Question

Solve the difference equation \(\Delta^{3} \varphi=0\) with the initial condition \(\varphi(0)=\) \(0,(\Delta \varphi)(0)=0,\left(\Delta^{2} \varphi\right)(0)=2\) for \(t\) a multiple of the step size \(h=1\)

Short answer

Question: Determine the function \(\varphi(t)\) for integer values of \(t\) that satisfies the third-order difference equation \(\Delta^{3} \varphi=0\) with the initial conditions: \(\varphi(0) = 0\), \((\Delta \varphi)(0) = 0\), and \((\Delta^2 \varphi)(0) = 2\). Answer: The function \(\varphi(t)\) is given by \(\varphi(t) = -e^t + e^{\frac{2\pi i}{3}t}\) for integer values of \(t\).

Step-by-step solution

Finding the general form of the solution for the third-order difference equation

We have the difference equation \(\Delta^{3} \varphi=0\). Let us denote the difference operator, i.e., \(\Delta\) by \(D\). Then, we can rewrite the equation as \(D^3 \varphi = 0\). The general form of the solution will consist of all possible linearly independent solutions. We assume a solution of the form \(\varphi(t) = e^{\lambda t}\). Substituting this into the difference equation, we get: \(D^3 e^{\lambda t} = e^{\lambda (t+3)} - 3e^{\lambda (t+2)} + 3e^{\lambda (t+1)} - e^{\lambda t} = 0\) Dividing both sides by \(e^{\lambda t}\), we get the characteristic equation: \(\lambda^3 = 1\) We solve the cubic equation for \(\lambda\). We know one root must be \(\lambda_1 = 1\). Now, we have to find the other roots, which are given by \(\lambda = e^{\frac{2\pi i}{3}}\) and \(\lambda = e^{\frac{4\pi i}{3}}\). Let us denote them as \(\lambda_2, \lambda_3\). As we have found three linearly independent solutions, the general form of the solution for \(\varphi(t)\) can be written as: \(\varphi(t) = A e^{\lambda_1 t} + B e^{\lambda_2 t} + C e^{\lambda_3 t}\)

Applying the initial conditions to find the coefficients

We apply the initial conditions to find the coefficients, i.e., \(A, B, C\): 1. \(\varphi(0) = 0\). 2. \((\Delta \varphi)(0) = 0\). 3. \((\Delta^2 \varphi)(0) = 2\). Applying the first initial condition: \(\varphi(0) = 0\): \(0 = A e^{0} + B e^{0} + C e^{0} \Rightarrow A+B+C = 0\) Applying the second initial condition: \((\Delta \varphi)(0) = 0\): \(0 = A(e^{\lambda_1}-1) + B(e^{\lambda_2}-1) + C(e^{\lambda_3}-1)\) Applying the third initial condition: \((\Delta^2 \varphi)(0) = 2\): \(2 = A(e^{2\lambda_1}-1) + B(e^{2\lambda_2}-1) + C(e^{2\lambda_3}-1)\) Now, we have a system of three linear equations with three unknowns. Solving this system, we get \(A=-1, B=1, C = 0\).

Writing down the final solution

Using the values for \(A, B, C\), we write down the final solution for \(\varphi(t)\): \(\varphi(t) = -e^{\lambda_1 t} + e^{\lambda_2 t}\) Hence, \(\varphi(t) = -e^t + e^{\frac{2\pi i}{3}t}\) for \(t\) being a multiple of the step size \(h=1\).

Key concepts

Initial Conditions

Initial conditions are vital in solving difference equations because they allow us to determine specific solutions from a general form. In our problem, we have three initial conditions: \( \varphi(0)=0 \), \( (\Delta \varphi)(0)=0 \), and \( (\Delta^2 \varphi)(0)=2 \). Each condition gives us information about the state of the function \( \varphi(t) \) at the initial time \( t=0 \).

• The first condition, \( \varphi(0) = 0 \), tells us the value of the function at time zero.
• The second condition, \( (\Delta \varphi)(0) = 0 \), involves the first difference, indicating a zero rate of change at that point.
• The third, \( (\Delta^2 \varphi)(0) = 2 \), involves the second difference, providing information about the acceleration or curvature at the initial point.

By applying these conditions, we can solve for constants in the general solution, giving us the specific function that satisfies the problem.

Characteristic Equation

The characteristic equation is a powerful tool for solving difference equations and helps find the form of solutions. It's derived from assuming the solution in an exponential form, like \( \varphi(t) = e^{\lambda t} \). Substituting this form into the difference equation allows us to find a relationship between lambda values.

For the equation \( \Delta^3 \varphi = 0 \), substituting leads us to the equation \( \lambda^3 = 1 \). This is the characteristic equation, and solving it reveals that the possible solutions are functions like \( e^{\lambda_1 t}, e^{\lambda_2 t}, e^{\lambda_3 t} \). These exponentials form the basis for constructing the general solution of the equation.

Linear Independence

In solving difference equations, the concept of linear independence ensures that our solutions are distinct and non-redundant. When we find linearly independent solutions, like \( e^{\lambda_1 t}, e^{\lambda_2 t}, e^{\lambda_3 t} \), they form a basis for the space of solutions.

• Each exponential term must not be expressible as a linear combination of the others.
• This ensures that every coefficient we solve for (\( A, B, C \)) among these terms contributes uniquely to the solution.

The condition of linear independence guarantees that the general solution spans all possible behaviors described by the differential equation, allowing us to cater it to specific initial conditions.

Cubic Equation

A cubic equation involves finding roots of a polynomial of degree three, as seen in our characteristic equation, \( \lambda^3 = 1 \). Here's how to handle it:

• Cubic equations always have three roots, which can be real or complex.
• In this problem, one known root is \( \lambda_1 = 1 \), while the other roots are found from complex solutions \( \lambda = e^{\frac{2\pi i}{3}} \) and \( \lambda = e^{\frac{4\pi i}{3}} \).
• These roots reflect symmetry and periodicity, characteristic of the cube root of unity.

Understanding cubic equations helps identify the underlying behavior patterns in the difference equation solution, revealing much about the system's dynamics.