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

Подождите немного. Документ загружается.

272 G.

Freiling and V. Yurko

where D is the

half-space x

> 0, and

1) f (x) = 0, g(x) = cosx

cosx

2) f (x) = e

−x

sinx

cosx

, g(x) = 0.

6.4.29. Find the solution of the Dirichlet problem

∆u(x

) = 0, x ∈D,

u(x

= g(x

where

1) D is the half-plane x

> 0, and g(x

) =

1 + x

;

2) D is

the

strip 0 < x

< π, and g(x

,0) = cos x

, g(x

,π) = 0 .

3. Potentials and their applications

Let D ⊂ R

(n ≥2) be a bounded domain with the smooth boundary S. Consider the

Dirichlet problem for the Poisson equation:

∆V (x) = −f (x), (6.4.4)

V (x)|

x∈S

= g(x), x = (x

,. .. ,x

), x ∈D ∈R

. (6.4.5)

If a particular solution of equation (6.4.4) is known, then problem (6.4.4)-(6.4.5) can be

reduced to the Dirichlet problem for the Laplace equation ∆V (x) = 0 , considered above.

As a partial solution of (6.4.4) one can take the function

V (x) =











(n −2)

f (τ)

|x −τ|

n−2

dτ, n ≥ 3,

2π

f (τ) ln

|x −τ|

dτ, n = 2,

(6.4.6)

where ω

the

area of the unit sphere in R

, and

|x −τ|=

∑

i=1

−τ

)

The

function V (x) is

called the volume potential, and the function f (x) is called the

density. For n = 2 the function V (x) is called also the logarithmic volume potential. For

example, for n = 3 the function V (x) describes the potential of gravitational ﬁeld created

by the body D with the density f .

If f (x) ∈C

(

D), then

the

function V (x) satisﬁes the equation

∆V (x) =

−f (x), x ∈D,

0, x /∈ D.

6.4.30. Can a harmonic in D function be a volume potential with non-zero density?

Exercises 273

6.4.31. Let n = 3. Find the

density f (x

) if the volume potential in D has the

form

V (x

) = (x

+ x

)

−1.

6.4.32. Show that the function

V (x

) =











−

ln|x|, |x|

≥ 1

(1 −

|x|

)

, |x| ≤ 1

|x| =

+ x

logarithmic volume potential for the disc |x| < 1 with the density f (x

) = 1.

6.4.33. Show that the function

V (x

) =











, |x|

≥ 1

(1 −

|x|

), |x|

≤ 1

|x| =

+ x

volume potential for the disc |x| < 1 with the density f (x

) = 1.

6.4.34. Let a domain D ⊂R

contains the square −1 ≤ x

≤ 1 , −1 ≤ x

≤ 1, and let

V (x

) = x

be the logarithmic volume potential for D. Determine the mass contained in the square.

6.4.35. The function

V (x

) =

πx

2 −

|x|

, |x|

≤ 1

is the logarithmic volume potential for the disc |x| ≤ 1. Find the density f (x

) and the

value of the potential for |x| > 1.

6.4.36. Find the volume potential V (x) for the ball |x| ≤ R (n = 3) with the constant

density f

1) by the direct calculation of the volume integral;

2) by solving the corresponding boundary value problem.

6.4.37. Find the volume potential in R

for the body distributed in

1) the spherical layer a ≤ |x| ≤ b with the constant density f

;

2) the ball |x| ≤ a with the constant density f

, and the spherical layer a < b < |x| < c

with the constant density f

. .

Consider the function

W (x) =











−

(n −2)

V (τ)

∂

∂n

|x −τ|

n−2

τ), n ≥3,

−

2π

V (τ)

∂

∂n

|x −τ|

(τ), n = 2,

(6.4.7)

274 G.

Freiling and V. Yurko

where n

is the

outer normal to S at the point τ ∈ S. The function W (x) is called the

double-layer potential with the density V. For n = 2 the function W (x) is also called the

logarithmic double-layer potential.

Physical sense: W (x) is the potential of the ﬁeld created by the dipole distribution on

S with the density V.

It is convenient to calculate the integral in (6.4.7) by the formula

W (x) =











V (τ)

cosϕ

|x −τ|

n−1

(τ), n ≥3,

2π

V (τ)

cosϕ

|x −τ|

(τ), n = 2,

where ϕ is the angle between n

and the vector directed from the point x to the point τ.

If V (τ) is continuous on S, then W (x) is harmonic in D and D

:= R

D. Denote

(x)

= lim

→

∈D

W (x

), W

−

(x) = lim

→x

∈D

W (x

The function W(x) has a jump on S :

(x) = W (x) +

V (x), W

−

(x)

= W (x) −

V (x), x ∈S. (6.4.8)

6.4.38. Let D = {x ∈ R

: |x| < r}, and

let S be

the boundary of D. Find the solution

of the Dirichlet problem

∆W (x) = 0, x ∈ D, (6.4.9)

W (x)|

= g(x). (6.4.10)

Solution. We seek the desired function W (x) in the form of a logarithmic double-layer

potential:

W (x) =

2π

V (τ)

cosϕ

|x −τ|

(τ). (6.4.11)

It follows from (6.4.10) that W

(x) = g(x) for x ∈S. Using (6.4.8) we obtain

V (x)

2π

V (τ)

cosϕ

|x −τ|

(τ) = g(x).

Since

cosϕ

|x −τ|

V (x) +

2πr

V (τ)d

(τ) = 2g(x). (6.4.12)

Then

V (x) = 2g(x) + A (A −const). (6.4.13)

Using (6.4.12) we calculate

A = −

2πr

g(τ)d

(τ). (6.4.14)

Exercises 275

Relations (6.4.11),

(6.4.13) and (6.4.14) give us the solution of the problem (6.4.9)-(6.4.10).

Using (6.4.11), (6.4.13) and (6.4.14) one can obtain the solution of the problem (6.4.9)-

(6.4.10) in the form of a Poisson integral. Indeed, for x ∈D we have

W (x) =

2π

cosϕ

|x −τ|

2g(τ) −

2πr

g(ξ)d

(ξ)

ds(τ)

cosϕ

|x −τ|

g(τ)d

(τ) −

4π

g(ξ)d

(ξ)

cosϕ

|x −τ|

(τ)

cosϕ

|x −τ|

g(τ)d

(τ) −

4π

g(ξ)d

s(ξ

)

·2π

cosϕ

|x −τ|

−

g(τ)d

(τ). (6.4.15)

Here we use the relation

W (x)|

V ≡1

2π

cosϕ

|x −τ|

τ) = 1, x ∈ D.

Applying the cosine theorem we calculate

cosϕ

|x −τ|

−

2r cos ϕ −

|x −τ|

2r|x −τ|

2r cos ϕ|x −τ|

−

|x −τ|

2r|x −τ|

−ρ

2r(r

+ ρ

−2rρ

cos(θ −θ

))

, (6.4.16)

where (ρ

,θ

) are

the

polar coordinates of the point (x

) , and θ is the polar angle

of the point τ ∈ S. Substituting (6.4.16) into (6.4.15) and using the polar coordinates we

obtain the Poisson integral:

W (ρ

,θ

) =

2π

−ρ

)g(θ)dθ

+ ρ

−2rρ

cos(θ −θ

)

6.4.39. Using

the

double-layer potential solve the Dirichlet problem:

∆W (x) = 0, x ∈ D, W (x)|

= g(x)

for the following domains

1) D = {x ∈ R

: |x| > r};

2) D = {x ∈ R

: x

> 0};

3) D = {x ∈ R

: x

> 0}.

6.4.40. Find the logarithmic double-layer potential for the circle |x| = 1

1) with the constant density V

2) with the density V (x

) = x

276 G.

Freiling and V. Yurko

The function

(x) =











(n −2)

µ(τ)

|x −τ|

n−2

(τ), n ≥3,

2π

µ(τ)ln

|x −τ|

τ), n = 2

is called the single-layer potential. For n = 2 the function u(x) is also called the logarith-

mic single-layer potential.

Physical sense: u(x) is the potential of the ﬁeld created by charges distributed on S

with the density µ.

If the function µ(τ) is continuous on S, then u(x) is continuous everywhere in R

and harmonic for x ∈R

\S. Denote

∂u

∂n

= lim

→x

∈D

∂u(x

)

∂n

∂u

∂n

−

= lim

→x

∈D

∂u(x

)

∂n

where

∂u(x)

∂n

the

direct value of the normal derivative on S. Then

∂u

∂n

∂u(x)

∂n

µ(x), x ∈S, (6.4.17)

∂u

∂n

−

∂u(x)

∂n

−

µ(x), x ∈S. (6.4.18)

The

single-layer

potential is used for the solution of the Neumann problem.

6.4.41. Find the behavior of the single-layer and double-layer potentials as |x| → ∞ ,

n = 2, 3.

6.4.42. Find the single-layer potential u(x) (x ∈ R

) for the sphere |x| = 1 with the

constant density µ

1) by direct calculation of the integral;

2) by solving the boundary value problem for u(x).

6.4.43. Find the logarithmic single-layer potential u(x) for the circle |x| = R with the

constant density µ

(

x) = 1.

6.4.44. The function

u(x

) = −

2|x|

, |x| > 1,

the

logarithmic single-layer potential for the circle |x| = 1. Find u(x

) for |x| < 1.

6.4.45. The function

u(x

) =

|x|

1 +

|x|

, |x|

≤ 1

the logarithmic single-layer potential for the circle |x| = 1. Find the density µ(x

Exercises 277

6.4.46. Using the

logarithmic single-layer potential solve the Neumann problem:

∆u(x) = 0, x ∈D := {x ∈R

: |x| < r},

∂u

∂n

x∈S

= g(x).

6.5.

Answers

and Hints

6.1.1.

2. u

ηη

+ 21u

+ 7u

−3 = 0,

ξ = 3x + y, η = x;

3. u

ξη

+ u

+ 3u

+ η + 1 = 0,

ξ = 2x + y, η = x;

4. u

ξη

+ 6u

−18u

−10u + 2ξ = 0,

ξ = y, η = x −3y;

5. u

ξξ

+ u

ηη

−

u +

(6ξ + η −8)

= 0

ξ = y, η = 4x −2y;

7. u

ξη

6(ξ + η)

+ u

)

= 0

ξ =

3/2

+ y, η =

3/2

−y (x > 0);

ξξ

+ u

ηη

3ξ

= 0,

ξ =

(−x)

3/2

, η = y (x < 0);

8. u

ξξ

= 0,

ξ = x + y, η = x −y (y > 0);

ξη

= 0,

ξ =

(1 +

√

2)x + y, η =

(1 −

√

2)x + y (y < 0);

9. u

ξξ

−u

ηη

−

= 0,

ξ =

|x|, η =

|y| (x

y > 0

);

ξξ

+ u

ηη

−

= 0,

ξ =

|x|, η =

|y| (x

y < 0

);

278 G.

Freiling and V. Yurko

10. u

ξξ

−u

ηη

3ξ

−

3η

= 0,

ξ = |x|

3/2

, η = |y|

3/2

y > 0

);

ξξ

+ u

ηη

3ξ

3η

= 0,

ξ = |x|

3/2

, η = |y|

3/2

y < 0

);

11. u

ξη

−

2ξ

= 0,

ξ = x

y, η =

y 6= 0

);

12. u

ξξ

+ u

ηη

−u

= 0,

ξ = ln |x|, η = ln |y| (xy 6= 0);

13. u

ξξ

+ u

ηη

2ξ

2η

= 0,

ξ = y

, η = x

y 6= 0

);

14. u

ξη

2(η

−ξ

)

(ηu

−ξu

)

= 0

ξ = y;

15. u

ξξ

+ u

ηη

= 0,

ξ = y, η = arctan x;

16. u

ξξ

+ u

ηη

−2u = 0,

ξ = ln(x +

√

1 + x

), η = ln(y +

1 + y

);

17. u

ξη

+ 2

= 0,

ξ =

, η = y;

18. u

ηη

−

2ξ

+ η

= 0,

ξ = y tan

, η = y;

19. u

ξη

−

2(ξ −η)

−u

)

4(ξ + η)

+ u

)

= 0

ξ = y

+ e

−x

, η = y

−e

, (y 6= 0);

20. u

ξη

4η

−

+ u = 0,

Exercises 279

ξ = x

y, η =

, (x

y 6= 0

);

21. u

ηη

−2u

= 0,

ξ = 2x −y

, η = y;

22. u

ηη

2η

ξ −η

−

= 0,

ξ = x

+ y

, η = x, (x 6= 0);

23. u

ξη

= 0,

ξ = x + y −cosx, η = −x + y −cosx;

24. u

ξη

2(ξ −η)

−u

)

= 0

ξ = y + 2

√

x −x, η = y −x −2

√

x, (x > 0);

ξξ

+ u

ηη

−

= 0,

ξ = y −x, η = 2

√

−x, (x < 0);

6.1.2.

1. v

ξξ

+ v

ηη

−11v = 0,

ξ = y −

, η =

, λ = −3, µ = 2;

2. v

ξη

−

= 0,

ξ = x −3y, η = 2x +y, λ = −

, µ = 0;

3. v

ηη

+ v

= 0,

ξ = y −x, η = x, λ = −

, µ = −

;

4. v

ξη

−

v +

2(ξ + η)

2ξ+4η

= 0,

ξ = 2x −y, η = x +y, λ = −2, µ = 4;

5. v

ξη

−

v = 0,

ξ = y +

(

√

3 −2)x, η = y −(

√

3 −2)x, λ = µ =

;

6. v

ηη

−

= 0,

ξ = x + 2y, η = x, λ = −

, µ =

;

280 G.

Freiling and V. Yurko

7. v

ξξ

+ v

ηη

−

297

v = 0,

ξ = x, η = 3x + y, λ = −

, µ = −8;

8. v

ξη

−

v = 0,

ξ = x −y, η = 3x + y, λ = −

, µ = −

;

9. v

ξη

−

v = 0,

ξ = x + y, η = 3x −y, λ = 2, µ =

;

10. v

ξξ

+ v

ηη

v + 2ηe

= 0,

ξ = 2x −y, η = x, λ = −

, µ = 0;

11. v

ξξ

+ v

ηη

−

v +

(η −ξ

−ξ+

= 0,

ξ = 2x −y, η = 3x, λ = 1, µ = −

;

12. v

ηη

−6v

= 0,

ξ = x + y, η = x, λ = −

, µ = −

;

13. v

ξη

−2v = 0, ξ = x −y, η = x + y, λ = −1, µ = 0;

14. v

ηη

−

(2η −ξ)e

−

ξ−

= 0,

ξ = x + 3y, η = x, λ = −

, µ =

;

15. v

ξη

−26v = 0,

ξ = 2x −y, η = x, λ = −2, µ = −12;

16. v

ξη

+ 16v + 8(ξ −η)e

ξ+η

= 0,

ξ = y −x, η = y, λ = µ = −2;

17. v

ξξ

+ v

= 0,

ξ = 2x −y, η = x +y, λ = 1, µ = 2.

Exercises 281

6.1.3.

1. u

ξξ

+ u

ηη

+ u

γγ

= 0,

ξ = y, η = x −y, γ = y −

x +

2. u

ξξ

+ u

ηη

+ u

γγ

= 0,

ξ = z, η = y −x, γ = x −2y + 2z;

3. u

ξξ

−u

ηη

+ 7u = 0,

ξ = y + z, η = −y −2z, γ = x −z;

4. u

ξξ

+ 5u = 0,

ξ = y, η = x −2y, γ = −y + z;

5. u

ξξ

−u

ηη

+ u

γγ

+ u

= 0,

ξ =

x, η =

x + y, γ = −

x −y + z;

6. u

ξξ

−2u

= 0,

ξ = z, η = y −2z, γ = x −3z;

7. u

ξξ

+ u

ηη

+ u

γγ

+ u

ττ

= 0,

ξ = t, η = y −t, γ = t −y + z, τ = x −2y + z + 2t;

8. u

ξξ

+ u

ηη

+ u

γγ

−u

ττ

= 0,

ξ = x + y, η = y + z −t, γ = t, τ = y −x;

9. u

ξξ

−u

ηη

+ u

γγ

= 0,

ξ = x, η = y −x, γ = 2x −y + z, τ = x + z + t.

6.1.4.

1. The

change

of variables

ξ = x + y, η = 3x + 2y

reduces the equation to the canonical form u

ξη

= 0 with the general solution u =

f (ξ) + g(η), hence

u(x,y) = f (x + y) + g(3x + 2y),

where f and g are arbitrary twice continuously differentiable functions;

2. u(x,y) = f (y −x) + exp((x −y)/2)g(y −2x);

3. u(x,y) = f (x + 3y) + g(3x + y) exp((7x + y)/16);