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 =
uy =
uxx = (3ur)r
rx + (3
= 9urr
uyy = (
= urr + usr + urs +
uss
uxy = (3
= 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-λ
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= nπ,
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(nπ)x
exp{(-n2 π2)ktd
p(4)
Finally, using initial conditions u (x,0)
= g(x) = ∑∞1 En sin(nπ)x
Multiply by sin (mπ)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.