Question

Prove that if \(u(x)\) and \(v(x)\) satisfy the general homogeneous boundary itions $$ \begin{aligned} &\alpha_{1} u(a)+\beta_{1} u^{\prime}(a)=0 \\ &\alpha_{2} u(b)+\beta_{2} u^{\prime}(b)=0 \end{aligned} $$ \(=a\) and \(x=b\), then $$ p(x)\left[u(x) v^{\prime}(x)-v(x) u^{\prime}(x)\right]_{x=a}^{x=b}=0 $$

Short answer

The proof starts with the given conditions of \(u(a)\) and \(u(b)\) being equal to zero, and by substituting these into the stated equation the result is zero, proving the equation is correct.

Step-by-step solution

Understand the given

The given boundary conditions equate to 0 for both \(u(x)\) at \(x=a\) and \(x=b\), with coefficients \(\alpha_1, \beta_1\) and \(\alpha_2, \beta_2\) respectively. The function that needs to be proven is \(p(x)[u(x) v^{\prime}(x) - v(x) u^{\prime}(x)]\) evaluated from \(x=a\) to \(x=b\) equals 0.

Apply Boundary Conditions

Replace \(u(a)\) and \(u'(a)\) with the provided boundary conditions: \(u(a) = -\frac{\beta_1}{\alpha_1}u'(a)\) and \(u(b) = -\frac{\beta_2}{\alpha_2}u'(b)\). Similarly, derive the values for \(u'(a)\) and \(u'(b)\) as \(u'(a) = -\frac{\alpha_1}{\beta_1}u(a)\) and \(u'(b) = -\frac{\alpha_2}{\beta_2}u(b)\)

Substitute the Values in Given Equation

Now substitute the above derived values for \(u(a), u(b)\) and \(u'(a), u'(b)\) in the given equation. Then perform the evaluations for \(x=a\) and \(x=b\). Since both \(u(a)\) and \(u(b)\) are equal to zero from the given, we find that the result of the evaluated equation is also zero

Key concepts

General Solution of Differential Equations

In the realm of mathematics, a differential equation represents a relationship involving a function and its derivatives. When we talk about the general solution of a differential equation, we are referring to a family of solutions that includes every possible solution to the differential equation. This encompasses all particular, singular, and arbitrary constant involving solutions.

To elucidate, suppose we have a linear differential equation of the second order. Its general solution is usually expressed in terms of two arbitrary constants because the order of the equation indicates the number of initial conditions necessary to reach a unique solution. For instance, the general solution to the equation \( y'' + p(x)y' + q(x)y = 0 \) might be represented as \( y = C_1y_1(x) + C_2y_2(x) \), where \( y_1(x) \) and \( y_2(x) \) are two linearly independent solutions to the homogeneous equation, and \( C_1 \) and \( C_2 \) are constants that will be defined by initial or boundary conditions.

In essence, the general solution reveals the broadest form of the solution before applying specific circumstances or conditions which include, but are not constrained to, boundary conditions typically seen in a boundary value problem.

Mathematical Proof

A mathematical proof is a rigorous argument that shows the truth of a given statement beyond a doubt within the framework of mathematical logic. Proofs can utilize a variety of techniques, such as direct proofs (where the statement is shown to follow from others by a straight chain of logical reasoning), proof by contradiction (where the assumption that the statement is false leads to a contradiction), and proof by induction (where the validity of a statement is established for all natural numbers).

In the context of our exercise, proving that \( p(x)(u(x)v'(x) - v(x)u'(x)) \) evaluated from \( x=a \) to \( x=b \) equals zero, would require us to construct a convincing argument using assumptions like the supplied boundary conditions and properties of differential equations. The step-by-step solution found in the exercise demonstrates that by applying these boundary conditions, we can show that the expressions are indeed zero, thereby establishing the proof.

It is crucial in a mathematical proof to communicate effectively, ensuring every step is clear and logically follows from the previous one. This establishes the certainty required for conclusive proof, allowing others to follow the argument and agree on its validity.

Boundary Value Problem

The term boundary value problem (BVP) holds a special place in differential equations and refers to finding a solution to a differential equation that must satisfy a set of conditions specified at the boundaries. These problems are pivotal in many areas of physics and engineering as they accurately describe systems that are constrained by conditions at the extremities.

For example, in heat transfer, one might seek the temperature distribution in a bar, which is governed by the heat equation—a partial differential equation. Here, the value of temperature at the ends of the bar (the boundaries) is typically known, and these known temperatures are the boundary conditions that the solution must satisfy. As another example, the simple harmonic motion of a string fixed at both ends can be modeled by a wave equation with boundary conditions that the displacement is zero at the fixed points.

Homogeneous boundary conditions, which are a focal point of our exercise, mean that the boundary conditions are set to zero, like \( \alpha_1 u(a) + \beta_1 u'(a) = 0 \) and \( \alpha_2 u(b) + \beta_2 u'(b) = 0 \). These specific types of boundary conditions often make the problem easier to solve and are common in problems exhibiting symmetry or a lack of external forces at the boundaries.