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Partial Differential Equations Introduction to . Partial Differential Equations The Heat Equation Laplace's Equation Peter Coutts & Jerad Meisner Harmonic Functions Spherical Harmonic . May 18, 2015 Mathematica Code Home Page Abstract Title Page Partial differential equations appear frequently in the study of physical phenomena including sound, heat, fluid dynamics, and elec- JJ II trostatics and also play a central roll in the study of quantum me- chanics. This paper will examine several partial differential equa- J I tions (PDEs) and how they apply to the real word. It will detail one method for solving PDEs known as separation of variables and take Page1 of 45 a brief look at what some of these solutions mean. Go Back Full Screen Close Quit 1. Introduction to Partial Differential Equa- tions Introduction to . A partial differential equation (PDE) is any differential equation that con- The Heat Equation tains unknown multivariable functions and their partial derivatives. They Laplace's Equation are commonly used to model the rates of change of physical quantities with Harmonic Functions respect to both space and time. Some examples are: the wave equation Spherical Harmonic . 2 Mathematica Code @ u 2 2 2 = c r u @t Home Page for modeling sound, light, and water waves; the Euler-Tricomi equation Title Page @2u @2u = x @x2 @y2 JJ II for studying transonic flows; and the Schr¨odingerequation J I @ i Ψ = H^ Ψ Page2 of 45 ~@t for describing how the quantum state of a physical system changes with Go Back respect to time. While many PDEs exist for describing a wide variety of Full Screen physical phenomena we will be focusing on two specific equations in this paper. Close Quit 2. The Heat Equation The one-dimensional heat equation Introduction to . @u @2u The Heat Equation = k (1) @t @x2 Laplace's Equation Harmonic Functions describes the temperature distribution in an object. The solution, u(x; t), Spherical Harmonic . represents the temperature at a given point x at time t, and k > 0 is a Mathematica Code constant value called the thermal conductivity and measures the ability of the object in question to conduct heat. Home Page Title Page 2.1. Method of Separation of Variables Many PDEs, including the heat equation, can be solved using a technique JJ II known as the method of separation of variables. We will use the one- dimensional heat equation as a simple example to explain how to perform J I separation of variables so that we may use it later to solve a more difficult Page3 of 45 equation. Go Back 2.1.1. General Solution Full Screen When using the method of separation of variables, we begin by making the assumption that our solution (in this case u(x; t)) is actually a product Close solution { a product of functions, each of which is dependent upon a single Quit variable [6]. In the case of the heat equation, we make the assumption that our solution will be of the form Introduction to . u(x; t) = F (x)G(t): The Heat Equation The reason for this, is that it will turn the PDE into two ordinary differential Laplace's Equation equations (ODEs) which should be much easier to solve. Now that we have Harmonic Functions defined the form for our final solution, we can plug it into equation (1) and Spherical Harmonic . differentiate Mathematica Code @ @2 [F (x)G(t)] = k [F (x)G(t)] @t @x2 Home Page dG d2F Title Page F (x) = kG(t) : dt dx2 JJ II In the step above, we factored out F (x) from the partial derivative on the left hand side (LHS), and G(t) from partial derivative on the right J I hand side (RHS), since they are not dependent upon the variable we are differentiating with respect to. Since we are then only differentiating single Page4 of 45 variable functions, we no longer have any partial derivatives to worry about. Now we need to move all expressions involving t to one side of the equation, Go Back and all expressions involving x to the other side. To do this we just divide both sides of the equation by F Gk and we are left with Full Screen 1 dG 1 d2F = : (2) Close kG dt F dx2 Quit Now, since we are dealing with an equation, we know that the left side and right side must be equal. We also know that in order for this equation to be true, both the LHS and RHS must be equal to a constant. If this were not Introduction to . the case, when we varied either t or x, only one side of the equation would The Heat Equation change and the equation would no longer be true. With this information in Laplace's Equation mind, we will let γ be a constant value known as the separation constant, Harmonic Functions and rewrite equation (2) as two equations Spherical Harmonic . 1 dG 1 d2F Mathematica Code = γ = γ kG dt F dx2 (3) Home Page dG d2F = γkG = γF (4) Title Page dt dx2 We now have two linear ODEs that we can solve using standard techniques. JJ II J I 2.1.2. Particular Solution with Initial and Boundary Conditions In order to find a particular solution we need to first understand the sit- Page5 of 45 uation we are modeling. In the case of the one-dimensional heat equation we can think of this as a thin bar of material on the x-axis and u(x; t) will Go Back represent the temperature of a cross-section of the material at position x at Full Screen time t (see Figure1)[6]. Close Quit Introduction to . The Heat Equation Laplace's Equation Harmonic Functions Spherical Harmonic . Figure 1: One-dimensional heat equation Mathematica Code Home Page Now let's define some initial and boundary conditions so that we may find a particular solution Title Page k = 2; 0 ≤ x ≤ π; t > 0 (5) JJ II u(0; t) = 0; u(π; t) = 0 (6) u(x; 0) = x; 0 ≤ x ≤ π: (7) J I Here, in Equation (5) we are defining the thermal conductivity of the object Page6 of 45 to be 2, and defining the domain of our function. In Equation (6) we are stating that the temperature at the boundaries is 0 at all t. Equation (7) is Go Back our initial condition and states that, at t = 0, the temperature at any given cross-section is equal to the x position of the cross-section [6]. We will now Full Screen rewrite our linear ODEs from Equation (4) using prime notation and the k Close Quit value we chose F 00 − γF = 0 (8) G0 − 2γG = 0: (9) Introduction to . The Heat Equation Now we can begin to solve these ODEs to satisfy our boundary conditions. Laplace's Equation We will start with Equation (8) and we are only looking for non-trivial Harmonic Functions solutions. To do this we need to recognize that we have three possibilities Spherical Harmonic . for γ that we will have to examine: γ = 0, γ < 0, and γ > 0, so we will need Mathematica Code to solve three times: once for each possibility. We will begin with γ = 0 and Equation (8) becomes Home Page F 00 = 0: Title Page Integrating twice yields F (x) = Ax + B: (10) JJ II Now, our boundary conditions from Equation (6) give: J I u(0; t) = u(π; t) = 0: Page7 of 45 Since u(x; t) = F (x)G(t) our boundary conditions will be satisfied if Go Back F (0) = F (π) = 0 Full Screen or G(t) = 0; for all t; Close Quit but since the latter will give us the trivial solution for all t, we are only in- terested in the former. Now, we can see here that, to satisfy these boundary conditions, both A and B must be 0. If we plug these values into Equa- Introduction to . tion (10) this will yield the trivial solution to our original PDE, so we will The Heat Equation ignore it. Still using Equation (8), we can now examine the situation where Laplace's Equation γ > 0. In this case we can find eigenvalues using the characteristic equation: Harmonic Functions λ2 − γ = 0 Spherical Harmonic . p λ = ± γ: Mathematica Code Since we have distinct real roots, we know that our solution will be a linear Home Page combination of exponentials of the eigenvalues Title Page p p F (x) = Ae γx + Be− γx: (11) JJ II Now we will use our boundary conditions to solve for A and B. We will start with F (0) = 0 J I F (0) = 0 Page8 of 45 Ae0 + Be0 = 0 Go Back A + B = 0: Full Screen We can see here that A = −B. Now we will use our second boundary condition, but this time, armed with this new knowledge, we will substitute Close Quit −A in for B F (π) = 0 p p Introduction to . γπ − γπ Ae − Ae = 0 The Heat Equation p p γπ − γπ A[e − e ] = 0: Laplace's Equation Harmonic Functions Here we can see that A must equal 0 to satisfy this equation, therefore, since Spherical Harmonic . A = −B, B is also equal to 0. If we plug these values into Equation (11), Mathematica Code we see that F (x) = 0 for all x, so, again, we have the trivial solution, and this situation is also of no interest to us. So lets examine the final case Home Page where γ < 0, but first we are going to let γ = −ρ2.
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