1, u(, t = c 2 for all t > 0; IC u(x, 0 = ( x c 1 + x c 2. (ii Use the superposition principle to solve the above IBVP for all initial conditions φ of the form: N ( nπx φ(x = u 0 (x, 0 + a n sin. et v be a solution of PDE v t v xx = 0 for (x, t Ω = [0, ] [0, ; BC v(0, t = v(, t = 0 for all t > 0; IC v(x, 0 = N a n sin ( nπx. The explicit formula of the lecture gives: v(x, t = N nπ a n e ( 2t ( nπ sin x. We claim that u(x, t = u 0 (x, t+v(x, t is the required solution (of the first IBVP. Indeed, u t u xx = (v t v xx + (u 0t u 0xx = 0, and u(x, 0 = u 0 (x, 0 + v(x, 0. Moreover, u(0, t = u 0 (0, t+v(0, t = c 1 +0 = c 1 and u(, t = u 0 (, t+v(, t = c = c 2. 11