1 Partial derivatives and Chain ruleguermond/M602_FALL_2020/all_exams.pdfMath 602 1 1 Partial...

Math 602 1

1 Partial derivatives and Chain rule

Question 1: Let u be a solution to the PDE ∂tu(x, t) +12∂xu

2(x, t) − ν∂xxu(x, t) = 0, x ∈(−∞,+∞), t > 0.(a) Let ψ(x, t) =

∫ x−∞ ∂tu(ξ, t)dξ +

12u

2(x, t)− ν∂xu(x, t). Compute ∂xψ(x, t).

Solution: The definition of ψ implies that

∂xψ(x, t) = ∂x(

∫ x−∞

∂tu(ξ, t)dx+1

2u2(x, t)− ν∂xu(x, t))

= ∂tu(x, t)x+1

2∂xu

2(x, t)− ∂xxu(x, t) = 0

i.e., ∂xψ(x, t) = 0. This means that ψ depends on t only.

(b) Let φ(x, t) := e−12ν

∫ x−∞ u(ξ,t)dξ. Compute ∂tφ, ∂xφ, and ∂xxφ.

Solution: The definition of φ, together with the chain rule, implies that

∂tφ(x, t) = ∂t

(− 1

2ν

∫ x−∞

u(ξ, t)dξ

)e−

12ν

∫ x−∞ u(ξ,t)dξ

=

(− 1

2ν

∫ x−∞

∂tu(ξ, t)dξ

)e−

12ν


and

∂xφ(x, t) = ∂x

(− 1

2ν

∫ x−∞

u(ξ, t)dξ

)e−

12ν


=

(− 1

2νu(x, t)

)e−

12ν


and

∂xxφ(x, t) =

(− 1

2ν∂xu(x, t)

)e−

12ν

∫ x−∞ u(ξ,t)dξ +

(− 1

2νu(x, t)

)2e−

12ν


(c) Compute ∂tφ− ν∂xxφ, assuming ψ(x, t) = 0.Solution: The above computations give

−ν∂xxφ(x, t) = −1

2ν

(−ν∂xu(x, t) +

1

2u2(x, t)

)e−

12ν


In conclusion

∂tφ− ν∂xxφ = −1

2ν

(∫ x−∞

∂tu(ξ, t)dξ +1

2u2(x, t)− ν∂xu(x, t)

)e−

12ν


= − 12νψ(x, t)e−

12ν

∫ x−∞ u(ξ,t)dξ.

This means ∂tφ− ν∂xxφ = 0.Question 2: Let d be positive integer and let u : Rd×R+ −→ R be the solution to the nonlinearPDE ∂tu(x, t)− a∆u(x, t) + b(∇u(x, t))·(∇u(x, t)) = 0, with u(x, 0) = f(x), where a > 0 andb 6= 0 are real numbers, and f(x) is a smooth function. Let v(x, t) = e− bau(x,t). (i) Compute∇v(x, t) and ∇·(∇v(x, t)) (Hint: compute ∂xiv(x, t) for i = 1, . . . , d, then

∑di=1 ∂xixiv(x, t). It

is the same question as question 1 in mdt1; it is worded differently.)

Solution: The chain rule gives

∂xiv(x, t) = −b

ae−

bau(x,t)∂xiu(x, t),

Math 602 2

and

d∑i=1

∂xixiv(x, t) = −b

ae−

bau(x,t)

d∑i=1

∂xixiu(x, t) +b2

a2e−

bau(x,t)

d∑i=1

∂xiu(x, t)∂xiu(x, t).

This implies that

∇v(x, t) = − ba

e−bau(x,t)∇u(x, t),

and

∆v(x, t) = − ba

e−bau(x,t)∆u(x, t) +

b2

a2e−

bau(x,t)(∇u(x, t))·(∇u(x, t)).

(ii) Compute ∂tv(x, t), ∂tv(x, t)− a∆v(x, t), and v(x, 0).Solution: Using again the chain rule we obtain

∂tv(x, t) = −b

ae−

bau(x,t)∂tu(x, t).

Combining the above results gives

∂tv(x, t)− a∆v(x, t) = −b

ae−

bau(x,t) (∂tu(x, t)− a∆u(x, t) + b(∇u(x, t))·(∇u(x, t))) = 0

i.e., v solves the heat equation ∂tv(x, t)− a∆v(x, t) = 0 with initial condition v(x, 0) = e−ba f(x).

(iii) Recall that the solution to the heat equation ∂tφ(x, t)−a∆φ(x, t) = 0 with initial conditionφ(x, 0) = ψ0(x) is given by ψ0(x)(4πat)

−d/2 ∫Rd e− ‖x−y‖

2

4at dy. Give the integral representationof u(x, t) in terms of a, b, and f .

Solution: Since ∂tv(x, t) − a∆v(x, t) = 0 with initial condition v(x, 0) = e−ba f(x), the integral

representation of the function u is then given by

u(x, t) = −ab

log(v(x, t)), with, v(x, t) = e−ba f(x)(4πat)−d/2

∫Rd

e−‖x−y‖2

4at dy

Question 3: Let u be a vector field in Rd, (d is the space dimension). Let p be a scalar fieldin Rd.(a) Using the product rule, express div(pu) in terms of ∇·u and ∇p. (You may use divu or ∇·uto denote the divergence of u and ∇p or gradp to denote the gradient of p. The space dimensionis d.)

Solution: Using the product rule we have

∇·(pu) =d∑i=1

∂i(pui) =

d∑i=1

ui∂ip+ p∂iui

= u·∇p+ p∇·u

In conclusion ∇·(pu) = u·∇p+ p∇·u.

(b) Let u be a smooth vector field in Ω ⊂ Rd with zero divergence and zero normal componentat the boundary of Ω. Let p be a smooth scalar field in Ω. Compute

∫Ωp(x)∇·u(x)dx and∫

Ωu(x)·∇p(x)dx.

Solution: (i) Using (a) and the fact that u is divergence free, we have∫Ω

u·∇pdx =∫

Ω

(∇·(pu)− p∇·u)dx =∫

Ω

∇·(pu)dx.

The divergence theorem (fundamental theorem of calculus in d space dimension) implies that∫Ω

u·∇pdx =∫∂Ω

pu·nds,

Math 602 3

where ∂Ω denotes the boundary of Ω. Since u·n = 0 at the boundary, we finally infer that∫Ωu·∇pdx = 0.

(ii) Note finally that∫

Ωp∇·udx = 0 since ∇·u = 0.

Question 4: Let φ be a smooth scalar field in Rd, (d is the space dimension). (a) Prove that∇φ·∂i(∇φ) = ‖∇φ‖∂i‖∇φ‖. Give all the details.Solution: Using the product rule, we have

∇φ·∂i(∇φ) =d∑j=1

∂jφ∂i(∂jφ) = ∂i(1

2

d∑j=1

(∂jφ)2) = ∂i(

1

2‖∇φ‖2) = ‖∇φ‖∂i‖∇φ‖.

(b) Let � > 0. Show that ∇·(√‖∇φ‖2 + �ei) = ∇φ√‖∇φ‖2+� ·∂i(∇φ) where ei is the unit vector in

the direction i.

Solution: Using the chain rule, we have

∇·(√‖∇φ‖2 + �ei) = ∂i(

√‖∇φ‖2 + �) = 1

2

1√‖∇φ‖2 + �

∂i(‖∇φ‖2 + �)

=‖∇φ‖√‖∇φ‖2 + �

∂i‖∇φ‖.

Then using (a) we infer that

∇·(√‖∇φ‖2 + �ei) =

∇φ√‖∇φ‖2 + �

·∂i(∇φ).

(c) Assume that φ(x) = 0 for all ‖x‖ ≥ R where R is a positive real number. Compute∫Rd

∇φ√‖∇φ‖2+�

·∂i(∇φ)dx, for all i = 1, . . . , d. Give all the details.

Solution: Using (b) we have∫Rd

∇φ√‖∇φ‖2 + �

·∂i(∇φ)dx =∫Rd∇·(√‖∇φ‖2 + �ei)dx,

where ei is the unit vector in the direction i. The divergence theorem together with the assumptionthat φ(x) = 0 for all ‖x‖ ≥ R implies that∫

Rd

∇φ√‖∇φ‖2 + �

·∂i(∇φ)dx = 0.

Question 5: Let y(x, t) = x cos(2t+ log(|x|)). Compute ∂tty + x2∂xxy − x∂xy + 6y.Solution: This exercise is meant to check whether you understand the notion of partial derivativesand the chain rule

∂tty(x, t) = −4x cos(2t+ log(|x|)) = −4y(x, t),

∂xy(x, t) = cos(2t+ log(|x|))− x sin(2t+ log(|x|))1

x= cos(2t+ log(|x|))− sin(2t+ log(|x|)),

∂xxy(x, t) = −1

xsin(2t+ log(|x|))− 1

xcos(2t+ log(|x|))

In conclusion

∂tty + x2∂xxy − x∂xy + 6y = −4x cos(2t+ log(|x|))− x sin(2t+ log(|x|))

− x cos(2t+ log(|x|))− x cos(2t+ log(|x|)) + x sin(2t+ log(|x|)) + 6x cos(2t+ log(|x|))= 0,

that is to say, y(x, t) solve the PDE ∂tty + x2∂y − x∂xy + 6y = 0.

Math 602 4

Question 6: Let ∇× be the curl operator acting on vector fields: i.e., let A = (A1, A2, A3) :R3 −→ R3 be a three-dimensional vector field over R3, then ∇×A = (∂2A3 − ∂3A2, ∂3A1 −∂1A3, ∂1A2−∂2A1). Accept as a fact that ∇·(A×B) = B·∇×A−A·∇×B for all smooth vectorfields A and B. Let Ω be a subset of R3 with a smooth boundary ∂Ω. Find an integration byparts formula for

∫ΩB·∇×Adx.

Solution: Using the divergence Theorem we infer that∫Ω

(B·∇×A−A·∇×B)dx =∫

Ω

∇·(A×B) =∫∂Ω

(A×B)·nds.

which implies that ∫Ω

B·∇×Adx =∫

Ω

A·∇×Bdx+∫∂Ω

(A×B)·nds.

Question 7: Let u, f : R −→ R be two functions of class C1. (a) Compute ∂xf(u(x)).Solution: Using the chain rule we obtain

∂xf(u(x)) = f′(u(x))∂xu.

where f ′ denotes the derive of f .

(b) Let ψ : R −→ R be functions of class C1. Let F : R −→ R be defined by F (v) =∫ v0f ′(t)ψ′(t)dt. Use (a) to compute ∂x(F (u(x))− ∂x(f(u(x)))ψ′(u(x)).

Solution: Using the chain rule we obtain

∂x(F (u(x)) = F′(u(x))∂xu(x) = f

′(u(x))ψ′(u(x))∂xu(x) = ∂x(f(u(x)))ψ′(u(x)).

This means that ∂x(F (u(x)) = ∂x(f(u(x)))ψ′(u(x)).

(c) Using the notation of (a) and (b), assume that u(±∞) = 0 and compute∫ +∞−∞ ∂x(f(u(x)))ψ

′(u(x))dx.

Solution: Using (b) and u(±∞) = 0 we have∫ +∞−∞

∂x(f(u(x)))ψ′(u(x))dx =

∫ +∞−∞

∂x(F (u(x)))dx = F (u(x))|+∞−∞ = F (0)− F (0) = 0.

Question 8: Let φ(x, y) = cos(x+ sin(x− y)). Compute ∂xφ(x, y) and ∂yφ(x, y). (Do not tryto simplify the results).

Solution: We apply the chain rule repeatedly

∂xφ(x, y) = − sin(x+ sin(x− y))(1 + cos(x− y)).

∂yφ(x, y) = − sin(x+ sin(x− y))(− cos(x− y)).

Question 9: Let φ = x4 − y4 (a) Compute ∆φ(x, y). (b) Consider the square Ω = [0, 1]×[0, 1]and let Γ be the boundary of Ω. Compute

∫Γ∂nφdΓ.

Solution: (a) The definition ∆φ = ∂xxφ+ ∂yyφ implies that

∆φ = ∂xxφ+ ∂yyφ = 12x2 − 12y2 = 12(x2 − y2).

(b)The definition ∆φ = div(∇φ) and the fundamental theorem of calculus (also known as thedivergence theorem) imply that∫

Γ

∂nφdΓ =

∫Γ

n·∇φdΓ =∫

Ω

div(∇φ)dΩ =∫

Ω

∆φdΩ =

∫Ω

12(x2 − y2)dxdy = 0.

Math 602 5

Question 10: Let φ = sin(x) − sin(y) (a) Compute ∆φ(x, y). (b) Consider the square Ω =[0, 1]×[0, 1] and let Γ be the boundary of Ω. Compute

∫Γ∂nφdΓ.


∆φ = ∂xxφ+ ∂yyφ = − sin(x) + sin(y).


Γ

∂nφdΓ =

∫Γ

n·∇φdΓ =∫

Ω

div(∇φ)dΩ =∫

Ω

∆φdΩ = −∫

Ω

sin(x)dxdy +

∫Ω

sin(x)dxdy = 0.

Question 11: Let φ(x, y) = log(2 + sin(x− y)). Compute ∂xφ(x, y) and ∂yφ(x, y). (Do not tryto simplify the results).

Solution: We apply the chain rule repeatedly

∂xφ(x, y) =1

2 + sin(x− y)cos(x− y)

∂yφ(x, y) = −1

2 + sin(x− y)cos(x− y).

Question 12: Let φ(x, y) =∫ sin(x)

0log(1+y2 +z2)dz. Given α, β, γ ∈ R, compute ∂xφ(α, α+β)

and ∂yφ(α− β, γ). (Do not try to simplify the results).Solution: Recalling the fundamental theorem of calculus

∂t

(∫ t0

f(z)dz

)= f(t),

we apply the chain rule repeatedly

∂xφ(x, y) = log(1 + y2 + sin(x)2) cos(x)

This means that∂xφ(α, α+ β) = log(1 + (α+ β)

2 + sin(α)2) cos(α)

Recalling that

∂t

∫ vu

f(s, t)ds =

∫ vu

∂tf(s, t)ds,

we apply the chain rule repeatedly

∂yφ(x, y) =

∫ sin(x)0

2y

1 + y2 + z2dz

This means that

∂yφ(α− β, γ) =∫ sin(α−β)

0

2γ

1 + γ2 + z2dz.

Question 13: Let φ = x2 + 2y2 (a) Compute ∆φ(x, y). (b) Consider the disk of radius 1centered at (0, 0) and let Γ be the boundary of Ω. Compute

∫Γ∂nφdΓ.


∆φ = ∂xxφ+ ∂yyφ = 2 + 4 = 6.


Γ

∂nφdΓ =

∫Γ

n·∇φdΓ =∫

Ω

div(∇φ)dΩ =∫

Ω

∆φdΩ = 6

∫Ω

dΩ = 6π,

because the surface of Ω,∫

ΩdΩ, is equal to π.

Math 602 6

2 Heat equation

Question 14: Let u be the solution of ∂tu = ∂xxu + 2, x ∈ (0, L), with ∂xu(0, t) = α,∂xu(L, t) = 3, u(x, 0) = f(x).

(a) Let α = 1. Compute∫ L

0u(x, t)dx as a function of t.

Solution: dt∫ L

0u(x, t)dx = ∂xu|L0 + 2L. That is dt

∫ L0u(x, t)dx = 2 + 2L. This implies∫ L

0u(x, t)dx = (2 + 2L)t+

∫ L0f(x)dx.

(b) Let α be an arbitrary number. For which value of α,∫ L

0u(x, t)dx does not depend on t?

Solution: The above computation yields∫ L

0u(x, t)dx = (3 − α + 2L)t +

∫ L0f(x)dx. This is

independent of t if (3− α+ 2L) = 0, meaning α = 3 + 2L.Question 15: Let u solve ∂tu − ∂xxu = 5, x ∈ (0, L), with ∂xu(0, t) = α, ∂xu(L, t) = −3,u(x, 0) = f(x).

(a) Let α = 1. Compute∫ L


Solution: Integrate the equation over (0, L):

dt

∫ L0

u(x, t)dx =

∫ L0

∂tu(x, t)dx =

∫ L0

∂xxudx+ 5L = ∂xu|L0 + 5L.

That is dt∫ L

0u(x, t)dx = −4 + 5L. This implies

∫ L0u(x, t)dx = (−4 + 5L)t+

∫ L0f(x)dx.




0u(x, t)dx = (−3 − α + 5L)t +


independent of t if (−3− α+ 5L) = 0, meaning α = −3 + 5L.Question 16: Let u solve ∂tu−∂x((2x+1)∂xu) = 3, x ∈ (0, L), with ∂xu(0, t) = α, ∂xu(L, t) =−1, u(x, 0) = f(x).(a) Let α = 1. Compute

∫ L0u(x, t)dx as a function of t.

Solution: Integrate the equation over the domain (0, L):

dt

∫ L0

u(x, t)dx =

∫ L0

∂tu(x, t)dx =

∫ L0

∂x((2x+ 1)∂xu)dx+ 3L

= (2L+ 1)∂xu(L, t)− ∂xu(0, t) + 3L = −(2L+ 1)− α+ 3L= L− 1− α

That is dt∫ L

0u(x, t)dx = L− 2. This implies

∫ L0u(x, t)dx = (L− 2)t+

∫ L0f(x)dx.




0u(x, t)dx = (L − 1 − α)t +


independent of t if L− 1− α = 0, meaning α = L− 1.Question 17: Let u solve ∂tu−∂x((3x+1)∂xu) = −3, x ∈ (0, L), with ∂xu(0, t) = 1, ∂xu(L, t) =α, u(x, 0) = f(x).

(a) Compute∫ L


Solution: Integrate the equation over the domain (0, L):

dt

∫ L0

u(x, t)dx =

∫ L0

∂tu(x, t)dx =

∫ L0

∂x((3x+ 1)∂xu)dx− 3L

= (3L+ 1)∂xu(L, t)− ∂xu(0, t)− 3L = (3L+ 1)α− 1− 3L= (3L+ 1)(α− 1)

That is dt∫ L

0u(x, t)dx = (3L+1)(α−1). This implies

∫ L0u(x, t)dx = (3L+1)(α−1)t+

∫ L0f(x)dx.

Math 602 7

(b) For which value of α the quantity∫ L



0u(x, t)dx = (3L + 1)(α − 1)t +


independent of t if (3L+ 1)(α− 1) = 0, meaning α = 1.

Question 18: Consider the differential equation −d2φdt2 = λφ, t ∈ (0, π), supplemented with the

boundary conditions φ(0) = 0, 3φ(π) = −φ′(π).(a) Prove that it is necessary that λ be positive for a non-zero solution to exist.

Solution: (i) Let φ be a non-zero solution to the problem. Multiply the equation by φ and integrateover the domain. ∫ π

0

(φ′(t))2dt− φ′(π)φ(π) + φ′(0)φ(0) = λ∫ π

0

φ2(t)dt.

Using the BCs, we infer ∫ π0

(φ′(t))2dt+ 3φ(π)2 = λ

∫ π0

φ2(t)dt,

which means that λ is non-negative since φ is non-zero.

(ii) If λ = 0, then∫ π

0(φ′(t))2dt = 0 and φ(π)2 = 0, which implies that φ′(t) = 0 and φ(π) = 0.

The fundamental theorem of calculus implies φ(t) = φ(π) +∫ tπφ′(τ)dτ = 0. Hence, φ is zero if

λ = 0. Since we want a nonzero solution, this implies that λ cannot be zero.

(iii) In conclusion, it is necessary that λ be positive for a nonzero solution to exist.

(b) Find the equation that λ must solve for the above problem to have a nonzero solution.

Solution: Since λ is positive, φ is of the following form

φ(t) = c1 cos(√λt) + c2 sin(

√λt).

The boundary condition φ(0) = 0 implies c1 = 0. The other boundary condition φ′(π) = −3φ(π)

implies√λc2 cos(

√λπ) = −3c2 sin(

√λπ). The constant c2 cannot be zero since we want φ to be

nonzero; as a result, λ must solve the following equation

√λ cos(

√λπ) + 3 sin(

√λπ) = 0,

for a nonzero solution φ to exist.

Question 19: Consider the Laplace equation −∆u = 0 in the rectangle {(x, y); x ∈ [0, L], y ∈[0, H]} with the boundary conditions u(0, y) = 0, u(L, y) = 3 cos( 52π

yH ), ∂yu(x, 0) = 0,

u(x,H) = 0. Solve the equation using the method of separation of variables. (Give all thedetails.)

Solution: Let u(x) = φ(x)ψ(y). Then, provided ψ and φ are non zero functions, this impliesφ”(x)φ(x) = −

ψ”(y)ψ(y) = λ. Observe that ψ

′(0) = 0 and ψ(H) = 0. The energy technique applied to

−ψ”(y) = λψ(y) gives∫ H0

−ψ”(y)ψ(y)dy =∫ H

0

ψ′(y)2dy − ψ′(H)ψ(H) + ψ′(0)ψ(0) = λ∫ H

0

ψ(y)2y,

which implies∫H

0ψ′(y)2dy = λ

∫H0ψ(y)2y since ψ′(0) = 0 and ψ(H) = 0. This in turn implies

that λ is nonnegative. Actually λ cannot be zero since it would mean that ψ = 0, which wouldcontradict the fact that the solution u is nonzero (λ = 0 ⇒ ψ′(y) = 0 ⇒ ψ(y) = ψ(H) = 0 forall y ∈ [0, H]). As a result λ is positive and ψ(y) = a cos(

√λy) + b sin(

√λy). The Neumann

condition at y = 0 gives b = 0. The Dirichlet condition at H implies cos(√λH) = 0, which implies√

λH = (n + 12 )π, where n is any integer. This means that ψ(y) = a cos((n +12 )π

yH ). The fact

that λ is positive implies φ(x) = c cosh(√λx) + d sinh(

√λx). The boundary condition at x = 0

implies c = 0. Then

u(x, y) = A cos((n+ 12 )πy

H) sinh((n+ 12 )π

x

H).

Math 602 8

The boundary condition at x = L gives

3 cos(5

2πy

H) = A cos((n+ 12 )π

y

H) sinh((n+ 12 )π

L

H),

which implies n = 2 and A = sinh−1(

52πL

H

), i.e.,

u(x, y) = 3sinh

(52πx

H

)sinh

(52πL

H

) cos(52πy

H).

Question 20: Let k : [−1,+1] −→ R be such that k(x) = 1, if x ∈ [−1, 0] and k(x) = 2if x ∈ (0, 1]. Solve the boundary value problem −∂x(k(x)∂xT (x)) = 0 with T (−1) = 0 and∂xT (1) = 1.(i) What should be the interface conditions at x = 0 for this problem to make sense?

Solution: The function T and the flux k(x)∂xT (x) must be continuous at x = 0. Let T− denotethe solution on [−1, 0] and T+ the solution on [0,+1]. One should have T−(0) = T+(0) andk−(0)∂xT

−(0) = k+(0)∂xT+(0), where k−(0) = 1 and k+(0) = 2.

(ii) Solve the problem, i.e., find T (x), x ∈ [−1,+1].Solution: On [−1, 0] we have k−(x) = 1, which implies ∂xxT−(x) = 0. This in turn impliesT−(x) = a+ bx. The Dirichlet boundary condition at x = −1 implies T−(−1) = 0 = a− b. Thisgives a = b and T−(x) = a(1 + x).

We proceed similarly on [0,+1] and we obtain T+(x) = c+ dx. The Neumann boundary conditionat x = +1 gives ∂xT

+(+1) = 1 = d and T+(x) = c+ x.

The interface conditions T−(0) = T+(0) and k−(0)∂xT−(0) = k+(0)∂xT

+(0) give

a = c, and a = 2.

In conclusion

T (x) =

{2(1 + x) if x ∈ [−1, 0],2 + x if x ∈ [0,+1].

Question 21: Let u solve ∂tu − ∂x((sin(x) + 2)∂xu) = g(x)e−t, x ∈ (0, L), with ∂xu(0, t) =sin(L) + 2, ∂xu(L, t) = 2, u(x, 0) = f(x), where f and g are two smooth functions.

(a) Compute ddt∫ L


Solution: Integrate the equation over the domain (0, L) and apply the fundamental Theorem ofcalculus:

d

dt

∫ L0

u(x, t)dx =

∫ L0

∂tu(x, t)dx =

∫ L0

∂x((sin(x) + 2)∂xu)dx+ e−t∫ L

0

g(x)dx

= (sin(L) + 2)∂xu(L)− (sin(0) + 2)∂xu(0) + e−t∫ L

0

g(x)dx

= (sin(L) + 2)2− 2(sin(L) + 2) + e−t∫ L

0

g(x)dx

= e−t∫ L

0

g(x)dx.

That isd

dt

∫ L0

u(x, t)dx = e−t∫ L

0

g(x)dx.

(b) Use (a) to compute∫ L


Math 602 9

Solution: Applying the fundamental Theorem of calculus again gives∫ L0

u(x, T )dx =

∫ L0

u(x, 0)dx+

∫ T0

d

dt

∫ L0

u(x, t)dxdt

=

∫ L0

f(x)dx+ (1− e−T )∫ L

0

g(x)dx.

(c) What is the limit of∫ L

0u(x, t)dx as t→ +∞?

Solution: The above formula gives

limT→+∞

∫ L0

u(x, T )dx =

∫ L0

f(x)dx+

∫ L0

g(x)dx.

Question 22: Let u solve ∂tu + ∂x(v(x, t)u − µ(x, t)∂xu

)= g(x)e−t, x ∈ (0, L), t > 0, with

µ(0, t)∂xu(0, t) = 1, µ(L, t)∂xu(L, t) = 1 + 2e−t, u(x, 0) = f(x), where v, µ > 0, f and g are

four smooth functions and v(0) = v(L) = 0.



Solution: Integrate the equation over the domain (0, L) and apply the Fundamental Theorem ofcalculus:

d

dt

∫ L0

u(x, t)dx =

∫ L0

∂tu(x, t)dx =

∫ L0

∂x(−v(x, t)u+ µ(x, t)∂xu)dx+ e−t∫ L

0

g(x)dx

= µ(L, t)∂xu(L)− µ(0, t)∂xu(0) + e−t∫ L

0

g(x)dx

= 1 + 2e−t − 1 + e−t∫ L

0

g(x)dx.

That isd

dt

∫ L0

u(x, t)dx = e−t(

∫ L0

g(x)dx+ 2).



Solution: Applying the Fundamental Theorem of calculus again (with respect to time this time)gives ∫ L

0

u(x, T )dx =

∫ L0

u(x, 0)dx+

∫ T0

e−tdt(

∫ L0

g(x)dx+ 2).

=

∫ L0

f(x)dx+ (1− e−T )(∫ L

0

g(x)dx+ 2).




limT→+∞

∫ L0

u(x, T )dx =

∫ L0

f(x)dx+

∫ L0

g(x)dx+ 2.

Question 23: Let u solve ∂tu−∂x(µ(x, t)∂xu) = g(x)e−t, x ∈ (0, L), t > 0, with µ(0, t)∂xu(0, t) =1, µ(L, t)∂xu(L, t) = 1+2e

−t, u(x, 0) = f(x), where µ > 0, f and g are three smooth functions.



Math 602 10

Solution: Integrate the equation over the domain (0, L) and apply the fundamental Theorem ofcalculus:

d

dt

∫ L0

u(x, t)dx =

∫ L0

∂tu(x, t)dx =

∫ L0

∂x(µ(x, t)∂xu)dx+ e−t∫ L

0

g(x)dx

= µ(L, t)∂xu(L)− µ(0, t)∂xu(0) + e−t∫ L

0

g(x)dx

= 1 + 2e−t − 1 + e−t∫ L

0

g(x)dx.

That isd

dt

∫ L0

u(x, t)dx = e−t(

∫ L0

g(x)dx+ 2).



Solution: Applying the fundamental Theorem of calculus again gives∫ L0

u(x, T )dx =

∫ L0

u(x, 0)dx+

∫ T0

e−tdt(

∫ L0

g(x)dx+ 2).

=

∫ L0

f(x)dx+ (1− e−T )(∫ L

0

g(x)dx+ 2).




limT→+∞

∫ L0

u(x, T )dx =

∫ L0

f(x)dx+

∫ L0

g(x)dx+ 2.

Question 24: Consider the heat equation ∂tT−k∂xxT = f(x), x ∈ [a, b], t > 0, with f(x) = kx,where k > 0. Compute the steady state solution (i.e., ∂tT = 0) assuming the boundaryconditions: −k∂nT (a) = 0, T (b) = 0 (∂n is the normal derivative)Solution: At steady state, T does not depend on t and we have ∂xxT (x) = −x, which implies∂xT (x) = α − 12x

2, and T (x) = β + αx − 16x3, where α, β ∈ R. The two constants α and β

are determined by the boundary conditions. 0 = −∂nT (a) = ∂xT (a) = α − 12a2 and 0 = T (b) =

β + αb− 16b3. We conclude that α = 12a

2 and β = −αb+ 16b3 = − 12a

2b+ 16b3. In conclusion

T (x) = −12a2b+

1

6b3 +

1

2a2x− 1

6x3 = −1

2a2(b− x) + 1

6(b3 − x3).

Question 25: Consider the equation ∂tc(x, t) − ∂xxc(x, t) = x, where x ∈ [0, L], t > 0, withc(x, 0) = f(x), −∂nc(0, t) = 6, −∂nc(L, t) = 5, (∂n is the normal derivative). Compute E(t) :=∫ L

0c(ξ, t)dξ.

Solution: We integrate the equation with respect to x over [0, L]∫ L0

∂tc(ξ, t)dξ −∫ L

0

∂ξξc(ξ, t)dξ =

∫ L0

ξdξ.

Using that∫ L

0∂tc(ξ, t)dξ = dt

∫ T0c(ξ, t)dξ together with the fundamental theorem of calculus, we

infer that

dtE(t)− ∂xc(L, t) + ∂xc(0, t) =1

2L2.

The boundary conditions ∂xc(0, t) = −∂nc(0, t) = 6, −∂xc(L, t) = −∂nc(L, t) = 5 give

dtE(t) + 5 + 6 =1

2L2.

Math 602 11

We now apply the fundamental theorem of calculus with respect to t

E(t)− E(0) =∫ t

0

∂τE(τ)dτ = (1

2L2 − 11)t.

In conclusion

E(t) =

∫ L0

f(ξ)dξ + (1

2L2 − 11)t.

Question 26: Consider the heat equation ∂tu(x, t)−3∂xxu(x, t) = 0, ∂xu(0, t) = 0, ∂xu(1, t) =0, u(x, 0) = u0(x), t > 0, x ∈ (0, 1). The general solution is u(x, t) =

∑∞n=0An cos(nπx)e

−3n2π2t.Compute the solution corresponding to the initial data u0(x) = 5 cos(4πx).

Solution: The solution contains one term only, corresponding to n = 4,

u(x, t) = 5 cos(4πx)e−48π2t.

Question 27: Consider the heat equation ∂tu(x, t) − 2∂xxu(x, t) = 0, u(0, t) = 0, u(1, t) = 0,u(x, 0) = u0(x), t > 0, x ∈ (0, 1). The general solution is u(x, t) =

∑∞n=0An sin(nπx)e

−2n2π2t.Compute the solution corresponding to the initial data u0(x) = 3 sin(4πx)− 5 sin(πx).Solution: The solution contains two terms only, corresponding to n = 1 and n = 4,

u(x, t) = −5 sin(πx)e−2π2t + 3 sin(4πx)e−32π

2t.

Question 28: Consider the equation ∂tc(x, t) − ∂xxc(x, t) = 6x/L2, where x ∈ [0, L], t > 0,with c(x, 0) = f(x), −∂nc(0, t) = 1, −∂nc(L, t) = 2, (∂n is the normal derivative). ComputeE(t) :=

∫ L0c(ξ, t)dξ.



0

∂ξξc(ξ, t)dξ =6

L2

∫ L0

ξdξ.

Using that∫ L



infer thatdtE(t)− ∂xc(L, t) + ∂xc(0, t) = 3.

The boundary conditions ∂xc(0, t) = −∂nc(0, t) = 1, −∂xc(L, t) = −∂nc(L, t) = 2 give

dtE(t) + 2 + 1 = 3.


E(t)− E(0) =∫ t

0

∂τE(τ)dτ = 0.

In conclusion

E(t) =

∫ L0

f(ξ)dξ, ∀t ≥ 0.

Question 29: Consider the equation ∂tc(x, t)−∂x((1+x2)∂xc(x, t)

)= 6x/L2, where x ∈ [0, L],

t > 0, with c(x, 0) = f(x), −∂nc(0, t) = 1, −∂nc(L, t) = 21+L2 , (∂n is the normal derivative).Compute E(t) :=

∫ L0c(ξ, t)dξ.



0

∂ξ((1 + ξ2)∂ξc(ξ, t)

)dξ =

6

L2

∫ L0

ξdξ.

Math 602 12

Using that∫ L



infer thatdtE(t)− (1 + L2)∂xc(L, t) + ∂xc(0, t) = 3.

The boundary conditions ∂xc(0, t) = −∂nc(0, t) = 1, −∂xc(L, t) = −∂nc(L, t) = 21+L2 give

dtE(t) + 2 + 1 = 3.


E(t)− E(0) =∫ t

0

∂τE(τ)dτ = 0.

In conclusion

E(t) =

∫ L0

f(ξ)dξ, ∀t ≥ 0.

Question 30: Consider the heat equation ∂tu(x, t) − 2∂xxu(x, t) = 0, u(0, t) = 0, u(1, t) = 0,u(x, 0) = u0(x), t > 0, x ∈ (0, 1). The general solution is u(x, t) =

∑∞n=0An sin(nπx)e

−2n2π2t.Compute the solution corresponding to the initial data u0(x) = 3 sin(4πx).

Solution: The solution contains one term only, corresponding to n = 4,

u(x, t) = 3 sin(4πx)e−32π2t.

Question 31: Consider the heat equation ∂tT −k∂xxT = f(x), x ∈ [a, b], t > 0, with f(x) = 0,where k > 0. Compute the steady state solution (i.e., ∂tT = 0) assuming the boundaryconditions: −k∂nT (a) = 1, T (b) = 0 (∂n is the normal derivative).Solution: At steady state, T does not depend on t and we have ∂xxT (x) = 0, which implies∂xT (x) = α, and T (x) = β + αx, where α, β ∈ R. The two constants α and β are determined bythe boundary conditions. 1 = −k∂nT (a) = k∂xT (a) = kα and 0 = T (b) = β + αb. We concludethat α = 1k and β = −αb = −

bk . In conclusion

T (x) =x− bk

.

Math 602 13

3 Laplace equation

Question 32: (a) Find a function U(x, y) = a + bx + cy + dxy, such that U(x, 0) = x,U(1, y) = 1 + y, U(x, 1) = 3x− 1, and U(0, y) = −y.Solution:

U(x, y) = x− y + 2xy

solves the problem.

(b) Use (a) to solve the PDE uxx + uyy = 0, ∀(x, y) ∈ (0, 1)×(0, 1), with the boundaryconditions u(x, 0) = 3 sin(πx) + x, u(1, y) = 1 + y, u(x, 1) = 3x− 1,and u(0, y) = sin(2πy)− y.Solution: By setting φ = u − U , we observe that φxx + φyy = 0 and at the boundary of thedomain we have

φ(x, 0) = 3 sin(πx), φ(1, y) = 0, φ(x, 1) = 0, and φ(0, y) = sin(2πy).

It is clear that

φ(x, y) = 3 sin(πx)sinh(π(1− y))

sinh(π)+

sinh(2π(1− x))sinh(2π)

sin(2πy)

Then,u(x, y) = φ(x, y) + U(x, y)

Question 33: Consider the Laplace equation ∆u = 0 in the rectangle x ∈ [0, L], y ∈ [0, H]with the boundary conditions u(0, y) = 0, u(L, y) = 0, u(x, 0) = 0, u(x,H) = f(x).(a) Is there any compatibility condition that f must satisfy for a smooth solution to exist?

Solution: f must be such that f(0) = 0 and f(L) = 0, otherwise u would not be continuous atthe two upper corners of the domain.

(b) Solve the Equation.

Solution: Use the separation of variable technique. Let u(x) = φ(x)ψ(y). Then, provided ψ and

φ are non zero functions, this implies φ′′

φ = −ψ′′

ψ = λ. Observe that φ(0) = φ(L) = 0. The usual

energy technique implies that λ is negative. That is to say φ(x) = a cos(√λx) + b sin(

√λx). The

boundary conditions imply a = 0 and√λL = nπ, i.e., φ(x) = b sin(nπx/L). The fact that λ is

negative implies ψ(y) = c cosh(√λy) + d sinh(

√λy). The boundary condition at y = 0 implies

c = 0. Then

u(xy) =

∞∑n=1

An sin(nπx

L) sinh(

nπy

L)

Question 34: Solve the PDE (note that the width and the height of the rectangle are notequal)

∂xxu+ ∂yyu = 0, 0 < x < 1, 0 < y < 2,

u(x, 0) = 8 sin(9πx), u(x, 2) = 0, 0 < x < 1,

u(0, y) = sin(2πy), u(1, y) = 0, 0 < y < 2.

Solution: The method of separation of variables tells us that the solution is a sum of termslike sin(nπx) sinh(nπ(y − 2)) and sin(mπy/2) sinh(mπ(x − 1)/2). By looking at the boundaryconditions we infer that there are two nonzero terms in the expansion: one corresponding to n = 9and one corresponding to m = 4. This gives

u(x, y) = 8 sin(9πx)sinh(9π(2− y))

sinh(18π))+ sin(2πy)

sinh(2π(1− x))sinh(2π))

Math 602 14

Question 35: Consider the Laplace equation ∆u = 0 in the rectangle x ∈ [0, L], y ∈ [0, H]with the boundary conditions u(0, y) = 0, u(L, y) = 0, u(x, 0) = 0, u(x,H) = f(x).(a) Is there any compatibility condition that f must satisfy for a smooth solution to exist?

Solution: f must be such that f(0) = 0 and f(L) = 0, otherwise u would not be continuous atthe two upper corners of the domain.

(b) Solve the Equation.

Solution: Use the separation of variable technique. Let u(x) = φ(x)ψ(y). Then, provided ψ and

φ are non zero functions, this implies φ′′

φ = −ψ′′

ψ = λ. Observe that φ(0) = φ(L) = 0. The usual

technique implies that λ is negative. That is to say φ(x) = a cos(√λx)+b sin(

√λx). The boundary

conditions imply a = 0 and√λL = nπ, i.e., φ(x) = b sin(nπx/L). The fact that λ is negative

implies ψ(y) = c cosh(√λy) + d sinh(

√λy). The boundary condition at y = 0 implies c = 0. Then

the ansatz is

u(x, y) =

∞∑n=1

An sin(nπx

L) sinh(

nπy

L),

and the usual computation gives

An =2

L sinh(nπHL )

∫ L0

f(ξ) sin(nπξ

L)dξ.

Question 36: Consider the Laplace equation ∆u = 0 in the rectangle x ∈ [0, L], y ∈ [0, H]with the boundary conditions u(0, y) = 0, ∂xu(L, y) = 0, u(x, 0) = 0, u(x,H) = sin(

32πx/L).

Solve the Equation using the method of separation of variables. (Give all the details.)

Solution: Let u(x) = φ(x)ψ(y). Then, provided ψ and φ are non zero functions, this impliesφ′′

φ = −ψ′′

ψ = λ. Observe that φ(0) = 0 and φ′(L) = 0. The usual energy technique implies that

λ is negative. That is to say φ(x) = a cos(√λx) + b sin(

√λx). The Dirichlet condition at x = 0

implies a = 0. The Neumann condition at L implies cos(√λL) = 0, which implies

√λL = (n+ 12 )π,

where n is an integer. This means that φ(x) = b sin((n + 12 )πx/L). The fact that λ is negative

implies ψ(y) = c cosh(√λy) + d sinh(

√λy). The boundary condition at y = 0 implies c = 0. Then

u(x, y) = A sin((n+ 12 )πx

L) sinh((n+ 12 )π

y

L).

The boundary condition at y = H gives

sin(3

2πx

L) = A sin((n+ 12 )π

x

L) sinh((n+ 12 )π

H

L),

which implies n = 1 and A = sinh−1(

32πH

L

), i.e.,

u(x, y) =sinh

(32πy

L

)sinh

(32πH

L

) sin(32πx/L)

Question 37: Consider the equation −∆u = 0 in the rectangle {(x, y); x ∈ [0, L], y ∈ [0, H]}with the boundary conditions u(0, y) = 0, u(L, y) = −5 cos( 32π

yH ), ∂yu(x, 0) = 0, u(x,H) = 0.

Solve the equation using the method of separation of variables. (Give all the details.)

Solution: Let u(x) = φ(x)ψ(y). Then, provided ψ and φ are non zero functions, this impliesφ”(x)φ(x) = −

ψ”(y)ψ(y) = λ. Observe that ψ

′(0) = 0 and ψ(H) = 0. The energy technique applied to

−ψ”(y) = λψ(y) gives∫ H0

−ψ”(y)ψ(y)dy =∫ H

0

ψ′(y)2dy − ψ′(H)ψ(H) + ψ′(0)ψ(0) = λ∫ H

0

ψ(y)2y,

which implies∫H

0ψ′(y)2dy = λ

∫H0ψ(y)2dy since ψ′(0) = 0 and ψ(H) = 0. This in turn implies

that λ is nonnegative. Actually λ cannot be zero since it would mean that ψ = 0, which would

Math 602 15

contradict the fact that the solution u is nonzero (λ = 0 ⇒ ψ′(y) = 0 ⇒ ψ(y) = ψ(H) = 0 for ally ∈ [0, H]). As a result λ is positive and

ψ(y) = a cos(√λy) + b sin(

√λy).

The Neumann condition at y = 0 gives b = 0. The Dirichlet condition at H implies cos(√λH) = 0,

which implies√λH = (n+ 12 )π, where n is any integer. This means that ψ(y) = a cos((n+

12 )π

yH ).

The fact that λ is positive implies φ(x) = c cosh(√λx) + d sinh(

√λx). The boundary condition at

x = 0 implies c = 0. Then

u(x, y) = A cos((n+ 12 )πy

H) sinh((n+ 12 )π

x

H).

The boundary condition at x = L gives

−5 cos(32πy

H) = A cos((n+ 12 )π

y

H) sinh((n+ 12 )π

L

H),

which, by identification, implies 1 = 2 and A = −5 sinh−1(

32πL

H

), i.e.,

u(x, y) = −5sinh

(32πx

H

)sinh

(32πL

H

) cos(32πy

H).

Question 38: Does any of the following expressions solve the Laplace equation inside therectangle 0 ≤ x ≤ L, 0 ≤ y ≤ H, with the following boundary conditions ∂xu(0, y) =15πH sin(

5πyH ) cosh(

5πLH ), ∂xu(L, y) = 0, u(x, 0) = 0, u(x,H) = 0? (justify clearly your answer):

u1(x, y) = 3 cos(5πy

H) cosh(

5π(x− L)H

), u2(x, y) = 3 sin(5πy

H) cosh(

5π(x− L)H

),


H) sinh(

5π(x− L)H

), u4(x, y) = 3 sin(5πy

H) sinh(

5π(x− L)H

).

Solution: From class, we know that all the above expressions solve the Laplace equation, hence wejust need to verify the boundary conditions. We observe that u1 and u3 do not satisfy the Dirichletboundary conditions u(x, 0) = 0, u(x,H) = 0; therefore u1 and u3 must be discarded.

Both u2 and u4 satify that Dirichlet conditions: u2(x, 0) = 0, u2(x,H) = 0, and u4(x, 0) = 0,u4(x,H) = 0. Now we need to check the Neumann conditions.

Note that u4 is such that ∂xu4(L, y) = 35πH sin(

5πyH ) cosh(0) 6= 0; a result u4 must be discarded as

well.

Finally u2 is such that ∂xu2(L, y) = 35πH sin(

5πyH ) sinh(0) = 0, but ∂xu2(0, y) = 3

5πH sin(

5πyH ) sinh(−

5πLH ),

which shows that u2 is not the solution to our problem either.

In conclusion, none of the proposed solutions solve the problem. The correct solution is

u(x, y) = 3cosh( 5πLH )

sinh(−5πLH )sin(

5πy

H) cosh(

5π(x− L)H

).

Question 39: Does any of the following expressions solve the Laplace equation inside therectangle 0 ≤ x ≤ L, 0 ≤ y ≤ H, with the following boundary conditions u(0, y) = 0, u(L, y) =3 sinh(5πLH ) sin(

5πyH ), u(x, 0) = 0, u(x,H) = 0? (justify clearly your answer):


H) cosh(

5π(x− L)H

), u2(x, y) = 3 sin(5πy

H) cosh(

5π(x− L)H

),


H) sinh(

5πx

H), u4(x, y) = 3 sin(

5πy

H) sinh(

5πx

H).

Math 602 16

Solution: From class, we know that all the above expressions solve the Laplace equation, hence wejust need to verify that the boundary conditions are met. We observe that u1 and u3 do not satisfythe Dirichlet boundary conditions u(x, 0) = 0, u(x,H) = 0; therefore u1 and u3 must be discarded.

Both u2 and u4 satisfy that Dirichlet conditions: u2(x, 0) = 0, u2(x,H) = 0, and u4(x, 0) = 0,u4(x,H) = 0. Now we need to check the Neumann conditions.

Note that u2 is such that u2(0, y) = 3 sin(5πyH ) cosh(

5π(−L)H ) 6= 0, which shows that u2 is not the

solution to our problem either.

Finally u4(0, y) = 3 sin(5πyH ) sinh(0) = 0 and u4(L, y) = 3 sin(

5πyH ) sinh(

5πLH ); a result u4 is the

solution.

3.1 Cylindrical coordinates

Question 40: Using cylindrical coordinates and the method of separation of variables, solvethe Laplace equation, 1r∂r(r∂ru) +

1r2 ∂θθu = 0, inside the domain D = {θ ∈ [0,

π2 ], r ∈ [0, 1]},

subject to the boundary conditions ∂θu(r, 0) = 0, u(r,π2 ) = 0, u(1, θ) = cos(3θ).

Solution: We set u(r, θ) = φ(θ)g(r). This means φ′′ = −λφ, with φ′(0) = 0 and φ(π2 ) = 0, andrdr(rdrg(r)) = λg(r). Then using integration by parts plus the boundary conditions we prove thatλ is non-negative. Then

φ(θ) = c1 cos(√λθ) + c2 sin(

√λθ).

The boundary condition φ′(0) = 0 implies c2 = 0. The boundary condition φ(π2 ) = 0 implies√

λπ2 = (2n + 1)π2 with n ∈ N. This means

√λ = (2n + 1). From class we know that g(r) is of

the form rα, α ≥ 0. The equality rdr(rdrrα) = λrα gives α2 = λ. The condition α ≥ 0 implies2n+ 1 = α. The boundary condition at r = 1 gives cos(3θ) = 12n+1 cos((2n+ 1)θ). This impliesn = 1. The solution to the problem is

u(r, θ) = r3 cos(3θ).

Question 41: Using cylindrical coordinates and the method of separation of variables, solvethe Laplace equation, 1r∂r(r∂ru) +


π2 ], r ∈ [0, 1]},

subject to the boundary conditions u(r, 0) = 0, u(r, π2 ) = 0, u(1, θ) = sin(2θ). (Give all thedetails.)

Solution: We set u(r, θ) = φ(θ)g(r). This means φ′′ = −λφ, with φ(0) = 0 and φ(π2 ) = 0,and rdr(rdrg(r)) = λg(r). The usual energy argument applied to the two-point boundary valueproblem

φ′′ = −λφ, φ(0) = 0, φ(π2

) = 0,

implies that λ is non-negative. If λ = 0, then φ(θ) = c1 + c2θ and the boundary conditions implyc1 = c2 = 0, i.e., φ = 0, which in turns gives u = 0 and this solution is incompatible with theboundary condition u(1, θ) = sin(2θ). Hence λ > 0 and


√λθ).

The boundary condition φ(0) = 0 implies c1 = 0. The boundary condition φ(π2 ) = 0 implies√

λπ2 = nπ with n ∈ N. This means√λ = 2n. From class we know that g(r) is of the form rα,

α ≥ 0. The equality rdr(rdrrα) = λrα gives α2 = λ. The condition α ≥ 0 implies 2n = α. Theboundary condition at r = 1 gives sin(2θ) = 12n sin(2nθ) for all θ ∈ [0, π2 ]. This implies n = 1.The solution to the problem is

u(r, θ) = r2 sin(2θ).

Question 42: Using cylindrical coordinates and the method of separation of variables, solvethe equation, 1r∂r(r∂ru) +

1r2 ∂θθu = 0, inside the domain D = {θ ∈ [0, π], r ∈ [0, 1]}, subject

to the boundary conditions u(r, 0) = 0, u(r, π) = 0, u(1, θ) = 2 sin(5θ). (Give all the details.)

Math 602 17

Solution: (1) We set u(r, θ) = φ(θ)g(r). This means φ′′ = −λφ, with φ(0) = 0 and φ(π) = 0,and r ddr (r

ddrg(r)) = λg(r).

(2) The usual energy argument applied to the two-point boundary value problem

φ′′ = −λφ, φ(0) = 0, φ(π) = 0,

implies that λ is non-negative. If λ = 0, then φ(θ) = c1 + c2θ and the boundary conditions implyc1 = c2 = 0, i.e., φ = 0, which in turns gives u = 0 and this solution is incompatible with theboundary condition u(1, θ) = 2 sin(5θ). Hence λ > 0 and


√λθ).

(3) The boundary condition φ(0) = 0 implies c1 = 0. The boundary condition φ(π) = 0 implies√λπ = nπ with n ∈ N \ {0}. This means

√λ = n, n = 1, 2, . . ..

(4) From class we know that g(r) is of the form rα, α ≥ 0. The equality r ddr (rddr r

α) = λrα

gives α2 = λ. The condition α ≥ 0 implies n = α. The boundary condition at r = 1 gives2 sin(5θ) = c21

n sin(nθ) for all θ ∈ [0, π]. This implies n = 5 and c2 = 2.

(5) Finally, the solution to the problem is

u(r, θ) = 2r5 sin(5θ).



32π], r ∈ [0, 3]}, subject

to the boundary conditions u(r, 0) = 0, u(r, 32π) = 0, u(3, θ) = 18 sin(2θ). (Give all the details.)

Solution: (1) We set u(r, θ) = φ(θ)g(r). This means φ′′ = −λφ, with φ(0) = 0 and φ( 32π) = 0,and r ddr (r

ddrg(r)) = λg(r).


φ′′ = −λφ, φ(0) = 0, φ(32π) = 0,



√λθ).

(3) The boundary condition φ(0) = 0 implies c1 = 0. The boundary condition φ(32π) = 0 implies√

λ 32π = nπ with n ∈ N \ {0}. This means√λ = 23n, n = 1, 2, . . ..


α) = λrα gives

α2 = λ. The condition α ≥ 0 implies 23n = α =√λ. The boundary condition at r = 3 gives

18 sin(2θ) = c2323n sin( 23nθ) for all θ ∈ [0,

32π]. This implies n = 3 and c2 = 2.


u(r, θ) = 2r2 sin(2θ).



32π], r ∈ [0, 3]}, subject

to the boundary conditions ∂θu(r, 0) = 0, u(r,32π) = 0, u(3, θ) = 9 cos(θ). (Give all the details

of all the steps.)

Solution: (1) We set u(r, θ) = φ(θ)g(r). This means φ′′ = −λφ, with φ′(0) = 0 and φ( 32π) = 0,and r ddr (r

ddrg(r)) = λg(r).

Math 602 18


φ′′ = −λφ, φ′(0) = 0, φ(32π) = 0,



√λθ).

(3) The boundary condition φ′(0) = 0 implies c2 = 0. The boundary condition φ(32π) = 0 implies

that cos(√λ 32π) = 0, i.e.,

√λ 32π = (2n + 1)

π2 with n ∈ N. This means

√λ = 13 (2n + 1),

n = 0, 1, 2, . . ..


α) = λrα gives

α2 = λ. The condition α ≥ 0 implies 13 (2n+ 1) = α =√λ. The boundary condition at r = 3 gives

9 cos(θ) = c1313 (2n+1) cos( 13 (2n+ 1)θ) for all θ ∈ [0,

32π]. This implies n = 1 and c1 = 3.


u(r, θ) = 3r cos(θ).

Question 45: The solution of the equation, 1r∂r(r∂ru) +1r2 ∂θθu = 0, inside the domain D =

{θ ∈ [0, π], r ∈ [0, 2]}, subject to the boundary conditions u(r, 0) = 0, u(r, π) = 0, u(2, θ) =g(θ) is u(r, θ) =

∑∞n=1 bnr

n sin(nθ). What is the solution corresponding to g(θ) = 5 sin(2θ) +2 sin(5θ)? (Give all the details.)

Solution: The only non-zero terms in the expansion are a2r2 sin(2θ) + b5r5 sin(5θ). The boundarycondition u(2, θ) = 5 sin(2θ) + 2 sin(5θ) = a22

2 sin(2θ) + a525 sin(5θ) is satisfied if a2 = 5/(2

2)and b5 = 2/(2

5), i.e.,

u(r, θ) = 5r2

22sin(2θ) + 2

r5

25sin(5θ).


{θ ∈ [0, π2 ], r ∈ [0, 2]}, subject to the boundary conditions ∂θu(r, 0) = 0, u(r,π2 ) = 0, u(2, θ) =

g(θ) is u(r, θ) =∑∞n=0 a2n+1r

(2n+1) cos((2n+1)θ). What is the solution corresponding to g(θ) =5 cos(3θ) + 2 cos(5θ)? (Give all the details.)

Solution: The only non-zero terms in the expansion are a3r3 cos(3θ)+a5r5 cos(5θ). The boundarycondition u(2, θ) = 5 cos(3θ) + 2 cos(5θ) = a32

3 cos(3θ) + a525 cos(5θ) is satisfied if a3 = 5/(2

3)and a5 = 2/(2

5), i.e.,

u(r, θ) = 5r3

23cos(3θ) + 2

r5

25cos(5θ).


{θ ∈ [0, π2 ], r ∈ [0, 3]}, subject to the boundary conditions ∂θu(r, 0) = 0, u(r,π2 ) = 0, u(3, θ) =

g(θ) is u(r, θ) =∑∞n=0 a2n+1r

(2n+1) cos((2n+1)θ). What is the solution corresponding to g(θ) =5 cos(θ) + 2 cos(3θ)? (Give all the details.)

Solution: The only non-zero terms in the expansion are a1r cos(θ) + a3r3 cos(3θ). The boundarycondition u(3, θ) = 5 cos(θ) + 2 cos(3θ) = a13

1 cos(3θ) + a333 cos(5θ) is satisfied if a1 = 5/3 and

a3 = 2/(33), i.e.,

u(r, θ) = 5r

3cos(θ) + 2

r3

33cos(3θ).

Math 602 19

3.2 Variable coefficients

Question 48: Let k : [−1,+1] −→ R be such that k(x) = 1, if x ∈ [−1, 0] and k(x) = 2 ifx ∈ (0, 1]. Solve the boundary value problem −∂x(k(x)∂xT (x)) = 0 with ∂xT (−1) = T (−1)and ∂xT (1) = 1.(i) What should be the interface conditions at x = 0 for this problem to make sense?


−(0) = k+(0)∂xT+(0), where k−(0) = 1 and k+(0) = 2.

(ii) Solve the problem, i.e., find T (x), x ∈ [−1,+1].Solution: On [−1, 0] we have k−(x) = 1, which implies ∂xxT−(x) = 0. This in turn impliesT−(x) = a+ bx. The Robin boundary condition at x = −1 implies ∂xT−1(−1)− T−(−1) = 0 =2b− a. This gives a = 2b and T−(x) = b(2 + x).

We proceed similarly on [0,+1] and we obtain T+(x) = c+ dx. The Neumann boundary conditionat x = +1 gives ∂xT

+(+1) = 1 = d and T+(x) = c+ x.


+(0) give

2b = c, and b = 2.

In conclusion

T (x) =

{2(2 + x) if x ∈ [−1, 0],4 + x if x ∈ [0,+1].

Question 49: Let k : [−1,+1] −→ R be such that k(x) = 6, if x ∈ [−1, 0] and k(x) = 3 if x ∈(0, 1]. Solve the boundary value problem −∂x(k(x)∂xT (x)) = 0 with 6∂xT (−1) = T (−1) + 13and T (1) = 5.(i) What should be the interface conditions at x = 0 for this problem to make sense?


−(0) = k+(0)∂xT+(0), where k−(0) = 6 and k+(0) = 3.

(ii) Solve the problem, i.e., find T (x), x ∈ [−1,+1].Solution: On [−1, 0] we have k−(x) = 1, which implies ∂xxT−(x) = 0. This in turn impliesT−(x) = a+ bx. The Robin boundary condition at x = −1 implies 6∂xT−(−1)− T−(−1) = 13 =6b− (a− b). This gives a = 7b− 13 and T−(x) = 7b− 13 + bx.

We proceed similarly on [0,+1] and we obtain T+(x) = c+ dx. The Dirichlet boundary conditionat x = +1 gives T+(1) = 5 = d+ c. This implies c = 5− d and T+(x) = 5− d+ dx.


+(0) give

7b− 13 = 5− d, and 6b = 3d.

This implies d = 4 and b = 2. In conclusion

T (x) =

{2x+ 1 if x ∈ [−1, 0],4x+ 1 if x ∈ [0,+1].

Question 50: Let k : [−1,+1] −→ R be such that k(x) = 2, if x ∈ [−1, 0] and k(x) = 3 ifx ∈ (0, 1]. Solve the boundary value problem −∂x(k(x)∂xT (x)) = 0 with ∂xT (−1) = T (−1) + 3and −∂xT (1) = T (1)− 7.(i) What should be the interface conditions at x = 0 for this problem to make sense?


−(0) = k+(0)∂xT+(0), where k−(0) = 2 and k+(0) = 3.

Math 602 20

(ii) Solve the problem, i.e., find T (x), x ∈ [−1,+1].Solution: On [−1, 0] we have k−(x) = 1, which implies ∂xxT−(x) = 0. This in turn impliesT−(x) = a + bx. The Robin boundary condition at x = −1 implies ∂xT−(−1) − T−(−1) = 3 =2b− a. This gives a = 2b− 3 and T−(x) = 2b− 3 + bx.

We proceed similarly on [0,+1] and we obtain T+(x) = c+ dx. The Robin boundary condition atx = +1 gives −∂xT+(+1) − T+(1) = −7 = −2d − c. This implies c = −2d + 7 and T+(x) =−2d+ 7 + dx.


+(0) give

2b− 3 = −2d+ 7, and 2b = 3d.

This implies d = 2 and b = 3. In conclusion

T (x) =

{3x+ 3 if x ∈ [−1, 0],2x+ 3 if x ∈ [0,+1].

Question 51: Let k, f : [−1,+1] −→ R be such that k(x) = 2, f(x) = 0 if x ∈ [−1, 0] andk(x) = 1, f(x) = 2 if x ∈ (0, 1]. Consider the boundary value problem −∂x(k(x)∂xT (x)) = f(x)with T (−1) = 0 and T (1) = 3.(a) What should be the interface conditions at x = 0 for this problem to make sense?


−(0) = k+(0)∂xT+(0), where k−(0) = 2 and k+(0) = 1.

Question 52: Let k, f : [−1,+1] −→ R be such that k(x) = 2 + x, f(x) = 0 if x ∈ [−1, 0] andk(x) = 1+2x, f(x) = 2 if x ∈ (0, 1]. Consider the boundary value problem −∂x(k(x)∂xT (x)) =f(x) with T (−1) = 5 and T (1) = −1.(a) What should be the interface conditions at x = 0 for this problem to make sense?


−(0) = k+(0)∂xT+(0), where k−(0) = 2 and k+(0) = 1, i.e., 2∂xT

−(0) = ∂xT+(0).

Question 53: Let k, f : [−1,+1] −→ R be such that k(x) = 2, f(x) = 0 if x ∈ [−1, 0] andk(x) = 1, f(x) = 2 if x ∈ (0, 1]. Consider the boundary value problem −∂x(k(x)∂xT (x)) = f(x)with T (−1) = −2 and T (1) = 2.(a) What should be the interface conditions at x = 0 for this problem to make sense?


−(0) = k+(0)∂xT+(0), where k−(0) = 2 and k+(0) = 1.

(b) Solve the problem, i.e., find T (x), x ∈ [−1,+1].Solution: On [−1, 0] we have k−(x) = 2 and f−(x) = 0 which implies −∂xxT−(x) = 0. Thisin turn implies T−(x) = ax+ b. The Dirichlet condition at x = −1 implies that T−(−1) = −2 =−a+ b. This gives a = b+ 2 and T−(x) = (b+ 2)x+ b.

We proceed similarly on [0,+1] and we obtain −∂xxT−(x) = 2, which implies that T+(x) =−x2 + cx + d. The Dirichlet condition at x = 1 implies T+(1) = 2 = −1 + c + d. This givesc = 3− d and T−(x) = −x2 + (3− d)x+ d.


+(0) give b = d and2(b+ 2) = 3− d, respectively. In conclusion b = − 13 , d = −

13 and

T (x) =

{53x−

13 if x ∈ [−1, 0],

−x2 + 103 x−13 if x ∈ [0, 1].

Question 54: Let k, f : [−1,+1] −→ R be such that k(x) = 2, f(x) = 0 if x ∈ [−1, 0] andk(x) = 1, f(x) = 2 if x ∈ (0, 1]. Consider the boundary value problem −∂x(k(x)∂xT (x)) = f(x)

Math 602 21

with T (−1) = 0 and T (1) = 2.(a) What should be the interface conditions at x = 0 for this problem to make sense?


−(0) = k+(0)∂xT+(0), where k−(0) = 2 and k+(0) = 1.

(b) Solve the problem, i.e., find T (x), x ∈ [−1,+1]. Give all the details.Solution: On [−1, 0] we have k−(x) = 2 and f−(x) = 0 which implies −∂xxT−(x) = 0. This inturn implies T−(x) = ax+b. The Dirichlet condition at x = −1 implies that T−(−1) = 0 = −a+b.This gives a = b and T−(x) = bx+ b.

We proceed similarly on [0,+1] and we obtain −∂xxT−(x) = 2, which implies that T+(x) =−x2 + cx + d. The Dirichlet condition at x = 1 implies T+(1) = 2 = −1 + c + d. This givesc = 3− d and T−(x) = −x2 + (3− d)x+ d.


+(0) give b = d and2b = 3− d, respectively. In conclusion b = 1, d = 1 and

T (x) =

{x+ 1 if x ∈ [−1, 0],−x2 + 2x+ 1 if x ∈ [0, 1].

Question 55: Let k, f : [−1,+1] −→ R be such that k(x) = 3, f(x) = −6 if x ∈ [−1, 0] andk(x) = 1, f(x) = 2 if x ∈ (0, 1]. Consider the boundary value problem −∂x(k(x)∂xT (x)) = f(x)with T (−1) = 1 and ∂xT (1) = 1.(a) What should be the interface conditions at x = 0 for this problem to make sense?


−(0) = k+(0)∂xT+(0), where k−(0) = 3 and k+(0) = 1.

(b) Solve the problem, i.e., find T (x), x ∈ [−1,+1]. Give all the details.Solution: On [−1, 0] we have k−(x) = 3 and f−(x) = −6 which implies −3∂xxT−(x) = −6.This in turn implies T−(x) = x2 + ax + b. The Dirichlet condition at x = −1 implies thatT−(−1) = 1 = 1− a+ b. This gives a = b and T−(x) = x2 + bx+ b.

We proceed similarly on [0,+1] and we obtain −∂xxT−(x) = 2, which implies that T+(x) =−x2 + cx + d. The Neumann condition at x = 1 implies T+(1) = 1 = −2 + c. This gives c = 3and T−(x) = −x2 + 3x+ d.


+(0) give b = d and3b = 3, respectively. In conclusion b = 1, d = 1 and

T (x) =

{x2 + x+ 1 if x ∈ [−1, 0],−x2 + 3x+ 1 if x ∈ [0, 1].

3.3 Maximum principle

Question 56: Consider the square D = (−1,+1)×(−1,+1). Let f(x, y) = x2 − y2 − 3. Letu ∈ C2(D) ∩ C0(D) solve −∆u = 0 in D and u|∂D = f . Compute min(x,y)∈D u(x, y) andmax(x,y)∈D u(x, y).

Solution: We use the maximum principle (u is harmonic and has the required regularity). Then

min(x,y)∈D

u(x, y) = min(x,y)∈∂D

f(x, y), and max(x,y)∈D

u(x, y) = max(x,y)∈∂D

f(x, y).

A point (x, y) is at the boundary of D if and only if x2 = 1 and y ∈ (−1, 1) or y2 = 1 andx ∈ (−1, 1). In the first case, x2 = 1 and y ∈ (−1, 1), we have

f(x, y) = 1− y2 − 3, y ∈ (−1, 1).

Math 602 22

The maximum is −2 and the minimum is −3. In the second case, y2 = 1 and x ∈ (−1, 1), we have

f(x, y) = x2 − 1− 3, x ∈ (−1, 1).

The maximum is −3 and the minimum is −4. We finally can conclude

min(x,y)∈∂D

f(x, y) = min−1≤x≤1

x2 − 4,= −4, max(x,y)∈∂D

f(x, y) = max−1≤y≤1

−2− y2 = −2.

In conclusionmin

(x,y)∈Du(x, y) = −4, max

(x,y)∈Du(x, y) = −2

Question 57: (i) Let Ω be an open connected set in R2. Let u be a real-valued nonconstantfunction continuous on Ω and harmonic on Ω. Assume that there exists x0 in Ω such∇u(x0) = 0.Do we have a minimum, a maximum, or a saddle point at x0? (explain)

Solution: The Maximum principle implies that u cannot be either minimum or maximum at x0.This point is a saddle point.

(ii) Let Ω = (0, 1) (note that Ω = [0, 1]), and let u : Ω −→ R be such that u(x) = 1 for allx ∈ Ω, u(0) = 0, and u(1) = −1. Is u harmonic on Ω? Find a point in Ω where u reaches itsmaximum? Does this example contradict the Maximum Principle? (explain)

Solution: Yes u is harmonic on Ω since u′′(x) = 0 for all x in Ω. Note however that u is notcontinuous on Ω; as a consequence, the hypotheses for the Maximum principle are not satisfied. Inother words, the above example does not contradict the Maximum principle .

Question 58: Let Ω be an open connected set in R2. Let u be a real-valued nonconstantfunction continuous on Ω and harmonic on Ω and of class C2 in Ω. Assume that there exists x0in Ω such ∇u(x0) = 0. Is the point x0 a minimum, a maximum, or a saddle point? (explain)Solution: The keyword here is that Ω is open, meaning that the points at the boundary of Ω arenot in Ω (it is not possible to punch a hole around the boundary points). The point x0 is in Ω,that is x0 is not at the boundary. Since u is continuous on Ω, harmonic on Ω and of class C2 in Ω,the maximum principle can be applied. The Maximum principle implies that u cannot have eithera minimum or a maximum at x0. This means that x0 is a saddle point.

Question 59: Let u be a continuous function on D where D is some open, connected set inR2. Explain why, if u is harmonic, it is generally a waste of time to locate a point where uachieves its maximum by solving ∂xu = 0 and ∂yu = 0 simultaneously.

Solution: From the Maximum Principle, we know that if u is not constant, the maximum of u isachieved at the boundary of D. The zero gradient condition does not apply for maximums at theboundary.

Question 60: Let D be the open disk of radius√

2 centered at (1, 2). Let u ∈ C2(D) ∩ C0(D)be a harmonic function in the disk D. Assume that u(x, y) = (x+ y)2 on the boundary of disk.Compute min(x,y)∈D u(x, y) and max(x,y)∈D u(x, y). You can give a geometric answer.

Solution: Since u is in C2(D) ∩ C0(D) and is harmonic, we can apply the maximum principle(Theorem). This theorem says that the maximum and minimum of u are attained at the boundaryof the disk. The problem then amounts to finding the maximum and minimum of (x + y)2 overthe circle of radius

√2 centered at (1, 2). The iso-values of the function (x + y)2 are parallel

lines of slope −1. These iso-line are perpendicular to the gradient of (x + y)2 which is the vector(2(x + y), 2(x + y)) = 2(x + y)(1, 1). We must find the two tangent lines to the circle that areperpendicular to the vector (1, 1). One easily verify that (0, 1) and (2, 3) are the tangent points(verify that the segment connecting these two points is parallel to the vector (1, 1) and passesthrough the center of the circle). As a result

min(x,y)∈D

u(x, y) = 1 and max(x,y)∈D

u(x, y) = 25.

Math 602 23

0 1 2 3 4 5

1

2

3

4

5

6

The dashed lines are iso-lines of (x+ y)2.

Question 61: Consider the disk D centered at (0, 0) of radius 1. Let f(x, y) = x2−y2 +4y−3.Let u ∈ C2(D) ∩ C0(D) solve −∆u = 0 in D and u|∂D = f . Compute min(x,y)∈D u(x, y) andmax(x,y)∈D u(x, y).


min(x,y)∈D




f(x, y).

A point (x, y) is at the boundary of D if and only if x2 + y2 = 1; as a result, the following holdsfor all (x, y) ∈ ∂D:

f(x, y) = x2 − y2 + 4y − 3 = 1− y2 − y2 + 4y − 3 = −2(1− y)2.

We obtain

min(x,y)∈∂D

f(x, y) = min−1≤y≤1

−2(1− y)2 = −8, max(x,y)∈∂D

f(x, y) = max−1≤y≤1

−2(1− y)2 = 0.

In conclusionmin

(x,y)∈Du(x, y) = −8, max

(x,y)∈Du(x, y) = 0

Question 62: Consider the square D = (−1,+1)×(−1,+1). Let f(x, y) = x2 − y2 − 3. Letu ∈ C2(D) ∩ C0(D) solve −∆u = 0 in D and u|∂D = f . Compute min(x,y)∈D u(x, y) andmax(x,y)∈D u(x, y).


min(x,y)∈D




f(x, y).

A point (x, y) is at the boundary of D if and only if x2 = 1 and y ∈ (−1, 1) or y2 = 1 andx ∈ (−1, 1). In the first case, x2 = 1 and y ∈ (−1, 1), we have

f(x, y) = 1− y2 − 3, y ∈ (−1, 1).

Math 602 24

The maximum is −2 and the minimum is −3. In the second case, y2 = 1 and x ∈ (−1, 1), we have

f(x, y) = x2 − 1− 3, x ∈ (−1, 1).

The maximum is −3 and the minimum is −4. We finally can conclude

min(x,y)∈∂D

f(x, y) = min−1≤x≤1

x2 − 4,= −4, max(x,y)∈∂D

f(x, y) = max−1≤y≤1

−2− y2 = −2.

In conclusionmin

(x,y)∈Du(x, y) = −4, max

(x,y)∈Du(x, y) = −2

Question 63: Consider the triangular domain D = {(x, y);x ≥ 0, y ≥ 0, 1 − x − y ≤ 0}. Letf(x, y) = x2 − y2 − 3. Let u ∈ C2(D) ∩ C0(D) solve −∆u = 0 in D and u|∂D = f . Computemin(x,y)∈D u(x, y) and max(x,y)∈D u(x, y).


min(x,y)∈D




f(x, y).

A point (x, y) is at the boundary of D if and only if {x = 0 and y ∈ [0, 1]} or {y = 0 and x ∈ [0, 1]},or {1− y − x = 0 and x ∈ [0, 1]}.(i) In the first case, x = 0 and y ∈ [0, 1], we have

f(x, y) = −y2 − 3, y ∈ [0, 1].

The maximum is −3 and the minimum is −4.(ii) In the second case, y = 0 and x ∈ [0, 1], we have

f(x, y) = x2 − 3, x ∈ [0, 1].

The maximum is −2 and the minimum is −3.(iii) In the third case, 1− x = y and x ∈ [0, 1], we have

f(x, y) = x2 − (1− x)2 − 3 = 2x− 4, x ∈ [0, 1].

The maximum is −2 and the minimum is −4.We finally can conclude

min(x,y)∈∂D

f(x, y) = −4, max(x,y)∈∂D

f(x, y) = −2.

In conclusionmin

(x,y)∈Du(x, y) = −4, max

(x,y)∈Du(x, y) = −2

Question 64: Consider the elliptic domain D = {(x, y);x2+2y2 ≤ 2}. Let f(x, y) = x2−y2−3.Let u ∈ C2(D) ∩ C0(D) solve −∆u = 0 in D and u|∂D = f . Compute min(x,y)∈D u(x, y) andmax(x,y)∈D u(x, y).


min(x,y)∈D




f(x, y).

A point (x, y) is at the boundary of D if and only if x2 + 2y2 = 2, meaning that x2 = 2− 2y2. Inconclusion, for any point (x, y) ∈ ∂D we have f(x, y) = x2−y2−3 = 2−2y2−y2−3 = −1−3y2and y ∈ [−1,+1].(1) The maximum of−1−3y2 over the interval [−1,+1] is reached for y = 0. Hence max(x,y)∈∂D f(x, y) =−1.(2) The minimum of−1−3y2 over the interval [−1,+1] is reached for y = ±1. Hence min(x,y)∈∂D f(x, y) =

Math 602 25

−4.We finally can conclude

min(x,y)∈∂D

f(x, y) = −4, max(x,y)∈∂D

f(x, y) = −1.

In conclusionmin

(x,y)∈Du(x, y) = −4, max

(x,y)∈Du(x, y) = −1.

Math 602 26

4 Eigenvalue problems

Question 65: Consider the differential equation d2φdt2 + λφ = 0, t ∈ (0, π), supplemented with

the boundary conditions φ(0) = 0, φ(π) = 0.(a) What is the sign of λ? Prove your answer.

Solution:

(b) Compute all the possible eigenvalues λ for this problem.

Solution:

Question 66: Consider the differential equation d2φdt2 + λφ = 0, t ∈ (0, π), supplemented with

the boundary conditions φ(0) = 0, φ′(π) = 0.(a) What is the sign of λ? Prove your answer.

Solution: Multiply the equation by φ and integrate over the domain.

−∫ π

0

(φ′(t))2dt+ φ′φ|π0 + λ∫ π

0

φ2(t)dt = 0.

Using the BCs, we infer

−∫ π

0

(φ′(t))2dt = λ

∫ π0

φ2(t)dt,

which means that λ is non-negative.

(b) Compute all the possible eigenvalues λ for this problem and compute φ.

Solution: There are two cases: Either λ > 0 or λ = 0. Assume first λ > 0, then


√λt).

The boundary condition φ(0) = 0 implies imply c1 = 0. The other BC implies c2 cos(√λπ) = 0.

c2 = 0 gives φ = 0, which is not a proper eigenfunction. The other possibility is√λπ = (n+ 12 )π,

n ∈ N. In conclusionφ(t) = c2 sin((n+

12 )t), n ∈ N.

The case λ = 0 gives φ(t) = c1 + c2t. The BCs imply c1 = c2 = 0, i.e. φ = 0, which is not aproper eigenfunction.

Question 67: Let Ω = (0, L) and let (λ, u) be an eigenpair of the Laplace equation over Ω withzero Dirichlet condition. Assume that λ ∈ C and that the function u(x) is complex-valued.(i) Write the PDE solved by u.

Solution: u and λ are such

−∂xxu(x) = λu(x), u(0) = 0, u(L) = 0.

(ii) Let ū be the complex conjugate of u. Write the PDE solved by ū (Hint: take the conjugateof (i)).

Solution: Take the complex conjugate of (i) we obtain

−∂xxū(x) = −∂xxu(x) = λu(x) = λ̄ū(x), ū(0) = 0, ū(L) = 0

which gives−∂xxū(x) = λ̄ū(x), ū(0) = 0, ū(L) = 0.

Note that we used the fact that ∂xu = ∂xū.

(iii) Prove that λ ∈ R (Hint: Use an energy argument with ū in (i) and an energy argumentwith u (ii) and conclude that λ = λ̄. Recall u 6= 0 and |z|2 = zz̄ for all z ∈ C).

Math 602 27

Solution: Multiply (i) by ū and integrate over Ω∫Ω

−∂xxu(x)ū(x)dx =∫

Ω

λu(x)ū(x)dx∫Ω

∂xu(x)∂xū(x)dx− [∂xu(x)ū(x)]L0 = λ∫

Ω

|u(x)|2dx∫Ω

∂xu(x)∂xu(x)dx = λ

∫Ω

|u(x)|2dx∫Ω

|∂xu(x)|2dx = λ∫

Ω

|u(x)|2dx.

Multiply (ii) by u and integrate over Ω∫Ω

−∂xxū(x)u(x)dx =∫

Ω

λ̄ū(x)u(x)dx∫Ω

∂xū(x)∂xu(x)dx− [∂xū(x)u(x)]L0 = λ̄∫

Ω

|u(x)|2dx∫Ω

∂xu(x)∂xu(x)dx = λ̄

∫Ω

|u(x)|2dx∫Ω

|∂xu(x)|2dx = λ̄∫

Ω

|u(x)|2dx.

In conclusion

λ

∫Ω

|u(x)|2dx = λ̄∫

Ω

|u(x)|2dx,

which means

(λ− λ̄)∫

Ω

|u(x)|2dx.

This in turn implies that λ = λ̄ since∫

Ω|u(x)|2dx is not zero (recall u 6= 0). In conclusion λ is

real. Note in passing that this also prove that λ ≥ 0.

Question 68: Consider the eigenvalue problem − ddt (t12ddtφ(t)) = λt

− 12φ(t), t ∈ (0, 1), supple-mented with the boundary condition φ(0) = 0, ∂tφ(1) = 0.(a) Prove that it is necessary that λ be positive for a non-zero smooth solution to exist.

Solution: (i) Let φ be a non-zero smooth solution to the problem. Multiply the equation by φ andintegrate over the domain. Use the Fundamental Theorem of calculus (i.e., integration by parts) toobtain ∫ 1

0

t12 (φ′(t))2dt− [t 12φ′(t)φ(t)]10 = λ

∫ 10

t−12φ2(t)dt.

Using the boundary conditions, we infer∫ 10

t12 (φ′(t))2dt = λ

∫ 10

t−12φ2(t)dt,


(ii) If λ = 0, then∫ 1

0t

12 (φ′(t))2dt = 0, which implies that φ′(t) = 0 for all t ∈ (0, 1]. This implies

that φ(t) is constant, and this constant is zero since φ(0) = 0. Hence, φ is zero if λ = 0. Since wewant a nonzero solution, this implies that λ cannot be zero.

(iii) In conclusion, it is necessary that λ be positive for a nonzero smooth solution to exist.

(b) The general solution to − ddt (t12ddtφ(t)) = λt

− 12φ(t) is φ(t) = c1 cos(2√t√λ) + c2 sin(2

√t√λ)

for λ ≥ 0. Find all the eigenvalues λ > 0 and the associated nonzero eigenfunctions.Solution: Since λ ≥ 0 by hypothesis, φ is of the following form

φ(t) = φ(t) = c1 cos(2√t√λ) + c2 sin(2

√t√λ).

Math 602 28

The boundary condition φ(0) = 0 implies c1 = 0. The other boundary condition implies ∂xφ(1) =0 = c2

√λ cos(2

√λ). The constant c2 cannot be zero since we want φ to be nonzero; as a result,

2√λ = (n+ 12 )π, n = 0, 2, . . .. In conclusion

λ = ((2n+ 1)π)2/16, n = 1, 2, . . . , φ(t) = c sin((n+1

2)π√t).


− 12φ(t), t ∈ (0, 1), supple-mented with the boundary condition φ(0) = 0, φ(1) = 0.(a) Prove that it is necessary that λ be positive for a non-zero smooth solution to exist.

Solution: (i) Let φ be a non-zero smooth solution to the problem. Multiply the equation by φ andintegrate over the domain. Use the fundamental Theorem of calculus (i.e., integration by parts) toobtain ∫ 1

0

t12 (φ′(t))2dt− [t 12φ′(t)φ(t)]10 = λ

∫ 10

t−12φ2(t)dt.


t12 (φ′(t))2dt = λ

∫ 10

t−12φ2(t)dt,



0t

12 (φ′(t))2dt = 0, which implies that φ′(t) = 0 for all t ∈ (0, 1]. The

fundamental theorem of calculus applied between t and 1 implies φ(t) = φ(1) +∫ t

1φ′(τ)dτ = 0

since φ(1) = 0 and φ′(τ) = 0 for all τ ∈ (t, 1]. Hence, φ is zero if λ = 0. Since we want a nonzerosolution, this implies that λ cannot be zero.




√t√λ)

for λ ≥ 0. Find all the eigenvalues λ ≥ 0 and the associated nonzero eigenfunctions.Solution: Since λ ≥ 0 by hypothesis, φ is of the following form

φ(t) = φ(t) = c1 cos(2√t√λ) + c2 sin(2

√t√λ).

The boundary condition φ(0) = 0 implies c1 = 0. The other boundary condition implies φ(1) =0 = c2 sin(2


2√λ = nπ, n = 1, 2, . . .. In conclusion

λ = (nπ)2/4, n = 1, 2, . . . , φ(t) = c sin(nπ√t).

Question 70: Consider the eigenvalue problem − d2

dt2φ(t) + 2ddtφ(t) = λφ(t), t ∈ (0, π), supple-

mented with the boundary condition φ(0) = 0, φ(π) = 0. (Hint: 2φ(t) ddtφ(t) =ddtφ

2(t).)(a) Prove that it is necessary that λ be positive for a non-zero solution to exist.

Solution: (i) Let φ be a non-zero solution to the problem. Multiply the equation by φ and integrateover the domain. Use the fundamental Theorem of calculus and use the hint to obtain∫ π

0

(φ′(t))2dt− φ′(π)φ(π) + φ′(0)φ(0) +∫ π

0

d

dt(φ2(t))dt = λ

∫ π0

φ2(t)dt∫ π0

(φ′(t))2dt− φ′(π)φ(π) + φ′(0)φ(0) + φ2(π)− φ2(0) = λ∫ π

0

φ2(t)dt

Using the boundary conditions, we infer∫ π0

(φ′(t))2dt = λ

∫ π0

φ2(t)dt,

Math 602 29


(ii) If λ = 0, then∫ π

0(φ′(t))2dt = 0 and φ(π)2 = 0, which implies that φ′(t) = 0. The fundamental

theorem of calculus implies φ(t) = φ(0) +∫ t

0φ′(τ)dτ = 0. Hence, φ is zero if λ = 0. Since we

want a nonzero solution, this implies that λ cannot be zero.

(iii) In conclusion, it is necessary that λ be positive for a nonzero solution to exist.

(b) The general solution to −φ′′ + 2φ′ = λφ is φ(t) = et(c1 cos(t√λ− 1) + c2 sin(t

√λ− 1)) for

λ ≥ 1. Find all the eigenvalues λ ≥ 1 and the associated nonzero eigenfunctions.Solution: Since λ ≥ 1 by hypothesis, φ is of the following form

φ(t) = et(c1 cos(√λ− 1t) + c2 sin(

√λ− 1t)).

The boundary condition φ(0) = 0 implies c1 = 0. The other boundary condition implies φ(π) =0 = eπc2 sin(

√λ− 1π). The constant c2 cannot be zero since we want φ to be nonzero; as a result,√

λ− 1 = n, n = 1, 2, . . .. In conclusion

λ = n2 + 1, n = 1, 2, . . . , φ(t) = cet sin(nt)


− 12φ(t), t ∈ (0, 1), supple-mented with the boundary condition φ(0) = 0, φ(1) = 0.(a) Prove that it is necessary that λ be positive for a non-zero smooth solution to exist.

Solution: (i) Let φ be a non-zero smooth solution to the problem. Multiply the equation by φ andintegrate over the domain. Use the fundamental Theorem of calculus (i.e., integration by parts) toobtain ∫ 1

0

t12 (φ′(t))2dt− [t 12φ′(t)φ(t)]10 = λ

∫ 10

t−12φ2(t)dt.


t12 (φ′(t))2dt = λ

∫ 10

t−12φ2(t)dt,



0t

12 (φ′(t))2dt = 0, which implies that φ′(t) = 0 for all t ∈ (0, 1]. The

fundamental theorem of calculus applied between t and 1 implies φ(t) = φ(1) +∫ t

1φ′(τ)dτ = 0

since φ(1) = 0 and φ′(τ) = 0 for all τ ∈ (t, 1]. Hence, φ is zero if λ = 0. Since we want a nonzerosolution, this implies that λ cannot be zero.




√t√λ)

for λ ≥ 0. Find all the eigenvalues λ ≥ 0 and the associated nonzero eigenfunctions.Solution: Since λ ≥ 0 by hypothesis, φ is of the following form

φ(t) = φ(t) = c1 cos(2√t√λ) + c2 sin(2

√t√λ).

The boundary condition φ(0) = 0 implies c1 = 0. The other boundary condition implies φ(1) =0 = c2 sin(2


2√λ = nπ, n = 1, 2, . . .. In conclusion

λ = (nπ)2/4, n = 1, 2, . . . , φ(t) = c sin(nπ√t).

Question 72: Consider the differential equation −d2φdt2 = λφ, t ∈ (0, π), supplemented with the

boundary conditions 2φ(0) = φ′(0), φ(π) = 0.(a) What should be the sign of λ for a non-zero solution to exist? Prove your answer.

Math 602 30

Solution: Let φ be a non-zero solution to the problem. Multiply the equation by φ and integrateover the domain. ∫ π

0

(φ′(t))2dt− φ′(π)φ(π) + φ′(0)φ(0) = λ∫ π

0

φ2(t)dt.

Using the BCs, we infer ∫ π0

(φ′(t))2dt+ 2φ(0)2 = λ

∫ π0

φ2(t)dt,


(b) Assume that 2 sin(√λπ)+

√λ cos(

√λπ) 6= 0. Prove that φ = 0 is the only possible solution.

Solution: Observe first that λ cannot be zero, otherwise we would have 2 sin(√λπ)+

√λ cos(

√λπ) =

0. As a result, (a) implies that λ is positive. Then φ is of the following form


√λt).

The boundary condition φ′(0) = 2φ(0) implies 2c1−√λc2 = 0. The other BC implies c1 cos(

√λπ)+

c2 sin(√λπ) = 0. The constants c1 and c2 solve the following linear system{

2c1 −√λc2 = 0

c1 cos(√λπ) + c2 sin(

√λπ) = 0

The determinant is equal to 2 sin(√λπ) +

√λ cos(

√λπ) and is non-zero by hypothesis. Then the

only solution is c1 = c2 = 0, which again gives φ = 0.

In conclusion, the only possible solution to the problem is φ = 0 if 2 sin(√λπ)+

√λ cos(

√λπ) 6= 0.


− 12φ(t), t ∈ (0, 1), supple-mented with the boundary condition φ(0) = 0, ∂tφ(1) = 0.(a) Prove that it is necessary that λ be positive for a non-zero smooth solution to exist.

Solution: (i) Let φ be a non-zero smooth solution to the problem. Multiply the equation by φ andintegrate over the domain. Use the Fundamental Theorem of calculus (i.e., integration by parts) toobtain ∫ 1

0

t12 (φ′(t))2dt− [t 12φ′(t)φ(t)]10 = λ

∫ 10

t−12φ2(t)dt.


t12 (φ′(t))2dt = λ

∫ 10

t−12φ2(t)dt,



0t

12 (φ′(t))2dt = 0, which implies that φ′(t) = 0 for all t ∈ (0, 1]. This implies

that φ(t) is constant, and this constant is zero since φ(0) = 0. Hence, φ is zero if λ = 0. Since wewant a nonzero solution, this implies that λ cannot be zero.




√t√λ)

for λ ≥ 0. Find all the eigenvalues λ > 0 and the associated nonzero eigenfunctions.Solution: Since λ ≥ 0 by hypothesis, φ is of the following form

φ(t) = φ(t) = c1 cos(2√t√λ) + c2 sin(2

√t√λ).

The boundary condition φ(0) = 0 implies c1 = 0. The other boundary condition implies ∂xφ(1) =0 = c2

√λ cos(2


2√λ = (n+ 12 )π, n = 0, 2, . . .. In conclusion

λ = ((2n+ 1)π)2/16, n = 1, 2, . . . , φ(t) = c sin((n+1

2)π√t).

Math 602 31

Question 74: Let p, q : [−1,+1] −→ R be smooth functions. Assume that p(x) ≥ 0 and q(x) ≥q0 for all x ∈ [−1,+1], where q0 ∈ R. Consider the eigenvalue problem −∂x(p(x)∂xφ(x)) +q(x)φ(x) = λφ(x), supplemented with the boundary conditions φ(−1) = 0 and φ(1) = 0.(a) Prove that it is necessary that λ ≥ q0 for a non-zero (smooth) solution, φ, to exist. (Hint:q0∫ +1−1 φ

2(x)dx ≤∫ +1−1 q(x)φ

2(x)dx.)

Solution: As usual we use the energy method. Let (φ, λ) be an eigenpair, then∫ +1−1

(−∂x(p(x)∂xφ(x))φ(x) + q(x)φ2(x))dx = λ∫ +1−1

φ2(x)dx.

After integration by parts and using the boundary conditions, we obtain∫ +1−1

(p(x)∂xφ(x)∂xφ(x) + q(x)φ2(x))dx = λ

∫ +1−1

φ2(x)dx.

which, using the hint, can also be re-written∫ +1−1

(p(x)∂xφ(x)∂xφ(x) + q0φ2(x))dx ≤ λ

∫ +1−1

φ2(x)dx.

Then ∫ +1−1

p(x)(∂xφ(x))2dx ≤ (λ− q0)

∫ +1−1

φ2(x)dx.

Assume that φ is non-zero, then

λ− q0 ≥∫ +1−1 p(x)(∂xφ(x))

2dx∫ +1−1 φ

2(x)dx≥ 0,

which proves that it is necessary that λ ≥ q0 for a non-zero (smooth) solution to exist.(b) Assume that p(x) ≥ p0 > 0 for all x ∈ [−1,+1] where p0 ∈ R+. Show that λ = q0 cannotbe an eigenvalue, i.e., prove that φ = 0 if λ = q0. (Hint: p0

∫ +1−1 ψ

2(x)dx ≤∫ +1−1 p(x)ψ

2(x)dx.)

Solution: Assume that λ = q0 is an eigenvalue. Then the above computation shows that

p0

∫ +1−1

(∂xφ(x))2dx ≤

∫ +1−1

p(x)(∂xφ(x))2dx = 0,

which means that∫ +1−1 (∂xφ(x))

2dx = 0 since p0 > 0. As a result ∂xφ = 0, i.e., φ(x) = c where cis a constant. The boundary conditions φ(−1) = 0 = φ(1) imply that c = 0. In conclusion φ = 0if λ = q0, thereby proving that (φ, q0) is not an eigenpair.

Question 75: Let φ be a non-zero solution to the eigenvalue problem −∂xxφ(x) = λφ(x),x ∈ (0, π), φ(0) = 0, φ(π) = 0. Determine the sign of λ using the energy method.Solution: Multiply the equation by φ, integrate over (0, π), and apply the fundamental theorem ofcalculus (i.e. integrate by parts):

−∫ π

0

φ(x)∂xxφ(x)dx = −∫ π

0

(∂x(φ(x)∂xφ(x))− (∂xφ(x))2)dx

= −φ(π)∂xφ(π) + φ(0)∂xφ(0) +∫ π

0

(∂xφ(x))2dx = λ

∫ π0

(φ(x))2dx.

In conclusion ∫ π0

(∂xφ(x))2dx = λ

∫ π0

(φ(x))2dx.

Math 602 32

Assuming that φ is nonzero, we obtain that λ =∫ π

0(∂xφ(x))

2dx/∫ π

0(φ(x))2dx ≥ 0, i.e. λ is

non-negative. If λ = 0 then ∂xφ = 0, which implies that φ is constant. The boundary conditionsimply that φ = 0 which contradicts our assumption that φ is non-zero. In conclusion λ is negative.

Question 76: Let φ be a non-zero solution to the eigenvalue problem −∂xxφ(x) = λφ(x),x ∈ (0, π), φ(0) = 0, ∂xφ(π) + φ(π) = 0. Determine the sign of λ using the energy method.Solution: Multiply the equation by φ, integrate over (0, π), and apply the fundamental theorem ofcalculus (i.e. integrate by parts):

λ

∫ π0

(φ(x))2dx = −∫ π

0

φ(x)∂xxφ(x)dx = −∫ π

0

(∂x(φ(x)∂xφ(x))− (∂xφ(x))2)dx

= −φ(π)∂xφ(π) + φ(0)∂xφ(0) +∫ π

0

(∂xφ(x))2dx

= (φ(π))2 +

∫ π0

(∂xφ(x))2dx.

In conclusion

(φ(π))2 +

∫ π0

(∂xφ(x))2dx = λ

∫ π0

(φ(x))2dx.

Assuming that φ is nonzero, we obtain that λ = ((φ(π))2 +∫ π

0(∂xφ(x))

2dx)/∫ π

0(φ(x))2dx ≥ 0,

i.e. λ is non-negative. If λ = 0 then φ(π) = 0 and ∂xφ = 0, which implies that φ is constant. Theother condition φ(π) = 0 implies that φ = 0 which contradicts our assumption that φ is non-zero.In conclusion λ is positive.

Question 77: Compute all the positive eigenvalues λ to the above eigenvalue problem.

Solution: Owing to λ being non-negative, the general solution to −∂xxφ(x) = λφ(x) is φ(x) =a cos(

√λx) + b sin(

√λx). The boundary conditions imply that φ(0) = 0 = a and φ(π) =

b sin(√λπ) = 0. As a result,

√λπ = nπ, where n ∈ N, i.e.

λ = n2, n ∈ N.

Question 78: Let φ be a non-zero solution to the eigenvalue problem −∂x((1 + x2)∂xφ(x)

)=

λφ(x), x ∈ (0, π), φ(π) = 0, −∂xφ(0) + φ(0) = 0. Determine the sign of λ using the energymethod.

Solution: Multiply the equation by φ, integrate over (0, π), and apply the fundamental theorem ofcalculus (i.e. integrate by parts):

λ

∫ π0

(φ(x))2dx = −∫ π

0

φ(x)∂x((1 + x2)∂xφ(x)

)dx = −

∫ π0

(∂x(φ(x)(1 + x2)∂xφ(x))− (1 + x2)(∂xφ(x))2)dx

= −(1 + π2)φ(π)∂xφ(π) + φ(0)∂xφ(0) +∫ π

0

(1 + x2)(∂xφ(x))2dx

= (φ(0))2 +

∫ π0

(1 + x2)(∂xφ(x))2dx.

In conclusion

(φ(0))2 +

∫ π0

(1 + x2)(∂xφ(x))2dx = λ

∫ π0

(φ(x))2dx.

Assuming that φ is nonzero, we obtain that λ = (φ2(0)+∫ π

0(1+x2)(∂xφ(x))

2dx)/∫ π

0(φ(x))2dx ≥

0, i.e. λ is non-negative. If λ = 0 then φ(0) = 0 and ∂xφ = 0, which implies that φ is constant. Theother condition φ(0) = 0 implies that φ = 0 which contradicts our assumption that φ is non-zero.In conclusion λ is positive.

Math 602 33

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