HW 23 Solutions
March 26, 2013
10.5.7 Find the solution of the heat conduction problem
100uxx = ut , 0 < x < 1, t > 0;
u(0, t) = 0, u(1, t) = 0, t > 0;
u(x, 0) = sin(2πx) − sin(5πx), 0 ≤ x ≤ 1
Solution: Note from equation 19 that the answer takes the form
u(x, t) =
∞
X
2 π2 t
cn e−100n
sin(nπx).
n=1
Substituting for t = 0 yields
u(x, 0) =
∞
X
cn sin(nπx) = sin(2πx) − sin(5πx).
n=1
Therefore, c2 = 1, c5 = −1, otherwise cj = 0. It follows immediately that
2
2
u(x, t) = e−400π t sin(2πx) − e−2500π t sin(5πx).
10.5.8 Find the solution of the heat conduction problem
uxx = 4ut , 0 < x < 2, t > 0;
u(0, t) = 0, u(2, t) = 0, t > 0;
u(x, 0) = 2 sin(πx/2) − sin(πx) + 4 sin(2πx), 0 ≤ x ≤ 2.
1
Solution: The answer must take the form
u(x, t) =
∞
X
2 π 2 t/16
cn e−n
sin(nπx/2).
n=1
Using the boundary conditions,
u(x, 0) =
∞
X
cn sin(nπx/2) = 2 sin(πx/2) − sin(πx) + 4 sin(2πx)
n=1
from which it follows that c1 = 2, c2 = −1, c4 = 4, else cj = 0.
Plugging in these values yields
u(x, t) = 2e−π
2 t/16
sin(πx/2) − e−π
2 t/4
2
2
sin(πx) + 4e−π t sin(2πx).