Myint Tyn U., Debnath L. Linear Partial Differential Equations for Scientists and Engineers

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4.5 Summary and Further Simpliﬁcation 111

Example 4.4.5. Use u = f (ξ), ξ =

√

4κt

to solve the parabolic system

= κu

, −∞ <x<∞,t>0, (4.4.7)

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

,x>0, (4.4.8)

where κ and u

are constant.

We use the given transformations to obtain

= f

′

(ξ) ξ

= −

√

4κt

′

(ξ) ,

∂

∂x

∂

∂x

′

(ξ) · ξ

4κt

′′

(ξ) .

Consequently, equation (4.4.7) becomes

′′

(ξ)+2ξf

′

(ξ)=0.

The solution of this equation is

′

(ξ)=A exp



−ξ



where A is a constant of integration. Integrating again gives

f (ξ)=A



−α

dα + B,

where B is an integrating constant.

Using the given conditions yields

0=A



−∞

−α

dα + B, u

= A



∞

−α

dα + B,

which give

A =

√

and B =

Thus, the ﬁnal solution is

u (x , t)=u



√



√

4κt

−α

dα +



4.5 Summary and Further Simpliﬁcation

We summarize the classiﬁcation of linear second-order partial diﬀerential

equations with constant coeﬃcients in two independent variables.

112 4 Classiﬁcation of Second-Order Linear Equations

hyperb oli c: u

= a

+ a

u + f

, (4.5.1)

− u

= a

∗

+ a

∗

+ a

∗

u + f

∗

, (4.5.2)

parabolic: u

= b

+ b

u + f

, (4.5.3)

elliptic: u

+ u

= c

+ c

u + f

, (4.5.4)

where r and s represent the new independent variables in the linear tran s-

formations

r = r (x, y) ,s= s (x, y) , (4.5.5)

and the Jacobian J =0.

To simplify equation (4.5.1) further, we introduce the new dependent

variable

v = ue

−(ar+bs)

, (4.5.6)

where a and b are undetermined coeﬃcients. Finding the derivatives, we

obtain

=(v

+ av) e

ar+bs

=(v

+ bv) e

ar+bs



+2av

+ a



ar+bs

=(v

+ av

+ bv

+ abv) e

ar+bs



+2bv

+ b



ar+bs

Substitution of these equation (4.5.1) yields

+(b − a

) v

+(a − a

) v

+(ab − a

a − a

b − a

) v = f

−(ar+bs)

In order that the ﬁrst derivatives vanish, we set

b = a

and a = a

Thus, the above equation becomes

=(a

+ a

) v + g

where g

= f

−(a

r+a

. In a similar manner, we can transform equa-

tions (4.5.2)–(4.5.4). Thus, we have the following transformed equations

corresponding to equations (4.5.1)–(4.5.4).

hyperb oli c: v

= h

v + g

− v

= h

∗

v + g

∗

, (4.5.7)

parabolic: v

= h

v + g

elliptic: v

+ v

= h

v + g

In the case of partial diﬀerential equations in several independent vari-

ables or in higher order, the classiﬁcation is considerably more complex.

For further reading, see Courant and Hilbert (1953, 1962).

4.6 Exercises 113

4.6 Exercises

1. Determine the region in which the given equation is hyperbolic, parabolic,

or elliptic, and transform the equation in the respective region to canon-

ical form.

(a) xu

+ u

= x

,(b)u

+ y

= y,

+ xyu

=0, (d)x

− 2xyu

+ y

= e

(e) u

+ u

− xu

=0, (f) e

+ e

= u,

(g) u

−

√





+2xu

− 3yu

+2u =exp



− 2y



y ≥ 0,

(h) u

−

√

+ xu

=cos



− 2y



, y ≥ 0,

(i) u

− yu

+ xu

+ yu

+ u =0,

(j) sin

+sin2xu

+cos

= x,

2. Obtain the general solution of the following equations:

(i) x

+2xyu

+ y

+ xyu

+ y

=0,

(ii) ru

− c

− 2c

=0,c = constant,

(iii) 4u

+12u

+9u

− 9u =9,

(iv) u

+ u

− 2u

− 3u

− 6u

=9(2x − y),

(v) yu

+3yu

+3u

=0, y =0.

(vi) u

+ u

=0,

(vii) 4 u

+ u

=0,

(viii) u

− 2 u

+ u

=0,

(ix) 2 u

+ u

=0,

(x) u

+4u

=0,

(xi) 3 u

+4u

−

=0.

114 4 Classiﬁcation of Second-Order Linear Equations

3. Find the characteristics and characteristic coordinates, and reduce the

following equations to canonical form:

(a) u

+2u

+3u

+4u

+5u

+ u = e

(b) 2u

− 4u

+2u

+3u =0,

+5u

+4u

+7u

=sinx,(d)u

+ u

+2u

+8u

+ u =0,

(e) u

+2u

+9u

+ u

=2, (f)6u

− u

+ u = y

(g) u

+ u

=3x,(h)u

− 9u

+7u

=cosy,

(i) x

− y

− u

=1+2y

,(j)u

+ yu

+4yu

=0,

(k) x

+2xyu

+ u

=0, (l)u

+ yu

=0.

4. Determine the general solutions of the following equations:

(i) u

−

=0,c= constant, (ii) u

+ u

=0,

(iii) u

xxxx

+2u

xxyy

+ u

yyyy

= 0, (iv) u

− 3u

+2u

=0,

(v) u

+ u

= 0, (vi) u

+10u

+9u

= y.

5. Transform the following equations to the form v

ξη

= cv, c = constant,

(i) u

− u

+3u

− 2u

+ u =0,

(ii) 3u

+7u

+2u

+ u

+ u =0,

by intr oducing the new variables v = ue

−(aξ+bη)

,wherea and b are

undetermined coeﬃcients.

6. Given the parabolic equation

= au

+ bu

+ cu + f,

where the coeﬃcients are constants, by the substitution u = ve

,for

the case c = −





, show that the given equation is redu ced to the

heat equation

= av

+ g, g = fe

−bx/2

7. Reduce the Tricomi equation

+ xu

=0,

4.6 Exercises 115

to the canonical form

(i) u

ξη

− [6 (ξ − η)]

−1

− u

)=0, for x<0,

(ii) u

αα

+ u

ββ

3β

=0,x>0.

Show that the characteristic curves for x<0 are cubic parabolas.

8. Use the polar coordinates r and θ (x = r cos θ, y = r sin θ) to tr ansform

the Laplace equation u

+ u

= 0 into the polar form

∇

u = u

θθ

=0.

9. (a) Using the cylindrical polar coordinates x = r cos θ, y = r sin θ, z = z,

transform the three-dimensional Laplace equation u

+ u

into the form

θθ

+ u

=0.

(b) Use the spherical polar coordinates (r, θ, φ)sothatx = r sin φ cos θ,

y = r sin φ sin θ, z = r cos φ to transform the three-dimensional Laplace

equation u

+ u

= 0 into the form

sin φ

(sin φu

)

sin

θθ

=0.

= κ (u

+ u

) ,

into the axisymmetric form

= κ





10. (a) Apply a linear transformation ξ = ax + by and η = cx + dy,to

transform the Euler equation

+2Bu

+ Cu

into canonical form, where a, b, c, d, A, B and C are constants .

(b) Show that the same transformation as in (a) can be used to trans-

form the nonhomogeneous Euler equation

+2Bu

+ Cu

= F (x, y, u, u

)

into canonical form.

116 4 Classiﬁcation of Second-Order Linear Equations

11. Obtain the solution of the Cauchy problem

+ u

=0,

u (x , 0) = f (x)andu

(x, 0) = g (x) .

12. Classify each of th e f ollowing equations and reduce it to canonical form:

(a) yu

− xu

=0,x>0,y>0; (b) u



sech



=0,

+ x

=0, (d)u

−



sech



=0,

(e) u

+6u

+9u

+3yu

=0,

(f) y

+2xy u

+2x

+ xu

=0,

(g) u

− (2 cos x) u



1+cos



+ u =0,

(h) u

+ (2 cosec y) u



cosec



=0.

(i) u

− 2 u

+ u

+3u

− u +1=0,

(j) u

− y

+ u

− u + x

=0,

(k) u

+ yu

− xu

+ y =0.

13. Transform the equation

+ yu

+sin(x + y)=0

into the canonical form. Use the canonical form to ﬁnd the general

solution.

14. Classify each of the following equations for u (x, t):

(a) u

=(pu

)

,(b)u

− c

+ αu =0,

)

+(au

)

=0, (d)u

− au

=0,

where p (x), c (x, t), a (x, t), and α (x) are given functions that take

only positive values in the (x, t) plane. Find the general solution of the

equation in (d).

The Cauchy Problem and Wave Equations

“Since a general soluti on must be judged impossible from want of analysis,

we must be content with the knowledge of some special cases, and that all

the more, since the development of various cases seems to be the only way

to bringing us at last to a more perfect knowledge.”

Leonhard Euler

“What would geometry be without Gauss, mathematical logic without

Boole, algebra without Hamilton, analysis without Cauchy?”

George Temple

5.1 The Cauchy Problem

In the theory of ordinary diﬀerential equations, by the initial-value problem

we mean the problem of ﬁnding the solutions of a given diﬀerential equation

with the appropriate number of initial conditions prescribed at an initial

point. For example, the second-order ordinary diﬀerential equation

= f



t, u,



and the initial conditions

u (t

)=α,





)=β,

constitute an initial-value problem.

An analogous problem can be deﬁned in the case of partial diﬀerential

equations. Here we shall state the problem involving second-order partial

diﬀerential equations in two independent variables.

118 5 The Cauchy Problem and Wave Equations

We consider a second-order partial diﬀerential equation for the function

u in the independent variables x and y, and suppose that this equation can

be solved explicitly for u

, and hence, can be represented in the from

= F (x, y, u, u

) . (5.1.1)

For some value y = y

, we prescribe the initial values of the unknown

function and of the derivative with respect to y

u (x , y

)=f (x) ,u

(x, y

)=g (x) . (5.1.2)

The problem of determining the solution of equation (5.1.1) satisfying

the initial conditions (5.1.2) is known as the initial-value problem. For in-

stance, the initial-value problem of a vibr ating string is the problem of

ﬁnding the solution of the wave equation

= c

satisfying the initial conditions

u (x , t

)=u

(x) ,u

(x, t

)=v

(x) ,

where u

(x) is the initial displacement and v

(x) is the initial velocity.

In initial-value p roblems, the initial values usually refer to the data

assigned at y = y

. It is not essential that these values b e given along

the line y = y

; they may very well be prescribed along some curve L

the xy plane. In such a context, the problem is called the Cauchy problem

instead of the initial-value problem, although the two names are actually

synonymous.

We consider the Euler equation

+ Bu

+ Cu

= F (x, y, u, u

) , (5.1.3)

where A, B, C are functions of x and y.Let(x

) denote points on a

smooth curve L

in the xy plane. Also let the parametric equations of this

curve L

= x

(λ) ,y

= y

(λ) , (5.1.4)

where λ is a parameter.

We supp ose that two functions f (λ)andg (λ) are prescribed alon g

the curve L

. The Cauchy problem is now one of determining the solution

u (x , y) of equation (5.1.3) in the neighborhood of the curve L

satisfying

the Cauchy conditions

u = f (λ) , (5.1.5a)

∂u

∂n

= g (λ) , (5.1.5b)

5.1 The Cauchy Problem 119

on the curve L

where n is the direction of the normal to L

which lies

to the left of L

in the counterclockwise direction of increasing arc length.

The function f (λ)andg (λ) are called the Cauchy data.

For every p oint on L

,thevalueofu is speciﬁed by equation (5.1.5a).

Thus, the curve L

represented by equation (5.1.4) with the condition

(5.1.5a) yields a twisted curve L in (x, y, u) space whose projection on

the xy planeisthecurveL

. Thus, the solution of the Cauchy problem is a

surface, called an integral surface,inthe(x, y, u) space passing through L

and satisfying the condition (5.1.5b), which represents a tangent plane to

the integral surface along L.

If the function f (λ) is diﬀerentiable, then along the curve L

,wehave

dλ

∂u

∂x

dλ

∂u

∂y

dλ

, (5. 1.6)

and

∂u

∂n

∂u

∂x

∂u

∂y

= g, (5.1.7)

but

= −

and

. (5.1.8)

Equation (5.1.7) may be written as

∂u

∂n

= −

∂u

∂x

∂u

∂y

= g. (5.1.9)

Since



dλ

−dy



(dx)

+(dy)

ds dλ

=0, (5.1.10)

it is possible to ﬁnd u

and u

on L

from the system of equations (5.1.6)

and (5.1.9). Since u

and u

are known on L

, we ﬁnd the higher derivatives

by ﬁrst diﬀerentiating u

and u

with respect to λ.Thus,wehave

∂

∂x

dλ

∂

∂x∂y

dλ



∂u

∂x



, (5.1.11)

∂

∂x∂y

dλ

∂

∂y

dλ



∂u

∂y



. (5.1.12)

From equation (5.1.3), we have

∂

∂x

+ B

∂

∂x∂y

+ C

∂

∂y

= F, (5.1.13)

120 5 The Cauchy Problem and Wave Equations

where F is known since u

and u

have been found. The system of equations

canbesolvedforu

, u

,andu

,if



dλ

ABC



= C



dλ



− B



dλ



dλ



+ A



dλ



=0. (5.1.14)

The equation





− B





+ C =0, (5.1.15)

is called the characteristic equation. It is then evident that the necessary

condition for obtaining the second derivatives is that the curve L

must not

be a characteristic curve.

If the coeﬃcients of equation (5.1.3) and the function (5.1.5) are ana-

lytic, then all the derivatives of higher orders can be computed by the above

process. The solution can then be represented in the form of a Taylor series:

u (x , y)=

∞



n=0

∞



k=0

k!(n − k)!

∂

∂x

∂y

n−k

(x − x

)

(y − y

)

n−k

, (5.1.16)

which can be shown to converge in the neighborhood of the curve L

.Thus,

we may state the famous Cauchy–Kowalewskaya theorem.

5.2 The Cauchy–Kowalewskaya Theorem

Let the partial diﬀerential equation be given in the form

= F (y, x

,...,x

,u,u

...,u

,...,u

) , (5.2.1)

and let the initial conditions

u = f (x

,...,x

) , (5.2.2)

= g (x

,...,x

) , (5.2.3)

be given on the noncharacteristic manifold y = y

If the function F is analytic in some neighborhood of the point



,...,x

,...



and if the functions f and g are analytic in

some neighborhood of the point



,...,x



, then the Cauchy prob-

lem has a unique analytic solution in some neighborhood of the point



,...,x

