Lecture 12

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Lecture 12. Partial Differential Equations Boundary Value Problems. Contents. Partial Differential Equations Boundary Value Problems Differential Operators Cylindrical and Spherical coordinate systems Examples of the Problems Solution of the heat equation Solution of the string equation.

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### Lecture 12

Partial Differential Equations

Boundary Value Problems

Lecture 12

Contents
• Partial Differential Equations
• Boundary Value Problems
• Differential Operators
• Cylindrical and Spherical coordinate systems
• Examples of the Problems
• Solution of the heat equation
• Solution of the string equation

Lecture 12

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)

Lecture 12

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.

Lecture 12

Inside the region

Along the Boundary

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.

Lecture 12

Differential Operators

In 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.

The sum of second derivatives of f(x,y,z), formally obtained as the scalar product of two gradient operators is called a Laplacian:

(2)

(3)

Lecture 12

Differential Operators 2

The 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.

(4)

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Lecture 12

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:

Cylindrical coordinates

Lecture 12

Other Coordinate Systems

Spherical coordinates

We can obtain in cylindrical coordinates

And in spherical:

(2)

Lecture 12

Laplacian in Cylindrical and Spherical systems

We can obtain in cylindrical coordinates

And in spherical:

(6)

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Lecture 12

Example: The Heat Equation
• 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:

y

x

Lecture 12

The Heat Equation

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:

(9)

Lecture 12

Examples of Physical Equations
• 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.

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Lecture 12

Solution of the Heat Equation

(12)

Consider 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.

Lecture 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).

(13)

Lecture 12

Solution of the Heat Equation 3

Boundary 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.

Lecture 12

Example 1

(14)

Solve

Given

Solution

The solution consists of functions:

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

(15)

(16)

Lecture 12

Example 1

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

(17)

Lecture 12

Solution of the String Equation

(18)

The vibrating string equation

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

into (14), we obtain

Lecture 12

Example 2 (1/3)

(19)

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:

(20)

Lecture 12

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)

Lecture 12

Example 2 (3/3)

Where

Lecture 12