Ma221 17F Hw11 SOLN - ADFASDFASDF PDF

Title Ma221 17F Hw11 SOLN - ADFASDFASDF
Author J Gray
Course Differential Equations
Institution Stevens Institute of Technology
Pages 2
File Size 55.7 KB
File Type PDF
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Ma 221

Homework 11 (Solutions)

Due: Dec 7, 2017

1. Complete solution to the IBVP for the Heat Equation. Consider the following initial-boundary value problem (IBVP) modeling heat flow in a wire. (PDE) (BC) (IC)

∂2u ∂u =k 2 , ∂t ∂x

for 0 < x < 2π ,

u (0, t ) = 0, u x (2π , t ) = 0, u(x, 0) = f (x) =

x , 2π

t >0

t >0

0 ≤ x ≤ 2π

(a) Assuming a product solution, u(x, t ) = X (x) · T (t ), use separation of variables to set up an eigenvalue problem for X (x) and a first-order ODE for T (t ). (b) Solve the eigenvalue problem and then construct a general solution u(x, t ) satisfying the PDE and BCs. (c) Calculate the appropriate Fourier Series for the initial data f (x) and use this to complete the solution to the IBVP. Write out the first three non-zero terms in the series solution for u(x, t ). Solution: X ′′(x ) T ′(t ) = −λ = kT (t ) X (x) (EVP) X ′′ + λX = 0, X (0) = 0, X ′(2π ) = 0; T ′ + kλT = 0.

(a) (PDE) X (x) · T ′(t ) = kX ′′(x) · T (t ) ⇒

  (b) The eigenvalues/functions: λn = (2n − 1)2 /16, Xn (x) = sin (2n − 1)x/4 , n = 1, 2, 3, . . . 2

T ′ + kλnT = 0 ⇒ Tn (t ) = e −kλn t = e −k(2n−1) t /16 ∞ Õ Õ   2 Cne −k(2n−1) t /16 sin (2n − 1)x/4 , ⇒ u(x, t ) = Cn Xn (x) · Tn (t ) = n=1

n

where the Cn are free constants.

(c) Applying the initial conditions, u(x, 0) = f (x) =

∞ Õ n=1

  Cn sin (2n − 1)x/4 .

It follows that the Cn are exactly the coefficients of the Fourier sine series for f (x) on (0, 2π ), 2 Cn = 2π



0



1 f (x) sin (2n − 1)x/4 dx = 2π 2 





0



  x sin (2n − 1)x/4 dx .

"  # 2π 2       4 1 4 − sin (2n − 1)x/4 Cn = x cos 2n − 1)x/4 + 2n − 1 2π 2 2n − 1 0 # "  2   n+1   8 · (−1) 4 1 4 (−1)n+1 − 0 = 2 − Cn = (0 − 0) + 2 2n − 1 2π π (2n − 1)2 2n − 1

∞ 8 Õ (−1)n+1 −k (2n−1)2 t /16   e · sin (2n − 1)x/4 2 π n=1 (2n − 1)2     o e −25kt/16 sin 5x/4  e −9kt/16 sin 3x/4 8 n −kt/16  sin x/4 − + u(x, t ) = 2 e − ··· 9 25 π

u(x, t ) =

2. Complete solution to the IBVP for the Wave Equation. Solve the following Initial Boundary Value Problem. (PDE)

u tt = 16 u xx ,

(BC)

for 0 < x < 2,

u(0, t ) = 0, u(2, t ) = 0,

t >0

t >0

sin(πx /2) sin(3πx /2) , 0...


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