Graded Homework 2 Solutions

Spring 2014

Do the problems in the order in which they are listed, write on only oneside of each page of your paper, and staple your pages. Papers are due at thebeginning of class. No late homework will be accepted. Study the PDE helpfile, "Two-Point Boundary Value Problems."

Note that

sin z = 0 if and only if z = kπ for some integer k,

and

cos z = 0 if and only if z = (k − 12)π for some integer k.

When k is an integer,coskπ = (−1)k

and

sin(k − 12)π = (−1)k+1.

1. Find 2× 2 matrices M and N such that

ϕ(a) + 2ϕ′(a) = 3ϕ(b)− ϕ′(b) and3ϕ(a)− ϕ′(a) = ϕ(b)

is equivalent to

M

�ϕ(a)ϕ′(a)

�+N

�ϕ(b)ϕ′(b)

�=

�00

�.

Solution. The given conditions are equivalent to

(1)ϕ(a) + (2)ϕ′(a) + (−3)ϕ(b) + (1)ϕ′(b) = 0 and

(3)ϕ(a) + (−1)ϕ′(a) + (−1)ϕ(b) + (0)ϕ′(b) = 0.

These are equivalent to

M

�ϕ(a)ϕ′(a)

�+N

�ϕ(b)ϕ′(b)

�=

�00

�.

where

M =

�1 23 −1

�and N =

�−3 1−1 0

�.

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2. Consider the following two-point boundary value problem in which L is apositive number.

−ϕ′′ = λϕ on [0, L],

ϕ′(0) = 0, and

ϕ(L) = 0.

(a) Show that all eigenvalues are positive.

Solution to part a. Suppose that λ is an eigenvalue and ϕ isa corresponding real valued eigenfunction. The Rayleigh quotientyields

λ =ϕ(0)ϕ′(0)− ϕ(L)ϕ′(L) +

�L

0(ϕ′(x))2 dx

�L

0(ϕ(x))2 dx

.

Using the boundary conditions, this becomes

λ =

� L0(ϕ′(x))2 dx

�L

0(ϕ(x))2 dx

.

Thus λ ≥ 0. The non zero constant functions do not satisfy thesecond boundary condition so

� L

0

(ϕ′(x))2dx > 0.

From this it follows that λ > 0.

(b) Find D(λ) and ∆(λ) in the case where λ > 0.

Solution to part b. The boundary conditions are equivalent to

M

�ϕ(a)ϕ′(a)

�+N

�ϕ(b)ϕ′(b)

�=

�00

�

where

M =

�0 10 0

�and N =

�0 01 0

�.

Remember that for this d.e. and λ > 0,we take

uλ(x) = cos√λx and vλ(x) = sin

√λx

so that

Φλ(x) =

�cos ρx sin ρx−ρ sin ρx ρ cos ρx

�

2

whereρ =

√λ.

Thus

D(λ) =MΦλ(0) +NΦλ(L) =

�0 ρ

cos ρL sin ρL

�

and∆(λ) = −ρ cos ρL

(c) Give a proper listing of eigenvalues and eigenfunctions for the prob-lem.

Solution for part c.

∆(λ) = 0 if and only if λ =

�(2k − 1)π2L

�2for some positive integer k

and when k is a positive integer

D

��(2k − 1)π2L

�2��c1c2

�=

�0(2k−1)π2L

0 (−1)k

��c1c2

�=

�00

�

if and only ifc2 = 0.

Thus a proper listing of eigenvalues and eigenfunctions is {λk}∞k=1and {ϕk}∞k=1 where

λk =

�(2k − 1)π2L

�2for k = 1, 2, . . .

and

ϕk(x) = cos

�(2k − 1)πx

2L

�for 0 ≤ x ≤ L and k = 1, 2, . . . .

3. Consider the following two-point boundary value problem.

−ϕ′′ = λϕ on [0, 1],

3ϕ(0) + ϕ′(0) = 0, and

ϕ(1) = 0.

(a) Find D(λ) and ∆(λ) in the case where λ < 0.

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Solution for part a.The boundary conditions are equivalent to

M

�ϕ(0)ϕ′(0)

�+N

�ϕ(1)ϕ′(1)

�=

�00

�

where

M =

�3 10 0

�and N =

�0 01 0

�.

Remember that for this d.e. and λ < 0 we take

uλ(x) = cosh√−λx and vλ(x) = sinh

√−λx

so that

Φλ(x) =

�cosh ρx sinh ρxρ sin ρx ρ cosh ρx

�

whereρ =

√−λ.

Thus

D(λ) =

�3 10 0

� �1 00 ρ

�+

�0 01 0

��cosh ρ sinh ρρ sinh ρ ρ cosh ρ

�.

So

D(λ) =

�3 ρ

cosh ρ sinh ρ

�

and∆(λ) = 3 sinh ρ− ρ cosh ρ.

(b) Show graphically that there is exactly one negative eigenvalue.

Solution for part b. Note that λ = −ρ2 is a negative eigenvalue ifand only if ρ > 0 and

3 sinh ρ− ρ cosh ρ = 0

or

tanh ρ =1

3ρ.

Analysis of the graph consisting of all (x, y) where x ≥ 0 and y = 1

3x

and the graph consisting of all (x, y) where x ≥ 0 and y = tanhxshows that there is exactly one such number ρ and it is near thenumber 3.

4

0 1 2 3 4 5 6 7 80.0

0.5

1.0

1.5

2.0

2.5

x

y

y = tanhx and y = x/3

.

(c) Give a numerical approximation for this eigenvalue.

Solution to pert c. Using Newton’s method with initial estimateof 3 or a graphing calculator, we find that the positive number x suchthat

tanhx− 13x = 0

is approximately2.984705

so the negative eigenvalue is approximately

−8.90846

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