DREAM IDEA PLAN IMPLEMENTATION 1 Present to Amirkabir

DREAM IDEA PLAN IMPLEMENTATION 1

Present to: Amirkabir University of Technology (Tehran Polytechnic) & Semnan University Dr. Kourosh Kiani Email: kkiani 2004@yahoo. com Email: Kourosh. kiani@aut. ac. ir Web: aut. ac. com 2

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What is a differential equation? 4

An equation with derivatives in it. 5

Examples 6

What was that last line? That is a partial differential equation (P. D. E. ) because it has partial derivatives. 7

Partial Derivatives? When u is a function of two variables, r and t, it has two partial first derivatives, one with respect to r, one with respect to t. 8

Find the partial derivative of u with respect to t by holding r constant and differentiating as usual. 9

Example Suppose Then the partial derivative of u with respect to t would be 10

Notation 11

Lingo Second partial derivative of u with respect to t Second mixed partial derivatives 12

Dealing with Differential Equations • Determine what the dependent variable is when you are presented with a differential equation. • Determine what the independent variable(s) is (are), too. 13

Example Dependent variable Independent variable 14

Why? In the equation is a function of is dependent on is the dependent variable, is the independent variable. 15

Example Dependent variable Independent variable 16

Why? In the equation are all functions of as evidenced by the right hand side of the equation. is the dependent variable, is the independent variable. 17

Example Dependent variable Independent variables 18

Why? is a function of and as evidenced by the partial derivatives and is the dependent variable, are the independent variables. 19

Why Does This Matter? • We want solutions to differential equations. • A solution to a differential equation is a function of the independent variable(s) which can successfully play the role of the dependent variable in the differential equation. 20

In Other Words The unknown in a differential equation is the dependent variable. It is the thing we want to find. It is the thing whose derivatives appear in the differential equation. It is a function expressed in terms of the independent variable(s). 21

Example is a solution to because 22

Example is a solution to because Check that this is true by calculating the derivatives! 23

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: (1) The higher order partial derivatives are defined recursively and include the mixed x, y derivatives: 24

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 17 th 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. 25

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. 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: 26

Differential Operators The divergence of a vector function (f 1(x, y, z), f 2(x, y, z), f 3(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 (f 1(x, y, z), f 2(x, y, z), f 3(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. 27

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 28

Other Coordinate Systems We can obtain in cylindrical coordinates And in spherical: (2) Spherical coordinates 29

Laplacian in Cylindrical and Spherical systems We can obtain in cylindrical coordinates And in spherical: 30

Example: The Heat Equation 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: y Consider a small square of size δ, shown on the figure. Its heat capacitance is δ 2·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: x 31

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: 32

Some examples with two independent variables (x y) or (x, t) : 1) ux + uy = 0 transport 2) ux + yuy = 0 transport 3) ux+uuy = 0 shock wave 4) uxx + uyy = 0 Laplace equation 5) utt –uxx + u 3 = 0 wave with interaction 6) ut + uux + uxxx = 0 dispersive wave 7) utt + uxxxx = 0 vibrating bar 8) ut + j uxx = 0 j = -1, quantum mechanics Examples 1 to 3 have order one. Examples 4, 5, & 8 have order two. Example 6 has order three. Example 7 has order four. Examples 3, 5, & 6 are not linear. 33

• Schrodinger equation which describe the nature of phenomena at the microscopic spatial scales • where (x, y, z, t) is the wave function of a particle of mass m, is Planck’s constant and V(x, y, z) is the potential at point (x, y, z). 34

• Elasticity • Wave equation • Euler-Bernoulli beam equation: 35

First-order Linear Partial Differential Equations When a, b and c are constant, the partial differential equation can be solve by rotating the axes. 36

Example 37

Replacing u and v by the substitutions in (2), we get We call such a solution a general solution for the PDF 38

Up to now we have stressed the similarity between firstorder ordinary and partial differential equations. We encounter a substantial difference when we consider initial value problems. The solution of an ODE is completely determined by prescribing a value for it at a single point. This is generally not true for PDE. In order to obtain a unique solution, it is usually necessary to prescribe values for the solution on an entire line. A suitable initial condition is z(x, 0)= z 0(x) for all x, where z 0 is a differentiable function. 39

Suppose we are given the initial condition Setting y=0 in (1) and substituting the initial condition for the lefthand side of (1), we have Replacing this function for G in (1) yields 40

Tangent planes, Normal Lines, and Gradients Let z = f(x, y) be a function of two variables, the graph of f ia a surface in R 3. More generally, the graph of the equation F(x, y, z) = 0 is called differentiable at point (x 0, y 0, z 0) if In R 2, a differentiable curve has a unique tangent line at each point. In R 3 a differentiable surface in R 3 has a unique tangent plane at each point at which 41

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Cauchy Problem A problem like that in above Example in which we require that the solution satisfy a line of initial conditions, is called a Cauchy problem or an initial value problem. The solution of a Cauchy problem can be visualized as a surface z = z(x, y) in three-dimensional Euclidean space. Thus we can use three-dimensional analytic geometry to increase our understanding of the solution. 43

We call a surface z = z(x, y) that is the solution of a Cauchy problem an integral surface. If we write the equation of an integral surface in the form F(x, y, z) = z(x, y)- z = 0 We can find the tangent plane to the integral surface by taking the total differential The vector point. is normal to the integral surface at any Rewriting (1) in the form Where g(x, y, z)=f(x, y)-c(x, y)z, 44

we see that the vector: Is orthogonal to N since Hence V is tangent to the integral surface and lies in the tangent plane at every point. Thus the first-order partial differential equation provides the geometric requirement that any integral surface through a given point be tangent to the vector V. This means that if we begin at some point specified by the initial condition (on the integral surface) and move in the direction of the known tangent vector V, we move along a curve lying entirely on the integral surface F(x, y, z) = 0. This curve is called a characteristic. 45

Points on a characteristic curve can be described parametrically by an expression of the form If we differentiate this with respect to t, we obtain a tangent vector. This vector must belong to the tangent plane at the given point. Furthermore, it must be proportional to V, since the characteristic is obtained by moving in the V direction. Thus the coordinates of and V are proportional; that is 46

or Equations (2) provide a pair of the first-order ordinary differential equations: or 47

Example a= 1, b=1, and g=z 48

y=0, we have x=c 1 49

Example a= 1, b=y, and g=2 xz 50

with 51

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Solution of a Cauchy Problem To solve the Cauchy problem Partial Differential equation Initial condition: By the method of characteristics, perform the following four steps: 53

Step 1 Write the differential equations 54

Step 2 Solve (if possible) the first differential equation For x in terms of y, obtaining x = x(y, c 1). Substitute this expression (if necessary) for x in the second differential equation And solve for z in terms of y, obtaining z = z(y, c 2). 55

Step 3 Set y=0 in the two solutions in step 2 and substitute the resulting expressions into the initial condition obtaining an equation involving only C 1 and C 2: 56

Step 4 Use the solutions in step 2 to eliminate the constants C 1 and C 2 in the final equation in step 3, The resulting equation in x, y, and z is the solution of the Cauchy problem 57

Questions? Discussion? Suggestions ? 58

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