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9.6 Other Heat Conduction Problems. We discussed in 9.5, the heat conduction equation α 2 u xx = u t , 0 < x < L, t > 0 , (1) the boundary conditions u (0 ,t )=0 , u ( L,t ) =0 , t > 0 , (2) and the initial condition u ( x, 0)= f ( x ) , 0 ≤ x ≤ L. (3)
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9.6 Other Heat Conduction Problems We discussed in 9.5, the heat conduction equation α2uxx= ut , 0 < x < L, t > 0, (1) the boundary conditions u(0,t)=0, u(L,t) =0, t > 0, (2)and the initial condition u(x,0)=f(x), 0 ≤ x ≤ L. (3) We now consider two other problems of one-dimensional heat conduction that can be handled by the method developed in Section 9.5. 1. Nonhomogeneous Boundary Conditions 2. Bar with Insulated Ends
Nonhomogeneous Boundary Conditions Suppose now that one end of the bar is held at a constant temperature T1 and the other is maintained at a constant temperature T2. Then the boundary conditions are u(0,t) = T1, u(L,t) = T2, t > 0. (4) The differential equation (1) and the initial condition (3) remain unchanged.
The Steady State Solution. After a long time—that is, as t →∞—we anticipate that a steady temperature distribution v(x) will be reached, which is independent of the time t and the initial conditions. Since v(x) must satisfy the equation of heat conduction (1), we have v''(x) = 0, 0 < x < L. (5) Hence the steady-state temperature distribution is a linear function of x. Further v(x) must satisfy the boundary conditions v(0) = T1, v(L) = T2, (6) which are valid even as t →∞. The solution of Eq. (5) satisfying Eqs. (6) is v(x) = (T2 − T1)x/L + T1. (7)
The Transient Solution. Returning to the original problem, Eqs. (1), (3), and (4), we will try to express u(x, t) as the sum of the steady-state temperature distribution v(x) and another (transient) temperature distribution w(x, t); thus we write u(x, t) = v(x) + w(x, t). (8) We obtain where
Example Consider the heat conduction problem uxx= ut , 0 < x < 30, t > 0, u(0, t) = 20, u(30, t) = 50, t > 0, u(x, 0) = 60 − 2x, 0 < x < 30. Find the steady-state temperature distribution and the boundary value problem that determines the transient distribution.
Bar with Insulated Ends A slightly different problem occurs if the ends of the bar are insulated so that there is no passage of heat through them. That is, α2uxx= ut , 0 < x < L, t > 0, the boundary conditions ux(0, t) = 0, ux(L, t) = 0, t > 0, and the initial condition u(x,0)=f(x), 0 ≤ x ≤ L.
Bar with Insulated Ends 1. Solve by the method of separation of variables to get the fundamental solutions for the problem 2. Solution to the heat conduction problem for a rod with insulated ends,
Example Find the temperature u(x, t) in a metal rod of length 25 cm that is insulated on the ends as well as on the sides and whose initial temperature distribution is u(x, 0) = x for 0 < x < 25.
More General Problems The method of separation of variables can also be used to solve heat conduction problems with other boundary conditions than those given by previous two types. For example, the left end of the bar might be held at a fixed temperature T, while the other end is insulated. In this case the boundary conditions are u(0, t) = T, ux(L, t) = 0, t > 0.
9.7 The Wave Equation: Vibrationsof an Elastic String a2uxx= utt Wave Equation Let u(x, t) denote the vertical displacement experienced by the string at the point x at time t. If damping effects, such as air resistance, are neglected, and if the amplitude of the motion is not too large, then u(x, t) satisfies the partial differential equation a2uxx= uttin the domain 0 < x < L, t > 0 where a is the velocity of propagation of waves along the string given by a2 = T/ρ for T is the tension (force) in the string, and ρ is the mass per unit length of the string material.
Boundary and Initial Conditions The ends are assumed to remain fixed, and therefore the boundary conditions are u(0,t) = 0, u(L,t) = 0, t ≥ 0. Initial Conditions: The initial position of the string u(x, 0) = f (x), 0 ≤ x ≤ L and its initial velocity ut (x, 0) = g(x), 0 ≤ x ≤ L, where f and g are given functions. For consistency, f (0) = f (L) = 0, g(0) = g(L) = 0.
Example: Elastic String with Nonzero Initial Displacement. First suppose that the string is disturbed from its equilibrium position and then released at time t = 0 with zero velocity to vibrate freely. Then the vertical displacement u(x, t) must satisfy the wave equation a2uxx= utt, 0 < x < L, t > 0; (1) the boundary conditions u(0, t) = 0, u(L, t) = 0, t ≥ 0; (3) and the initial conditions u(x, 0) = f (x), ut(x, 0) = 0, 0 ≤ x ≤ L, (9) where f is a given function describing the configuration of the string at t = 0.
Solve by method of separation of variables The formal solution of the problem is Where The quantities nπa/L for n = 1, 2, . . . are the natural frequencies of the string. The factor sin (nπx/L) represents the displacement pattern occurring in the string when it is executing vibrations of the given frequency. Each displacement pattern is called a natural mode of vibration and is periodic in the space variable x; the spatial period 2L/n is called the wavelength of the mode of frequency nπa/L.
Example Consider a vibrating string of length L = 30 that satisfies the wave equation 4uxx= utt, 0 < x < 30, t > 0. Assume that the ends of the string are fixed and that the string is set in motion with no initial velocity from the initial position u(x, 0) = f (x) ={x/10, 0 ≤ x ≤ 10, {(30 − x)/20, 10 < x ≤ 30. Find the displacement u(x, t) of the string and describe its motion through one period.
Answer where
General Problem for the Elastic String. Let us modify the problem just considered by supposing that the string is set in motion from its equilibrium position with a given velocity. Then the vertical displacement u(x,t) must satisfy the wave equation (1) a2uxx= utt, 0 < x < L, t > 0; the boundary conditions (3) u(0, t) = 0, u(L, t) = 0, t ≥ 0; and the initial conditions u(x, 0) = 0, ut(x, 0) = g(x), 0 ≤ x ≤ L, (31) where g(x) is the initial velocity at the point x of the string.
Formal solution Where Use of the principle of superposition. To solve the wave equation with the general initial conditions u(x,0) = f(x), ut(x,0)=g(x), 0<x<L, you can solve instead the somewhat simpler problems with the initial conditions (9) and (31) discussed before, respectively, and then add together the two solutions.
9.8 Laplace’s Equation Laplace’s equation in two dimensions is uxx+ uyy= 0, and in three dimensions uxx+ uyy + uzz= 0. The potential function of a particle in free space acted on only by gravitational forces satisfies the same equations. So, Laplace’s equation is also called potential equation. The problem of finding a solution of Laplace’s equation that takes on given boundary values is known as a Dirichlet problem (the first boundary value problem of potential theory). If the values of the normal derivative are prescribed on the boundary, the problem is said to be a Neumann problem (the second boundary value problem of potential theory).
Dirichlet Problem for a Rectangle The mathematical problem of finding the function u satisfying Laplace’s equation (1) uxx+ uyy= 0, in the rectangle 0 < x < a, 0 < y < b, and also satisfying the boundary conditions u(x, 0) = 0, u(x, b) = 0, 0 < x < a, u(0, y) = 0, u(a, y) = f (y), 0 ≤ y ≤ b where f is a given function on 0 ≤ y ≤ b.
Solution To solve this problem, we construct a fundamental set of solutions satisfying the partial differential equation and the homogeneous boundary conditions; then we superpose these solutions so as to satisfy the remaining boundary condition. Solution where
Example In above problem, let a = 3, b = 2, and f(y)={ y, 0 ≤ y ≤ 1, {2 − y, 1 ≤ y ≤ 2. Answer We find that Plots and level curves are in the figures.
Dirichlet Problem for a Circle Consider the problem of solving Laplace’s equation in a circular region r < a subject to the boundary condition u(a, θ) = f (θ), where f is a given function on 0 ≤ θ < 2π (see Figure). In polar coordinates Laplace’s equation has the form u must be periodic in θ with period 2π. Moreover u(r, θ) is bounded for r ≤ a.
Apply the method of separation of variables to solve this problem Above equation represents the solution of the boundary value problem where
Section 9.2 Fourier Series & Section 9.3 The Fourier Convergence Theorem
Section 9.5 Separation of Variables; Heat Conduction in a Rod Section 9.6 Other Heat Conduction Problems The classical equation for unidirectional heat conduction in a rod of finite length L is α2uxx= ut , 0 < x < L, t > 0, where u(x, t) represents the temperature at time t and position x along the rod with initial and boundary conditions discussed. Method of separation of variables to solve the heat equationis discussed.
Section 9.7 The Wave Equation: Vibrations of an Elastic String The classical wave equation that describes the vibrations of an elastic string of length L is a2uxx= utt, 0 < x < L, t > 0, where u(x, t) represents the string’s displacement from equilibrium at time t and position x along the string. Initial and boundary conditions are given. Solved by method of separation of variables. Defined modal shape and natural frequency.
