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

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242 G.

Freiling and V. Yurko

hav

e the general solutions

(t) = e

−t

cos

n +

t + B

sin

n +

, n ≥0,

where A

and B

are

arbitrary

constants. Hence,

(x,t) = e

−t

cos

n +

t + B

sin

n +

cos

n +

x .

seek the solution of the mixed problem of the form

u(x,t) =

∞

∑

n=0

−t

cos

n +

t + B

sin

n +

cos

n +

and

ﬁnd A

and B

from the initial conditions:

= 0, B

8(−1)

(2n + 1)

−

π(2n + 1)

This

yields

(x,t) = e

−t

∞

∑

n=0

sin

n +

t cos

n +

x .

The

method

of separation of variables allows us to construct solutions of mixed prob-

lems for non-homogeneous equations with boundary conditions.

Consider the problem of forced oscillations of a non-homogeneous string under the

inﬂuence of an exterior force of density f :

ρ(x)u

= (k(x)u

)

−q(x)u + f (x,t), (6.2.7)

ρ(x) > 0, k(x) > 0, 0 < x < l, t > 0,

t=0

= ϕ(x), u

t=0

= ψ(x), (6.2.8)

+ hu)|

x=0

= µ(t), (H

+ Hu)|

x=l

= ν(t). (6.2.9)

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

of problem (6.2.7)-(6.2.9) in the form of a series

u(x,t) =

∞

∑

n=1

(t)y

(x), (6.2.10)

with respect to the eigenfunctions of problem L . Substituting (6.2.10) into (6.2.7) and

(6.2.8) we obtain the following Cauchy problem for T

(t) :

(t) + λ

(t) = f

(t),

(0) = ϕ

, T

(0) = ψ

Exercises 243

where f

(t), ϕ

, ψ

are the

Fourier coefﬁcients for the functions

ρ(x)

f (x,t), ϕ(x), ψ(x),

respecti

ely:

(t) =

f (x,t)y

(x)dx,

ρ(x)ϕ(x)y

(x)dx,

ρ(x)ψ(x)y

(x)dx.

Solving the Cauchy problem we arrive at

(t) = ϕ

cos

t + ψ

sin

√

sin

√

(t −τ)

√

(τ)dτ.

Problem

(6.2.7)-(6.2.9)

with non-homogeneous boundary conditions can be reduced to

a problem with homogeneous boundary conditions using of the following replacement of

the unknown function u(x,t) = V (x,t) +W (x,t), where

W (x,t) = (a

+ a

x + a

)µ(t) + (a

+ a

x + a

)ν(t),

and where the coefﬁcients a

i j

are chosen such that the function W (x,t) satisﬁes the bound-

ary conditions (6.2.9). For example, for the boundary conditions u(0,t) = µ(t), u(l,t) =

ν(t) one can take

W (x,t) =

1 −

µ(t)

ν(t).

6.2.17. Solv

the problem of forced oscillations of a homogeneous string (0 < x < l)

with ﬁxed end-points under the inﬂuence of the exterior force of density f (x,t) = Asin ωt

and with zero initial conditions. Study the resonance effect and ﬁnd the solution in the case

of resonance.

6.2.18. Solve the problem of forced oscillations of a homogeneous string (0 < x < l)

under the inﬂuence of the exterior force of density f (x,t) and with zero initial conditions

provided that:

1. f (x,t) = f

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

2. f (x,t) = f

, u

(0,t) = u

(l,t) = 0;

3. f (x,t) = cos

πt

, u(0,t)

= u

(l,t) = 0;

4. f (x,t) = Axe

−t

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

5. f (x,t) = Ae

−t

sin

x, u(0,t)

= u

(l,t) = 0;

244 G.

Freiling and V. Yurko

6. f (x,t) = x, u

(0,t) = u

(l,t) = 0;

7. f (x,t) = x(l −x), u(0,t) = u(l,t) = 0.

Solution of Problem 1. We have to solve the following mixed problem

= a

+ f

, 0 < x < l, t > 0,

|x=0

= u

|x=l

= 0,

|t=0

= u

t|t=0

= 0.

We seek the solution of the form

u(x,t) =

∞

∑

n=1

(t) sin

nπ

have

∞

∑

n=1

sin

nπ

where

2 f

sin

nπ

x =

2 f

nπ

(1 −(−1)

) .

The

functions u

(t) are

solutions of the Cauchy problems

¨u

(t) +

anπ

(t)

= F

) = ˙u

(0) = 0.

Hence

(t) =

anπ

sin

anπ

(t −τ) dτ

(anπ)

1 −cos

anπ

and

consequently

u(x,t) =

∞

∑

n=1

2 f

(1 −(−1)

)

1 −cos

anπ

sin

nπ

x .

6.2.19. Solv

the problem of forced transverse oscillations of a homogeneous string

(0 < x < l) ﬁxed at the end-point x = 0 under the inﬂuence of the force of density Asin ωt

applied to the end-point x = l provided that the initial displacement and velocity are equal

to zero.

6.2.20. Solve the problem of forced oscillations of a homogeneous string (0 < x < l)

under the inﬂuence of the exterior force of density f (x,t) provided that:

1. u(x,0) = u

(x,0) = 0, u(0,t) = µ

, u(l,t) = ν

, f (x,t) = 0;

Exercises 245

2. u(x,0) = u

(x,0

) = 0, u

(0,t) = µ

, u

(l,t) = ν

, f (x,t) = 0;

3. u(x,0) = u

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

4. u(x,0) = 0, u

(x,0) = 1, u

(0,t) = sin 2t, u

(l,t) = 0,

f (x,t) = sin 2t;

5. u(x,0) = sin

πx

, u

(x,0)

= 0

, u(0,t) = t

, u(l,t) = t

, f (x,t) = 0;

6. u(x,0) = 0, u

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

(l,t) = 1, f (x,t) = f

Solution of Problem 1. We have to solve the following mixed problem

= a

, 0 < x < l, t > 0,

|x=0

= µ

, u

|x=l

= ν

|t=0

= u

t|t=0

= 0.

Denote

w(x) =

1 −

The

replacement u

(x,t) = v(x,t) + w(x) yields the following mixed problem with respect

to v(x,t) :

= a

, 0 < x < l, t > 0,

|x=0

= 0, v

|x=l

= 0,

|t=0

= −w(x), v

t|t=0

= 0.

Applying the method of separation of variables we ﬁnd

v(x,t) =

∞

∑

n=1

cos

anπ

t sin

nπ

x ,

where

= −

w(x)sin

nπ

x =

nπ

(ν

(−1)

−µ

Hence

u(x,t)

1 −

∞

∑

n=1

nπ

(ν

(−1)

−µ

)cos

anπ

t sin

nπ

x .

6.2.21. Solv

the problem of free oscillations of a homogeneous string (0 < x < l)

with ﬁxed end-points in a medium whose resistance is proportional to the ﬁrst degree of the

velocity.

6.2.22. Solve the following problems:

1. u

= u

, 0 < x < 1,

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

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

+ 2;

246 G.

Freiling and V. Yurko

2. u

= u

+ u, 0 < x < 2,

u(x,0) = 0, u

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

3. u

+ u

= u

, 0 < x < 1,

u(x,0) = 0, u

(x,0) = 1 −x, u(0,t) = t, u(1,t) = 0;

4. u

−7u

= u

+ 2u

−2t −7x −e

−x

sin3x, 0 < x < π,

u(x,0) = 0, u

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

5. u

= u

+ 4u + 2sin

x, 0 < x < π,

u(x,0) = u

(x,0) = 0, u

(0,t) = u

(π,t) = 0;

6. u

−3u

= u

+ u −x(4 + t) + cos

, 0 < x < π,

u(x,0)

= u

(x,0

) = x, u

(0,t) = t + 1, u(π,t) = π(t + 1);

7. u

−3u

= u

+ 2u

−3x + 2t, 0 < x < π,

u(x,0) = e

−x

sinx, u

(x,0) = x, u(0,t) = 0, u(π,t) = πt.

Solution of Problem 2. Denote

w(x,t) = t(2 −x).

The replacement u(x,t) = v(x,t)+w(x,t) yields the following mixed problem with respect

to v(x,t) :

= v

+ v + w, 0 < x < 2, t > 0,

|x=0

= 0, v

|x=2

= 0,

|t=0

= 0, v

t|t=0

= x −2.

We seek the solution of this problem of the form

v(x,t) =

∞

∑

n=1

(t) sin

nπ

have

x −2 =

∞

∑

n=1

sin

nπ

x ,

where

(x −2)sin

nπ

x =

nπ

The

functions v

(t) are

solutions of the Cauchy problems

¨v

(t) =

nπ

−1

(t)

= −tγ

) = 0, ˙v

(0) = γ

Exercises 247

Hence

(t) = γ

sinµ

−

sinµ

(t −τ)

τγ

dτ

nπ

sinµ

nπ

−1,

and

consequently

u(x,t) = t(2 −x) +

∞

∑

n=1

nπ

sinµ

sin

nπ

The

Riemann method

6.2.23. Using the Riemann method (see Section 2.4) solve the following problems in

the domain −∞ < x < ∞, t > 0 :

1. u

−u

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

(x,0) = x;

2. u

−u

= 2, u(x,0) = sin x, u

(x,0) = x

;

3. u

−u

+ 2u

+ u = e

−x

, u(x, 0) = x, u

(x,0) = 1;

4. u

−u

+ 2u

−u = e

, u(x, 0) = x, u

(x,0) = 0.

Solution of Problem 1. The Riemann function satisﬁes the conditions

ξξ

−v

ηη

= 0 in ∆

PQM

v = 1 on MP,

v = 1 on MQ.

Consequently, v ≡1. Using the Riemann formula we get

u(x

) =

−t

x −

−t+t

t−t

= −

)

−(x

−t

)

−

2t(t

−t) d

= x

−

= x

−

= x

−

Hence,

u(x,t)

= x

t −

248 G.

Freiling and V. Yurko

6.2.24. Find the

Riemann function for the operator

L u = u

−a

, a = const,

and solve the Cauchy problem

= a

+ f (x,t), −∞ < x < ∞, t > 0,

u(x,0) = ϕ(x), u

(x,0) = ψ(x).

6.2.25. Solve the Cauchy problem

−y

= 0, −∞ < x < ∞, 1 < y < +∞,

y=1

= ϕ(x), u

y=1

= ψ(x).

6.3. Parabolic Partial Differential Equations

Parabolic partial differential equations usually describe various diffusion processes. The

most important equation of parabolic type is the heat (conduction) equation or diffusion

equation. In this section we suggest exercises for this type of equations.

1. The Cauchy problem for the heat equation

6.3.1. Using the Poisson formula (3.2.6) solve the following problems:

1. u

, u

(x,0) = e

2x−x

, −∞ < x < ∞, t ≥ 0;

2. u

, u

(x,0) = e

−x

sinx, −∞ < x < ∞, t ≥ 0;

3. u

= u

, u(x, 0) = xe

−x

, −∞ < x < ∞, t ≥ 0.

6.3.2. Prove that the non-homogeneous equation

= a

+ f (x,t), −∞ < x < ∞, t > 0

with the initial condition

u(x,0) = 0

has the following solution:

u(x,t) =

∞

−∞

f (ξ, τ)

π(t −τ)

−

(ξ −x)

(t −τ)

dξdτ.

Hint. Apply

the

method from Chapter 2, Section 2.1 for the non-homogeneous wave

equation.

6.3.3. Solve the following problems in the domain −∞ < x < ∞, t ≥ 0 :

Exercises 249

1. u

= 4u

+t + e

, u

(x,0) = 2;

2. u

= u

+ 3t

, u(x, 0) = sin x;

3. u

= u

+ e

−t

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

4. u

= u

+ e

sinx, u(x,0) = sin x;

5. u

= u

+ sint, u(x, 0) = e

−x

2. The mixed problem for the heat equation

Consider the following problem for the heat equation

ρ(x)u

= (k(x)u

)

−q(x)u, (6.3.1)

ρ(x) > 0, k(x) > 0, 0 < x < l, t > 0,

with the boundary conditions

+ hu)|

x=0

= (H

+ Hu)|

x=l

= 0. (6.3.2)

and with the initial condition

|t=0

= ϕ(x). (6.3.3)

For solving problem (6.3.1)-(6.3.3) we apply the method of separation of variables. We

will seek non-trivial (i.e. not identically equal to zero) solutions of equation (6.3.1) satis-

fying the boundary conditions (6.3.2) and having the form u(x,t) = T (t)y(x) . Substituting

this into (6.3.1) and (6.3.2) we obtain the equation

T (t) + λT (t) = 0

for the function T (t) , and the Sturm-Liouville problem L

−(k(x)y

(x))

+ q(x)y(x) = λρ(x)y(x), 0 ≤x ≤l,

(0) + hy(0) = H

(l) + Hy(l) = 0,

for the function y(x). Let {λ

} and {y

(x)} be eigenvalues and eigenfunctions of the

Sturm-Liouville problem respectively. Then the functions

(x,t) = A

exp(−λ

t)y

(x)

satisfy (6.3.1) and (6.3.2) for any A

We will seek the solution of the mixed problem (6.3.1)-(6.3.3) in the form

u(x,t) =

∞

∑

n=1

−λ

(x).

250 G.

Freiling and V. Yurko

Using the

initial condition (6.3.3) for ﬁnding the coefﬁcients A

, we get

ϕ(x) =

∞

∑

n=1

(x),

and consequently,

ρ(x)ϕ(x)y

(x)dx.

6.3.4. 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 temperature zero, and the initial temperature u(x,0) is given:

1. u(x,0) = u

;

2. u(x,0) = u

x(l −x);

3. u(x,0) = x

0 ≤ x ≤

u(x,0)

= l −x

≤ x ≤ l

;

4. u(x,0)

= sin

6.3.5. Find

the

temperature distribution u(x,t) in a slender homogeneous bar (0 < x <

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

the initial temperature u(x,0) is given:

1. u(x,0) = u

;

2. u(x,0) = u

0 ≤ x ≤

, u(x, 0)

= 0

≤ x ≤ l

;

3. u(x,0)

= sin

4. u(x,0)

= x

0 ≤ x ≤

, u(x, 0)

= l −x

≤ x ≤ l

Solution

Problem 3. We have to solve the following mixed problem

= a

, 0 < x < l, t > 0,

x|x=0

= u

x|x=l

= 0,

|t=0

= sin

Exercises 251

First we

seek nontrivial particular solutions of the equation of the form u(x,t) = Y (x)T (t),

which satisfy the boundary conditions. Repeating the arguments from section 3.1 we obtain

for the function T (t) the ordinary differential equation

T (t) + a

λT (t) = 0,

and for the function Y(x) we get the Sturm-Liouville boundary value problem

(x) + λY (x) = 0, Y

(0) = Y

(l) = 0.

Here λ is the spectral parameter. The eigenvalues and the eigenfunctions of this boundary

value problem are

nπ

, Y

(x)

= cos

x, n ≥0.

The

corresponding

equations

(t) + a

(t) = 0, ,n ≥0,

have the general solutions

(t) = A

exp

−

anπ

, n ≥0,

where A

are

arbitrary

constants. Hence,

(x,t) = A

exp

−

anπ

cos

nπ

x, n ≥0.

seek the solution of the mixed problem of the form

u(x,t) =

∞

∑

n=0

exp

−

anπ

cos

nπ

and

ﬁnd A

from the initial conditions:

sin

x =

∞

∑

n=0

cos

nπ

x ,

hence

sin

x cos

nπ

x n ≥1,

We calculate

, A

2n+1

= 0, A

π(1 −4n

)

and

consequently

u(x,t) =

∞

∑

n=1

π(1 −4n

)

−

anπ

cos

2nπ