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MAE 384. Advanced Mathematical Methods for Engineers. Homework Assignment 4. 1. (15 pts.) Classify the following PDE’s as elliptic, hyperbolic, or parabolic or the conditions under which it would change between types if it could be multiple types:

MAE 384. Advanced Mathematical Methods for Engineers. Homework Assignment 4. 1. (15 pts.) Classify the following PDE’s as elliptic, hyperbolic, or parabolic or the conditions under which it would change between types if it could be multiple types: (a) 9 πœ• 2𝑒 πœ•π‘₯ 2 βˆ’ 6 πœ•π‘’ πœ•π‘₯ + 5 πœ• 2𝑒 πœ•π‘¦2 = 0 (b) 𝑒 πœ• 2𝑒 πœ•π‘₯ 2 + 6 πœ• 2𝑒 πœ•π‘‘ 2 = 0 (c) πœ• 2𝑒 πœ•π‘₯ 2 βˆ’ 5 πœ• 2𝑒 πœ•π‘₯πœ•π‘¦ + 4 πœ• 2𝑒 πœ•π‘¦2 = 0 (d) 1 𝑐 πœ• 2𝑒 πœ•π‘₯ 2 = πœ•π‘’ πœ•π‘‘ + 4 πœ• 2𝑒 πœ•π‘¦2 (e) 17 = πœ•π‘’ πœ•π‘‘ + 𝑑 πœ• 2𝑒 πœ•π‘¦2 + 𝑑 2 πœ• 2𝑒 πœ•π‘₯ 2 + 20 πœ• 2𝑒 πœ•π‘₯πœ•π‘¦ 2. (15 pts.) Assume that the wavelength of acoustic waves in an organ pipe is long relative to the width of the pipe so that the acoustic waves are one-dimensional (they travel only lengthwise in the pipe). Therefore, the equation governing the pressure in the wave is πœ• 2𝑝 πœ•π‘‘ 2 βˆ’ 𝑐 2 πœ• 2𝑝 πœ•π‘₯ 2 = 0 Where 𝑝 is the acoustic pressure (disturbance). At the closed end, air velocity (𝑒) must be zero. Note that βˆ’πœŒ πœ•π‘’ πœ•π‘‘ = πœ•π‘ πœ•π‘₯ So that if 𝑒 = 0, then πœ•π‘β„πœ•π‘₯ = 0. At an open end, 𝑝 β‰ˆ 0 Find the acoustic modes in a pipe of length 𝑙 with (a) Two closed ends (b) Two open ends (c) One closed end and one open end. 3. (30 pts.) Solve the heat equation (Equation 1) for each of the following two set of conditions: πœ•π‘‡ πœ•π‘‘ = π‘˜ πœ• 2𝑇 πœ•π‘₯ 2 (1) (a) 𝑇(π‘₯, 0) = sin πœ‹π‘₯ 𝑙 𝑇(0,𝑑) = 100 ; 𝑇(𝑙,𝑑) = 0 (b) 𝑇(π‘₯, 0) = 0 𝑇(0,𝑑) = 100 ; πœ•π‘‡ πœ•π‘₯ (𝑙,𝑑) = 0 4. (15 pts.) Under the assumptions of incompressible (constant density), irrotational (no rotation, βˆ‡ Γ— 𝒖 = 0) and inviscid (no viscous dissipation), the fluid-dynamic equations are governed by the Laplace equation βˆ‡ 2πœ“ = 0 Where πœ“ is called the stream function and is defined such that 𝑒 = πœ•πœ“ πœ•π‘₯ and 𝑣 = πœ•πœ“ πœ•π‘¦ where 𝑒 and 𝑣 are the x and y components of velocity, respectively. This means that if we know the stream function, we can figure out the velocity components at every point. While we normally think of this in terms of Cartesian coordinate system, we can also do all of this in terms of the polar coordinate system. This will be a more relevant coordinate system for the situation we will be looking at, the flow around a circular cylinder. The Laplace equation in cylindrical coordinates is 1 π‘Ÿ πœ• πœ•π‘Ÿ (π‘Ÿ πœ•πœ“ πœ•π‘Ÿ ) + 1 π‘Ÿ 2 πœ• 2πœ“ πœ•πœƒ 2 = 0 The boundary conditions are that the velocity must be zero on the cylinder surface so that πœ“(π‘Ž, πœƒ) = 0 and πœ“ β‰ˆ π‘ˆπ‘Ÿ sin πœƒ (for π‘Ÿ β†’ ∞) Where π‘Ž is the radius of the cylinder and π‘ˆ is the free-stream velocity (the velocity a long way away from the cylinder). What is the solution for the stream function, πœ“(π‘Ÿ, πœƒ)? Plot the streamlines (lines of constant πœ“) for various values of πœ“ for π‘ˆ = 25 m/s and π‘Ž = 0.75 m (you may want to create matrices of radius and angle values for various points in the domain and then evaluate the solution at these points and use the contour plotting command to plot the streamlines since each contour is a line of constant πœ“). Note: The solution to a second order ODE of the form π‘Ÿ 2 𝑑 2𝑓(π‘Ÿ) π‘‘π‘Ÿ 2 + π‘Ÿ 𝑑𝑓(π‘Ÿ) π‘‘π‘Ÿ βˆ’ 𝑛 2𝑓(π‘Ÿ) = 0 is 𝑓(π‘Ÿ) = 𝑐1π‘Ÿ 𝑛 + 𝑐2π‘Ÿ βˆ’π‘› where the c values are constants that are determined by the boundary conditions.

 
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