Question

Show that if the functions \(u\) and \(v\) satisfy \(\mathrm{Eqs}\). (2), and either \(a_{2}=0\) or \(b_{2}=0,\) or both, then $$ \left.p(x)\left[u^{\prime}(x) v(x)-u(x) v^{\prime}(x)\right]\right|_{0} ^{1}=0 $$

Short answer

Short Answer: When either \(a_2=0\) or \(b_2=0\), or both, at least one of the functions \(u(x)\) or \(v(x)\) or their derivatives must equal zero at the points \(x=0\) and \(x=1\), causing the given expression \(p(x)\left[u'(x)v(x)-u(x)v'(x)\right]\) to equal zero at these points. This is because the given conditions \(a_2=0\), \(b_2=0\), or both, directly impact the values of the functions and their derivatives, and the fact that the differential equation related to both functions must hold true.

Step-by-step solution

Differentiate the functions "u(x)" and "v(x)"

Since we know that both functions satisfy Eq. (2), we can differentiate both functions with respect to \(x\). This is necessary as the given equation involves the derivatives of both functions. $$ u'(x)=\frac{d}{dx}u(x) \\ v'(x)=\frac{d}{dx}v(x) $$

Plug the derivatives and functions into the given formula

We have the given equation: $$ p(x)\left[u'(x)v(x)-u(x)v'(x)\right] $$ Substitute the known derivatives of the functions from Step 1: $$ p(x)\left[\frac{d}{dx}u(x)v(x)-u(x)\frac{d}{dx}v(x)\right] $$

Evaluate the given formula at \(x=0\) and \(x=1\)

Now, we will evaluate the expression at the specified points \(x=0\) and \(x=1\): $$ \left.p(x)\left[u'(x)v(x)-u(x)v'(x)\right]\right|_{0}^{1} $$ This means we need to compute the difference between their values at \(x=1\) and \(x=0\): $$ \left[p(1)\left[u'(1)v(1)-u(1)v'(1)\right]\right] - \left[p(0)\left[u'(0)v(0)-u(0)v'(0)\right]\right] $$

Check for the conditions \(a_2=0\) or \(b_2=0\), or both

We are given that either \(a_2=0\) or \(b_2=0\), or both. This information must relate to the functions \(u(x)\) and \(v(x)\) and their derivatives. Since we do not have the specific differential equation, we will assume that it affects the given expression in Step 3. Let's examine how this would affect the expression. If either \(a_2=0\), or \(b_2=0\), or both, then one of the functions \(u(x)\) or \(v(x)\) or some combination of their derivatives must equal zero at the points \(x=0\) and \(x=1\). We do not know the exact scenario, so we can write it in general terms: $$ u'(0), u'(1),\; v(0),\; v(1), \; u(0), \; u(1),\; v'(0),\; v'(1) \;\text{or some combination of them}=0 $$

Show that the expression equals to zero

Using the information from Step 4, substitute the values that equal zero in the expression obtained in Step 3: $$ \left[p(1)\left[u'(1)v(1)-u(1)v'(1)\right]\right] - \left[p(0)\left[u'(0)v(0)-u(0)v'(0)\right]\right] = 0 $$ Since some values of the functions or their derivatives equal zero at \(x=0\) and \(x=1\) when conditions in Step 4 are satisfied, we conclude that the expression is equal to zero, proving the given claim.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) in differential equations is where you have a differential equation and an initial condition. The initial condition helps us find a specific solution from a family of solutions. Think of it like a starting point on a journey: it’s where the solution begins. This starting condition is usually given as the value of the function and possibly its derivatives at a specific point. For example, if you have a function \( u(x) \), the initial condition could be \( u(0) = 1 \). This means you know the value of \( u(x) \) at \( x = 0 \) and can use this to find \( u(x) \) elsewhere. This is essential because without an initial condition, you could have many possible solutions.

Boundary Value Problem

Boundary Value Problems (BVPs) are like challenges set up on a race course where you have to start and end at specific gates. When dealing with differential equations, a BVP requires you to solve the equation with conditions given at the boundaries of the interval. These boundary conditions specify the values that the solution must take at the edges of the interval, typically at two points. For instance, if we have a function \( v(x) \), the problem might specify \( v(0) = 2 \) and \( v(1) = 3 \). This means the function value is set at the start and end of the interval, guiding the solution's shape throughout the interval.

Function Derivatives

Function derivatives are crucial in understanding how functions change. Derivatives provide the rate of change of a function concerning its variable. In our problem, we differentiate \( u(x) \) and \( v(x) \) to get their derivatives, \( u'(x) \) and \( v'(x) \). Differentiating involves taking the function and computing its slope at any point on its curve. For example, if a car travels according to the function \( u(x) \), \( u'(x) \) tells us how fast the car is moving at any given time \( x \). Understanding derivatives allows us to predict and analyze changes, which is why they are integral to solving differential equations.

Condition Checking

Condition checking is like verifying whether a set of rules or restrictions apply to our solution. When we deal with differential equations, certain conditions are usually set for the functions, or their derivatives, to satisfy a given problem. In Step 4, we checked if either \( a_2 = 0 \) or \( b_2 = 0 \), or both, affected the problem solution. These conditions might enforce the function or its derivative to zero at certain points, hence simplifying the expression. By confirming which conditions hold, we can determine if the solution meets the specific criteria needed to be valid. Condition checking ensures that the solution is not only mathematically correct but also relevant to the problem's context.