MA1506 Tutorial 11 Solutions

1. Let S(t) be the number of Spartans, P (t) the number of Persians. We have

dS

dt= −P

dP

dt= −11, 111, 111.1S

so

dSdt

dPdt

=

[0 − 1

−11, 111, 111.1 0

] [SP

]

Eigenvalues λ2 − 11, 111, 111.1 = 0 ⇒ λ = ±3333.3333

Eigenvectors

[±3333.3333 −1

−11, 111, 111.1 ±3333.3333

] [1α

]= 0

⇒ α = ±3333.3333.

Since we are interested only in the first quadrant, take +. We have a SADDLE.

The trajectory that passes

through the origin is a straight

line parallel to the eigenvector(1

3333.3333

). That straight

line has equation P = 3333.3333S.

We want to start at the point on this straight line that has P = 1000000. So the

initial value of S has to be 10000003333.3333 = 300.

So Leonidas should take 299 soldiers, assuming that he intends to fight (and die)

himself.

1

2. Differentiating

u(x, y) = F (y − 3x) (∗)

w.r.t. x and y separately using chain rule, we get

ux = −3F ′(y − 3x) and uy = F ′(y − 3x).

Therefore,

ux + 3uy =[−3F ′(y − 3x)

]+ 3

[F ′(y − 3x)

]= 0.

i.e. the p.d.e. is satisfied and (∗) is a general solution.

To find the particular solution to the p.d.e., we need to find the specific single variable function

F (t) that satisfies the given initial condition.

(a) Substituting x = 0 in (∗), we get u(0, y) = F (y). Comparing with the given condition,

we have

F (y) = 4 sin(y) (†).

Substituting (†) into (∗), we get the particular solution u(x, y) = 4 sin(y − 3x).

(b) Substituting y = 0 in (∗), we get u(x, 0) = F (−3x). Comparing with the given condition,

we have

F (−3x) = ex+1 (△).

We set t = −3x to get x = −t/3. So equation (△) can be rewritten as

F (t) = e(−t3+1) (♢).

Substituting (♢) into (∗), we get the particular solution u(x, y) = e

(− (y−3x)

3+1

).

3. According to the lecture notes, we just need to extend f(x) = x to an odd function on the

domain (−π, π) [which is immediate, because y = x is already odd!], and then turn that into

a periodic function of period 2π. The result is a discontinuous function with jumps; at those

jumps, the average value is zero [which is also the value of y(0, π)]. Next, we just need the

Fourier sine coefficients of y = x, that is,

bn =2

π

∫ π

0xsin(nx)dx =

2

π

[− xcos(nx)

n+

sin(nx)

n2

]π

0

=−2

ncos(nπ) =

2

n(−1)n+1.

So again according to the lecture notes, the solution of the wave equation in this case is

y(t, x) =

∞∑n=1

2

n(−1)n+1sin(nx)cos(nct).

Notice that y(0, π) from this formula is zero, as it should be.

4. (a) Setting u(x, y) = X(x)Y (y), the p.d.e. yux − xuy = 0 becomes yX ′Y − xXY ′ = 0.

Dividing by XY gives

yX ′

X− x

Y ′

Y= 0

i.e.1

x· X

′

X=

1

y· Y

′

Y= k1 (constant)

This gives two separable o.d.e., the first of which is1

x· X

′

X= k1. Integrating this yields

ln |X| =1

2k1x

2 + c1

i.e. X = ±ec1ecx2,where c =

1

2k1

Similarly, integrating the second separable o.d.e.1

y· Y

′

Y= k1 gives Y = ±ec2ecy

2.

Thus,

u(x, y) = XY = ±ec1ecx2ec2ecy

2

= kec(x2+y2).

(b) Setting u(x, y) = X(x)Y (y), the p.d.e. ux = yuy becomes X ′Y = yXY ′. Dividing by

XY givesX ′

X= y

Y ′

Y= c (constant)

This gives two separable o.d.e., the first of which isX ′

X= c. Integrating this yields

ln |X| = cx+ k1

i.e. X = ±ek1ecx.

Similarly, integrating the second separable o.d.e.Y ′

Y=

c

ygives

ln |Y | = c ln |y|+ k2

i.e. Y = ±ek2yc.

Thus,

u(x, y) = XY = ±ek1ecxek2yc

= kycecx.

(c) Setting u(x, y) = X(x)Y (y), the p.d.e. uxy = u becomes X ′Y ′ = XY . Dividing by

XY givesX ′

X· Y

′

Y= 1.

2

This implies that bothX ′

Xand

Y ′

Yare nonzero constants and we set

X ′

X= c and

Y ′

Y=

1

c.

This gives two separable o.d.e., the first of which isX ′

X= c. Integrating this yields

ln |X| = cx+ k1

i.e. X = ±ek1ecx.

Similarly, integrating the second separable o.d.e.Y ′

Y=

1

cgives

ln |Y | =y

c+ k2

i.e. Y = ±ek2ey/c.

Thus,

u(x, y) = XY = ±ek1ecxek2ey/c

= kecx+y/c.

(d) Setting u(x, y) = X(x)Y (y), the p.d.e. xuxy+2yu = 0 becomes xX ′Y ′+2yXY = 0.

Dividing by −2yXY gives (xX ′

X

)(− 1

2y· Y

′

Y

)= 1.

This implies that both xX ′

Xand

−1

2y· Y

′

Yare nonzero constants. We set x

X ′

X= c and

−1

2y· Y

′

Y=

1

c, which respectively give two separable o.d.e.

X ′

X=

c

xand

Y ′

Y= − 2y

c.

Integrating the first o.d.e.X ′

X=

c

xyields

ln |X| = c lnx+ k1

i.e. X = ±ek1xc.

Similarly, integrating the second separable o.d.e.Y ′

Y= − 2y

cgives

ln |Y | = − y2

c+ k2

i.e. Y = ±ek2e−y2/c.

Thus,

u(x, y) = XY = ±ek1xcek2e−y2/c

= kxce−y2/c.

3

Question 5

We want to solvec2yxx = ytt,

subject to the four conditions

y(t, 0) = y(t, π) = 0, y(0, x) = f(x), yt(0, x) = 0,

where f(x) is a given function which is zero at 0 and π, and which we extend whennecessary to be an odd periodic function of period 2π. D’Alembert claimed that thefollowing object satisfies all of the above:

y(t, x) =1

2

[f(x+ ct) + f(x− ct)

].

First, note that if y(t,x) is given by this formula, then

ytt =c2

2

[f ′′(x+ ct) + f ′′(x− ct)

],

while

yxx =1

2

[f ′′(x+ ct) + f ′′(x− ct)

].

So it does satisfy the equation. We also see that

y(t, 0) =1

2

[f(ct) + f(−ct)

],

which is zero because f(x) is odd. Then

y(t, π) =1

2

[f(π + ct) + f(π − ct)

]=

1

2

[f(π + ct) + f(−π − ct)

],

because f(x) is periodic with period 2π, and this too vanishes because f(x) is odd. Next,clearly

y(0, x) =1

2

[f(x) + f(x)

]= f(x),

and finally

yt(0, x) =1

2

[cf ′(x) − cf ′(x)

]= 0.

So D’Alembert was right.

Question 6

If you take the wave equationc2yxx = ytt,

and change the variables from (x,y) to (ξ, τ), where ξ = ax, τ = at, for any constant a,then the equation continues to hold because you get a factor involving a2 coming out onboth sides of the equation. We can use this here, because suppose that x ranges from 0to L: then define

ξ =πx

L, τ =

πt

L.

Then the range of ξ is 0 to π, so all our familiar formulae for the wave equation hold whenwe use ξ and τ . Thus for example the function f(x) giving the initial shape of the string,when expressed in terms of ξ, will have a Fourier sine series with coefficients of the form

bn =2

π

∫ π

0f(ξ)sin(nξ)dξ,

which, returning to the variable x, can be written [since dξ = πdx/L]

bn =2

L

∫ L

0f(x)sin(

nxπ

L)dx.

The solution of the wave equation likewise takes the form

∞∑n=1

bnsin(nξ)cos(ncτ),

which means that it is ∞∑n=1

bnsin(nxπ

L)cos(

nctπ

L),

and this is the answer to our question. In summary: all you have to do is to put in factorsof π/L, BUT you have to remember to do it for both x AND t.

Question 7

To solveytt = c2yxx − byt,

set y(t,x) = T(t)X(x), and get

T ′′X = c2TX ′′ − bT ′X.

Divide throughout by TX:T ′′

T= c2

X ′′

X− b

T ′

T,

from which we getT ′′

T+ b

T ′

T= c2

X ′′

X.

As usual, both sides have to be equal to some constant −K, and so we get a pair ofordinary differential equations:

T ′′ + bT ′ +KT = 0,

X ′′ + c−2KX = 0,

and so we have successfully separated the variables. To get y(t,0) = y(t,π) = 0, we wantX(0) = X(π) = 0, and, as usual, that means that we want trigonometric solutions forX(x); so from the second equation we see that K is positive. No doubt you now recognisethe first equation as the equation of motion of the damped simple harmonic oscillator.[Remember that, in the ODE for the damped harmonic oscillator, all of the coefficientswere positive, and that was very important.] So the solutions for T(t) will be [if b is smallenough] products of exponentials with trigonometric functions, as in Chapter 2.

2

Question 8

Here we use the following useful observation: the Laplace equation is completely sym-metric between x and y. That is, if you switch x and y in the Laplace equation, nothingchanges. Furthermore, the boundary conditions we have here [with u(x,y) equal to zeroon all sides of the square except the left vertical side, ie along the y axis] are obtained byflipping x and y in the problem in the lecture notes [the one at the very end of Chapter8, where u(x,y) is equal to zero on all sides of the square except the bottom, ie along thex axis]. Let’s call solution of that problem in the lecture notes u1(x,y), so

u1(x, y) =∞∑n=1

cnsin(nx)sinh[n(π − y)],

where

cn =2π

∫ π0 f(x)sin(nx)dx

sinh(nπ).

Exchanging x with y we get the solution of our problem here, which I will call u2(x,y):

u2(x, y) =∞∑n=1

dnsin(ny)sinh[n(π − x)],

where

dn =2π

∫ π0 g(y)sin(ny)dy

sinh(nπ).

[If you don’t like this symmetry argument, you can do this problem by just following thesteps in the lecture notes, making the necessary changes at each step.]

For the second part of the problem, I claim that the answer is simply u(x,y) ≡ u1(x,y)+ u2(x,y). Since each of these is a solution, the linearity of the Laplace equation ensuresthat their sum is likewise a solution. But now we have to check the boundary conditionsto see that they agree with the ones we want. [It might be helpful here to draw twosquares, marking on the edges of each one the values of u1(x,y) and u2(x,y) there, so youcan take the sums easily.]

u(x, 0) = u1(x, 0) + u2(x, 0) = f(x) + 0 = f(x);

u(x, π) = u1(x, π) + u2(x, π) = 0 + 0 = 0;

u(π, y) = u1(π, y) + u2(π, y) = 0 + 0 = 0;

u(0, y) = u1(0, y) + u2(0, y) = 0 + g(y) = g(y),

all of which are correct. So this is indeed the solution.

3