University:
Whitman College
Course:
MATH-367 | Introduction to Partial Differential Equations
Academic year:

2023
Views:

291

Pages:

2
Author:

Grace Wallace

Extra Practice: Exercise 4.4.2 In 4.2, it was shown that the displacement u of a nonuniform string satisfies the PDE: ρ0 utt = T0 uxx + Q where Q represents the vertical component of body force per unit length. If Q = 0, the PDE is homogeneous. A slightly different homogeneous equation occurs if Q = αu. 1. Show that, if α < 0, the body force is restoring (towards u = 0), and if α > 0, the force tends to push the string farther away from u = 0. Extra HINT: Look at the how the acceleration and u are related by considering ρ0 utt = αu SOLUTION: If α < 0 and u > 0, then utt is negative- Therefore, u is acting in the opposite direction, and is a restoring force. If α > 0 and u > 0, then utt is also positive, and the string is pushed away from 0. 2. Assume that ρ0 and α are functions of x, but that T0 is constant, and separate variables. Analyze the time-dependent ODE. SOLUTION: Let u = XT , and ρ0 (x)XT 00 = T0 X 00 T + α(x)XT Divide both sides by ρ0 (x)XT and we get: T0 X 00 α(x) T 00 = + = −λ T ρ0 (x) X ρ0 (x) Therefore, the time dependent ODE is T 00 + λT = 0. The solution to this, as usual, depends on the solutions to the characteristic equation, which depends on λ: √ r2 + λ = 0 ⇒ r = ± −λ The three possible outcomes are: • λ = 0: T (t) = c1 + c2 t √ √ • λ < 0: T (t) = c1 cosh( −λ t) + c2 sinh( −λ t) NOTE: It is OK to use exponential here, but we should try to get used to using hyperbolic sine and cosine- it makes some things a lot easier. √ √ • λ > 0: T (t) = c1 cos( λ t) + c2 sin( λ t) 3. Specialize the previous part and make α, ρ0 , T0 all constant with the conditions: BCs ICs u(0, t) = 0 u(L, t) = 0 u(x, 0) = 0 ut (x, 0) = f (x) SOLUTION: Go back and analyze the ODE for X, since we have the full set of boundary conditions for it: X(0) = 0, X(L) = 0. α T0 00 X + X = −λX ρ0 ρ0 1 Simplifying a bit, we get the following, and we’ll make a substitution to make our work a little clearer: ρ0 α 00 X =− + λ X ⇒ X 00 = −βX ⇒ X 00 + βX = 0 X(0) = X(L) = 0 T0 ρ0 Therefore, our only nontrivial solution for X is the periodic solution, with eigenvalues and eigenfunctions: p nπ nπ 2 Xn (x) = sin βn x = sin x βn = L L Now going back to T , we have to see if restricting βn ≥ 0 puts a restriction on λn : ρ0 α nπ 2 βn = + λn = T0 ρ0 L Now, solving for λn , we get: λn = − T0 nπ 2 α + ρ0 ρ0 L The problem said to assume α < 0, and every other constant is positive, so therefore, λn > 0. Now it’s easy to solve the time-dependent ODE: p p T 00 + λn T = 0 ⇒ T (t) = An cos( λn t) + Bn sin( λn t) Taking the superposition: u(x, t) = ∞ X (An cos( n=1 nπ p p λn t) + Bn sin( λn t)) sin x L To find An , Bn , we use the initial position and velocity: u(x, 0) = 0 = ∞ X An sin n=1 nπ x L ⇒ An = 2 L Z L 0 · sin 0 nπ x dx = 0 L And differentiating, we get ut (x, t) = ∞ X Bn p λn cos( p λn t) sin n=1 nπ x L so that, with ut (x, 0) = f (x), we get the Bn ’s through the sine series for f : ∞ X nπ p f (x) = Bn λn sin x L n=1 ⇒ 2 Bn = √ λn L 4. What are the frequencies of vibration? SOLUTION: The nth (circular) frequency is s p λn = − 2 α T0 nπ 2 + ρ0 ρ0 L Z L f (x) sin 0 nπ x dx L

Math 367 Extra Practice Exercise 4.4.2

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