Question

Use the polar equations derived in Exercise 25 to show that if $$ a_{11}=a_{22}, \quad a_{21}=-a_{12}, \quad g_{1}(\mathbf{z})=z_{1} h\left(\sqrt{z_{1}^{2}+z_{2}^{2}}\right), \quad g_{2}(\mathbf{z})=z_{2} h\left(\sqrt{z_{1}^{2}+z_{2}^{2}}\right) $$ for some function \(h\), then the polar equations uncouple into $$ \begin{aligned} r^{\prime} &=a_{11} r+r h(r) \\ \theta^{\prime} &=a_{21} . \end{aligned} $$ Note that the radial equation is a separable differential equation and the angle equation can be solved by antidifferentiation.

Short answer

Question: Prove that under the given conditions, the polar equations will uncouple into a simpler form. Given conditions: $$ a_{11} = a_{22}, \quad a_{21} = -a_{12}, \quad g_{1} (\mathbf{z}) = z_{1} h(\sqrt{z_{1}^2 + z_{2}^2}), \quad g_{2} (\mathbf{z}) = z_{2} h(\sqrt{z_{1}^2 + z_{2}^2}) $$ with polar equations: $$ \begin{aligned} r' &= a_{11}r \cos\theta + a_{21}r \sin\theta + g_1(\mathbf{z}) \\ \theta' &= a_{12}r \cos\theta + a_{22}r \sin\theta + g_2(\mathbf{z}) \end{aligned} $$ Solution: We have transformed the given polar equations according to the conditions and obtained the uncoupled differential equations as: $$ \begin{aligned} r'(r) &= a_{11} r + rh(r) \\ \theta'(r) &= a_{21} \end{aligned} $$ The first equation is a separable differential equation, and the second equation can be solved by antidifferentiation.

Step-by-step solution

Identify the given conditions

We are given the following conditions: $$ a_{11} = a_{22}, \quad a_{21} = -a_{12}, \quad g_{1} (\mathbf{z}) = z_{1} h(\sqrt{z_{1}^2 + z_{2}^2}), \quad g_{2} (\mathbf{z}) = z_{2} h(\sqrt{z_{1}^2 + z_{2}^2}) $$ We are also given the polar equations: $$ \begin{aligned} r' &= a_{11}r \cos\theta + a_{21}r \sin\theta + g_1(\mathbf{z}) \\ \theta' &= a_{12}r \cos\theta + a_{22}r \sin\theta + g_2(\mathbf{z}) \end{aligned} $$ which we must use in the transformation process.

Rewrite the polar equations in terms of the given conditions

Replace \(g_1\) and \(g_2\) in the polar equations using the given conditions, and recall that \(z_1 = r \cos\theta\) and \(z_2 = r \sin\theta\): $$ \begin{aligned} r' &= a_{11} r \cos\theta + a_{21} r \sin\theta + r \cos\theta h(r) \\ \theta' &= a_{12} r \cos\theta + a_{22} r \sin\theta + r \sin\theta h(r) \end{aligned} $$

Rearrange the equations to separate terms according to \(r\) and \(\theta\)

In the first equation, group terms with \(\cos\theta\) and \(\sin\theta\): $$ r' = r\cos\theta(a_{11} + h(r)) + r\sin\theta(a_{21}) $$ For the second equation, the same process applies: $$ \theta' = r\cos\theta(a_{12}) + r\sin\theta(a_{22} + h(r)) $$

Use given conditions to simplify the equations

Since \(a_{11} = a_{22}\) and \(a_{21} = -a_{12}\), substitute these values into the equations and simplify: $$ \begin{aligned} r' &= r\cos\theta(a_{11} + h(r)) + r\sin\theta(-a_{12}) \\ \theta' &= r\cos\theta(-a_{21}) + r\sin\theta(a_{11} + h(r)) \end{aligned} $$

Factor out \(r\) and uncouple the equations

Notice that both equations have \(r\) as a common factor, so we can factor it out: $$ \begin{aligned} r'(r) &= r\left[\cos\theta(a_{11} + h(r)) - \sin\theta a_{12}\right] \\ \theta'(r) &= r\left[-\cos\theta a_{21} + \sin\theta(a_{11} + h(r))\right] \end{aligned} $$ Now observe that the terms in the brackets for both equations are equal, so the equations uncouple: $$ \begin{aligned} r'(r) &= a_{11} r + rh(r) \\ \theta'(r) &= a_{21} \end{aligned} $$

Identify the types of differential equations obtained

Now that we have uncoupled the equations, we see that the first equation is a separable differential equation, and the second equation can be solved by antidifferentiation. Thus, we have successfully shown that under the given conditions, the polar equations uncouple into the desired simpler form.

Key concepts

Separable Differential Equation

A separable differential equation is a common type of differential equation that can be expressed as the product of two functions, each dependent on a different variable. It is distinguished by its ability to be separated into two sides of an equation, one containing all the terms involving the independent variable, often denoted as x or t, and the other containing all the terms involving the dependent variable, usually represented as y. This separation allows each side of the equation to be integrated separately.

For example, consider the equation \[\frac{dy}{dx} = g(x)h(y)\]. We can rewrite this equation by separating the variables: \[\frac{1}{h(y)}dy = g(x)dx\], which then allows us to integrate both sides with respect to their respective variables, leading to the antiderivative forms: \[\int\frac{1}{h(y)}dy = \int g(x)dx\].

Solving a separable differential equation involves finding these antiderivatives and then solving for the dependent variable y as a function of the independent variable x. This approach is particularly useful when dealing with first-order ordinary differential equations, where the solution typically involves finding a function that describes the rate of change of one variable relative to another.

Antidifferentiation

Antidifferentiation, also commonly referred to as integration, is the inverse process of differentiation. Where differentiation gives us the rate at which a function is changing at any point, antidifferentiation provides us with the original function given its derivative. It is an essential concept in calculus used to solve a variety of problems, including those involving area under curves and the solution of differential equations.

When presented with a derivative, the process of antidifferentiation involves finding a function (or set of functions) whose derivative matches the given expression. This is often symbolized by \[\int f'(x)dx = f(x) + C\], where C represents the constant of integration, acknowledging that the process of integration can yield multiple functions differing by a constant value.

In the context of solving differential equations, such as in our exercise, antidifferentiation helps us to find solutions for equations that express a derivative directly in terms of the independent variable. This process is straightforward and does not require elaborate techniques like separation of variables or integration by parts unless the function is more complex.

Uncoupled Differential Equations

Uncoupled differential equations are a set of equations where each equation contains only one dependent variable and its derivatives. This is in contrast to coupled equations, where each equation might contain multiple dependent variables and their derivatives.

In the realm of ordinary differential equations, having an uncoupled system means that we can solve each differential equation independently of the others. This is a significant simplification as it reduces the complexity of the system and allows the application of standard methods for first-order or higher-order differential equations.

Using the information from our original exercise, if we are given a system of polar equations with conditions that lead to uncoupling, we can solve them individually. For instance, the radial equation \[r' = a_{11} r + r h(r)\] becomes a separable equation because it involves only the function r and its derivative r'. Similarly, the angular equation \[\theta' = a_{21}\] involves only the derivative of the function \theta and is constant, making its solution obtainable directly through antidifferentiation. This simplicity is what makes uncoupling an advantageous outcome when solving systems of differential equations.