sample final exam solutions - mcmaster...

LAST (family) NAME:

FIRST (given) NAME:

ID # :

MATHEMATICS 3FF3

McMaster University Final Examination Dr. J.-P. GabardoDay ClassDuration of Examination: 3 hours

THIS EXAMINATION PAPER INCLUDES 22 PAGES AND 9 QUESTIONS. YOUARE RESPONSIBLE FOR ENSURING THAT YOUR COPY OF THE PAPER ISCOMPLETE. BRING ANY DISCREPANCY TO THE ATTENTION OF YOURINVIGILATOR.

Instructions:

• The total of marks is 100.

• No calculator or other aid are allowed.

• Pages 18–21 are for scratch or overflow work. Please indicate clearly that you are continuingyour work on one of these pages if you decide to do so.

• There is a formula sheet on the last page.

SAMPLE FINAL EXAM SOLUTIONS

Continued. . . of 22

Final Exam / Math 3FF3

NAME: ID #:

Provide all details and fully justify your answer in order to receive credit.

1. (10 pts.) Use the method of characteristics the solve the first-order partial differential equation(for u = u(x, y)):

ux −ex

1 + eyuy = 0, −∞ < x <∞, y > 0,

with initial condition u(x, 0) = e2x.

Solution. The equation for the characteristics curves is

dy

dx= − ex

1 + ey

which is a separable ODE. We have

(1 + ey)dy

dx= −ex and thus

∫(1 + ey) dy = −

∫ex dx

which yield y + ey = −ex + C.

The characteristics curves are thus the curves with equation

ey + y + ex = C, C arbitary constant.

The general solution of the PDE is this

u(x, y) = F (ey + y + ex),

where F (s) is an arbitrary function of 1 variable. Using the initial condition we have

u(x, 0) = e2x = F (e0 + 0 + ex) = F (1 + ex)

Letting s = 1 + ex, we have e2x = (ex)2 = (s− 1)2 and thus F (s) = (s− 1)2.

The solution is thereforeu(x, y) = (ey + y + ex − 1)2.



NAME: ID #:

2. (8 pts.) Consider the solution u(x, t) of the diffusion equationut = uxx, 0 < x < l, t > 0,

ux(0, t) = ux(l, t) = 0, t > 0,

u(x, 0) = φ(x), 0 < x < l.

Show that the function F (t) =

∫ l

0

eu(x,t) dx is decreasing for t ≥ 0.

Solution. We compute

F ′(t) =d

dt

{∫ l

0

eu(x,t) dx

}=

∫ l

0

d

dt

{eu(x,t)

}dx =

∫ l

0

ut(x, t) eu(x,t) dx

=

∫ l

0

uxx(x, t) eu(x,t) dx (using the PDE)

=[ux(x, t) e

u(x,t)]l0−∫ l

0

ux(x, t)d

dx

{eu(x,t)

}dx (using integration by parts)

= ux(l, t)︸︷︷︸=0

eu(l,t) − ux(0, t)︸︷︷︸=0

eu(0,t) −∫ l

0

(ux(x, t))2 eu(x,t)︸︷︷︸

≥0

dx ≤ 0

Since F ′(t) ≤ 0 for all t > 0, it follows that F (t)is decreasing for t ≥ 0.



NAME: ID #:

3. Consider the eigenvalue problem{X ′′(x) + λX(x) = 0, 0 < x < 1,

X(0) = 0, X(1) = −X ′(1).

(a) (4 pts.) Show that the boundary conditions for the problem above are symmetric.(Youshould assume that all functions involved are complex-valued.) What can you conclude aboutthe eigenvalues and eigenfunctions of the problem above?

Solution. If f(x) and g(x) both satisfy the boundary conditions, we have[f ′(x) g(x)− f(x) g′(x)

]10

= f ′(1)g(1)− f(1) g′(1)− f ′(0) g(0)︸︷︷︸=0

− f(0)︸︷︷︸=0

g′(0)

= −f(1)g(1) + f(1) g(1) = 0.

It follows that all the eigenvalues are real

and that eigenfunctions associated with different eigenvalues are orthogonal to each other.

(b) (4 pts.) State a simple condition that can be used to check that there are no negativeeigenvalues and use it to show that this is the case for the problem above. Check directly thatthey are indeed no negative eigenvalues.

Solution. If the boundary conditions are symmetric and we have

[f(x) f ′(x)]10 ≤ 0

for any (real-valued) function f(x) satisfying the boundary conditions, then all then eigenvaluesare non-negative. For the boundary conditions in part (a), we have

[f(x) f ′(x)]10 = f(1) f ′(1)− f(0)︸︷︷︸

=0

f ′(0) = −f(1)2 ≤ 0

and thus that condition holds. By direct computation, letting λ = −β2 where β > 0, ifX ′′(x)− β2X(x) = 0, then X(x) = C1 sinh(βx) + C2 cosh(βx). If X(0) = 0, then C2 = 0 andthe condition X(1) = −X ′(1) implies that C1 sinh(β) = −C1 β cosh(β). If C1 6= 0, we wouldneed

tanh(β) = −β

which has no solution as tanh(β) > 0 and −β < 0 for β > 0.



NAME: ID #:

(c) (2 pts.) Is zero an eigenvalue? If so find the corresponding eigenfunction.

Solution. If X(x) is a corresponding eigenfunction, then X ′′(x) = 0, so X(x) = C1 x + C2.The condition X(0) = 0 implies C2 = 0 and the condition X(1) = −X ′(1) implies C1 = −C1

or C1 = 0. Thus there is non non-trivial solution and 0 is not an eigenvalue.

(d) (4 pts.) Use the graphical method to show that there are infinitely many eigenvaluesλn > 0 and find the corresponding eigenfunctions.

Solution. If λ = β2 where β > 0 and X ′′(x) + β2X(x) = 0, then X(x) = C1 sin(βx) +C2 cos(βx). If X(0) = 0, then C2 = 0 and the condition X(1) = −X ′(1) implies thatC1 sin(β) = −C1 β cos(β). If C1 6= 0, we need

tan(β) = −β

We can see from the graph above that the previous equation has infinitely many solutionsβn for n ≥ 1, where (n − 1/2)π < βn < nπ and thus infinitely many corresponding positiveeigenvalues λ = β2

n.

and eigenfunctions Xn(x) = sin(βnx).



NAME: ID #:

4. Let D = {(x, y), x2 + y2 < 1} be the unit disk centered at the origin in the x, y plane and letC = {(x, y), x2 + y2 = 1} be the unit circle bounding D. Find the solution u(x, y) of Laplaceequation

uxx + uyy = 0, (x, y) ∈ D,

which satisfies the boundary condition

u(x, y) = 4x3, (x, y) ∈ C.

(a) (8 pts.) Solve this problem by passing to polar coordinates. Note that the Laplaceoperator is written as

∂2

∂r2+

1

r

∂

∂r+

1

r2∂2

∂θ2

in polar coordinates.

(Hint. cos3 θ = 14

cos(3θ) + 34

cos(θ).)

Solution. We look first for non-trivial solutions of the PDE in polar coordinates of the formu(r, θ) = R(r) Θ(θ) where Θ(θ) is 2π-periodic. We have

R′′(r) Θ(θ) +1

rR′(r) Θ(θ) +

1

r2R(r) Θ′′(θ) = 0 or

r2R′′(r) + r R′(r)

R(r)= −Θ′′(θ)

Θ(θ)= λ.

This leads to the eigenvalue problem for Θ(θ):

Θ′′(θ) + λΘ(θ) = 0, Θ(−π) = Θ(π), Θ′(−π) = Θ′(π),

The eigenvalues are λn = n2, n ≥ 0, with corresponding eigenfunctions Θ0(θ) = 1 and,forn ≥ 1, Θn(θ) = An cos(nθ) +Bn sin(nθ).

For each n ≥ 0, we need to solve the DE

r2R′′(r) + r R′(r)− n2R(r) = 0

which is of Cauchy-Euler type. Letting R(r) = rβ we have β (β − 1) + β − n2 = 0 or β2 = n2.If n = 0, the solution is R0(r) = C + D ln r and and we need D = 0 to ensure the continuityof the solution at r = 0. Thus R0(r) = C. If n ≥ 1, the solution is of the form

Rn(r) = Cn rn +Dn r

−n.

Again, to ensure the continuity of the solution at r = 0, we need to take Dn = 0. We obtainthus particular solutions of the form C0 and (An cos(nθ) + Bn sin(nθ)) rn for n ≥ 1. By thesuperposition principle, the general solution has the form

u(r, θ) =A0

2+∞∑n=1

(An cos(nθ) +Bn sin(nθ)) rn

where A0

2= C0 for convenience.



NAME: ID #:

If x2 + y2 = 1, we have x = cos θ, y = sin θ, and 4x3 = 4 cos3 θ = cos(3θ) + 3 cos(θ). To satisfythe boundary condition, we thus need

u(1, θ) =A0

2+∞∑n=1

(An cos(nθ) +Bn sin(nθ)) = cos(3θ) + 3 cos(θ).

By inspection, we see immediately that A1 = 3, A3 = 1 and An = Bn = 0 otherwise. Thesolution is thus

u(r, θ) = cos(3θ) r3 + 3 cos(θ) r.

(b) (4 pts.) Express the solution found in part (a) in rectangular coordinates and verify thatit is a solution of the Laplace equation satisfying the given boundary condition.

Solution. Letting x+ i y = r(cos θ + i sin θ) = r ei θ, we have r cos θ = x = Re(x+ iy) and

r3 cos(3θ) = Re(r3 e3 i θ) = Re((r ei θ)3) = Re((x+ i y)3)

= Re(x3 + 3 i x2 y − 3x y2 − i y3) = x3 − 3x y2

In cartesian coordinates, the solution is thus given by

u(x, y) = x3 − 3x y2 + 3x.

We check that this is indeed the correct solution:

uxx + uyy = 6x− 6x = 0

and if x2 + y2 = 1, we have

x3 − 3x y2 + 3x = x3 − 3x (1− x2) + 3 x = 4x3.



NAME: ID #:

5. Consider the wave equation

utt = 4uxx, 0 < x < π, t > 0, (1)

together with the boundary conditions

u(0, t) = u(π, t) = 0, t > 0, (2)

and the initial conditions

u(x, 0) = 0, ut(x, 0) = 1, 0 < x < π. (3)

(a) (4 pts.) Using the method of separation of variables, find all non-trivial solutions of theequations (1) and (2) of the form u(x, t) = X(x)T (t).

Solution. Letting u(x, t) = X(x)T (t), we have using (1)

X(x)T ′′(t) = 4X ′′(x)T (t) orT ′′(t)

4T (t)=X ′′(x)

X(x)= −λ.

and, using (2),

X(0)T (t) = X(π)T (t) = 0 which implies X(0) = X(π) = 0.

We obtain thus the eigenvalue problem for X(x):{X ′′(x) + λX(x) = 0,

X(0) = X(π) = 0

The eigenvalues are λn = n2 for n ≥ 1 with corresponding eigenfunctions Xn(x) = sin(nx).

For each n, the solution of the ODE

T ′′(t) + 4n2 T (t) = 0

isTn(t) = An cos(2nt) +Bn sin(2nt).

The non-trivial solutions obtained are thus

un(x, t) = (An cos(2nt) +Bn sin(2nt)) sin(nx), n ≥ 1.



NAME: ID #:

(b) (8 pts.) Use the superposition principle to find the general form of a solution of the PDEsatisfying the given boundary conditions and compute the unique solution u(x, t) satisfying thegiven initial conditions (3).

Solution. By the superposition principle, the general solution of (1) and (2) is of the form

u(x, t) =∞∑n=1

(An cos(2nt) +Bn sin(2nt)) sin(nx), 0 < x < π, t > 0.

for appropriate constant An and Bn, n ≥ 1. The first condition in (3) yields

u(x, 0) = 0 =∞∑n=1

An sin(nx), 0 < x < π,

and thus that An = 0 for all n ≥ 1

(by uniqueness of the coefficients in a Fourier sine series). The first condition in (3) yields

ut(x, 0) = 1 =∞∑n=1

2nBn sin(nx), 0 < x < π.

It follows that

2nBn =2

π

∫ π

0

sin(nx) dx =2

π

[− cos(nx)

n

]π0

=2

π

1− (−1)n

n

and thus

Bn =1

π

1− (−1)n

n2, n ≥ 1.

The solution is thus

u(x, t) =1

π

∞∑n=1

1− (−1)n

n2sin(2nt) sin(nx), 0 < x < π, t > 0,

or

u(x, t) =2

π

∞∑k=0

1

(1 + 2k)2sin(2(1 + 2k)t) sin((1 + 2k)x), 0 < x < π, t > 0.



NAME: ID #:

6. (a) (5 pts.) Show that if u(x, t) is solution of the diffusion equation with variable dissipation{ut − k uxx + φ(t)u = 0, −∞ < x <∞, t > 0,

u(x, 0) = f(x), −∞ < x <∞,

and g(t) is solution of the differential equation

g′(t) = φ(t) g(t), g(0) = 1,

then v(x, t) = g(t)u(x, t) is solution of the diffusion equation{vt − k vxx = 0, −∞ < x <∞, t > 0,

v(x, 0) = f(x), −∞ < x <∞,

Solution. We havevt(x, t) = g′(t)u(x, t) + g(t)ut(x, t)

andvxx(x, t) = g(t)uxx(x, t).

Thus

vt(x, t)− k vxx(x, t) = g′(t)u(x, t) + g(t)ut(x, t)− k g(t)uxx(x, t)

= φ(t) g(t)u(x, t) + g(t)ut(x, t)− k g(t)uxx(x, t)

= g(t) (φ(t)u(x, t) + ut(x, t)− k uxx(x, t)) = 0

andv(x, 0) = g(0)u(x, 0) = 1 · f(x) = f(x).



NAME: ID #:

(b) (5 pts.) Solve the diffusion equation with variable dissipation{ut − k uxx + cos(t)u = 0, −∞ < x <∞, t > 0,

u(x, 0) = e−x, −∞ < x <∞,

(Hint. See part (a).)

Solution. Let v(x, t) be the solution of the diffusion equation{vt − k vxx = 0, −∞ < x <∞, t > 0,

v(x, 0) = e−x, −∞ < x <∞,

Then,

v(x, t) =1√

4πkt

∫ ∞−∞

e−(x−y)2/4kt e−y dy

=1√

4πkt

∫ ∞−∞

e−(x2−2 (x−2 kt) y+y2)/4kt dy

=1√

4πkt

∫ ∞−∞

e−(x−2 kt−y)2/4kt e(−4ktx+4 k2 t2)/4kt dy

= ekt−x1√

4πkt

∫ ∞−∞

e−(x−2 kt−y)2/4kt

After making the change of variable, (−x+ 2 kt+ y)/√

4kt = s with dy =√

4kt ds, we obtain

v(x, t) = ekt−x1√π

∫ ∞−∞

e−s2

ds︸︷︷︸=1

= ekt−x.

Using part (a), we first need to find g(t) the solution of g′(t) − cos(t) g(t) = 0 with g(0) = 1.The integrating factor is exp(

∫− cos t dt) = e− sin t. We have thus

(g(t) e− sin t)′ = 0 or g(t) e− sin t = C, a constant.

Since g(0) = 1, C = 1 and g(t) = esin t. By part (a), the solution is

u(x, t) =v(x, t)

g(t)= e− sin t ekt−x = e− sin t+k t−x.



NAME: ID #:

7. (a) (8 pts.) Use the method of separation of variables together with the superposition principleto find the solution u = u(x, y) of the boundary-value problem for the Laplace equation

uxx + uyy = 0, 0 < x < π, 0 < y < 1,

ux(0, y) = ux(π, y) = 0, 0 < y < 1,

u(x, 0) = 0, u(x, 1) = cos x, 0 < x < π.

Solution. We first look for solutions of the PDE that satisfy the boundary conditions and areof the form u(x, y) = X(x)Y (y). We have

X ′′(x)Y (y) +X(x)Y ′′(y) = 0 orX ′′(x)

X(x)= −Y

′′(y)

Y (y)= −λ

andX ′(0)Y (y) = X ′(π)Y (y) = 0 which yields X ′(0) = X ′(π) = 0.

We obtain thus the eigenvalue problem for X(x):{X ′′(x) + λX(x) = 0,

X ′(0) = X ′(π) = 0

The eigenvalues are λn = n2 for n ≥ 0 with corresponding eigenfunctions X0(x) = 1 andXn = cos(nx) for n ≥ 1.

For each n ≥ 0, the DE for Y (y) is Y ′′(y) − n2 Y (y) = 0. If n = 0, the solution is Y0(y) =A0 + B0 y and for n ≥ 1, the solution is Yn(y) = An cosh(ny) + Bn sinh(ny). We obtain thusparticular solutions of the form

A0 +B0 y and (An cosh(ny) +Bn sinh(ny)) cos(nx) n ≥ 1.

By the superposition principle, the general solution has the form

u(x, y) = A0 +B0 y +∞∑n=1

(An cosh(ny) +Bn sinh(ny)) cos(nx), 0 < x < π, 0 < y < 1.

The boundary condition u(x, 0) = 0 yields

0 = A0 +∞∑n=1

An cos(nx), 0 < x < π

and implies that An = 0 for all n ≥ 0.

The boundary condition u(x, 1) = cos x then yields

cosx = B0 +∞∑n=1

Bn sinh(n) cos(nx), 0 < x < π.



NAME: ID #:

By inspection, B1 sinh(1) = 1 or B1 = 1/ sinh(1) and Bn = 0 otherwise.

The solution is thus

u(x, y) =sinh y

sinh 1cos(x).

(b) (4 pts.) Find the maximum value of the solution u(x, y) in part (a) on the rectangle{(x, y), 0 ≤ x ≤ π, 0 ≤ y ≤ 1}. How does this relate to the “maximum principle”?

Solution. If 0 ≤ x ≤ π, cos x ≤ 1 = cos 0 and if 0 ≤ y ≤ 1, sinh y ≤ sinh 1 as sinh y is anincreasing function. It follows that

u(x, y) =sinh y

sinh 1cos(x) ≤ sinh 1

sinh 1cos(0) = 1 = u(0, 1).

The maximum value of u(x, y) on the rectangle {(x, y), 0 ≤ x ≤ π, 0 ≤ y ≤ 1} is thus 1 and isreached at the point (0, 1) which is on the boundary of the rectangle and not an interior pointin accordance with the maximum principle.



NAME: ID #:

8. (10 pts.) Solve the Poisson equation{uxx + uyy = x2 + y2, x2 + y2 < 1,

u(x, y) = 0, x2 + y2 = 1.

by passing to polar coordinates (See Problem 4). Express your final answer in rectangularcoordinates.

(Hint. Look for a solution that depends only on r.)

Solution. Let x = r cos θ and y = r sin θ and suppose that u(x, y) = v(r), for some functionv. Then,

v′′(r) +1

rv′(r) = r2 and v(1) = 0.

The function w = v′ satisfies the first-order linear ODE

w′(r) +1

rw(r) = r2

The integrating factor is e∫1/r dr = eln r = r.

We have thusr w′(r) + w(r) = (r w(r))′ = r3

Integrating yields

r w(r) =r4

4+ C or w(r) =

r3

4+C

r.

Since w(r) = v′(r) should be continuous at r = 0, we need C = 0. We have thus

v′(r) = w(r) =r3

4and thus after integrating, v(r) =

r4

16+D.

The condition v(1) = 0 yields D = − 116

. We have thus

v(r) =r4 − 1

16

and the solution is given in rectangular coordinates as

u(x, y) =(x2 + y2)2 − 1

16.



NAME: ID #:



NAME: ID #:

9. (a) (5 pts.) Compute the Fourier series of the function

f(x) = ex, −π < x < π,

in its complex form (i.e. ex =∑∞

n=−∞ cn einx, −π < x < π).

Solution. We have

cn =1

2π

∫ π

−πex e−inx dx

=1

2π

∫ π

−πe(1−in)x dx

=1

2π

[e(1−in)x

1− in

]π−π

=1

2π

e(1−in)π − e−(1−in)π

1− in=

(−1)n

2 π

eπ − e−π

1− in

The Fourier series expansion of ex is thus given by

ex =∞∑

n=−∞

(−1)n

2π

(eπ − e−π

1− in

)einx, −π < x < π.



NAME: ID #:

(b) (7 pts.) Apply Parseval’s identity to the function in part (a) and the complete orthogonalsystem {einx}n∈Z on the interval (−π, π) to prove the formula

∞∑n=−∞

1

1 + n2= π

eπ + e−π

eπ − e−π.

Solution. Letting Xn(x) = einx for n ∈ Z and f(x) = ex, we have using (a),

(f,Xn) =

∫ π

−πex e−inx dx = (−1)n

(eπ − e−π

1− in

), n ∈ Z,

‖Xn‖2 = (Xn, Xn) =

∫ π

−πeinx e−inx dx =

∫ π

−π1 dx = 2π,

and

‖f‖2 = (f, f) =

∫ π

−π|f(x)|2 dx =

∫ π

−π(ex)2 dx =

∫ π

−πe2x dx =

[e2x

2

]π−π

=e2π − e−2π

2.

Since

|(f,Xn)|2 =

∣∣∣∣(−1)n(eπ − e−π

1− in

)∣∣∣∣2 =(eπ − e−π)2

1 + n2,

Parseval’s identity yields

‖f‖2 =e2π − e−2π

2=

∞∑n=−∞

|(f,Xn)|2

‖Xn‖2=

∞∑n=−∞

(eπ − e−π)2

2π (1 + n2)

which shows that

∞∑n=−∞

1

1 + n2= π

e2π − e−2π

(eπ − e−π)2= π

(eπ + e−π) (eπ − e−π)

(eπ − e−π)2= π

eπ + e−π

eπ − e−π.



NAME: ID #:

SCRATCH



NAME: ID #:

Some formulas you may use:

f(x) ∼ a02

+∞∑n=1

an cos(nπx

l

), an =

2

l

∫ l

0

f(x) cos(nπx

l

)dx, 0 < x < l.

f(x) ∼∞∑n=1

bn sin(nπx

l

), bn =

2

l

∫ l

0

f(x) sin(nπx

l

)dx, 0 < x < l.

f(x) ∼ a02

+∞∑n=1

an cos(nπx

l

)+ bn sin

(nπxl

), −l < x < l,

where an =1

l

∫ l

−lf(x) cos

(nπxl

)dx, n ≥ 0, bn =

1

l

∫ l

−lf(x) sin

(nπxl

)dx, n ≥ 1.

f(x) ∼∞∑

n=−∞

cn eπinx/l, −l < x < l, with cn =

1

2 l

∫ l

−lf(x) e−πinx/l dx.

If {Xn(x)} is a complete orthogonal system on (a, b),

f(x) ∼∑n

(f,Xn)

‖Xn‖2Xn(x), ‖f‖2 =

∑n

|(f,Xn)|2

‖Xn‖2.

{X ′′(x) + λX(x) = 0,

X(0) = X(l) = 0,λn =

(nπl

)2, Xn(x) = sin(nπx/l), n ≥ 1.

{X ′′(x) + λX(x) = 0,

X ′(0) = X ′(l) = 0,λn =

(nπl

)2, Xn(x) = cos(nπx/l), n ≥ 0.

{X ′′(x) + λX(x) = 0,

X(l) = X(−l), X ′(−l) = X ′(l) = 0,λn =

(nπl

)2, n ≥ 0

withX0(x) = 1, Xn(x) = An cos(nπx/l) +Bn sin(nπx/l), n ≥ 1.

u(x, t) =1√

4πkt

∫ ∞−∞

e−(x−y)2/4kt φ(y) dy.

*** THE END*** of 22

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