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Chapter 11: Problem 17

Consider the problem $$ y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0, \quad y(0)=a, \quad y(1)=b $$ Let \(y=u+v,\) where \(v\) is any twice differentiable function satisfying the boundary conditions (but not necessarily the differential equation). Show that \(u\) is a solution of the problem $$ u^{\prime \prime}+p(x) u^{\prime}+q(x) u=g(x), \quad u(0)=0, \quad u(1)=0 $$ where \(g(x)=-\left[v^{\prime \prime}+p(x) v^{\prime}+q(x) v\right],\) and is known once \(v\) is chosen. Thus nonhomogeneities can be transferred from the boundary conditions to the differential equation. Find a function \(v\) for this problem.

Short Answer

Expert verified
Question: Show that you can transfer the nonhomogeneities from the boundary conditions to a second-order linear differential equation, simplifying the problem. Answer: By expressing y as the sum of two functions, u and v, we can substitute this expression into the given differential equation to obtain a new equation for u. We can then choose a suitable function v that satisfies the given boundary conditions, allowing us to transform the problem into a new problem with homogeneous boundary conditions and a known function g(x).

Step by step solution

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01

Substitute y = u + v in the given differential equation

Given the function y as the sum of u and v, we have: $$ y = u + v $$ Now, differentiate the equation above with respect to x to obtain the first and second derivatives of y in terms of u and v: $$ y' = u' + v' $$ $$ y'' = u'' + v'' $$ Substitute these expressions for y', y'', and y in the given differential equation to get: $$ (u'' + v'') + p(x)(u' + v') + q(x)(u + v) = 0 $$
02

Rearrange the differential equation for u

Now we need to find an equation for u. To do that, first take v'' and v' terms to the other side of the equation: $$ u'' + p(x)u' + q(x)u = -v'' - p(x)v' - q(x)v $$ Let's define g(x) as: $$ g(x) = -v'' - p(x)v' - q(x)v $$ Now the equation for u becomes: $$ u'' + p(x)u' + q(x)u = g(x) $$ with boundary conditions: $$ u(0) = 0 $$ $$ u(1) = 0 $$
03

Find a function v

The next step is to find a function v that satisfies the given boundary conditions. Since this is an arbitrary function, we can choose any function that satisfies y(0) = a and y(1) = b. A simple choice would be a straight line connecting the points (0, a) and (1, b). Thus, the function v can be given as: $$ v(x) = a(1-x) + bx $$ This function satisfies the given boundary conditions, i.e., v(0) = a and v(1) = b. Now, we have shown that the given problem can be transformed into a new problem for u with homogeneous boundary conditions and a known function g(x), which depends on the chosen function v.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boundary Value Problems
Boundary value problems (BVPs) are essential in the realm of differential equations. They involve finding a solution to a differential equation subject to specific conditions at the boundaries of the interval. In our example, we are tasked with a second-order differential equation with boundary conditions. These conditions are given as \( y(0) = a \) and \( y(1) = b \).

To solve a boundary value problem, a function must satisfy not only the differential equation itself but also the boundary conditions. This makes BVPs slightly more challenging than initial value problems, where conditions are specified only at the start of the interval.

Here, we use the sum \( y = u + v \) to break down our problem. By choosing \( v \) to specifically satisfy the boundary conditions, we transform the original problem into a more manageable form that involves only the function \( u \), which satisfies a new differential equation with homogeneous boundary conditions.
Differential Equation Transformation
Differential equation transformation is a powerful technique used to modify a problem into a form that is easier to tackle. This transformation often involves changing the form of the boundary conditions or the differential equation itself.

In our problem, we leverage the transformation \( y = u + v \), where \( v \) is a function meeting the boundary conditions but not necessarily the differential equation. By doing this substitution, we manipulate the equation to produce a new differential equation solely in terms of \( u \), which then has homogeneous boundary conditions.

This transformation is key as it shifts all complexities related to boundary conditions into a new function \( g(x) \), defined as:
  • \( g(x) = -[v'' + p(x)v' + q(x)v] \)
Now, instead of solving the original boundary value problem, we handle a fresh problem where the equation \( u'' + p(x)u' + q(x)u = g(x) \) must be addressed, with simpler, homogeneous conditions for \( u \).
Homogeneous Boundary Conditions
Homogeneous boundary conditions are conditions where the function is set to zero at the boundary points. In mathematical terms, the boundary conditions are \( u(0) = 0 \) and \( u(1) = 0 \). This is a simpler scenario to solve, as the solutions can often be more straightforward to determine.

In this exercise, transforming the problem into one with homogeneous boundary conditions simplifies the solution process considerably. With the function \( v \) chosen to manage initial conditions, the task of solving \( u \) with boundary values set to zero is made easier.

This strategy of zeroing out boundary conditions allows us to focus purely on the differential equation part. As such, our main task becomes finding a function \( u \) that satisfies the differential equation with a given \( g(x) \). This method of simplifying complex boundary requirements into homogeneous conditions is a common technique in mathematical physics and engineering.

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Most popular questions from this chapter

Indicate a way of solving nonhomogeneous boundary value problems that is analogous to using the inverse matrix for a system of linear algebraic equations. The Green's function plays a part similar to the inverse of the matrix of coefficients. This method leads to solutions expressed as definite integrals rather than as infinite series. Except in Problem 35 we will assume that \(\mu=0\) for simplicity. (a) Show by the method of variation of parameters that the general solution of the differential equation $$ -y^{\prime \prime}=f(x) $$ can be written in the form $$ y=\phi(x)=c_{1}+c_{2} x-\int_{0}^{x}(x-s) f(s) d s $$ where \(c_{1}\) and \(c_{2}\) are arbitrary constants. (b) Let \(y=\phi(x)\) also be required to satisfy the boundary conditions \(y(0)=0, y(1)=0\) Show that in this case $$ c_{1}=0, \quad c_{2}=\int_{0}^{1}(1-s) f(s) d s $$ (c) Show that, under the conditions of parts (a) and (b), \(\phi(x)\) can be written in the form $$ \phi(x)=\int_{0}^{x} s(1-x) f(s) d s+\int_{x}^{1} x(1-s) f(s) d s $$ (d) Defining $$ G(x, s)=\left\\{\begin{array}{ll}{s(1-x),} & {0 \leq s \leq x} \\ {x(1-s),} & {x \leq s \leq 1}\end{array}\right. $$ show that the solution takes the form $$ \phi(x)=\int_{0}^{1} G(x, s) f(s) d s $$ The function \(G(x, s)\) appearing under the integral sign is a Green's function. The usefulness of a Green's function solution rests on the fact that the Green's function is independent of the nonhomogencous term in the differential equation. Thus, once the Green's function is determined, the solution of the boundary value problem for any nonhomogeneous term \(f(x)\) is obtained by a single integration. Note further that no determination of arbitrary constants is required, since \(\phi(x)\) as given by the Green's function integral formula automatically satisfies the boundary conditions.

In this problem we indicate a proof that the eigenfunctions of the Sturm- Liouville problem \((1),(2)\) are real. (a) Let \(\lambda\) be an eigenvalue and \(\phi\) a corresponding eigenfunction. Let \(\phi(x)=U(x)+\) \(i V(x),\) and show that \(U\) and \(V\) are also eigenfunctions corresponding to \(\lambda .\) (b) Using Theorem \(11.2 .3,\) or the result of Problem \(20,\) show that \(U\) and \(V\) are linearly dependent. (c) Show that \(\phi\) must be real, apart from an arbitrary multiplicative constant that may be complex.

Using the method of Problem 17 , transform the problem $$ y^{\prime \prime}+2 y=2-4 x, \quad y(0)=1, \quad y(1)+y^{\prime}(1)=-2 $$ into a new problem in which the boundary conditions are homogeneous. Solve the latter problem by reference to Example 1 of the text.

Determine whether there is any value of the constant \(a\) for which the problem has a solution. Find the solution for each such value. $$ y^{\prime \prime}+\pi^{2} y=a, \quad y^{\prime}(0)=0, \quad y^{\prime}(1)=0 $$

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