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Math 6B Section, Week 9March 5, 2015

Exercise 1 (11.2.10,15) Find the Fourier series:

(a) f(x) =

x 4 :4 x 0x+ 4 : 0< x 4

(b) f(x) =

x : x

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utt(x, t) =F(x)G(t), uxx(x, t) =F

(x)G(t). Setting these equal we have F(x)G(t) =F(x)G(t) and we divide to find

G

G =

F

F .

Since the left hand side depends only ont whereas the right depends only onx, we have

G

G =

F

F =k

for some constantk .(i) First we solve for F. We have that F kF = 0. If k = 0 then F = 0 and

hence F is of the form ax+b for constants a, b. The boundary conditions implythat u(0, t) = bG(t) = 0. Since we assume G = 0, we find that b = 0. Similarly,u(1, t) = (a+b)G(t) = aG(t) = 0 and we find a= 0. Hence, the case k = 0 is ofno interest.If k > 0 then, setting k = p2, we find F p2F = 0. The general solution tothis equation is F(x) =aepx +bepx. The conditions F(0) =F(1) = 0 imply that

a+b= aep +be

p = 0. The system of equations1 1ep ep

ab

=

00

has only the trivial solution a = b = 0 since the matrix

1 1ep ep

has nonzero

determinantep ep (recall that p = 0). Hence, this case is also of no interest.Finally, we see that we must have k = p2

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must match the coefficients of the sines on the right. Hence, we getA1 = 1, A2= 12

and An= 0 for n >2.We calculate ut(x, t),

ut(x, t) =

n=1

sin(nx)(nAnsin(nt) +nBncos(nt)).

Hence, since our initial velocity ut(x, 0) is 0, we find that

ut(x, 0) =n=1

sin(nx)(nBn) = 0.

Since this is the Fourier sine series for 0, we must have nBn = 0 and thereforeBn= 0 for each n.

Finally then, our solution is

u(x, t) =

n=1

sin(nx)(Ancos(nt)) = sin(x)cos(t) 1

2

sin(2x) cos(2t).

The last equality just follows from substituting for An.(b) Now we resort to the general equations derived in class. We have that u is equal to

n=1

sin(n

Lx)(Ancos(

cn

L t) +Bnsin(

cn

L t))

where

An= 2

L

L0

f(x)sin(n

Lx)dx, Bn=

2

cn

L0

g(x)sin(n

Lx)dx

whereu(x, 0) =f(x) and ut(x, 0) =g(x).

Here we have c= 1, L= 1, f(x) =x(1 x) and g(x) = 0. We have right off the batthat Bn = 0. For An we calculate

An=2

1

10

x(1 x)sin(n

1 x)dx= 2

10

x(1 x)sin(nx)dx.

After calculating the antiderivative, we find that this is equal to

2

2

n33x sin(nx) +

1

n22sin(nx)

2

n33cos(nx)

1

nx(1 x)cos(nx)

|10

and evaluating we find

An= 2 2

n33cos(n) + 2

n33

= 4n33

((1)n 1) = 4n33

((1)n+1 + 1).

Therefore our solution is

u(x, t) =n=1

4

n33((1)n+1 + 1) sin(nx) cos(nt).

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Exercise 5 (12.3.14) Find the deflection u(x, t) of the string of length L= and c2 = 1for zero initial displacement and triangular initial velocity,

ut(x, 0) =

x : 0 x 2 x :

2x

.

Solution. As stated, we have L = , c = 1, f(x) = 0, and g(x) the triangular functiondescribed. Since f= 0, we have An= 0 for every n. On the other hand,

Bn= 2

1n

0

g(x)sin(n

x)dx=

2

n

/20

x sin(nx)dx+

/2

( x)sin(nx)dx

.

This is equal to

2

n

1

nx cos(nx) +

1

n2sin(nx)

|/20 +

1

n cos(nx) +

1

nx cos(nx)

1

n2sin(nx)

|/2

.

This comes out to be 4n3

sin(n2

). Hence, u is given by

n=1 sin(

n

x)(0 +

4

n3sin(n

2 )sin(

n

t)) =

n=1

4

n3sin(n

2 )sin(nx)sin(nt).

This can be simplified a bit more by noticing that

sin(n

2) =

0 :n is even(1)(n1)/2 :n is odd

.

Namely, we have

u(x, t) =k=0

(1)k 4

(2k+ 1)3sin((2k+ 1)x) sin((2k+ 1)t).

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