math for cslecture 121 partial differential equations boundary value problems

Math for CS Lecture 12 1

Lecture 12

Partial Differential Equations

Boundary Value Problems


Contents

1. Partial Differential Equations

2. Boundary Value Problems

3. Differential Operators

4. Cylindrical and Spherical coordinate systems

5. Examples of the Problems

6. Solution of the heat equation

7. Solution of the string equation


Partial Derivatives

Consider a function of two or more variables e.g. f(x,y). We can talk about derivatives of

such a function with respect to each of its variables:

The higher order partial derivatives are defined recursively and include the mixed x,y

derivatives:

(1)

2

),(),(lim

),(0

yxfyxf

x

yxff x

2

),(),(lim

),(0

yxfyxf

y

yxff y

2

),(),(lim

0

yxfyxf

x

f

y

ff xxyxxy


Partial Differential Equations

Partial differential equation (PDE) is an equation containing an unknown function of two or

more variables and its partial derivatives.

Invention of PDE’s by Newton and Leibniz in 17th century mark the beginning of modern

science. PDE’s arise in the physical problems, both in classical physics and quantum

mechanics.

Orbits of the planets or spaceships, flow of the liquid around a submarine or air around an

airplane, electrical currents in the circuit or processor, and actually majority of the physics

and engineering inspired problems are described by partial differential equations.


Boundary Value Problem Consider the shape of the soap film stretched on approximately horizontal frame. Let

z=h(x,y) be the description of the shape (height) of this film. The tension force, acting on the

unit piece of the surface is proportional to

And equals zero for an equilibrium solution. Let h(x(t),y(t))=g(t) be the parametric

description of the frame. Thus, the following differential equation defines the shape of the

soap film:

This is the Laplace equation with Dirichlet boundary conditions.

2

2

2

2

y

f

x

f

)()(),(

02

2

2

2

tgtytxf

y

f

x

f

Inside the region

Along the Boundary


Differential OperatorsIn the differential equations, there are several derivatives that occur very often. For example

the vector of first derivatives or gradient of the function:

To clarify the notation of PDE’s and facilitate the calculations, the notation of differential

operators was invented.

Thus, nabla stands for gradient:

The sum of second derivatives of f(x,y,z), formally obtained as the scalar product of two

gradient operators is called a Laplacian:

z

f

y

f

x

ff ,,

(3)

zyx

,,

2

2

2

2

2

22 ,,,,,

zyxzyxzyx

(2)


Differential Operators 2The divergence of a vector function (f1(x,y,z), f2(x,y,z), f3(x,y,z)) is the sum of its first

derivatives, or a scalar product of the function with nabla:

The rotor of a vector function (f1(x,y,z), f2(x,y,z), f3(x,y,z)) is the vector product of the

function with nabla:

Several identities can be derived for the operators of gradient, divergence, rotor and

Laplacian.

z

f

y

f

x

fff

321,div

y

f

x

f

z

f

x

f

z

f

y

f

xfxfxf

zyx

kji

ff 121323

321

,,

)()()(

rot

(4)

(5)


Other Coordinate Systems We defined the differential operators in Euclidian

coordinates. However, it is sometimes more convenient

to use other systems, like spherical coordinates (r,φ,θ)

for spherically symmetric problems or cylindrical

(ρ,φ,z) for cylindrically symmetrical problems.

Using the identities:

...

...

2

2

x

f

x

f

x

r

r

f

x

f

x

f

x

f

x

f

x

r

r

f

x

ff

xxxx

x

Cylindrical coordinates


Other Coordinate SystemsWe can obtain in cylindrical coordinates

And in spherical:

z

ff

fff

fff

z

y

x

cos1

sin

sin1

cos

(2)

Spherical coordinates

f

rr

ff

r

f

r

f

r

ff

r

f

r

f

r

ff

z

y

x

sincos

sin

cossincossinsin

sin

sincoscossincos


Laplacian in Cylindrical and Spherical systemsWe can obtain in cylindrical coordinates

And in spherical:

2

2

2

2

22

2 11

z

fffff

2

2

2222

2 sin

1sin

sin

11

f

r

f

rr

fr

rrf

(6)

(7)


1. The heat equation, describing the temperature in solid u(x,y,z,t) as a function of

position (x,y,z) and time t:

This equation is derived as follows:

Consider a small square of size δ, shown on

the figure. Its heat capacitance is δ2·q, where

q is the heat capacitance per unit area. The

heat flow inside this square is the difference

of the flows through its four walls. The heat

flow through each wall is:

Example: The Heat Equation

2/0

xx

u

y

xx

uW

2/0

xx

u

2/0

yy

u


Here δ is the size of the square, µ is the heat conductivity

of the body and is the temperature

gradient. The change of the temperature

of the body is the total thermal flow

divided by its heat capacitance:

the last expression is actually the definition of the second derivative, therefore:

The Heat Equation

2/1

0

xx

uF

q

yu

yu

q

xu

xu

q

FFFF

t

u yyxx

2

2/2/

2

2/2/

24321 0000

2/2

0

xx

uF

2/

3

0

yy

uF

2/

4

0

yy

uF

uqt

u

y

u

x

u

qt

u

2

2

2

2

(9)

t

u


Examples of Physical Equations

2. The vibrating string equation, describing the deviation y(x,t) of the taut string from its

equilibrium y=0 position:

The derivation of this equation is somewhat similar to the heat equation: we consider a small

piece of the string; the force acting on this piece is ; it causes the acceleration of

the piece which is .

3. The Schrödinger equation. This equation defines the wave function of the particle in the

static field, and used, for example to calculate the electron orbits of the atoms.

22

2)(

mVE

2

22

2

2

x

ya

t

y

2

2

x

ykF

2

2

t

y

(11)

(10)


Solution of the Heat EquationConsider again the Heat Equation:

Let u=XT, where X(x) and T(t); then

In the last equation the left part is function of t, while the right part is the function of x.

Therefore, this equation can only be valid if both parts are constant, say –λ2. Then:

We have chosen a negative constant in order to obtain a bounded solution.

2

2

x

uc

t

u

X

X

T

T

c

TXcTX

1

tceT

TcT2

2

(12)


Solution of the Heat Equation 2

Thus, the bounded solution of (12) is a linear combination of the functions from the

parametric family

the solution ax+b corresponds to λ=0.

The specific solution of equation (12) with initial conditions f(0,x) and f(t,x1) f(t,x2) is found

via decomposition of f(0,x) in the basis (13), satisfying f(t,x1) f(t,x2).

xbxaX

XX

sincos

2

baxxbxae tc ;)sincos(2

(13)


Solution of the Heat Equation 3Boundary and initial conditions restrict the family of these functions:

The limitation of the domain to [-a,a] restricts the functions from a parametric family to a

countable set of 2a periodic functions:

The symmetry (odd or even) of the initial conditions further restricts the basis to sin(..) or

cos(..).

Similarly, the Dirichlet (constant value) or Neuman (zero derivative) boundary conditions

restrict the basis to the functions, satisfying the conditions.

)sincos(

2

xa

xnbx

a

xnae

ta

nc


Example 1Solve

Given

Solution

The solution consists of functions:

The condition u(0,t)=u(3,t)=0 is fulfilled by

xxxxu

tutu

10sin28sin34sin5)0,(

0),3(),0(

0,30,2 2

2

txx

u

t

u

baxxbxae t ;)sincos(22

(14)

(15)

3sin

2

32 xn

ebt

n

n

(16)


Example 1

We need only n=12, 24 and 30 in order to fit f(x,0). The solution is

3sin

2

32 xn

ebt

n

n

xexexetxf ttt 10sin28sin34sin5),(222 20012832 (17)


Solution of the String EquationThe vibrating string equation

Can be solved in the way similar to solution of the heat equation. Substituting

into (14), we obtain

baxtcbtcaxbxa ;sincossincos 2211

(18)2

22

2

2

x

yc

t

y

XTy


Example 2 (1/3)The taut string equation is fixed at points x=-1 and x=1; f(-1,t)=f(1,t)=0; Its equation of

motion is

Initially it is pulled at the middle, so that

Find out the motion of the string.

Solution:

The solution of (19) is comprised of the functions, obtaining zero values at x=-1 and x=1:

},0{,

sincossin

2

)12(

2

)12(

2

)12(cos

1

Nmn

tmbtmaxma

tn

btn

axn

mmn

nn

(19)2

2

2

2

x

yc

t

y

10,1

01,1)0,(

xx

xxxy

(20)


Example 2 (2/3)Solution (continued):

Moreover, since the initial condition

is symmetric, only the cos(…x) remains in the solution.

The coefficients bn in (20) are zeros, since

Therefore, the solution has the form

(21)

10,1

01,1)0,(

xx

xxxy

0)0,(

t

xy

tnxn

atxyn

n cos

2

)12(cos),(

0


Example 2 (3/3)Where

dx

n

dxxn

an

1

0

2

1

0

212

cos2

)1(2

12cos2

math for cslecture 121 partial differential equations boundary value problems

Documents

y derivatives

boundary slide

vector function f

gradient operators

operators of gradient

derivatives of fx

string equation slide

unknown function