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

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42 G.

Freiling and V. Yurko

Therefore, equation

(2.2.67) is transformed to the following equation with respect to Z(x) :

−Z

(x) + a(x)Y (x) = λr(x)Z(x), 0 < x < l, (2.2.69)

where

a(x) = q

(x) −

(x)

The

boundary

conditions (2.2.35) takes the form

(0) −h

Z(0) = 0,

(l) + H

Z(l) = 0,

)

(2.2.70)

where

= h −

p(0)

, H

= H +

p(l)

Step

2. Denote

R(x)

r(x) > 0, T =

R(τ)dτ.

make the replacement

ξ =

R(τ)dτ, z(ξ) =

R(x)Z(x). (2.2.71)

follo

ws from (2.2.71) that

dξ

R(x)

. Hence

dξ

R(x)

(x)

2R(x)

R(x)

Z(x),

dξ

R(x)

(x)

R(x)

−

3(R

(x))

(x)

R(x)

Z(x).

Therefore,

the

boundary value problem (2.2.69)-(2.2.70) is transformed to the following

boundary value problem with respect to z(ξ) :

−

z(ξ)

dξ

+ P(ξ)z(ξ)

= λ

z(ξ), 0 < ξ < T, (2.2.72)

(0) −αz(0) = 0, z

(T ) + βz(T ) = 0, (2.2.73)

where

P(ξ) =

(x)

−

3(R

(x))

(x)

a(x)

(x)

−

5(r

(x))

16r

(x)

a(x)

r(x)

α =

R(0)

(0)

, β =

R(l)

−

(l)

Hyperbolic P

artial Differential Equations 43

Step 3. W

e make the substitution

η =

πξ

, y(η)

= z

Then

the

boundary value problem (2.2.72)-(2.2.73) is transformed to the following bound-

ary value problem with respect to y(η) :

−

y(η)

dη

+ Q(η)y(η)

= µy

(η), 0 < η < π,

(0) −α

y(0) = 0, y

(π) + β

y(T ) = 0,

where

Q(η) =

, µ =

λ,

α, β

β.

Thus,

have reduced the Sturm-Liouville problem (2.2.34)-(2.2.35) to the form

(2.2.36). Using the results obtained above we conclude that the following theorem holds.

Theorem 2.2.8. (1) The boundary value problem (2.2.34) −(2.2.35) has a countable

set of eigenvalues {λ

}

n≥0

. They are real and simple ( i.e. λ

6= λ

for n 6= k) , and

nπ

+ O

, n →∞, (2.2.74)

wher

T =

ρ(τ)

k(τ)

dτ.

each eigenvalue λ

, there exists only one ( up to a multiplicative constant ) eigenfunc-

tion Y

(x) . For n → ∞ ,

(x) =

(k(x)ρ(x))

1/4

cos

πn

ρ(s)

k(s)

+ O

uniformly

x ∈ [0, l].

(2) The eigenfunctions related to different eigenvalues are orthogonal in

2,ρ

(0,l) , i.e.

(x)Y

(x)ρ(x)dx = 0 for n 6= k.

The system of eigenfunctions {Y

(x)}

n≥0

is complete in L

2,ρ

(0,l) .

(3) Let f (x), x ∈[0,l] be an absolutely continuous function. Then

f (x) =

∞

∑

n=0

(x),

44 G.

Freiling and V. Yurko

where

f (x)Y

(x)ρ(x)d

x, α

(x)ρ

(x)dx.

and the series converges uniformly on [0,l].

4.3. Since nontrivial solutions of problem (2.2.34)-(2.2.35) exist only for the eigen-

values λ = λ

of the form (2.2.74), it makes sense to consider equation (2.2.33) only for

λ = λ

(t) + ρ

(t) = 0, n ≥ 0, λ

= ρ

The general solution of this equation has the form

(t) = A

sinρ

+ B

cosρ

t, n ≥0,

where A

and B

are

arbitrary

constants. Thus, the solutions of the auxiliary problem have

the form

(x,t) =

sinρ

+ B

cosρ

(x), n = 0,1,2,

.. (2.2.75)

We will seek the solution of the mixed problem (2.2.27)-(2.2.29) by superposition of

standing waves (2.2.75), i.e. as a formal series solution:

u(x,t) =

∞

∑

n=0

sinρ

+ B

cosρ

(x). (2.2.76)

construction

the function u(x,t) satisﬁes formally equation (2.2.27) and the boundary

conditions (2.2.28) for all A

and B

. Choose A

and B

such that u(x,t) satisﬁes also

the initial conditions (2.2.29). Substituting (2.2.76) into (2.2.29) we calculate

Φ(x) =

∞

∑

n=0

(x), Ψ(x) =

∞

∑

n=0

(x),

and consequently,

Φ(x)Y

(x)ρ(x)d

Ψ(x)Y

(x)ρ(x)d











(

2.2.77)

Thus, the solution of the mixed problem (2.2.27)-(2.2.29) has the form (2.2.76), where the

coefﬁcients A

and B

is taken from (2.2.77).

We note that the series in (2.2.76) converges uniformly in D := {(x,t) : 0 ≤ x ≤ l, t ≥

0}. Indeed, let

Y (x) := −(k(x)Y

(x))

+ q(x)Y (x).

Hyperbolic P

artial Differential Equations 45

Since `

(x) = λ

ρ(x)Y

(x), it

follows from (2.2.77) that

Ψ(x)`

(x)d

= −

Ψ(x)

(k(x)Y

(x))

x +

Ψ(x)q(x)Y

(x)d

Inte

grating by parts twice the ﬁrst integral we get

k(x)(Ψ

(x)Y

(x) −Ψ(x)Y

(x))

−

(k(x)Ψ

(x))

(x)d

x +

Ψ(x)q(x)Y

(x)d

The

substitution vanishes since the functions Ψ(x) and Y

(x) satisfy the boundary condi-

tions (2.2.35). Therefore,

(x)Y

(x)d

where Ψ

(

x) := `

Ψ(x). Similarly,

(x)Y

(x)d

where Φ

(x) :

= `

Φ(x). Taking the asymptotics for λ

and Y

(x) into account (see The-

orem 2.2.8) we conclude that the series (2.2.76) converges uniformly in D.

2.3. The Goursat Problem

The Goursat problem is the problem of solving a hyperbolic equation with given data on

characteristics. Therefore, this problem is also called the problem on characteristics. It

is convenient for us to consider the canonical form of hyperbolic equations for which the

characteristics are parallel to the coordinate axes. For simplicity, we conﬁne ourselves to

linear equations. So, we consider the following Goursat problem:

= a(x,y)u

+ b(x,y)u

+ c(x,y)u + f (x,y), (x,y) ∈ Π, (2.3.1)

|x=x

= ϕ(y), u

|y=y

= ψ(x). (2.3.2)

Here x,y are the independent variables, u(x, y) is an unknown function and Π = {(x, y) :

≤ x ≤ x

, y

≤ y ≤ y

} is a rectangle (see ﬁg. 2.3.1). Equation (2.3.1) has two families

of characteristics x = const and y = const. . Thus, the conditions (2.3.2) are conditions on

the characteristics x = x

and y = y

46 G.

Freiling and V. Yurko

Figure

2.3.1.

Deﬁnition

2.3.1. The function u(x, y) is called a solution of problem (2.3.1)-(2.3.2)

if u(x, y) is deﬁned and continuous with its derivatives u

in the rectangle Π and

satisﬁes (2.3.1) and (2.3.2).

Theorem 2.3.1. Let the functions a,b,c, f be continuous in Π, and let the functions

ϕ and ψ be continuously differentiable with ϕ(y

) = ψ(x

). Then the solution of problem

(2.3.1) −(2.3.2) exists and is unique.

Proof. 1) We reduce the Goursat problem (2.3.1)-(2.3.2) to an equivalent system of

integral equations. Suppose that the solution u(x,t) of problem (2.3.1)-(2.3.2) exists. Put

v = u

, w = u

Then

= av + bw + cu + f , w

= av + bw + cu + f , u

= w. (2.3.3)

Integrating (2.3.3) and taking (2.3.2) into account we obtain

v(x,y) = ψ

(x) +

(av + bw + cu + f )(x,η) dη,

w(x,y) = ϕ

(y) +

(av + bw + cu + f )(ξ,y)dξ,

u(x,y) = ψ(x) +

w(x,η) dη.











(2.3.4)

Thus, if u is a solution of problem (2.3.1)-(2.3.2), then the triple u,v,w is a solution of the

system (2.3.4).

The inverse assertion is also valid. Indeed, let the triple u,v,w deﬁne a solution of the

system (2.3.4). Differentiating (2.3.4) we deduce that the equalities (2.3.3) hold. Moreover,

(x,y) = ψ

(x) +

(x,η) dη

= ψ

(x) +

(av + bw + cu + f )(x,η) dη = v(x,y).

Hyperbolic P

artial Differential Equations 47

Thus u

= v. Together

with (2.3.3) this yields that the function u(x,t) is a solution of

equation (2.3.1). Furthermore, it follows from (2.3.4) for x = x

and y = y

that

u(x,y

) = ψ(x),

u(x

,y) = ψ(x

) +

w(x

,η) dη = ψ(x

) +

(η)dη = ϕ(y),

i.e. u(x,y) satisﬁes the conditions (2.3.2). Thus, the Goursat problem (2.3.1)-(2.3.2) is

equivalent to the system (2.3.4).

2) We will solve the system (2.3.4) by the method of successive approximations. Put

(x,y) = ψ

(x) +

f (x, η)dη,

(x,y) = ϕ

(y) +

f (ξ, y)dξ,

(x,y) = ψ(x),

n+1

(x,y) =

(av

+ bw

+ cu

)(x,η) dη,

n+1

(x,y) =

(av

+ bw

+ cu

)(ξ,y)dξ,

n+1

(x,y) =

(x,η) dη.











(2.3.5)

Take the constants M ≥ 0 and K ≥ 1 such that

|,|v

|,|w

| ≤ M, |a|+ |b|+ |c| ≤ K.

Using (2.3.5), by induction one gets the estimates:

(x,y)|, |w

(x,y)|, |u

(x,y)| ≤ MK

(x + y −x

−y

)

. (2.3.6)

Indeed,

for n = 0

these estimates are obvious. Fix N ≥ 0 and suppose that estimates

(2.3.6) are valid for n =

0,N. Then,

using

(2.3.5) and (2.3.6) we obtain

N+1

(x,y)| ≤

(|a(x,η)|+ |b(x,η)|

+|c(x,η)|)

(x + η −x

−y

)

dη,

and

consequently

N+1

(x,y)| ≤ MK

N+1

(x + η −x

−y

)

dη

= M

N+1

(x + η −x

−y

)

N+1

(N + 1)!

≤ M

N+1

(x + y −x

−y

)

N+1

(N + 1)!

48 G.

Freiling and V. Yurko

For w

and u

guments are similar.

By virtue of (2.3.6), the series

u(x,y) =

∞

∑

n=0

(x,y),

v(x,y) =

∞

∑

n=0

(x,y),

w(x,y) =

∞

∑

n=0

(x,y)











(2.3.7)

converge absolutely and uniformly in Π (since they are majorized by the convergent nu-

merical series M

∞

∑

n=0

+ y

−x

−y

)

and

|v(x,y

)|,|w(x,y)|,|u(x,y)| ≤ M exp(K(x

+ y

−x

−y

)). (2.3.8)

Obviously, the triple u, v,w, constructed by (2.3.7), solves system (2.3.4).

3) Let us prove the uniqueness. Let the triples (u,v, w) and ( ˜u, ˜v, ˜w) be solutions

of system (2.3.4). Then the functions u

∗

= u − ˜u , v

∗

= v − ˜v , w

∗

= w − ˜w satisfy the

homogeneous system

∗

(x,y) =

(av

∗

+ bw

∗

+ cu

∗

)(x,η) dη,

∗

(x,y) =

(av

∗

+ bw

∗

+ cu

∗

)(ξ,y)dξ,

∗

(x,y) =

∗

(x,η) dη.











Since the functions u

∗

are continuous in Π, there exists a constant M

> 0 such

that |u

∗

|,|v

∗

|,|w

∗

| ≤ M

. Repeating the previous arguments, by induction we obtain the

estimate

∗

(x,y)|, |w

∗

(x,y)|, |u

∗

(x,y)| ≤ M

(x + y −x

−y

)

As n → ∞ this

yields u

∗

(x,y

) = v

∗

(x,y) = w

∗

(x,y) = 0. Theorem 2.3.1 is proved. 2

Let us study the stability of the solution of the Goursat problem.

Deﬁnition 2.3.2. The solution of the Goursat problem is called stable if for each

ε > 0 there exists δ = δ(ε) > 0 such that if |ϕ

(ν)

(y) −

(ν)

(y)| ≤ δ , |ψ

(ν)

(x) −

(ν)

(x)| ≤

δ , ν = 0,1 , x ∈ [x

] , y ∈ [y

], then |u(x,y) − ˜u(x,y)| ≤ ε , |u

(x,y) − ˜u

(x,y)| ≤

ε , |u

(x,y) − ˜u

(x,y)| ≤ ε for all (x,y) ∈ Π. Here ˜u(x, y) is the solution of the Goursat

problem for the data

ϕ,

ψ.

Let us show that the solution of problem (2.3.1)-(2.3.2) is stable. Indeed, denote u

∗

u − ˜u , v

∗

= v − ˜v , w

∗

= w − ˜w , ϕ

∗

= ϕ −

ϕ , ψ

∗

= ψ −

ψ , f

∗

= 0. By virtue of (2.3.8),

∗

(x,y)|, |w

∗

(x,y)|, |u

∗

(x,y)| ≤ M

∗

exp(K(x

+ y

−x

−y

)),

Hyperbolic P

artial Differential Equations 49

where

∗

= max(max

|ψ

∗

(x)|,max

|ψ

∗,0

(x)|,max

|ϕ

∗,0

(y)|) ≤ δ.

Choosing δ = εexp(

−K(x

+ y

−x

−y

)) and using the relation M

∗

≤ δ, we get

∗

(x,y)|, |v

∗

(x,y)|, |w

∗

(x,y)| ≤ ε. Thus, the Goursat problem (2.3.1)-(2.3.2) is well-posed.

Now we reformulate the results obtained above for another canonical form of hyperbolic

equations. We consider the following Goursat problem

−u

+ a(x,t)u

+ b(x,t)u

+ c(x,t)u = f (x,t), (2.3.9)

(x,t) ∈ ∆(x

u(x,x −x

) = ϕ(x), u(x,−x + x

) = ψ(x), (2.3.10)

where

∆(x

) = {(x,t)) : t −t

+ x

≤ x ≤ −t +t

+ x

, 0 ≤t ≤ t

}

is the characteristic triangular (see ﬁg. 2.4.1). Equation (2.3.9) has two families of charac-

teristics x + t = const and x −t = const. Thus, the conditions (2.3.10) are conditions on

the characteristics I

: t = x −x

+ t

and I

: t = −x + x

+ t

. Problem (2.3.9)-(2.310)

can be reduced to a problem of the form (2.3.1)-(2.3.2) by the replacement:

x = ξ −η + t

, t = ξ + η −x

This yields

ξ =

t + x

−t

, η =

t −x

Indeed,

denote

˜u

(ξ,η) = u(ξ −η +t

,ξ + η −x

) = u(x,t).

Then u

= ( ˜u

− ˜u

)/2 , u

= ( ˜u

+ ˜u

)/2 , u

−u

= −˜u

ξη

, and consequently, equation

(2.3.9) takes the form

˜u

ξη

= ˜a(ξ,η) ˜u

b(ξ,η) ˜u

+ ˜c(ξ,η) ˜u +

f (ξ, η), (2.3.11)

where

˜a =

a + b

b = −

a −b

, ˜c = c,

f = −f .

The

characteristics I

and I

equation (2.3.9) are transformed into characteristics of

equation (2.3.11) η = t

and ξ = x

, respectively, and conditions (2.3.10) become

˜u(ξ,t

) =

ϕ(ξ), ˜u(x

,η) =

ψ(η),

where

ϕ(ξ) = ϕ(ξ) ,

ψ(η) = ψ(−η+x

). Therefore, the following theorem is a corol-

lary of Theorem 2.3.1.

Theorem 2.3.2. Let the functions a, b,c, f be continuous in ∆(x

), and let the

functions ϕ and ψ be continuously differentiable with ϕ(x

) = ψ(x

). Then the solution

of the Goursat problem (2.3.9) −(2.3.10) exists and is unique.

50 G.

Freiling and V. Yurko

2.4. The

Riemann Method

Riemann’s method is a classical technique for solving the Cauchy problem for hyperbolic

linear equations in two variables; in particular it provides information for domains of

dependence and inﬂuence for solutions.

We consider the following Cauchy problem

−u

+ a(x,t)u

+ b(x,t)u

+ c(x,t)u = f (x,t), (2.4.1)

−∞ < x < ∞, t > 0,

|t=0

= ϕ(x), u

t|t=0

= ψ(x). (2.4.2)

Denote D = {(x,t) : −∞ < x < ∞, t ≥ 0}. Suppose that a,b ∈C

(D) , c, f ∈C(D) , ϕ ∈

(R), ψ ∈C

(R).

Deﬁnition 2.4.1. A function u(x,t) is called a solution of problem (2.4.1)-(2.4.2) if

u(x,t) ∈C

(D), and u(x,t) satisﬁes (2.4.1) and (2.4.2).

Derivation of the Riemann formula. Denote

L u = u

−u

+ au

+ bu

+ cu,

∗

v = v

−v

−(av)

−(bu)

+ cv.

Fix a point (x

) and consider the characteristic triangular ∆(x

) = {(x,t) : t −t

≤ x ≤ −t + t

+ x

, 0 ≤ t ≤ t

} (see ﬁg. 2.4.1) with vertices at the points M = (x

) ,

P = (x

−t

,0) and Q = (x

,0). The boundary ∂∆ of ∆ consists of three segments:

∂∆ = I

∪I

, where

P :

= {(x,t) : t = x −x

, x

−t

≤ x ≤ x

QM := {(x,t) : t = −x + x

, x

≤ x ≤ x

PQ := {(x,t) : t = 0, x

−t

≤ x ≤ x

M =

)

∆

)

P = (x

−t

,0) Q = (x

,0)I

ﬁg.

2.4.1

Hyperbolic P

artial Differential Equations 51

Suppose that

the solution u(x,t) of problem (2.4.1)-(2.4.2) exists. Let v(x,t) ∈C

(D)

be an arbitrary function. Since

vL u −uL

∗

v = (vu

−uv

+ auv)

−(vu

−uv

−buv)

it follows from Green’s formula that

∆

(vL u −uL

∗

v)dxdt

∂∆

(vu

−uv

−buv)dx + (vu

−uv

+ auv)dt (2.4.3)

(with counterclockwise orientation of ∂∆ ). Let us transform the integral along the bound-

ary. We have

∂∆

1) On I

: t = x −x

, dt = dx. Denote

(x) = u(x, x −x

), β

(x) = v(x, x −x

Then α

(x) = (u

+ u

)

|t=x−x

, β

(x) = (v

+ v

)

|t=x−x

, and consequently,

(v(u

+ u

) −u(v

+ v

) + (a −b)uv)(x,x −x

)dx

(β

(x)α

(x) −α

(x)β

(x)

+(a −b)(x,x −x

)α

(x)β

(x))dx.

Integrating by parts the ﬁrst term we get

(x)α

(x)

−

(x)

2β

(x) −(a −b)(x,x −x

)β

(x)

dx.

Impose a ﬁrst condition on the function v(x,t), namely:

2β

(x) −(a −b)(x,x −x

)β

(x) = 0.

Solving this ordinary differential equation we calculate

v(x,x −x

) = exp

(a −b)

(ξ

,ξ −x

)dξ

. (2.4.4)

Then

= u(P)v(P) −u(M)v(M). (2.4.5)