Question

State a condition on \(F\) such that the boundary value problem $$ y^{\prime \prime}=F(x), \quad y(0)=0, \quad y(1)-y^{\prime}(1)=0 $$ has a solution, and find all solutions.

Short answer

In conclusion, a solution exists for the given second-order linear ordinary differential equation with a source term \(F(x)\) and the given boundary conditions if the derivatives of \(F(x)\) exist continuously. The general solution is given by: $$ y(x) = F_2(x) + (F_2(0) - F_2(1) + F_1(1))x - F_2(0), $$ where \(F_1(x)\) and \(F_2(x)\) are the continuous antiderivatives of \(F(x)\).

Step-by-step solution

Solve the ODE for a general \(F(x)\)#

To solve the given ODE, we will use the following general form for a second-order linear ODE with a source term \(F(x)\): $$ y^{\prime \prime} = F(x) $$ Let's denote an antiderivative of \(F(x)\) by \(F_1(x)\), and let's integrate the ODE once to find \(y'(x)\): $$ y^{\prime}(x) = \int F(x) dx = F_1(x) + C_1 $$ Now, let's integrate it again to find \(y(x)\): $$ y(x) = \int (F_1(x) + C_1) dx = F_2(x) + C_1x + C_2 $$ where \(F_1(x)\) and \(F_2(x)\) are the antiderivatives of \(F(x)\), and \(C_1\) and \(C_2\) are constants of integration.

Apply the first boundary condition \(y(0)=0\)#

Now we have the general solution \(y(x) = F_2(x) + C_1x + C_2\). We apply the first boundary condition \(y(0) = 0\): $$ 0 = F_2(0) + C_1 \cdot 0 + C_2 $$ We can see that \(C_2 = -F_2(0)\). So, the solution becomes: $$ y(x) = F_2(x) + C_1x - F_2(0) $$

Apply the second boundary condition \(y(1)-y'(1)=0\)#

Now we apply the second boundary condition \(y(1) - y'(1) = 0\). First, we find the expression for \(y'(1)\) using the relation we derived in Step 1, \(y^{\prime}(x) = F_1(x) + C_1\): $$ y^\prime(1) = F_1(1) + C_1 $$ Next, we apply the boundary condition: $$ y(1) - y'(1) = (F_2(1) + C_1 - F_2(0)) - (F_1(1) + C_1) = 0 $$ Upon solving the equation above, we find that $$ C_1 = F_2(0) - F_2(1) + F_1(1) $$ Now, we have found the specific values of the constants of integration, \(C_1\) and \(C_2\). Thus, the solution for \(y(x)\), which satisfies the boundary conditions, is: $$ y(x) = F_2(x) + (F_2(0) - F_2(1) + F_1(1))x - F_2(0) $$

State the condition on \(F(x)\) for the existence of a solution#

Based on the above analysis, a solution exists for any given function \(F(x)\) as long as its derivatives exist continuously. The general solution is $$ y(x) = F_2(x) + (F_2(0) - F_2(1) + F_1(1))x - F_2(0), $$ with the condition that \(F_1(x)\) and \(F_2(x)\) are the continuous antiderivatives of \(F(x)\).

Key concepts

Second-Order Linear ODE

A second-order linear ordinary differential equation (ODE) is a key concept in mathematical analysis and forms the foundation of many physical laws and phenomena. In its general form, it is represented as \( y'' = f(x, y, y') \), where \( y'' \) denotes the second derivative of the unknown function \( y(x) \) with respect to the independent variable \( x \). For a differential equation to be linear, it should not contain any products or nonlinear functions of \( y \) or \( y' \).

The equation in our task, \( y'' = F(x) \) is a specific case where the equation is linear and the function \( F(x) \) acts as a source term which can be any function of \( x \) alone. These types of second-order linear ODEs are incredibly important as they can describe a wide range of physical systems such as oscillations, waves, and dynamic systems.

To solve such equations, one typically seeks two integrations to go from the second derivative to the original function, applying relevant boundary conditions to determine the constants of integration.

Antiderivative

The antiderivative, commonly known as the indefinite integral, is fundamentally associated with the concept of integration in calculus. When presented with a function \( F(x) \), its antiderivative is a function \( G(x) \) such that \( G'(x) = F(x) \). In the context of the given problem, finding a solution to the second-order ODE involves the process of integration to determine \( y(x) \) first from \( y'' \) and then from \( y' \) to \( y \) itself.

The process requires integrating the source term \( F(x) \) twice, leading to the general solution form involving two antiderivatives, denoted \( F_1(x) \) and \( F_2(x) \) for each layer of integration. These antiderivatives embody the accumulated effect of the source term \( F(x) \) on the solution \( y(x) \) and must be determined to proceed with the solution.

Boundary Conditions

Boundary conditions are constraints that are necessary to uniquely determine the solution to a differential equation. They are typically given as values or relationships that the solution must satisfy at specific points, which, in many real-world applications, may represent the state of a system at the boundaries of its domain.

In our exercise, we have two boundary conditions. The first, \( y(0) = 0 \), specifies the value of the unknown function \( y \) at the point \( x=0 \). The second, \( y(1) - y'(1) = 0 \) provides a relationship between the value of the function and its first derivative at \( x=1 \). These conditions are key to determining the constants of integration, \( C_1 \) and \( C_2 \), resulting in a complete and unique solution to the boundary value problem.

Integration Constants

Integration constants are introduced during the process of solving differential equations when finding antiderivatives. Because there are an infinite number of antiderivatives for a given function, each differing by a constant, these constants of integration, \( C_1 \) and \( C_2 \) in our case, serve to represent that family of solutions.

Once boundary conditions are applied, the values of these constants can be determined, providing a specific solution fitting the parameters of the given problem. As shown in our exercise, the integration constants are found by substituting the boundary conditions into the general solution, which ultimately leads to a particular solution that satisfies both the differential equation and the boundary conditions.