Question

Find the Green's function for the boundary value problem $$ x^{2} y^{\prime \prime}+x y^{\prime}-y=F(x), \quad y(1)-2 y^{\prime}(1)=0, \quad y^{\prime}(2)=0 $$ given that \(\\{x, 1 / x\\}\) is a fundamental set of solutions of the complementary equation. Then use the Green's function to solve \((\mathrm{A})\) with (a) \(F(x)=1,\) (b) \(F(x)=x^{2},\) and \((\mathbf{c}) F(x)=x^{3}\).

Short answer

For case (a), with \(F(x) = 1\), the solution is as follows: (a) $$y(x) = \int_1^2 \frac{x_<}{x} x_>dx'$$ We break it into two parts: For \(1 \leq x' < x\), we have \(x_< = x'\) and \(x_> = x\): $$\int_1^x \frac{x'}{x} xdx' = \int_1^x x' dx' = \frac{x^3}{6} - \frac{1}{2}$$ For \(x < x' \leq 2\), we have \(x_< = x\) and \(x_> = x'\): $$\int_x^2 \frac{x}{x} x'dx' = \int_x^2 x' dx' = 2x^2 - \frac{7}{3}x + 1$$ So, the complete solution y(x) in the whole interval is given by: $$ y(x) = \begin{cases} \frac{x^3}{6} - \frac{1}{2} & 1 \leq x \leq 2 \\\\ 2x^2 - \frac{7}{3}x + 1 & x < x' \end{cases}$$ For case (b), with \(F(x) = x^2\), the solution is as follows: (b) $$y(x) = \int_1^2 \frac{x_<}{x} x_>(x'^2)dx'$$ We break it into two parts as before: For \(1 \leq x' < x\), we have \(x_< = x'\) and \(x_> = x\): $$\int_1^x \frac{x'}{x}(xx'^2) dx' = \int_1^x x'^3 dx' = \frac{x^4}{16} - \frac{1}{4}$$ For \(x < x' \leq 2\), we have \(x_< = x\) and \(x_> = x'\): $$\int_x^2 \frac{x}{x}(x'^2 x') dx'= \int_x^2 x'^3 dx' = \frac{15}{16}x^4 - 2x^3 + \frac{25}{12}x^2 - 3x + 1$$ So, the complete solution y(x) in the whole interval is given by: $$ y(x) = \begin{cases} \frac{x^4}{16} - \frac{1}{4} & 1 \leq x \leq 2 \\\\ \frac{15}{16}x^4 - 2x^3 + \frac{25}{12}x^2 - 3x + 1 & x < x' \end{cases}$$ For case (c), with \(F(x) = x^3\), the solution is as follows: (c) $$y(x) = \int_1^2 \frac{x_<}{x} x_>(x'^3)dx'$$ We break it into two parts as before: For \(1 \leq x' < x\), we have \(x_< = x'\) and \(x_> = x\): $$\int_1^x \frac{x'}{x}(xx'^3) dx' = \int_1^x x'^4 dx' = x^5/20 - 1/5$$ For \(x < x' \leq 2\), we have \(x_< = x\) and \(x_> = x'\): $$\int_x^2 \frac{x}{x}(x'^3 x') dx' = \int_x^2 x'^4 dx' = \frac{161}{30}x^5 - \frac{35}{4}x^4 + \frac{169}{12}x^3 - 7x^2 + 4x - 1$$ So, the complete solution y(x) in the whole interval is given by: $$ y(x) = \begin{cases} x^5/20 - 1/5 & 1 \leq x \leq 2 \\\\ \frac{161}{30}x^5 - \frac{35}{4}x^4 + \frac{169}{12}x^3 - 7x^2 + 4x - 1 & x < x' \end{cases}$$ In conclusion, we have found the solutions to the boundary value problem for different values of \(F(x)\), specifically \(F(x) = 1\), \(F(x) = x^2\), and \(F(x) = x^3\).

Step-by-step solution

Verify the fundamental set

First, the complementary equation corresponding to the given differential equation is: $$ x^{2} y^{\prime \prime}+x y^{\prime}-y=0 $$ We are given the fundamental set of solutions \(\\{x, 1 / x\\}\) for this complementary equation. It is important to remember that to find the Green's function we need this fundamental set for the given complementary equation.

Compute the Wronskian

To compute the Wronskian of the two functions \(u(x) = x\) and \(v(x) = \frac{1}{x}\), we can use the formula: $$W(u, v) = uv' - vu'$$ Differentiating \(u(x)\) and \(v(x)\) gives: $$u'(x) = 1, \quad v'(x) = -\frac{1}{x^2}$$ Then, the Wronskian is: $$W(u, v) = x\left(-\frac{1}{x^2}\right) - \frac{1}{x}(1) = -\frac{x}{x^2} - \frac{1}{x} = -\frac{1+x}{x}$$

Construct the Green's function

Now that we have the Wronskian and the fundamental set, we can construct the Green's function for the given boundary value problem as follows: $$G(x, x') = \begin{cases} \frac{x_<}{x} x_> & 1 \leq x',x \leq 2 \\\\ 0 & \text{otherwise} \end{cases}$$ where \(x_<\) and \(x_>\) indicate the smaller and larger values between \(x\) and \(x'\), respectively.

Apply the Green's function

In order to solve the given boundary value problem, we need to compute the convolution of the Green's function and the forcing function \(F(x)\): $$y(x) = \int_1^2 G(x, x')F(x')dx'$$ For the (a), (b) and (c) cases, we have: (a) \(F(x) = 1\) $$y(x) = \int_1^2 \frac{x_<}{x} x_>dx'$$ (b) \(F(x) = x^2\) $$y(x) = \int_1^2 \frac{x_<}{x} x_>(x'^2)dx'$$ (c) \(F(x) = x^3\) $$y(x) = \int_1^2 \frac{x_<}{x} x_>(x'^3)dx'$$ By solving each of these integrals, we will find the solutions corresponding to \(F(x)=1\), \(F(x)=x^2\), and \(F(x)=x^3\).

Key concepts

Wronskian

Understanding the Wronskian is crucial when working with differential equations. It's a determinant used to determine if a set of functions are linearly independent, which is vital when looking for a solution to a differential equation. Imagine you have two functions, let's call them u(x) and v(x), which are solutions to a homogeneous linear differential equation. To check their linear independence, you calculate the Wronskian, represented as:
\[ W(u, v) = uv' - vu' \]
Here, u' and v' are the derivatives of u(x) and v(x), respectively. If the Wronskian is not zero, the functions are linearly independent.
In the problem at hand, we calculated the Wronskian for the functions in the fundamental set, which helped us solidify that the given functions form a legitimate basis for the solution space and set the stage for constructing the Green's function.

Fundamental Set of Solutions

A 'fundamental set of solutions' refers to a collection of solutions to a homogeneous linear differential equation that is linearly independent and spans the solution space. Think of it like finding the right pieces that fit together to complete a puzzle. Each piece (solution) has to be different (linearly independent), and when you combine them, you should be able to see the whole picture (the entire solution space).
For the given boundary value problem, the fundamental set is \( \{x, 1/x\} \). These two functions are the 'pieces' that can construct any solution to the equation by forming a linear combination, which is why they're so important. They're the building blocks for the Green's function and ultimately for solving the entire equation.

Differential Equations

Differential equations are mathematical expressions that describe relationships involving rates of change and initial conditions. They are indispensable in modelling real-world systems where variables depend on one another and change over time, like physics, engineering, and economics.
In our case, we're dealing with a second-order linear ordinary differential equation (ODE), which means our equation involves functions of a single variable and their derivatives. ODEs like these often pop up in classical mechanics when dealing with motion and forces. The goal is to find a function y(x) that satisfies the equation for a given F(x), which represents the external force or input in our system. Solving differential equations involves finding the fundamental set of solutions for the complementary homogeneous equation and then using that to construct the particular solution to the inhomogeneous equation.

Convolution Integral

The convolution integral is a powerful tool in the analysis of systems. It essentially calculates the output of a system when you know the input signal and the system's impulse response. In the context of differential equations, the impulse response is the Green's function, and the input signal is a forcing function, F(x).
Mathematically, the convolution integral combines two functions, the Green's function G(x, x'), which encodes information about the system itself and the forcing function F(x). The resulting integral:
\[ y(x) = \int G(x, x') F(x') dx' \]
This process 'blends' the influence of F(x') over the interval of integration, weighted by the Green's function, to produce the desired solution, y(x), to our differential equation. The convolution integral is a versatile and profound concept, used not just in differential equations, but also in signal processing, probability, and other fields.