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

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262 G.

Freiling and V. Yurko

Exterior Diric

hlet problem:

∆u = 0 (x ∈ D

= g, u(∞) = 0 (u(x) = O(1), |x| → ∞ for n = 2).

Exterior Neumann problem:

∆u = 0 (x ∈ D

∂u

∂n

= g, u(∞)

= 0

, (u(x) = O(1), |x| → ∞ for n = 2).

One can consider the Dirichlet and Neumann problems also in other unbounded re-

gions (a sector, a strip, a half-strip, a half-plane, etc.). Analogously one can formulate the

Dirichlet and Neumann problems for the Poisson equation

∆u(x) = −f (x).

The Dirichlet problem has a unique solution. If n ≥ 3 then the exterior Neumann

problem also has a unique solution. The solution of the interior Neumann problem (for

n = 2 also of the exterior Neumann problem) is deﬁned up to an additive constant. Note

that

g(x)ds = 0

is a necessary condition for the solvability of the Neumann problem.

6.4.13. Find the solution of the Dirichlet problem

∆u(x

) = 0, x ∈D,

u(x

= g(x

where D = {x : |x| < l} ⊂ R

is the disc of radius l around the origin, and

1. g(x

) = a;

2. g(x

) = ax

+ bx

;

3. g(x

) = x

;

4. g(ρ,ϕ) = a + b sin ϕ;

5. g(ρ,ϕ) = a sin

ϕ + bcos

ϕ.

Solution of Problem 4. Let us go on from polar coordinates to cartesian ones. If

) ∈ S then x

= l cos ϕ , x

= l sin ϕ, hence

g(x

) = a +

Exercises 263

Since the

function A + Bx

is harmonic it follows that the solution has the form

u(x

) = a +

u(ρ,ϕ)

= a +

ρsin ϕ.

Solution

Problem 2. Let us go on to polar coordinates. Then

g(ρ,ϕ) = al cos ϕ + bl sin ϕ.

We seek a solution of the form

u(ρ,ϕ) = A(ρ)cos k

ϕ + B(ρ)sink

ϕ.

Then the functions A(ρ) and B(ρ) satisfy the Euler equations

(ρ) + ρA

(ρ) −k

A(ρ) = 0,

(ρ) + ρB

(ρ) −k

B(ρ) = 0.

The general solutions of these equations have the form

A(ρ) = α

−|k

+ β

B(ρ) = α

−|k

+ β

In our case we have k

= k

= 1. Since

|u(x

)| < ∞, u(l, ϕ) = g(l,ϕ), 0 ≤ϕ ≤ 2π,

we get

= β

= 0, α

−1

= al, α

−1

= bl.

Consequently, the solution has the form

u(ρ,ϕ) =

cosϕ +

sinϕ.

6.4.14. Find

the

solution of the Dirichlet problem

∆u(x

) = 0, x ∈D,

u(x

= g(x

where D = {x : |x| > l} ⊂ R

is the exterior of the disc of radius l around the origin, and

the function g(x

) is taken from Problem 6.4.13 (1.-5.)

6.4.15. Find the solution of the Dirichlet problem

∆u(x

) = 0, x ∈D,

264 G.

Freiling and V. Yurko

u(x

|x|=l

= a, u(x

|x|=l

= b,

where D = {x : l

< |x| < l

} ⊂ R

a ring.

6.4.16. Find the solution of the Dirichlet problem for the Poisson equation

∆u(x

) = 1,

u(x

= 0,

where D = {x : |x| < l}⊂ R

is the disc of radius l around the origin.

6.4.17. Find the solution of the Dirichlet problem

∆u(x

) = 0, x ∈ D,

u(x

= a,

where D = {x : |x| < l}⊂ R

is the ball of radius l around the origin.

6.4.18. Find the solution of the Dirichlet problem

∆u(x

) = 0, x ∈ D,

u(x

= a,

where D = {x : |x| > l}⊂ R

is the exterior of the ball of radius l .

6.4.19. Find the solution of the Neumann problem

∆u(x

) = 0, x ∈D,

∂u(x

)

∂n

= g(x

where D = {x : |x| < l}

⊂ R

the disc of radius l around the origin, and

1. g(x

) = a;

2. g(x

) = ax

;

3. g(x

) = a(x

−x

);

4. g(ρ,ϕ) = a cosϕ + b;

5. g(ρ,ϕ) = a sinϕ + b sin

ϕ.

Solution of Problem 2. First we check the condition

g(x)ds = 0.

Since

ds = al

2π

cosϕ dϕ = 0,

Exercises 265

it follo

ws that the problem is solvable. We seek the solution in the form

u(x

) = αx

and we deﬁne α from the boundary condition. Let us introduce polar coordinates, then

u(ρ,ϕ) = αρcos ϕ.

The differentiation with respect to the normal coincides with the differentiation with respect

to ρ, hence

∂u

∂n

= al cos ϕ = g = α cosϕ.

Therefore α = al, and

consequently

u(ρ,ϕ) = alρcos ϕ +C

u(x

) = alx

+C.

Let D ⊂R

be a bounded domain with the piecewise smooth boundary S. Fix y ∈D.

The function G(x,y) , x ∈

D is

called

the Green’s function for D if

G(x,y) =

4π|x −y|

+ v(x,y),

where

the

function v(x,y) is harmonic for x ∈ D and continuous for x ∈

D, and

such

that

G(x,y)|

x∈S

= 0.

If D is unbounded then there is the additional condition v(x,y) → 0 as |x| → ∞. Using

the Green’s function one can write the solution of the Dirichlet problem

∆u(x) = −f (x), u(x)|

= g(x)

in the form

u(x) = −

∂G(x,y)

∂n

g(y)d

y +

G(x,y

) f (y)dy. (6.4.3)

For constructing the Green’s function for some symmetrical domains one can use the so-

called reﬂection method. In this method we seek the function v(x, y) as a sum of terms

of the form

|x −y

where

the

points y

are chosen outside D and such that the Green’s

function satisﬁes the boundary condition.

For n = 2 the Green’s function is deﬁned analogously, namely:

(i) G(x,y) =

2π

|x −y|

+ v(x,y);

(ii) G(x,y)|

x∈S

= 0; x =

) ∈

D, y =

) ∈ D.

Note

that the Green’s function G(x, y) is symmetrical with respect to x and y :

G(x,y) = G(y, x).

266 G.

Freiling and V. Yurko

6.4.20. Construct the

Green’s function for the following domains:

1) the ball of radius R around the origin, n = 3 ;

2) the disc of radius R around the origin, n = 2.

Solution of Problem 1. Choose the point y

= (y

), symmetrically to the point

y = (y

) with respect to the sphere S (i.e. |y|·|y

| = R

). Then

|y|

seek the function v(x,y) in the form

v(x,y) = −

4π|x −y

The

Green’

s function is deﬁned by the formula

G(x,y) =

4π|x −y|

−

4π|x −y

where α is

constant.

O |y

| |y

|x −y

Figure 6.4.1.

Let x ∈S (see ﬁg. 6.4.1). Since the triangles Oxy

and Oxy are similar it follows that

|y|

|x −y

|x −y|

and

consequently

4π|x −y|

4π

|y|

|x −y

Hence,

α =

|y|

then

the

boundary condition for the Green’s function is fulﬁlled, and consequently,

G(x,y) =

4π|x −y|

−

4π|y||x −y

Exercises 267

6.4.21. Find the

Green’s function for the following domains:

1) the half-space x

> l , n = 3 , x = (x

) ;

2) the half-plane x

> l , n = 2 , x = (x

Solution of Problem 2. Choose the point y

symmetrically to y with respect to the line

= l (see ﬁg. 6.4.2).

= l

y =

(

)

= (y

,2l −y

)

x = (x

)

|x −y|

|x −y

Figure 6.4.2.

We seek the function v(x, y) in the form

v(x,y) = −

2π

|x −y

;

where y

l −y

). If x ∈S, then

|x −y|

|x −y

Hence,

if α = 1

, then the boundary condition for the Green’s function is fulﬁlled, and

consequently,

G(x,y) =

2π

|x −y|

−

2π

|x −y

note that the construction of the point y

in Problems 6.4.20-6.4.21 is called the reﬂec-

tion of the point y with respect to the ball and the plane, respectively.

6.4.22. Construct the Green’s function for the following domains in R

1) the dihedral angle x

> 0 , x

> 0 ;

2) the dihedral angle 0 < ϕ <

;

octant x

> 0

, x

> 0 , x

> 0 ;

4) the half-ball of radius R around the origin, x

> 0 ;

5) the quarter of the ball of radius R , x

> 0 , x

> 0 ;

6) the part of the ball of radius R , with x

> 0 , x

> 0 ;

268 G.

Freiling and V. Yurko

7) the

layer between the two planes x

= 0 and x

= l ;

8) the half of the layer 0 < x

< l , x

> 0 .

Solution of Problem 6. Let us make reﬂections with respect to the parts of the boundary

of the domain. First we reﬂect the point y with respect to the sphere, i.e. we construct the

point

|y|

1,0,0

Figure

6.4.3.

The

function

(x,y) =

4π|x −y|

−

4π|y||x −y

satisﬁes

the

boundary condition on the sphere, but G

(x,y) is not equal to zero on the other

parts of the boundary. Let us now reﬂect the points y and y

with respect to the plane

= 0. Let y

1,0,0

and y

1,0,0

be the symmetrical points to y and y

respectively (see ﬁg.

6.4.3). The function

(x,y) =

4π|x −y|

−

4π|y||x −y

−

4π|x −y

1,0,0

4π|y

1,0,0

|x −y

1,0,0

equal to zero on the sphere and on the plane x

= 0. Analogously we make reﬂections of

the points y , y

, y

1,0,0

, y

1,0,0

with respect to the plane x

= 0, and then we make reﬂec-

tions of the obtained points with respect to the plane x

= 0. Thus, the Green’s function

has the form

G(x,y) =

4π

∑

m,n,k=0

(−1)

m+n+k

|x −y

m,n,k

−

|y|

|x −y

m,n,k

where

0,0,0

= y =

), y

m,n,k

= ((−1)

,(−1)

Exercises 269

m,n,k

= y

m,n,k

6.4.23. Construct

the

Green’s function for the following domains in R

1) the right angle x

> 0, x

> 0 ;

2) the angle 0 < ϕ <

(in

the

polar coordinates);

3) the half-disc of radius R around the origin, x

> 0 ;

4) the quarter of the disc of radius R , x

> 0 , x

> 0 ;

5) the strip 0 < x

< π ;

6) the half-strip 0 < x

< l , x

> 0 .

Solution of Problem 2. Let us make reﬂections with respect to the parts of the boundary

of the domain. It is convenient to use the polar coordinates x = (ρ, ϕ) , y = (µ,ψ). First

we reﬂect the point y with respect to the line ϕ =

(see

ﬁg.

6.4.4), i.e. we construct the

point ˜y

= (µ,

2π

−ψ).

˜y

(

µ,

2π

−ψ)

n−1

(

µ,

2π(n−1)

+ ψ)

˜y

(

µ,−ψ)

y = (µ,ψ)

ψ =

Figure

6.4.4.

The

function

(x,y) =

2π

|x −y|

−

2π

|x − ˜y

equal

to zero on the line ϕ =

reﬂect now the points y and ˜y

with respect to the line ϕ = 0, and we obtain the

function

(x,y) =

2π

|x −y|

−

2π

|x − ˜y

−

2π

|x − ˜y

2π

|x −y

n−1

where

˜y

(µ

,−ψ), y

n−1

µ,−

2π

+ ψ

µ,

2π(n −1)

+ ψ

270 G.

Freiling and V. Yurko

Furthermore, we

again make a reﬂection with respect to the line ϕ =

viously, we

should reﬂect only the points ˜y

and y

n−1

, since the points y and ˜y

lie symmetrically

with respect to this line. We continue this procedure, and after the n -th reﬂection we obtain

the Green’s function of the form

G(x,y) =

2π

n−1

∑

k=0

|x −y

−ln

|x −

where

µ,

2πk

+ ψ

, ˜y

µ,

2πk

−ψ

, y

= y, k =

0,n −1.

The

construction

of the Green’s function for a plane domain D ⊂ R

connects with

the problem of the conformal mapping of D onto the unit disc. Let D ⊂ R

be a simply

connected domain with a sufﬁciently smooth boundary S , z = x

+ix

∈

D , ζ = y

+iy

∈

D ,

and

let ω(z,ζ) be the function realizing the conformal mapping of D onto the unit disc,

and ω(ζ,ζ) = 0. Then the Green’s function for D has the form

G(z,ζ) =

2π

|ω(z,ζ)|

6.4.24. Construct

the

Green’s function for the following domains in R

1) the half-plane ℑz > 0 ;

2) the quarter of the plane 0 < argz <

;

the

angle 0 < arg z <

;

the

half-disc |z| < R , ℑz > 0 ;

5) the quarter of the disc |z| < 1 , 0 < arg z <

;

the

strip 0 < ℑz < π ;

7) the half-strip 0 < ℑz < π , ℜz > 0 .

Solution of Problem 6. Let us ﬁnd the function realizing the conformal mapping of the

strip into the unit disc. Let ζ be a ﬁxed point of the strip. Using the function z

= e

we map the strip 0 < ℑz < π into the upper half-plane. The point ζ is mapped into the

point ζ

= e

. Then we map the upper half-plane into the unit disc such that the point ζ

is mapped into the origin. This is made by the linear-fractional function

ω(z

) = e

iα

(z −ζ

)

(z −ζ

)

Thus,

the

desired function has the form

ω(z,ζ) = e

iα

−e

Exercises 271

and we

construct the Green’s function

G(z,ζ) =

2π

−

−e

6.4.25. Using

the

Green’s function ﬁnd the solution of the Dirichlet problem

∆u(x

) = 0, x ∈ D,

= g(x

where D = {x : |x| < R} is the ball of radius R.

Solution. The Green’s function for the ball is constructed in Problem 4.20. Let us

calculate the normal derivative on the sphere

∂G(x,y)

∂n

. The

dif

ferentiation with respect to

the normal coincides with the differentiation with respect to the radius |y| = r. Therefore,

we introduce spherical coordinates. We get

|x −y|

= |x|

+ |y|

+ 2|x||y|cosγ;

|x −y

= |x|

+ |y

+ 2|x||y

|cos γ; |y

| :=

|y|

;

∂G(x,y)

∂n

4π

∂

∂r

|x|

+ r

−2|x|r cos γ

−

|x|

+ R

−2|x|r cos γ

r=R

|x|

−R

4πR(R

+ |x|

−2R|x|cosγ)

3/2

|x|

−R

4πR(|x −y|

y∈S

Thus,

the

solution of our Dirichlet problem is given by the formula (the Poisson formula):

u(x) =

4πR

−

|x|

|x −y|

g(y)d

Analogously

, for the disc ( n = 2 ) the Poisson formula has the form

u(x) =

2πR

−

|x|

|x −y|

g(y)d

6.4.26. Using

the Poisson formula solve Problems 6.4.13 and 6.4.17.

6.4.27. Using (6.4.3) solve Problems 6.4.15 and 6.4.16.

6.4.28. Find the solution of the Dirichlet problem

∆u(x

) = −f (x

), x ∈D,

u(x

= g(x