Solutions to Partial Differential Equations

Solution
Q1)
∂ u
∂ t = ∂2 u
∂ x2
u(x, t) = (C1cos(px) + C2sin(px))e− p2 C2 t
= (C1cos(px) + C2sin(px))e− p2 π 2 t
As given u(0, t) = 0= C1e− p2 πt
 C1 = 0
u( π ,t ¿=¿C2sin(p π))e− p2 π 2 t
sin(pπ ¿=0
p π=nπ
p= n
u(x, t) = bn*e−n2 πt sin (nx )
where bn = C2
u(x, t) = ∑ b n∗e−n2 πt sin (nx )
= ∑
n =1
∞
bnsin(nx)
Here,
bn = 2
π ∫
0
π
¿ ¿
= 2
π [3 ( sin ( 5−n ) x
2(5−n) − sin ( ( n+5 ) x )
2(n+5) )−11 ( sin ( 4−n ) x
2(4−n) − sin ( n+4 ) x
2(n+ 4) ) ] at limits of π ∧0
= 2
π [3 ( sin ( 5−n ) π
2(5−n) − sin ( ( n+5 ) π )
2(n+ 5) )−11 ( sin ( 4−n ) π
2(4−n) − sin ( n+ 4 ) π
2(n+ 4) ) ]
bn = 6
π [2 ( 5−n )∗2 ( n+5 ) ]− 22
π [2 ( 4−n )∗2 ( n+ 4 ) ]

bn = 6
π (25−n2)− 22
π (16−n2)
take n = 2m-1
u(x, t) = ∑
m=1
∞
¿∗e− (2 m−1 )2 πt∗sin (2 m−1)x ¿ ¿
Q2)
let u= XT
XT11 = kX11T
T11
T =k X 11
X
Let m = 0, T11 = 0=> T = a t + b
X11 = ux => x = Ae √ux+ B e− √ux
u(x, t) = (at + b)( Ae √ ux+ B e− √ ux ¿
u(x, 0) = 0
a = 0
u(0, t) = 0 => b(A + B) = 0
u(l, t) = 0 => b(Ae √ul+ B e− √ul ¿= 0
 b(Ae √ul+ B e− √ul ¿=2bAsinn√ul = 0
therefore
u(x, t) = 0
m ≠ 0
let m >0, m= λ2
T11 = λ2 T , T= A eλt + B e−λt
X11 = (λ2+u ¿ x
X = C+ D e− √ λ2 +ux
u(x, t) = ( A e λt + B e−λt )(C e √λ2 +ux+ D e− √ λ2+ux )
u(0, t) = 0
= > ( A e λt + B e−λt )(C + D) = 0
= > C – D
U(l, t) = ( A e λt + B e−λt )(C e √λ2 +ul+ D e− √λ2 +ul) = 0

= > ( C e √ λ2 +ul+ D e− √ λ2+ ul ) =0
= > c (e √λ2+ul +e− √λ2+ul )=0
= cSin(n√ λ2 +ul) = 0 , c= 0=D=0
General solution
u(x, t) = BDsin√ λ2−uxsinλt
general solution
u ( x , t ) =∑
n=1
∞
Ansin(nπx) sin√ λ2−uxsinλt
U(x, t) = f(x)
U(x, t) = ∑
n =1
∞
√n2 π2 +u2
Ansinn πxcos √ n2 π 2+u2 t
u(x, 0) = f(x)
f ( x )=∑
n=1
∞
An √n2 π2 +u2sinnπx
∑
n =1
∞
An √n2 π2 +u2sinnπxsinm πx=f ( x ) sinmπx
Am√m2 π2 +u2
∫
0
1
sin2 mπx=∫
0
1
f ( x ) sinmπx
= > Am
2 √ m2 π2 +u2
∫
0
1
1−cos2 mπx=∫
0
1
f ( x ) sinmπx
= > Am
2 √ m2 π2 +u2=∫
0
1
f ( x ) sinmπx
Am = 2
√ m2 π 2+u2 ∫
0
1
f ( x ) sinmπx
So,
u(x, t) = ∑
n =1
∞
Ansinnπxsin √m2 π2 +u2 t

b)
u(x, t) = (C1cos(px) + C2sin(px))e− p2 C2 t
= (C1cos(px) + C2sin(px))e− p2 π 2 t
As given u(0, t) = 0= C1e− p2 πt
 C1 = 0
u( π ,t ¿=¿C2sin(p π))e− p2 π 2 t
sin(pπ ¿=0
p π=nπ
p= n
u(x, t) = bn*e−n2 πt sin (nx )
where bn = C2
u(x, t) = ∑ b n∗e−n2 πt sin (nx )
= ∑
n =1
∞
bnsin(nx)
Here,
bn = 2
π ∫
0
π
¿ ¿
=
−2
π [ ( n2−6 n ) cos ( n+ 6 ) x +(10 n2−360)(cos ( nx )+ ( n2 +6 n ) c 0 s ( n−6 ) x )
4 n(n2−36) ]at thelimits between π∧0
= - 2
π [ ( n2−6 n ) cos ( n+6 ) π +(10 n2−360)(cos ( nπ ) + ( n2 +6 n ) cos ( n−6 ) π )
4 n (n2−36) ] at the limitsbetween π ∧0
bn = −2
π ( n2−6 n+10 n2−360+ n2 +6 n
4 n(n2 −36) )
take n = 2m-1