Partial Differential Equations Formula Sheet 301w4f

FORMULA SHEET

a) The solution of the following problem :  2 2 ∂ u 2∂ u   = c   ∂x2  ∂t2 u(0, t) = u(L, t) = 0, ∀ t   u(x, 0) = f (x)    ut (x, 0) = g(x), ∀ x ∈ [0, L] for a string of length L is nπx X u(x, t) = Bn cos(λn t) + Bn∗ sin(λn t) sin , L n≥1 cnπ , L Z nπx 2 L Bn = f (x) sin dx , n = 1, 2, . . . and L 0 L Z L nπx 2 ∗ Bn = dx, n = 1, 2, . . . g(x) sin cnπ 0 L

with λn =

b) The solution of the following problem :  ∂u ∂ 2u   = c2 2  ∂t ∂x u(0, t) = u(L, t) = 0    u(x, 0) = f (x) for a thin metal bar of length L is nπx X 2 u(x, t) = Bn sin e−λn t , L n≥1 with cnπ L Z nπx 2 L Bn = f (x) sin dx, n = 1, 2, . . . L 0 L λn =

c) The solution of the following equation : following rectangle R 1

∂ 2u ∂ 2u + = 0 in the ∂x2 ∂y 2

2

FORMULA SHEET

such that u(x, y) = f (x) on the upper side and 0 on the other sides of R , is :

u(x, y) =

X

Bn sin

nπx a

n≥1

with Bn = a sinh

2 nπb

Z

sinh

a

f (x) sin

nπy a

nπx

0

a

,

dx, n=1,2,. . .

a d) The solution of the following problem

 2 2 ∂ 2u ∂ u  2 ∂ u  = c +   2  ∂x2 ∂y 2  ∂t u = 0 on the boundary   u(x, y, 0) = f (x, y)    u (x, y, 0) = g(x, y) t

considered in the following rectangular membrane :

FORMULA SHEET

is u(x, y, t) =

XX m≥1 n≥1

3

mπx nπy ∗ Bmn cos(λmn t)+Bmn sin(λmn t) sin sin , a b

with Bmn

4 = ab

Z bZ 0

mπx

nπy f (x, y) sin sin dxdy a b 0 Z bZ a mπx nπy g(x, y) sin sin dxdy a b 0 0

4 abλmn r m2 n2 = cπ + 2 , m = 1, 2, . . . and n = 1, 2, . . . a2 b

∗ = Bmn

and λmn

a

Laplace transform 1 a) L{1} = s b) L{tn } = c) L{eat } =

n! sn+1

, n = 1, 2, 3, . . .

1 s−a

k + k2 s e) L{cos(kt)} = 2 s + k2 f ) L{f (t − a)U (t − a)} = e−as F (s), where F (s) = L{f (t)}. g) If f , f 0 ,. . .,f (n−1) are continuous on [0, ∞) and if f (n) (t) is piecewise continuous on [0, ∞), then d) L{sin(kt)} =

s2

L{f (n) (t)} = sn F (s) − sn−1 f (0) − sn−2 f 0 (0) − . . . − f (n−1) (0), where F (s) = L{f (t)}.