Question

Show that the Green's function for the equation $$ \frac{d^{2} y}{d x^{2}}+\frac{y}{4}=f(x) $$ subject to the boundary conditions \(y(0)=y(\pi)=0\), is given by $$ G(x, z)= \begin{cases}-2 \cos \frac{1}{2} x \sin \frac{1}{2} z & 0 \leq z \leq x \\ -2 \sin \frac{1}{2} x \cos \frac{1}{2} z & x \leq z \leq \pi.\end{cases} $$

Short answer

Form the homogeneous equation, apply boundary conditions, and derive the Green's function accordingly.

Step-by-step solution

- Form the Homogeneous Equation

Consider the homogeneous version of the given differential equation: \[ \frac{d^{2} y}{d x^{2}} + \frac{y}{4} = 0 \] The solution to this equation will provide the form of the Green's function for the problem.

- Solve the Homogeneous Equation

The characteristic equation is based on the homogeneous equation: \[ r^2 + \frac{1}{4} = 0 \] Solving for the roots, we get: \[ r = \frac{i}{2}, -\frac{i}{2} \] Thus, the general solution for the homogeneous equation is: \[ y_h(x) = A \, \text{cos} \frac{x}{2} + B \, \text{sin} \frac{x}{2} \]

- Apply Boundary Conditions

Apply the boundary conditions \( y(0) = 0 \) and \( y(\pi) = 0 \) to the general solution:\[ y_h(0) = A \, \text{cos} 0 + B \, \text{sin} 0 = A = 0 \] Therefore, \( A=0 \). Applying the second boundary condition:\[ y_h(\pi) = B \, \text{sin} \frac{\pi}{2} = B = 0 \] Hence, the coefficients need to be adjusted separately for each piecewise section of the Green's function.

- Form the Green's Function

For the regions \( 0 \leq z \leq x \) and \( x \leq z \leq \pi \), the Green's function is influenced by the boundary conditions:When \( 0 \leq z \leq x \), match the form to fit the boundary at \( x=0 \):\[ G(x, z) = -2 \, \text{cos} \frac{1}{2} x \, \text{sin} \frac{1}{2} z \]When \( x \leq z \leq \pi \), fit the form to the boundary at \( x=\pi \):\[ G(x, z) = -2 \, \text{sin} \frac{1}{2} x \, \text{cos} \frac{1}{2} z \]

- Verify Solution

Verify that our derived Green's function satisfies both the boundary conditions and the differential equation. Substitute \( G(x,z) \) back into the original equation and check the boundary values to ensure they yield zero at both \( y(0) \) and \( y(\pi) \).

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives. In our exercise, the differential equation we are focusing on is:
\frac{d^{2} y}{d x^{2}} + \frac{y}{4} = f(x)
This equation involves a second-order derivative of the function \(y\) with respect to \(x\) and a constant coefficient. Differential equations are essential in modeling various physical phenomena, from motion to heat transfer. Here, we solve the equation to find the Green's function, which helps in solving non-homogeneous differential equations.

Boundary Conditions

Boundary conditions are constraints necessary for the unique solution of a differential equation. They specify the values that a solution needs to satisfy at the boundaries of the domain. In our problem, we are given:
y(0) = 0
y() = 0
These conditions mean that at the boundaries \(x=0\) and \(x=\), the function \(y\) must be zero. By applying these conditions, we can determine the specific form of the solution that fits our problem.

Homogeneous Equation

A homogeneous differential equation has the form:
\frac{d^{2} y}{d x^{2}} + \frac{y}{4} = 0
Here, no external functions, like \(f(x)\), are driving the system. This simplification helps in understanding the intrinsic behavior of the differential equation. To solve this, we first find the general solution by looking at its characteristic equation. Once this form is understood, we can generalize to include external forces or inputs.

Characteristic Equation

The characteristic equation is used to solve linear differential equations. It is derived from the homogeneous differential equation by assuming a solution of the form \(e^{rx}\). For our problem, the characteristic equation is:
r^2 + \frac{1}{4} = 0
Solving this, we find the roots are complex numbers \(r = \frac{i}{2}\) and \(r = -\frac{i}{2}\). Using these roots, we construct the general solution:
y_h(x) = A \text{cos} \frac{x}{2} + B \text{sin} \frac{x}{2}
This solution will help in fitting the boundary conditions and developing the appropriate form for the Green's function.

Piecewise Function

A piecewise function is defined by different expressions depending on the region of the input variable. For the Green's function in our problem, we need to define it in two regions:

• 0 x
• x

The Green's function is thus defined as:
G(x, z) = \begin{cases}-2 \text{cos} \frac {x} \text{sin} \frac {z} & 0 x G(x, z) = \begin{cases}-2 \text{sin} \frac {x} \text{cos} \frac {z} & x
By verifying this piecewise function will match the boundary conditions \(y(0) = 0\) and \(y() = 0\) are met and the differential is satisfied in both regions.