Question

Show that the \(n\) th prolongation of \(\mathbf{v}=X(x, u) \partial_{x}+U(x, u) \partial_{u}\) for an ordinary DE of \(n\) th order is $$\mathrm{pr}^{(n)} \mathbf{v}=\mathbf{v}+\sum_{k=1}^{n} U^{[k]} \frac{\partial}{\partial u^{(k)}},$$ where $$u^{(k)} \equiv \frac{\partial^{k} u}{\partial x^{k}} \quad \text { and } \quad U^{[k]}=D_{x}^{k}\left(U-X u_{x}\right)+X u^{(k+1)}$$

Short answer

The \( n \)th prolongation of \( \mathbf{v} \) can be proven using the chain rule for differentiation in multiple variables, and the concepts of the total derivative and the series summation. The proof requires careful manipulation of the total derivative and summing up over the series.

Step-by-step solution

Understanding the formula

The given formula for the \( n \)th prolongation of \( \mathbf{v} \) can be rewritten as \[ \mathrm{pr}^{(n)} \mathbf{v}=\mathbf{v}+\sum_{k=1}^{n} U^{[k]} \frac{\partial}{\partial u^{(k)}} \]. This formula is suggesting a summation operation over a series of expressions involving the variable \( k \) from 1 to \( n \). Each term in the summation involves the hyper function \( U^{[k]} \) and a derivative.

Understanding \( u^{(k)} \) and \( U^{[k]} \)

Here \( u^{(k)} \) is the \( k \)th derivative of \( u \) with respect to \( x \). Essentially, it represents the rate of change of the function \( u \) after \( k \) derivations. \( U^{[k]} \) is a special notation used to denote the function \( D_{x}^{k}\left(U-X u_{x}\right)+X u^{(k+1)} \). As you can see, \( U^{[k]} \) is defined in terms of the derivative of \( u \), and the difference between the functions \( U \) and \( X \times u_x \), all taken to the \( k \)th power. Plus, the multiplication of \( X \) times the \( (k+1) \)th derivative of \( u \).

Proving the formula

The proof will heavily utilize the concept of total derivative. In general, the total derivative of a function can be expanded using the chain rule. Here, we will need to differentiate \( U^{[k]} \), keeping in mind that it is a function of \( x \) and \( u \). When we carry out this differentiation process over the series (from \( k = 1 \) to \( n \)), and then add the vector \( \mathbf{v} \), we will be able to show that the result matches the provided formula.

Key concepts

Prolongation

Prolongation in the context of differential equations refers to the process of extending or continuing the vector field to higher-order derivatives of the dependent variable. It provides a way to analyze complex ordinary differential equations (ODEs) beyond just first-order terms.
A vector field is often represented with operators acting on a space of functions. In our case,

• \( \mathbf{v} = X(x, u) \partial_{x} + U(x, u) \partial_{u} \)
• This indicates \( \mathbf{v} \) is acting as a directional derivative on the space of functions of \( x \) and \( u \).

The \( n \) th prolongation, \( \mathrm{pr}^{(n)} \mathbf{v} \), includes all necessary higher-order derivative components to fully describe the effects of \( \mathbf{v} \) on derivatives up to \( n \) th order. Extending a vector field through prolongation helps in solving differential equations by encompassing further derivative terms needed for complete transformation.

Derivatives

Derivatives are fundamental to calculus and differential equations. They describe how a function changes as its input variables change. Here, we focus on two important aspects: derivatives as rates of change and higher-order derivatives (successive derivatives).
Consider a function \( u(x) \). The ordinary derivative \( u_x \) denotes the instantaneous rate of change of \( u \) with respect to \( x \). Higher-order derivatives, such as \( u^{(k)} \), represent the function's nth derivative concerning \( x \). This detailed analysis is crucial for differential equations, especially when understanding prolongation, because:

• It enables us to examine not just the immediate behavior of \( u \), but also how it changes across multiple dimensions - in terms of rates and acceleration.
• The derivatives describe not only linear behavior but also how curvature and higher-order effects influence solutions.

Chain Rule

The chain rule is a powerful tool in calculus used to differentiate composite functions. It allows us to find the derivative of functions that are composed of other functions, which is crucial when dealing with implicit differentiation.
In the formula provided for the nth prolongation,

• We often need to differentiate terms that involve several variables, such as \( U^{[k]}=D_{x}^{k}(U-X u_{x})+X u^{(k+1)} \).
• The chain rule helps us manage these computations effectively.
  For example, if you have a composite function \( h(x) = f(g(x)) \), the chain rule states that \( h'(x) = f'(g(x)) \cdot g'(x) \).

By applying the chain rule, we can simplify differentiation of complex expressions encountered in the prolongation process. It ensures every component of a vector field's effect is accounted for, enhancing the ability to solve higher-order ODEs efficiently.