Math for CSTutorial 121 Heat Equation. Laplace Equation.

Math for CSTutorial 122 Contents 1.Heat Equation 2.Solution of the Heat Equation 3.Examples of the Physical Equations

Math for CSTutorial 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 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: Example: The Heat Equation y x

Math for CSTutorial 124 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

Math for CSTutorial 125 Solution of the Heat Equation 1/5 A square plate [0,1] 2 has the sides kept at u(0,y,t)=u(1,y,t)=u(x,0,t)=u(x,1,t)=0, and initial temperature u(x,y,0)=f(x,y). Determine u(x,y,t). Solution: Consider the solution in the form Then (1) becomes (2) (1)

Math for CSTutorial 126 Solution of the Heat Equation 2/5 Since the left side depends only on t and the right side depends only on x and y, each side must be equal to a constant, -λ 2, where the sign is chosen for convergence: The last equation can be rewritten as, which shows that both sides are constants, say -µ 2 (negative to have bounded solutions): (2) (3)

Math for CSTutorial 127 Solution of the Heat Equation 3/5 The solutions to equations (2) and (3) are: therefore, the solution to (1) is given by: From the boundary conditions that u(0,y,t)=0 and u(x,0,t)=0 we get a 1 =a 2 =0; Therefore the solution is already limited to the form: where B=b 1 b 2. (4)

Math for CSTutorial 128 Solution of the Heat Equation 4/5 From the boundary conditions u(1,y,t)=0 and u(x,1,t)=0, we see that the sin(..x) and sin(..y) are zero at the boundary, therefore: Therefore, the solution satisfying boundary conditions is given by: Since the equation (1) is linear, any linear combination of these functions is also a solution. Letting t=0, and using an initial condition u(x,y,0)=f(x,y), we obtain: (5)

Math for CSTutorial 129 Solution of the Heat Equation 5/5 The solution to (1) is given by Where And therefore

Math for CSTutorial 1210 Laplace Equation 1/3 Three sides of the square plate are kept at zero temperature u(0,y)=u(1,y)=u(x,0)=0, the fourth side, is kept at temperature u 1 : u(x,1)= u 1. Determine the steady state temperature of the plate. To solve, suppose u(x,y)=X(x)Y(y): or Setting each side to –λ 2 we obtain: From which we obtain:

Math for CSTutorial 1211 Laplace Equation 2/3 Where The boundary conditions u(0,y)=u(x,0)=0 imply a 1 =a 2 =0. The condition u(1,y)=0 implies λ=mπ and therefore, the general form of the solution is: From the condition u(x,1)=u 1 we have: And therefore

Math for CSTutorial 1212 Laplace Equation 3/3 The solution is This problem, which is the solution of Laplace equation Inside the region R when u(x,y) is specified an the boundary of R is called a Dirichlet problem. The boundary conditions, when the function is specified around the boundary is called Dirichlet boundary conditions.

Math for CSTutorial 1213 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. (11) (10)

Math for CSTutorial 1214 Example 1 Solve Given Solution The solution consists of functions: The condition u(0,t)=u(3,t)=0 is fulfilled by (14) (15) (16)

Math for CSTutorial 1215 Example 1 We need only n=12, 24 and 30 in order to fit f(x,0). The solution is (17)

Math for CSTutorial 1216 Solution of the String Equation The vibrating string equation Can be solved in the way similar to solution of the heat equation. Substituting into (14), we obtain (18)

Math for CSTutorial 1217 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: (19) (20)

Math for CSTutorial 1218 Example 2 (2/3) Solution (continued): Moreover, since the initial condition is symmetric, only the cos(…x) remains in the solution. The coefficients b n in (20) are zeros, since Therefore, the solution has the form (21)

Math for CSTutorial 1219 Example 2 (3/3) Where