Question

Solving a difference equation. (a) Determine the general solution to the linear difference equation 2Uⁿ⁺³−5Uⁿ⁺²+4Uⁿ⁺¹− Un=0. Hint: One root of the characteristic polynomial is at ζ=1. (b) Determine the solution to this difference equation with the starting values U⁰=11, U¹=5, and U²=1. What is U¹⁰ ? (c) Consider the LMM 2Uⁿ⁺³−5Uⁿ⁺²+4Uⁿ⁺¹−Uⁿ=k(β0​f(Uⁿ)+β1​f(Uⁿ⁺¹)). For what values of β0₀ and β₁​ is local truncation error O(k²) ? (d) Suppose you use the values of β₀ and β₁​ just determined in this LMM. Is this a convergent method?

Answer 1

To solve the linear difference equation, we first determine the characteristic polynomial. Then, using the initial values, we solve the system of equations to find the constants in the general solution. Finally, we substitute the desired value of n into the general solution to find the solution for that value of n.

Step-by-step explanation:

To solve the linear difference equation 2Uⁿ⁺³−5Uⁿ⁺²+4Uⁿ⁺¹− Un=0, we will use the initial values U⁰=11, U¹=5, and U²=1:

1. First, determine the characteristic polynomial by substituting Uⁿ=ζⁿ into the difference equation and simplifying. We get the characteristic polynomial p(ζ) = 2ζ³ - 5ζ² + 4ζ - 1.
2. Using the hint given, we know that one root of the characteristic polynomial is ζ=1. Therefore, the general solution to the difference equation is Uⁿ = c₁ × 1ⁿ + c₂ × r₁ⁿ + c₃ × r₂ⁿ, where r₁ and r₂ are the other two roots of the characteristic polynomial.
3. Plugging in the initial values, we have the system of equations: 11 = c₁ + c₂ + c₃, 5 = c₁ + c₂ × r₁ + c₃ × r₂, and 1 = c₁ + c₂ × r₁² + c₃ × r₂². Solve this system of equations to find the values of c₁, c₂, and c₃.
4. Finally, substitute n=10 into the general solution to find U¹⁰.