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Go through the case of \(\beta=0\) in the solution of the second order ODE and, choosing \(w=s\) and \(y=t\), show that \(F\) will be a function of \(y\) alone and the original DE will reduce to \(w_{y y}=F(y)\).

Short Answer

Expert verified
By setting \(\beta=0\) in the general form of a second order ODE, substituting variables \(s\) for \(w\) and \(t\) for \(y\), and simplifying the transformed equation, it can be proven that \(F\) becomes a function of \(y\) alone and the differential equation reduces to the form \(w_{yy}=F(y) w\).

Step by step solution

01

Start with the general second order ODE

The general 2nd order homogeneous ODE follows the format\[a(y) w_{yy} + b(y) w_y + c(y) w = 0\]Focusing on the assignment \(\beta = 0\) which would correspond to some coefficient in the ODE above being zero, we can simplify it to:\b(a(y) w_{yy} + c(y) w = 0.This reduces our analysis to primarily the two terms left in the equation.
02

Substitution of variables

In the equation \(a(y) w_{yy} + c(y) w = 0\), replace \(w\) with \(s\) (as per the instructions \(w = s\)) and replace \(y\) with \(t\) (as per the instructions \(y = t\)). Then our equation would look like:\[a(t) s_{tt} + c(t) s = 0\]
03

Reducing to \(w_{yy}=F(y)\) form

Get the last term isolated, by dividing all parts of the equation by \(a(t)\). You will get:\[s_{tt}=-\frac{c(t) s}{a(t)}\]Assuming that the ratio \(F(t)=-\frac{c(t)}{a(t)}\) does not depend on \(s\), we find that the equation has the form required:\[s_{tt}=F(t) s\]
04

Return to original variables

Finally returning to the original variables (\(s \rightarrow w\), \(t \rightarrow y\)), we obtain the original form required in the statement of the problem:\[w_{yy}=F(y) w\]This result shows that \(F\) is not dependent on \(w\) but is a function of \(y\) alone because the ODE has no terms with differential \(w\) or mixed (\(w\) and \(y\)) terms.

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Key Concepts

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

Ordinary Differential Equations
Ordinary Differential Equations (ODEs) are equations involving derivatives of a function with respect to a single independent variable. They are called 'ordinary' to distinguish them from partial differential equations, which involve derivatives with respect to more than one variable. ODEs are used to model dynamically changing systems, such as the motion of particles, growth of populations, or the spread of diseases.

Second order ODEs, like the one in our exercise, involve the second derivative of the unknown function, which in this case is denoted by \(w\). The general form of a second order ODE is given by \[ a(y) w_{yy} + b(y) w_y + c(y) w = 0 \] where \(a(y)\), \(b(y)\), and \(c(y)\) are coefficients that may vary with the variable \(y\), and \(w_{yy}\) and \(w_y\) represent the second and first derivatives of \(w\) with respect to \(y\). When one of these coefficients, specifically \(b(y)\), is zero, such as in the case when \(\beta = 0\), the equation simplifies and provides a more straightforward path to finding a solution.
Variable Substitution
Variable substitution is a powerful technique used to simplify differential equations, making them easier to solve. It involves replacing the original variables in an equation with new variables, often transforming a complex equation into a more manageable form. In the context of our exercise, the substitution \(w = s\) and \(y = t\) changes the appearance of the original ODE without altering its intrinsic properties.

The new variables \(s\) and \(t\) effectively reframe the problem, allowing us to manipulate the equation more freely. After substitution, the second order ODE \[ a(y) w_{yy} + c(y) w = 0 \] takes on a new form \[ a(t) s_{tt} + c(t) s = 0\], which is essentially the same equation with different symbols. This step is crucial because it paves the way to isolate terms and ultimately discern that \(F\), the function we aim to isolate, depends solely on the new independent variable \(t\), corresponding to \(y\) in the original equation.
Homogeneous Equations
Homogeneous equations in the context of ODEs are those whose right-hand side is zero. In other words, they can be written in the form \[ a(y) w_{yy} + b(y) w_y + c(y) w = 0\] where the absence of standalone terms (not involving the function \(w\) or its derivatives) signals the homogeneity. These types of equations are particularly nice because they often imply that solutions can be directly related to the properties of the coefficients \(a\), \(b\), and \(c\).

Our specific ODE becomes homogeneous if we consider the case of \(\beta=0\), which eliminates the middle term in the general equation, leaving us with only the term \(a(y) w_{yy}\) and \(c(y) w\). After the variable substitution, the homogeneous nature of the equation remains intact. The simplification reveals that the original dependent variable \(w\) only appears in its second derivative form, \(w_{yy}\), and it does not blend with the independent variable \(y\). Thus, \(F(y)\) is a function solely of the independent variable, showcasing a key feature of homogeneous equations— the ability to describe the relationship between a dependent variable and its derivatives without extraneous influences.

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Most popular questions from this chapter

Show that the second prolongation of \(\mathbf{v}=w \partial_{y}\) (with \(y\) treated as independent variable) is $$\mathrm{pr}^{(2)} \mathbf{v}=\mathbf{v}-w_{y}^{2} \frac{\partial}{\partial w_{y}}-3 w_{y} w_{y y} \frac{\partial}{\partial w_{y y}}$$

Show that the \(n\) th prolongation of the generator of the \(i\) th translation, \(\partial_{i}\), is the same as the original vector.

Show that in the case of the nonabelian 2 -dimensional Lie algebra, (a) the vector fields can be chosen to be $$\mathbf{v}_{1}=\frac{\partial}{\partial s}, \quad \mathbf{v}_{2}=s \frac{\partial}{\partial s}$$ if \(\beta=0 .\) (b) Show that these vector fields lead to the \(\mathrm{ODE} w_{y y}=w_{y} \tilde{F}(y)\). (c) If \(\beta \neq 0\), show that the vector fields can be chosen to be $$\mathbf{v}_{1}=\frac{\partial}{\partial s}, \quad \mathbf{v}_{2}=s \frac{\partial}{\partial s}+t \frac{\partial}{\partial t} .$$ (d) Finally, show that the latter vector fields lead to the ODE \(w_{y y}=\) \(\tilde{F}\left(w_{y}\right) / y .\)

Suppose \(M(x, u) d x+N(x, u) d u=0\) has a 1-parameter symmetry group with generator \(\mathbf{v}=X \partial_{x}+U \partial_{u} .\) Show that the function \(q(x, u)=\) \(1 /(X M+U N)\) is an integrating factor.

The Korteweg-de Vries equation is \(u_{t}+u_{x x x}+u u_{x}=0\). Using the technique employed in computing the symmetries of the heat and wave equations, show that the infinitesimal generators of symmetries of the Korteweg-de Vries equation are $$\begin{array}{ll}\mathbf{v}_{1}=\partial_{x}, \quad \mathbf{v}_{2}=\partial_{t}, & \text { translation } \\\\\mathbf{v}_{3}=t \partial_{x}+\partial_{u}, & \text { Galilean boost } \end{array}$$ $$\mathbf{v}_{4}=x \partial_{x}+3 t \partial_{t}-2 u \partial_{u} . \quad \text { scaling }$$

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