KING FAHD UNIVERSITY OF PETROLEUM & MINERALS

Department of Mathematical Sciences

 

Mathematical Methods for Engineers              MATH 513

 

 

First Major Exam                            Second Semester 2003/04

 

 

Name                                                                 ID #             

 

 

Q No.

Grade

1

/9

2

/8

3

/8

Total

/25

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Please write clearly and show all work.

 

 


p(1)

Q 1)

(a) Consider the partial differential equation

 

= 0

Use  r = 3x + y  and s = y  to reduce this PDE into a suitable form and write the general solution.

 

We are given

 

r = 3 x + y ;  so  rx = 3; ry = 1

s = y          ;  so  sx  = 0;  sy = 1

 

By chain rule

 

ux =  ur rx + usx = 3 ur

uy = ur ry  + us sy = ur + us

uxx = (3ur)r rx + (3 ur)s sx

      = 9urr

uyy = (ur + us)r ry + (ur + us)s sy

      = urr + usr + urs + uss

uxy = (3 ur )r ry + (3 ur)ssy

      = 3 urr + 3 urs

 

Put these in the PDE

 

9 urr – 6(3 urr + 3 urs) + 9(urr + 2usr + uss) = 0

 

This simplifies to

 

uss = 0.    Integrating with respect to s gives

us = f(r).  Integrating again

u(r,s) = s f (r) + g ( r) .

 

Thus, u(x,y) = y f(3x+y) + g(3x+y).


p(2)

Q 1)

(b) Apply separation of variables to solve the PDE

 

x2 uxx + y u y = 0

 

Assume u(x,y) = F(x) G(y)

So that,

uxx = F//(x) G(y)

uy  = F(x) G/(y)

PDE

so that we get

x2 F//(x) – λ F(x) = 0     (1)

y G/(y) + λ G(y) = 0        (2)

 

We can solve (1) as Cauchy Euler equation by putting t = ln x (x=et)

dF/dx=(dF/dt)(dt/dx) = (1/x)(dF/dt)

d2F/dx2 =(-1/x2)(dF/dt)+ (1/x2)(d2F/dt2)

The Diffequation (1) gives

d2 F/dt2 dF/dt – λ F =0, which gives the auxiliary equation

m2 – m – λ = 0.

If m= m1,2 are the two roots then after using t= ln x

F = C1 xm1 + C2 xm2

 

The DE (2) has solution G= y

Thus u(x,y) = E1 y – λ xm1 + E2 y – λ xm2


p(3)

Q 2)

Solve the following problem.

 

u xx = (1/k ) u t , 0 < x< 1, t > 0

u (0,t) = 0, u (1) = 0, t > 0,

u(x,0) = g(x).

 

Assume u(x,y) = F(x) G(y)

PDE gives

 

 

BC F(0) = 0, F(1) = 0.

λ= 0 gives F(x) = A x+ B. Using F(0) = 0 and F (1) = 0, we get A= B =0, the trivial solution.

λ > 0Take λ = p2 . This leads to solution

F(x) = A epx + Be-px

F(0) = 0 = A+ B

F(1) = 0 = Aep +Be-p

For nontrivial solution we put det(coeff) = 0

This leads to p = 0, which is above case.

λ < 0 

Put λ = - p2

This gives  m2 + p2 = 0, which gives m = ± i p

The general solution is

F(x) = A cos px + B sin px

F(0) = 0 gives A = 0.

F(1) = 0 gives B sin pl = 0

Which for non zero B gives sin p = 0 or p= , n = 1,2,3

We can use label n and write

Fn (x) = Bn sin nπx

Solving the G- equation we obtain

Gn(t) = Cn e- λt where λ = n2 π2

Using principle of superposition

U(x,t) = ∑1 En sin()x exp{(-n2 π2)ktd

p(4)

 

Finally, using initial conditions u (x,0) = g(x) = ∑1 En sin()x

Multiply by sin ()x and integrate from 0 to 1

En = 2 0i1g(x) sin(n π)x dx.

This completes the solution.

 

 

 

 


p(5)

Q 3)

(a) Find the Fourier series for the following function in interval [-π, π]

 

f(x) =   1, -π < x< 0

f(x) =  -1 , 0< x < π

 

Given function is odd so a n = 0

 

bn = 1/π -πiπ f(x) sin nx dx  = 2/π 0iπ (-1) sin nx dx 

 

= (2/π)[cos(nx)/n ] = (2/n π)[(-1) – 1]

This will be non zero for odd n. So

f(x) = (-4/π) ∑1 (1/(2n-1)) sin(2n-1)x.

 

 


p(6)

Q 3)

(b)

 

Find the Fourier cosine series in [0, π ] for   f(x) = ex.

Use this series to write The Fourier cosine series for f(x) = e-x.

Compare the two series and tell reason for your observation.

 

f(x) = a0/2 +∑1 an cosnx

 

a0 = (2/π) 0iπf(x) dx = (2/π) 0iπ e xdx = (2/π)[eπ – 1]

 

a n =(2/x) 0iπ e x cosnxdx = (2/π) (1/(1+n2) [eπ(-1)n – 1]

This can be put in the Fourier cosine series formula to get

f(x) = (1/π)[eπ – 1] + ∑1 (2/π) (1/(1+n2) [eπ(-1)n – 1] cosnx.

We notice that to get series for e-x we need to change x into –x. But as x appears only in cosine so we get the same series.

Reason: When we write cosine series, the function is treated like an even function. So f(x) = f(-x). And so the series remains un-changed.