Homework #04 Answers and Hints (MATH4052 Partial Differential Equations)

Transcription

1 Homework #4 Answers and Hints (MATH452 Partia Differentia Equations) Probem 1 (Page 89, Q2) Consider a meta rod ( < x < ), insuated aong its sides but not at its ends, which is initiay at temperature = 1 Suddeny both ends are punged into a bath of temperature = Write the differentia equation, boundary conditions, and initia condition Write the formua for the temperature u(x, t) at ater times In this probem, assume the infinite series expansion 1 = 4 ( sin πx + 1 3πx sin + 1 ) 5πx sin + (1) π 3 5 Soution In ideaized cases, we have DE: u t = ku xx ( < x <, < t < ), BC: u(, t) = u(, t) = ( t < ), IC: u(x, ) = 1 ( < x < ) It is a (homogeneous) Dirichet probem To sove it, we separate the variabes u(x, t) = T (t)(x) and derive T kt = = λ = constant Therefore, T (t) satisfies the equation T = λkt, whose soution is T (t) = Ae λkt For simpicity, we set A = 1, so that the initia condition is directy satisfied by (x) Furthermore, = λ in < x < with () = () = Using the boundary conditions, we have λ = n2 π 2, and u(x, t) = A n e ( 2 ) 2 kt sin x, This is the second order inear ODE we have taught about (see Tutoria notes 1&2 for detais) 1

2 where the coefficients are given by the initia condition u(x, ) = Appying the initia condition, we have which is for even n and 4 the given formua In summary, we have A n = A n sin x x u(x, ) sin sin2 x = 2 x sin 2x 1 cos = 2 = 2 sin x cos x = 2 1 cos(), for odd n Aternativey this can be obtained using u(x, t) = n odd 4 n 2 π 2 kt e 2 sin x Probem 2 (Page 89, Q4) Consider waves in a resistant medium that satisfy the probem u tt = c 2 u xx ru t for < x <, (2) u = at both ends, (3) u(x, ) = φ(x), (4) u t (x, ) = ψ(x), (5) where r is a constant, < r < 2πc/ Write down the series expansion of the soution Soution The equation is very simiar to the wave equation, so we can try whether separation of variabes sti work Let u(x, t) = T (t)(x), we have from the same argument For (x), we have T c 2 T + r T c 2 T = = λ = constant = λ in < x < with () = () = 2

3 Just ike before, from the boundary conditions, we have λ n = n2 π 2 so that 2 nontrivia soution exists Let n (x) sove the ODE = λ n and T n (t) sove the ODE T + rt + c 2 λ n T =, then n (x) = sin x, and where s (1) n s (2) Note that if s (1) n = s (2) n Since < r < 2πc/, T n (t) = A n e s(1) n t + B n e s(2) n t, n C are two roots of the quadratic equation s 2 +rs+c 2 λ n = = s n, that is, when n = r 2πc, T n (t) = A n e snt + B n te snt r 2πc N, and the soution can be written as u(x, t) = T n (t) n (x) = A n e s(1) n t + B n e s(2) n where the coefficients are given by the initia condition and u(x, ) = φ(x) = u t (x, ) = ψ(x) = A n + B n sin x, t sin x, A n s (1) n + B n s (2) n sin x Given that φ(x) = P n sin x and ψ(x) = Q n sin x, the coefficients sove a inear system In I n An Pn =, S 1 S 2 B n where S 1, S 2 are diagona matrices with the roots above Since s (1) n s (2) n, the coefficient matrix is of fu rank and the system admits unique soution Q n Remark 1 On the other hand, if m := r 2πc N, u(x, t) = A m e smt + B m te smt sin mπx A n e s(1) n where the coefficients are given by the initia condition u(x, ) = φ(x) = A m sin mπx t + B n e s(2) n A n + B n sin x, t sin x, 3

4 and u t (x, ) = ψ(x) = s m A m + B m sin mπx A n s (1) n + B n s (2) n sin x Now, if s m 1, the system matrix is aso of fu rank, and the soution is unique Actuay, since s m = r 2 <, this is the ony possibe case Probem 3 (Page 92, Q2) Consider the equation u tt = c 2 u xx for < x <, with the boundary conditions u x (, t) =, u(, t) = (Neumann at the eft, Dirichet at the right) 1 Show that the eigenfunctions are cos n + 1 2) πx/ 2 Write the series expansion for a soution u(x, t) Soution Separation of variabe u(x, t) = (x)t (t) yieds T c 2 T = = λ = constant As before, using boundary vaue, we know ony for λ = n)π/2, admits nonzero soution, (x) = C n cos n)πx/ := n(x), where here C n is a arbitrary constant, n =, 1, 2, have, T n (t) = A n cos n)cπt/ + B n sin n)cπt/ Therefore, soution can be written as, u(x, t) = Correspondingy, we C n cos n)πx/{a n cos n)cπt/ + B n sin n)cπt/} n=

5 Probem 4 (Page 92, Q3) Sove the Schrdinger equation u t = iku xx for rea k in the interva < x < with the boundary conditions u x (, t) =, u(, t) = Soution Separation of variabes u(x, t) = (x)t (t) eads to the equation T ikt = = λ = constant, so that T (t) = e ikλt and (using resuts from the previous probem) the eigenvaues are ( ) 2 2n 1 π λ n =, 2 with eigenfunctions n (x) = cos n + 1 ) πx 2 Then the soution can be written as u(x, t) = A n T n (t) n (x) = A n e ikλt cos n + 1 ) πx, 2 with the coefficients given by series expansion of initia conditions u(x, ) = φ(x) = A n cos n + 1 ) πx 2 5