Question

In Exercises 26-30 find necessary and sufficient conditions on \(\alpha, \beta, \rho,\) and \(\delta\) for the boundary value problem to have a unique solution for every continuous \(F,\) and find the Green's function. $$ y^{\prime \prime}-2 y^{\prime}+2 y=F(x), \quad \alpha y(0)+\beta y^{\prime}(0)=0, \quad \rho y(\pi)+\delta y^{\prime}(\pi)=0 $$

Short answer

Based on the given second-order linear differential equation and non-homogeneous boundary conditions, we found that the necessary and sufficient conditions on the coefficients \(\alpha\), \(\beta\), \(\rho\), and \(\delta\) for the boundary value problem to have a unique solution for every continuous function \(F(x)\) are \(\alpha^2 + \beta^2 > 0\) and \(\rho^2 + \delta^2 > 0\). The Green's function for this problem is given by: $$ G(x, t) = \frac{(\delta - \rho\pi)(y_1(x)y_2(t) - y_2(x)y_1(t))}{\alpha y_1(\pi)y_2(0) - \alpha y_2(\pi)y_1(0)}, $$ where \(y_1(x) = e^x \cos(x)\) and \(y_2(x) = e^x \sin(x)\).

Step-by-step solution

Find the homogeneous solutions

We start by finding the homogeneous solutions of the given differential equation. The equation is: $$ y^{\prime \prime}-2 y^{\prime}+2 y=0 $$ To find the homogeneous solutions, we assume a solution of the form \(y(x)=e^{rx}\) and substitute it into the equation to obtain: $$ (r^2-2 r+2)e^{rx}=0 $$ This gives the characteristic equation as follows: $$ r^2-2r+2=0 $$ Solving this quadratic equation for r, we get: $$ r_1=1+i, \quad r_2=1-i $$ Hence, the homogeneous solutions are: $$ y_1(x)=e^{x}\cos(x), \quad y_2(x)=e^{x}\sin(x) $$

Apply boundary conditions

Now we apply the given boundary conditions to the general solution: $$ y(x) = C_1 y_1(x) + C_2 y_2(x) $$ The boundary conditions are: $$ \alpha y(0)+\beta y^{\prime}(0)=0 \\ \rho y(\pi)+\delta y^{\prime}(\pi)=0 $$ Applying the first boundary condition: $$ \alpha (C_1\cdot1+C_2\cdot0)+\beta (C_1\cdot1+C_2\cdot1)=0 $$ This simplifies to: $$ \alpha C_1+\beta(C_1+C_2)=0 $$ Applying the second boundary condition: $$ \rho (C_1e^{\pi}\cos(\pi)+C_2e^{\pi}\sin(\pi))+\delta (C_1e^{\pi}\cos(\pi)-C_1e^{\pi}\sin(\pi)+C_2e^{\pi}\cos(\pi)+C_2e^{\pi}\sin(\pi))=0 $$ This simplifies to: $$ (C_1+C_2)(e^{\pi}(\rho\cos(\pi) + \delta(\cos(\pi)-\sin(\pi)))=0 $$

Find necessary and sufficient conditions

To have a unique solution for every continuous \(F,\) we require: $$ \begin{cases} \alpha C_1+\beta(C_1+C_2)=0 \\ (C_1+C_2)(e^{\pi}(\rho\cos(\pi) + \delta(\cos(\pi)-\sin(\pi)))=0 \end{cases} $$ These two equations must hold true for nontrivial solutions. The necessary and sufficient conditions for that are: $$ \alpha^2 + \beta^2 > 0 \\ \rho^2 + \delta^2 > 0 $$

Find Green's function

The Green's function \(G(x, t)\) satisfies the following differential equation and boundary conditions: $$ \frac{\partial^2 G}{\partial x^2} - 2 \frac{\partial G}{\partial x} + 2 G = \delta(x-t) \\ \alpha G(0, t) + \beta \frac{\partial G(0, t)}{\partial x} = 0 \\ \rho G(\pi, t) + \delta \frac{\partial G(\pi, t)}{\partial x} = 0 $$ The Green's function has the form: $$ G(x, t) = C_1(t) y_1(x) + C_2(t) y_2(x) $$ Applying the boundary conditions on the Green's function and solving for \(C_1(t)\) and \(C_2(t)\), we obtain: $$ C_1(t) = \frac{(\delta - \rho\pi)y_2(t)}{\alpha y_1(\pi)y_2(0) - \alpha y_2(\pi)y_1(0)} \\ C_2(t) = -\frac{(\delta - \rho\pi)y_1(t)}{\alpha y_1(\pi)y_2(0) - \alpha y_2(\pi)y_1(0)} $$ Thus, the Green's function is given by: $$ G(x, t) = \frac{(\delta - \rho\pi)(y_1(x)y_2(t) - y_2(x)y_1(t))}{\alpha y_1(\pi)y_2(0) - \alpha y_2(\pi)y_1(0)} $$

Key concepts

Homogeneous Solutions of Differential Equations

When we deal with differential equations, a fundamental aspect is to understand homogeneous solutions. These are solutions to the equation when it is set equal to zero, i.e., the right-hand side is absent. For the differential equation
\[ y'' - 2y' + 2y = 0 \]
we look for solutions that satisfy this equation without the forcing function, which in this case is represented by the general term \(F(x)\). By assuming a solution of the form \(y(x) = e^{rx}\), we transform the problem into an algebraic equation called the characteristic equation.
\[ r^2 - 2r + 2 = 0 \]
By solving the characteristic equation, we get values for \(r\) that give us the basis functions for the general homogeneous solution. In our case, the roots are complex, \(1+i\) and \(1-i\), leading to solutions involving both exponential and trigonometric functions due to Euler's formula, resulting in:
\[ y_1(x) = e^{x}\cos(x), \quad y_2(x) = e^{x}\sin(x) \]
These functions form the complementary solution to the non-homogeneous differential equation, providing the foundation upon which we build the general solution by adding a particular solution that depends on \(F(x)\).

Characteristic Equation

The characteristic equation plays a pivotal role in solving homogeneous linear differential equations. It is derived by substituting a trial solution, usually of the form \(y(x)=e^{rx}\), into the homogeneous differential equation and simplifying it to an algebraic equation.
In our exercise, the substitution resulted in the characteristic equation:
\[ r^2 - 2r + 2 = 0 \]
This equation is crucial because its roots determine the nature of the homogeneous solutions. Real roots lead to exponential functions, while complex roots, as in our case, lead to a combination of exponential and trigonometric functions. The roots \(1+i\) and \(1-i\) produce the fundamental set of solutions for our second-order differential equation, allowing us to express the homogeneous solution in terms of exponential and sine or cosine functions.

Boundary Conditions

Boundary conditions are crucial for uniquely determining the solution of a differential equation within a given domain. In the context of our boundary value problem, they are specified at the endpoints of the interval, \(x=0\) and \(x=\pi\):
\[ \alpha y(0) + \beta y'(0) = 0, \quad \rho y(\pi) + \delta y'(\pi) = 0 \]
Applying these conditions to our general solution aids in finding constants \(C_1\) and \(C_2\), thereby specifying a unique solution fitting our problem. The coefficients \(\alpha, \beta, \rho,\) and \(\delta\) control the type and strength of the boundary conditions, which can be Dirichlet (values of the function), Neumann (values of the function's derivative), or a mix. In this case, the boundary conditions involve both the functions and their derivatives, making it a mixed or Robin boundary condition. These conditions must be non-trivial to ensure we have a unique solution, which implies that neither pair of \(\alpha, \beta\) nor \(\rho, \delta\) can be simultaneously zero.

Unique Solution Conditions

For the existence of a unique solution for the boundary value problem under any continuous function \(F\), certain conditions must be met. These conditions stem from the theory behind linear differential equations and ensure that our boundary value problem won't have multiple or no solutions.
Specifically, the coefficients in the boundary conditions must satisfy:
\[ \alpha^2 + \beta^2 > 0, \quad \rho^2 + \delta^2 > 0 \]
These inequalities ensure that both boundary conditions contribute meaningfully to the solution. If they do not hold—say if \(\alpha = \beta = 0\) or \(\rho = \delta = 0\)—we would get trivial boundary conditions that would not impose any restriction on \(C_1\) or \(C_2\). Thus, these conditions on \(\alpha, \beta, \rho,\) and \(\delta\) are crucial for narrowing down the possibilities to a single, unique solution that fits within the constraints of the given differential equation and boundary values.

Differential Equations

In the realm of mathematics, differential equations are powerful tools for modeling dynamic systems and phenomena. The equation provided in our exercise,
\[ y'' - 2y' + 2y = F(x) \]
is a second-order linear non-homogeneous differential equation. Such equations are integral in a multitude of fields, including physics, engineering, and economics. The solutions to these equations describe the behavior of systems under various forces or influences. In essence, the differential equation relates a function with its derivatives, representing rates of change. The non-homogeneous term, \(F(x)\), acts as an external influence or forcing function that drives the system. Solving the equation involves finding a function \(y(x)\) that satisfies the relation for any given \(F(x)\). Our steps to solve it involve finding the homogeneous solution, applying boundary conditions, ensuring uniqueness conditions are met, and if necessary, finding a particular solution or constructing a Green's function to account for the non-homogeneous term.