MTH203: Assignment-14

1. (a) Using separation of variables, we get

u(x, t) =∞∑

n=1

an exp(−λ2nt) sin(nπx/2), λn = nπ

Initial condition gives

∞∑n=1

an sin(nπx/2) = sinπx

2+ 3 sin

5πx

2

Using Fourier methods or by inspection we find a1 = 1, a5 = 3 and rest of the an

are zero.

Thus,

u(x, t) = sin(πx/2) exp(−π2t) + 3 sin(5πx/2) exp(−25π2t)

(b) The boundary conditions suggest u(x, t) to be of the form

u(x, t) =∞∑

n=1

un(t) sin(nπx/10)

Substituting into the governing equation we get

u′n = [1− (nπ/10)2]un =⇒ un(t) = an exp([1− (nπ/10)2]t)

Thus

u(x, t) =∞∑

n=1

an exp([1− (nπ/10)2]t) sin(nπx/10)

Using initial condition,

∞∑n=1

an sin(nπx/10) = 3 sin(2πx)− 7 sin(4πx)

Comparing a20 = 3, a40 = −7 and the rest of the an’s are zero. Thus

u(x, t) = 3 exp([1− (2π)2]t) sin(2πx)− 7 exp([1− (4π)2]t) sin(4πx)

(c) Using separation of variables, the solution of the heat equation satisfying the BCs

is

u(x, t) =a0

2+

∞∑n=1

an exp[−(nπ/2)2t] cos(nπx/2)

Now

a0

2+

∞∑n=1

an cos(nπx/2) = u(x, 0) =⇒ a0 = 1, an = − 2

nπsin(nπ/2), n ≥ 1

Thus,

u(x, t) =1

2− 2

π

∞∑n=1

exp[−(nπ/2)2t]

ncos(nπx/2)

(d) Using separation of variables, the solution of the heat equation satisfying the BCs

is

u(x, t) =a0

2+

∞∑n=1

an exp[−(nπ/2)2t] cos(nπx/2)

Nowa0

2+

∞∑n=1

an cos(nπx/2) = u(x, 0) =1

2+

cos 2x

2

Comparing a0 = 1, a4 = 1/2 and the rest of the an’s are zero. Thus,

u(x, t) =1

2+

1

2exp[−4π2t] cos(2πx)

(e) Using separation of variables, the solution of the heat equation satisfying the BCs

is

u(x, t) =a0

2+

∞∑n=1

an exp[−(nπ/2)2t] cos(nπx/2)

Now

a0

2+

∞∑n=1

an cos(nπx/2) = u(x, 0) =⇒ a0 = 1, an =16 cos(nπ/2) sin2(nπ/4)

n2π2, n ≥ 1

Thus,

u(x, t) =1

2+

16

π2

∞∑n=1

cos(nπ/2) sin2(nπ/4)

n2exp[−(nπ/2)2t] cos(nπx/2)

2. Let w(x, t) = u(x, t)− (1− x/L)T1 − (x/L)T2. Then w(x, t) satisfies

wt = wxx, 0 < x < L

subject to w(0, t) = w(L, t) = 0 and w(x, 0) = f(x) − (1 − x/L)T1 − (x/L)T2 ≡ F (x).

Proceeding as before, we get

w(x, t) =∞∑

n=1

an exp[−(nπ/L)2t] sin(nπx/L),

where

an =2

L

∫ L

0

F (x) sin(nπx/L) dx

Thus

u(x, t) = (1− x/L)T1 + (x/L)T2 +∞∑

n=1

an exp[−(nπ/L)2t] sin(nπx/L).