Assigned: January 13, 2011Due: January 26, 2011

MATH 480: Homework 1SPRING 2011

NOTE: For each homework assignment observe the following guidelines:

• Include a cover page and an assignment sheet.

• Always clearly label all plots (title, x-label, y-label, and legend).

• Use the subplot command from MATLAB when comparing 2 or more plots to makecomparisons easier and to save paper.

1. Solve each of the following ODEs for y(x):

(a) y′′ + 16y = 0 with y(0) = 1 and y′(0) = 0.

(b) y′′ + 6y′ + 9y = 0 with y(0) = 1 and y′(0) = 0.

(c) y′′′ − y′′ + y′ − y = 0 with y(0) = 1 and y′(0) = y′′(0) = 0.

2. Integration by parts allows us to transfer the derivative from one part of the integrandto another: ∫ b

af(x) g′(x) dx =

[f(x) g(x)

]ba−∫ b

af ′(x) g(x) dx.

Prove that the above statement is true. (HINT: think product rule!)

3. A linear PDE can be written in differential operator notation L(u) = f , where Lis the linear differential operator, u is the unknown function, and f is the right-hand side function. For each of the following PDEs, determine the linear operatorand the right-hand side function, the order of the PDE, and whether the PDE ishomogeneous or nonhomogeneous:

(a) uxxx + uyyy − u = 0

(b) utt − uxx + uyy + uzz = xyz

(c) x2uxx − y2uy = cos(x)− sin(y)

(d) y2uxx − x2uy = cos(y)− sin(x)

(e) ut − cos(xt)uxxx − t5 = t2u

4. The following convection-diffusion-decay equation appears in many physical appli-cations:

ut = Duxx − cux − λu .

Show that this equation can be transformed into a heat equation for w(x, t) byapplying the transformation

u(x, t) = w(x, t)eαx−βt .

HINT: You will only obtain a heat equation for w(x, t) with an appropiate choicefor the constants α and β in terms of the constants D, c, and λ. Determine thechoice for α and β that produces a heat equation for w(x, t).

1

Assigned: January 13, 2011Due: January 26, 2011

5. Consider the heat equation:

ut =(K0(x)ux

)x

u(0, t) = 0

u(1, t) = 1 ,

where K0(x) = K0(x) = ex/ cos(x).

(a) Determine the steady-state solution.

(b) Plot the steady-state solution in MATLAB. Always clearly label all plots.

6. Consider the heat equation:

ut =(K0(x)ux

)x

+Q(t)

u(1, t) = 0

u(2, t) = 1 ,

where K0(x) = x2.

(a) Under what condition on the heat source Q(t) does a steady-state solution existfor this problem? Clearly explain your answer.

(b) Under this condition, determine the steady-state solution.

7. Consider the function:u(x, t) = sin(4πx) e−πt .

(a) Plot this function in MATLAB over the domain (x, t) ∈ [0, 1]× [0, 1] using themesh command. Always clearly label all plots. (HINT: see page 4 of the“Introduction to Plotting with MATLAB” guide.)

(b) Explain what you observe.

8. If L is a linear operator, prove that L(∑M

n=1 cnun

)=(∑M

n=1 cnLun)

.

(HINT: use induction to prove this result.)

9. Evaluate (be careful if n = m)∫ L

0sin(nπxL

)sin(mπx

L

)dx

where n, m are both integers with n > 0 and m > 0. Use the trigonometric identity

2 sin a sin b = cos(a− b)− cos(a+ b).

10. Separation of Variables. By using u(x, t) = X(x)T (t) or u(x, y, t) = X(x)Y (y)T (t),separate the following PDEs into two or three ODEs for X and T or X, Y , and T .The parameters c and k are constants. You do not need to solve the equations.NOTE: one of the equations cannot be separated. Indicate this when you discoverthat equation.

2

Assigned: January 13, 2011Due: January 26, 2011

(a) utt = (xux)x

(b) utt = c2uxx

(c) ut = k(uxx + uyy)

(d) ut = k(yux + uy)

(e) ut + cux = kuxx

(f) ut = k(yux + xuy)

3

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Assigned: January 13, 2011Due: January 26, 2011

Problem # 5 (b)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% HW#1: Problem 5 (b)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

clear all; clf;

x=0:0.01:1;

w=exp(1)*(1+exp(-x).*(sin(x)-cos(x)))/(exp(1)+sin(1)-cos(1));

plot(x,w)

xlabel(’x’)

ylabel(’w(x)’)

title(’Problem 5(b): steady-state solution w(x) vs. x’)

print -depsc2 hw1_pr5b.eps

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

w(x

)

Problem 5(b): steady−state solution w(x) vs. x

Figure 1: Steady-state solution w(x) in problem # 5 (b)

8

f

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.u/+.tfq -ztkt*i=l

Kok)= t t-

fuz '-z6z>dr 69'{22.1)r>z ufk)=,hIn'%/r/A ) 147 "L/- T4'n

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u/l,r/ =9 4rl4_4 -.!

/z lzA'

Assigned: January 13, 2011Due: January 26, 2011

Problem # 7 (a)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% HW # 1: Problem 7 (a)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

clear all;

x=linspace(0,1,50);

t=linspace(0,1,50);

for i=1:50

for j=1:50

u(i,j)=sin(4*pi*x(i))*exp(-pi*t(j));

end

end

mesh(x,t,u);

xlabel (’x ’)

ylabel (’t’)

zlabel(’u(x, t)’)

title(’Problem 7 (a): u(x,t)’)

print -depsc2 hw1_pr7a.eps

00.2

0.40.6

0.81

0

0.2

0.4

0.6

0.8

1−1

−0.5

0

0.5

1

x

Problem 7 (a): u(x,t)

t

u(x,

t)

Figure 2: Plot of u(x, t) from problem # 7 (a)

Problem # 7 (b)

At t = 0, u = sin(4πx). This sine wave rapidly decays to 0 as t→∞, i.e. limt→∞

u(x, t) = 0.

10

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