Question

Let \(L\) be a second order linear differential operator. Show that the solution \(y=\phi(x)\) of the problem $$ \begin{array}{c}{L[y]=f(x)} \\ {a_{1} y(0)+a_{2} y^{\prime}(0)=\alpha, \quad b_{1} y(1)+b_{2} y^{\prime}(1)=\beta}\end{array} $$ can be written as \(y=u+v,\) where \(u=\phi_{1}(x)\) and \(v=\phi_{2}(x)\) are solutions of the problems $$ \begin{aligned} L[u]=0 & \\ a_{1} u(0)+a_{2} u^{\prime}(0)=\alpha, & b_{1} u(1)+b_{2} u^{\prime}(1)=\beta \end{aligned} $$ and $$ \begin{aligned} L_{L}[v] &=f(x) \\ a_{1} v(0)+a_{2} v^{\prime}(0) &=0, \quad b_{1} v(1)+b_{2} v^{\prime}(1)=0 \end{aligned} $$ respectively.

Short answer

Explain the superposition principle in the context of solving linear differential equations with boundary value problems. The superposition principle is a technique used to find the solution of a linear differential equation with boundary value problems by dividing the main problem into two separate sub-problems, one homogeneous and the other non-homogeneous, while maintaining the original boundary conditions. The principle states that the main solution can be found by adding the individual solutions of the two sub-problems. The first solution, \(u\), solves the homogeneous equation with the same boundary conditions as the main problem. The second solution, \(v\), solves the non-homogeneous equation, but with homogeneous boundary conditions. By summing the solutions \(u\) and \(v\), we obtain the solution \(y\) that satisfies the main problem's differential equation and boundary conditions.

Step-by-step solution

Identify the main problem

The main problem is a second order linear differential equation: $$L[y] = f(x),$$ where \(L\) is a second order linear differential operator, and \(y = \phi(x)\) is the unknown function we wish to find. In addition, the boundary conditions are given by: $$a_{1} y(0) + a_{2} y^{\prime}(0) = \alpha,$$ and $$b_{1} y(1) + b_{2} y^{\prime}(1) = \beta.$$

Define the first sub-problem

We want to find a function \(u = \phi_{1}(x)\) that satisfies the homogeneous equation: $$L[u] = 0,$$ with the same boundary conditions as the main problem: $$a_{1} u(0) + a_{2} u^{\prime}(0) = \alpha,$$ and $$b_{1} u(1) + b_{2} u^{\prime}(1) = \beta.$$

Define the second sub-problem

We also want to find a function \(v = \phi_{2}(x)\) that satisfies the non-homogeneous equation: $$L[v] = f(x),$$ but with homogeneous boundary conditions: $$a_{1} v(0) + a_{2} v^{\prime}(0) = 0,$$ and $$b_{1} v(1) + b_{2} v^{\prime}(1) = 0.$$

Show that the sum of the two solutions is a solution for the main problem

Now, we want to show that the sum of the two sub-problems' solutions, \(y = u + v\), is a solution for the main problem. First, notice that: $$L[y] = L[u+v] = L[u]+L[v].$$ Since we defined \(L[u]=0\) and \(L[v]=f(x)\), it is clear that \(L[y]=f(x)\). Hence, \(y\) is a solution of the main differential equation. Now let's check the boundary conditions: $$a_1 y(0) + a_2 y'(0) = a_1(u(0)+v(0)) + a_2(u'(0)+v'(0))$$ $$= (a_1u(0)+a_2u'(0))+(a_1v(0)+a_2v'(0)) = \alpha + 0 = \alpha.$$ And for the second boundary condition: $$b_1 y(1) + b_2 y'(1) = b_1(u(1)+v(1)) + b_2(u'(1)+v'(1))$$ $$= (b_1u(1)+b_2u'(1))+(b_1v(1)+b_2v'(1)) = \beta + 0 = \beta.$$ Thus, \(y=u+v\) satisfies both the main differential equation and the boundary conditions, and \(y=\phi(x)=u+v\) , where \(y=u+v\), \(u=\phi_{1}(x)\) and \(v=\phi_{2}(x)\) are the solutions for the given problems.

Key concepts

Boundary Value Problems

When solving differential equations, a Boundary Value Problem (BVP) is a scenario where we are given a differential equation along with certain conditions at fixed boundaries. In simpler terms, it’s like being told not only what a journey involves (the differential equation), but also where you start and finish (the boundary values). These problems are essential in physics and engineering, because they describe how things behave at the edges or limits.

For a second order linear differential equation, these boundary conditions are typically specified at two different points and are crucial because they allow us to find unique solutions. Remember that without these conditions, we may end up with an infinite number of possible solutions! To visualize, imagine you have a string that is pinned at two points (these are your boundary conditions). The way the string stretches in between those points can be described by a BVP, where the shape of the string is the solution to the equation.

The exercise provided is an example of a BVP, where the boundary conditions are given at the points x=0 and x=1. Solving this problem involves finding a function that not only satisfies the differential equation but also fulfills these given conditions at the boundaries.

Homogeneous Equations

Homogeneous equations are a special subset of differential equations. Imagine a perfectly balanced seesaw without any external forces — this represents a homogeneous equation where all terms are in balance, resulting in a zero right-hand side of the equation. In mathematical terms, if you have an equation in the form L[y] = 0, it’s homogeneous, because there’s nothing (0) on the other side of the equation.

Homogeneous equations have the delightful property that their solutions can be added together, and the result will still be a solution. This characteristic makes them foundational building blocks when constructing more complex solutions. In our exercise, once we find the solution u = \(\phi_{1}(x)\) that satisfies the homogeneous equation with the specified boundary conditions, we have figured out one part of the overall behavior of our system. Think of it as finding the undisturbed state of the system before any external influence is introduced.

Non-Homogeneous Equations

On the flip side, we have non-homogeneous equations. These are the equations that do have an external force or influence — in the seesaw analogy, this is like having an extra weight on one side. Mathematically, a non-homogeneous equation takes the form L[y] = f(x), where f(x) represents this external influence and is not equal to zero.

The goal here is not only to balance the seesaw, but to do so considering this added weight. In practice, this involves finding a particular solution that can accommodate the influence of f(x). The beauty of our problem lies in splitting the main problem into a homogeneous and a non-homogeneous part, solving them separately, and combining their solutions. By finding a function v = \(\phi_{2}(x)\) that satisfies the non-homogeneous but simplifies the boundary conditions to be homogeneous, we grasp the impact of the external force while keeping the constraints simple. Combining u and v gives us the complete picture, as they jointly respect the original boundary conditions and account for the external forces described in the original equation.