Freiling G., Yurko V. Lectures on Differential Equations of Mathematical Physics: A First Course

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252 G.

Freiling and V. Yurko

6.3.6. Solve

the following problems applying the method of separation of variables:

1. u

= u

, 0 < x < 1,

(0,t) = u(1,t) = 0, u(x,0) = x

−1;

2. u

= u

+ u, 0 < x < l,

u(0,t) = u(l,t) = 0, u(x, 0) = u

;

3. u

= u

−4u, 0 < x < π,

u(0,t) = u(π,t) = 0, u(x,0) = x

−πx;

4. u

= a

−βu, 0 < x < l,

u(0,t) = u

(l,t) = 0, u(x, 0) = sin

πx

;

Solution

Problem 2. We seek the solution in the form

u(x,t) =

∞

∑

n=1

−λ

(x),

where the numbers λ

are the eigenvalues, and the functions y

(x) are the eigenfunctions

of the boundary value problem

+ (λ −1)y = 0, y(0) = y(l) = 0.

Solving this boundary value problem we get

πn

+ 1,

(x)

= sin

πn

x, n ≥1.

Therefore,

u(x,t)

∞

∑

−

1+(

πn

)

sin

πn

Using

the

initial conditions we obtain

∞

∑

n=1

sin

πn

x = u

and

consequently

πn

(1 +

(−1

)

n+1

The solution of Problem 2 has the form

u(x,t) =

∞

∑

n=1

πn

(1 +

(−1

)

n+1

−

1+(

πn

)

sin

πn

Exercises 253

Using the

method of separation of variables one can solve also problems for non-

homogeneous equations and boundary conditions. Consider the following mixed problem

ρ(x)u

= (k(x)u

)

−q(x)u + f (x,t), 0 < x < l, t > 0, (6.3.4)

+ hu)|

x=0

= µ(t), (H

+ Hu)|

x=l

= ν(t), (6.3.5)

t=0

= ϕ(x). (6.3.6)

In the case of homogeneous boundary conditions (µ(t) = ν(t) = 0) we seek the solution of

(6.3.4)-(6.3.6) in the form of a series

u(x,t) =

∞

∑

n=1

(t)y

(x)

with respect to the eigenfunctions of the corresponding problem L . Substituting this into

(6.3.4) and (6.3.5) we obtain a Cauchy problem for T

(t) :

(t) + λ

(t) = f

(t), T

(0) = ϕ

, (6.3.7)

where f

(t) and ϕ

are the Fourier coefﬁcients for the functions

ρ(x)

f (x,t) and ϕ(x),

respecti

ely:

(t) =

f (x,t)y

(x)dx, ϕ

ρ(x)ϕ(x)y

(x)dx.

Solving the Cauchy problem (6.3.7) we calculate

(t) = ϕ

−λ

(τ)e

−λ

(t−τ)

dτ.

Problem (6.3.4)-(6.3.6) with non-homogeneous boundary conditions can be reduced to the

problem with homogeneous boundary conditions with the help of the replacement of the

unknown function u(x,t) = v(x,t)+w(x,t), where the function w(x,t) is chosen such that

it satisﬁes the boundary conditions (6.3.5).

6.3.7. Find the temperature distribution u(x,t) in a slender homogeneous bar (0 < x <

l) with isolated lateral surface provided that the end-points x = 0 and x = l are maintained

at the constant temperatures u(0,t) = µ

, u(l,t) = ν

, and where the initial temperature is

u(x,0) = ϕ(x).

6.3.8. The heat radiation takes place from the lateral surface of a slender homogeneous

bar (0 < x < l) to the surrounding medium of zero temperature. Find the temperature

distribution u(x,t) if the end-point s x = 0 and x = l are maintained at the constant tem-

peratures u(0,t) = µ

, u(l,t) = ν

, and if the initial temperature is u(x,0) = ϕ(x).

254 G.

Freiling and V. Yurko

6.3.9. Solve

the following problems applying the method of separation of variables:

1. u

= a

, 0 < x < l,

u(0,t) = 0, u(l,t) = At, u(x,0) = 0;

2. u

= u

, 0 < x < l,

(0,t) = 1, u(l,t) = 0, u(x, 0) = 0;

3. u

= u

, 0 < x < l,

(0,t) = u

(l,t) = u

, u(x, 0) = Ax;

4. u

= u

, 0 < x < l,

(0,t) = At, u

(l,t) = B, u(x, 0) = 0;

5. u

= a

−βu, 0 < x < l,

u(0,t) = µ

, u

(l,t) = v

, u(x, 0) = 0;

6. u

= a

−βu + sin

πx

, 0 < x < l,

u(0,t)

= u

(l,t) = 0, u(x, 0) = 0;

7. u

= u

+ u −x + 2sin 2xcos x, 0 < x <

u(0,t)

= 0

, u

(

,t)

= 1

, u(x, 0) = x;

8. u

= u

+ u + xt(2 −t) + 2 cost, 0 < x < π,

(0,t) = u

(π,t) = t

, u(x, 0) = cos 2x;

9. u

= u

−2u

+ u + e

sinx −t, 0 < x < π,

u(0,t) = u(π,t) = 1 +t, u(x,0) = 1 + e

sin2x;

10. u

= u

−2u

+ x + 2t, 0 < x < 1,

u(0,t) = u(1,t) = t, u(x,0) = e

sinπx.

Solution of Problem 1. By the replacement

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

reduce our problem to the following mixed problem with respect to v(x,t) :

= a

−

, 0 < x < l, t > 0,

v(0,t)

= v

(l,t) = 0, v(x, 0) = 0.

Exercises 255

seek the solution of this problem in the form

v(x,t) =

∞

∑

n=1

(t) sin

πn

have

∞

∑

n=1

sin

πn

where

x sin

πn

x =

nπ

(−1)

n−1

The

functions v

(t) are

solutions of the Cauchy problems

˙v

(t) +

anπ

(t)

(−1)

nπ

(0)

= 0

Hence

(t) = (−1)

2Al

1 −e

−

anπ

and

consequently

u(x,t) =

t +

∞

∑

n=1

(−1

)

2Al

1 −e

−

anπ

sin

πn

Solution

Problem 5. Denote

w(x) =

1 −

Clearly

, w(0

) = µ

, w

(l) = ν

. The replacement u(x,t) = v(x,t) + w(x) yields the fol-

lowing mixed problem with respect to v(x,t) :

= a

−βv + f (x), 0 < x < l, t > 0,

v(0,t) = v

(l,t) = 0, v(x, 0) = −w(x),

where f (x) = a

(x) −βw(x) . We seek the solution of this problem of the form

v(x,t) =

∞

∑

n=0

(t)sin

n +

The

functions v

(t) are

solutions of the Cauchy problems

˙v

(t) +

a(2n + 1)π

(t) −βv

(t)

= f

) = −w

256 G.

Freiling and V. Yurko

where

f (x) sin

n +

x, w

w(x)sin

n +

Hence

(t) =

−

xp(−λ

t),

and

consequently,

u(x,t) = w(x) +

∞

∑

n=0

−

xp(−λ

sin

n +

where

a(2n + 1)π

−β.

Solution

Problem 10. By the replacement u(x,t) = v(x,t)+t we reduce our problem

to the following mixed problem with respect to v(x,t) :

= v

−2v

+ x + 2t −1, 0 < x < 1, t > 0,

v(0,t) = v(1,t) = 0, v(x,0) = e

sinπx.

We seek the solution of this problem of the form

v(x,t) = e

∞

∑

n=1

(t) sin πnx.

The functions v

(t) are solutions of the Cauchy problems

˙v

(t) +

(nπ)

+ 1

(t) = f

(t),

(0) = γ

where

(t) = 2

−x

(x + 2t −1)sin πnxdx, n ≥1,

= 1, γ

= 0, n ≥ 2.

Hence

(t) = γ

exp(−(n

+ 1)t) +

(τ)exp(−(n

+ 1)(t −τ))dτ.

Exercises 257

6.4. Elliptic

Partial Differential Equations

Elliptic equations usually describe stationary ﬁelds, for example, gravitational, electro-

statical and temperature ﬁelds. The most important equations of elliptic type are the Laplace

equation ∆u = 0 and the Poisson equation ∆u = f (x), where x = (x

,. .. ,x

) are spatial

variables, u(x) is an unknown function,

∆u :=

∑

k=1

∂

∂x

the

Laplace operator (or Laplacian), and f (x) is a given function. In this section we

suggest exercises for boundary value problems for elliptic partial differential equations.

We will need the expressions of the Laplace operator in curvilinear coordinates:

i) polar coordinates (n = 2, x

= ρ cos ϕ, x

= ρ sin ϕ) :

∆u(ρ,ϕ) =

∂

∂ρ

∂u

∂ρ

∂

∂ρ

;

ii)

ylindrical coordinates (n = 3, x

= ρ cos ϕ, x

= ρ sin ϕ, x

= z) :

∆u(ρ,ϕ,z) =

∂

∂ρ

∂u

∂ρ

∂

∂ϕ

∂

∂z

;

iii)

spherical

coordinates (n = 3, x

= r cosϕsinθ,x

= r sinϕcosθ,z = r cos θ) :

∆u(r,ϕ, θ) =

∂

∂r

∂u

∂r

sinθ

∂

∂θ

sinθ

∂u

∂θ

sin

∂

∂ϕ

Harmonic

functions and their properties

A function u(x) = u(x

,. .. ,x

) is called harmonic in a domain D ⊂ R

, if u(x) ∈

(D) (i.e. twice continuously differentiable in D ) and ∆u = 0 in D. If D is unbounded

then there is the additional condition: u(x) is bounded for n = 2, and u(x) → 0 for n ≥ 3

as |x| :=

∑

i=1

→ ∞ .

6.4.1. Check

that

the following functions are harmonic:

1. u

) = ln

|x −y|

= ln

−y

)

−y

)

n = 2

, x = (x

), y = (y

), x 6= y;

2. u

) =

|x −y|

−y

)

−y

)

−y

)

n = 3, x = (x

), y = (y

), x 6= y,

258 G.

Freiling and V. Yurko

where y is a

ﬁxed point.

The function u

is called the fundamental solution of the Laplace equation on the plane.

The function u

is called the fundamental solution of the Laplace equation in the space R

The importance of the fundamental solution connects with the isotropy of the space when

the physical picture depends only on the distance from the source of energy but not on

the direction. For example, the function u

represents the potential of the gravitational

(electro-statical) ﬁeld created by a point unit mass (point unit charge). Similar sense has the

fundamental solution of the Laplace equation on the plane: this is the potential of the ﬁeld

produced by a charge of constant linear density q = 1, distributed uniformly along the line

= y

, x

= y

6.4.2. Find all harmonic functions, deﬁned in a bounded domain D ⊂ R

, and having

the following form:

1. u(x

) is a polynomial of the second degree;

2. u(x

) depends only on one variable;

3. u(x

) depends only on the distance ρ =

+ x

;

4. u(x

) depends

only

on the angle ϕ between the x

-axis and the vector connect-

ing the origin and the point x = (x

6.4.3. Find all harmonic functions, deﬁned in a bounded domain D ⊂ R

, and having

the following form:

1. u(x

) is a polynomial of the second degree;

2. u(x

) depends only on the distance r =

+ x

;

3. u(x

) depends

only

on the angle θ between the coordinate x

-plane (i.e.

= 0 ) and the vector connecting the origin and the point x = (x

6.4.4. Find the value of the constant k, for which the following functions are harmonic

in a bounded domain D ⊂ R

1. u(x

) = x

+ kx

;

2. u(x

) = e

coshkx

;

3. u(x

) = sin 3x

coshkx

6.4.5. Let the function u(x

) be harmonic in a bounded domain D ⊂ R

. Check

which of these functions are harmonic:

∂u(x

)

∂x

∂u(x

)

∂x

;

2. x

∂u(x

)

∂x

−x

∂u(x

)

∂x

;

Exercises 259

3. x

∂u(x

)

∂x

−x

∂u(x

)

∂x

6.4.6. Let

the

function u(x

) be harmonic in a bounded domain D ⊂R

. Check

which of these functions are harmonic:

1. u(ax

,ax

), where a is a constant;

2. u(x

+ h

) , where h

are constants;

∂u(x

)

∂x

∂u(x

)

∂x

6.4.7. Can

non-constant harmonic function in a bounded domain D have a closed

level curve?

Hint. Use the maximum principle for harmonic functions: Let u(x) , x ∈ R

be har-

monic in a domain D ⊂ R

and continuous in

D. Then u(x) attains

its

maximum and

minimum on the boundary of D.

The theory of harmonic functions of two variables has a deep connection with the the-

ory of analytic functions. Let n = 2, i.e. x = (x

), and let z = x

+ ix

. If the function

f (z) = u(x

)+iv(x

) is analytic in D, then by virtue of the Cauchy-Riemann differ-

ential equations

∂u

∂x

∂v

∂x

∂u

∂x

= −

∂v

∂x

the

functions u and v are

harmonic in D (and they are called conjugate harmonic func-

tions). Conversely, if u(x

) is harmonic in the simply connected domain D, then there

exists a conjugate harmonic function v(x

) such that the function f = u + iv is analytic

in D, and

v(x

) =

)

−

∂u

∂x

∂

∂x

+C, (6

.4.1)

where (y

) is a ﬁxed point in D (the integral does not depend on the way of the inte-

gration since under the integral we have the total differential of v ).

Another way of recovering of an analytic function from its real part is given by the

Goursat formula:

f (z) = 2u

−u(0,0)

+ i

C, z = x

+ ix

, (0,0) ∈ D. (6.4.2)

6.4.8. Let the function u(x

) be harmonic. Show that the function

f (z) =

∂u(x

)

∂x

−i

∂u(x

)

∂x

, z = x

+ ix

analytic.

6.4.9. Let

the function u(x

) be harmonic in D with (u

)

+ (u

)

6= 0 in D .

Show that the function

)

260 G.

Freiling and V. Yurko

is also

harmonic in D .

6.4.10. Recover function f (z) , which is analytic in a simply connected domain D ,

from its real part:

1. u(x

) = x

−3x

;

2. u(x

) = e

sinx

;

3. u(x

) = sin x

coshx

;

4. u(x

) = x

−x

;

5. u(x

) = x

+ x

Solution of Problem 2. We construct the conjugate function v(x

) by (6.4.1). As

path of integration we choose the rectilinear segments connecting the points (y

),(x

)

and (x

),(x

) (we suppose that these segments lie in D ). Then

v(x

) = −

cosy

dt +

sint dt +C

= e

cosy

−e

cosx

+C.

Hence

f (z) = e

sinx

−ie

cosx

+ i(e

cosy

+C).

6.4.11. Show that if two harmonic functions in D coincide in a domain D

⊂ D , then

they coincide in the whole domain D.

6.4.12. Let u(x

) be a harmonic function. Calculate the integrals:

u(x

−x

,2x

)ds, S = {(x

) : x

+ x

= 1};

2π

(sincos ϕ)(coshsin ϕ),(cos cosϕ)(sinh sinϕ)

dϕ;

2π

∑

k=1

coskϕ,

∑

k=1

sinkϕ

dϕ.

Solution of Problem 3. We use the following important property of harmonic functions:

If the function u(x

) is harmonic in D and

z = g(ζ) = ν(ξ,η) + iµ(ξ,η)

is an analytic function in a domain ∆ having values in D, then the composite function

u(ν(ξ,η),µ(ξ,η)) = U(ξ,η)

is harmonic in D with respect to (ξ,η).

Exercises 261

introduce the function

U(ξ,η) = u

ℜ

∑

k=1

,ℑ

∑

k=1

, ζ = ξ + iη.

Clearly, U is harmonic. Consider this function on the circle ξ

+ η

= 1. Let us introduce

the polar coordinates ξ = cos ϕ,η = sin ϕ. Then ζ

= cos kϕ + i sinkϕ ,

ℜ

∑

k=1

∑

k=1

coskϕ, ℑ

∑

k=1

∑

k=1

sinkϕ.

Therefore,

2π

U(cosϕ,sinϕ)dϕ =

2π

∑

k=1

coskϕ,

∑

k=1

sinkϕ

dϕ.

By the mean-value theorem for harmonic functions,

2π

U(cosϕ,sinϕ)dϕ = 2πU(0,0).

Since

∑

k=1

ζ=0

= 0, we have U(0,0) = u(0,0). Hence the desired integral is equal to

2πu(0,0).

2. Boundary value problems for elliptic equations.

The Green’s function

Let D ⊂R

be a bounded domain with a piecewise smooth boundary S.

Dirichlet problem. Let a continuous function g(x) be given on S. Find a function u(x)

which is harmonic in D and continuous in

D and

has

on S assigned values:

∆u = 0 (x ∈ D),

= g.

Neumann problem. Let a continuous function g(x) be given on S. Find a function u(x)

which is harmonic in D and u(x) ∈C

(

D) ,

∂u

∂n

= g ,

where n is

the outer normal to S :

∆u = 0 (x ∈ D),

∂u

∂n

= g.

Then

the

Dirichlet and Neumann problems can be considered also in unbounded do-

mains. In this case we need the additional condition u(∞) = 0 (for n = 2 the additional

condition has the form u(x) = O(1), |x| → ∞ ). For example, let D ⊂ R

be a bounded set,

and D

:= R

D. Then

the

Dirichlet and Neumann problems in the domain D

are called

exterior problems.