Question

Rectify the following fields: 1) \(s_{1} \partial / \partial x_{1}+2 x_{2} \partial / \partial x_{2}\) for \(x_{1}>0\) 2) \(\partial / \partial x_{1}+\sin x_{1} \partial / \partial x_{2}\) 3) \(x_{1} \partial / \partial x_{1}+\left(1-x_{1}^{2}\right) \partial / \partial x_{2}\) for \(x_{1}^{2}<1\)

Short answer

Answer: The rectified form of the given vector fields is in the coordinates (u, v) for all three cases.

Step-by-step solution

Identify the given vector field

The given vector field is \(s_{1} \partial / \partial x_{1}+2 x_{2} \partial / \partial x_{2}\) for \(x_{1}>0\).

Find the differential equations for the coordinates

To do this, equate the given vector components with the rectified vector field's components - a constant field. In this case, we choose a constant field \(s_{1}\partial / \partial u + s_{2} \partial / \partial v\), where \(s_{1}\) and \(s_{2}\) are constants. So, \(\frac{\partial u}{\partial x_1} = s_{1}\), \(\frac{\partial v}{\partial x_2} = 2 x_2\).

Solve the differential equations

First, \(\int \frac{\partial u}{\partial x_1} dx_1 = \int s_{1} dx_1 \implies u = s_{1}x_1 + g(v).\) Next, \(\int \frac{\partial v}{\partial x_2} dx_2 = \int 2x_2 dx_2 \implies v = x_{2}^{2} + h(u).\)

Combine the equations to find the new coordinates

Now, we create a system of equations: \(u = s_{1}x_1 + g(v)\), \(v = x_{2}^{2} + h(u)\). Rearranging and inserting \(v\)'s expression into \(u\)'s equation, we have \(x_1 = \frac{1}{s_1}(u - g(x_2^2 + h(u)))\). Similarly, rearranging and inserting \(u\)'s expression into \(v\)'s equation, we have \(x_2 = \sqrt{v - h(s_1x_1 + g(v))}\). The vector field is rectified in the coordinates \((u, v)\). For the second vector field:

Identify the given vector field

The given vector field is \(\partial / \partial x_{1}+\sin x_{1} \partial / \partial x_{2}\).

Find the differential equations for the coordinates

Again, equate the given vector components with the constant field: \(u\partial / \partial x_1 + v \partial / \partial x_2\). So, \(\frac{\partial u}{\partial x_1} = 1\), \(\frac{\partial v}{\partial x_2} = \sin x_1\).

Solve the differential equations

First, \(\int \frac{\partial u}{\partial x_1} dx_1 = \int dx_1 \implies u = x_1 + g(v).\) Next, \(\int \frac{\partial v}{\partial x_2} dx_2 = \int \sin x_1 dx_2 \implies v = x_2\sin x_1 + h(u).\)

Combine the equations to find the new coordinates

Now, we create a system of equations: \(u = x_1 + g(v)\), \(v = x_2\sin x_1 + h(u)\). Rearranging and inserting \(v\)'s expression into \(u\)'s equation, we have \(x_1 = u - g(x_2\sin u + h(u))\). Similarly, rearranging and inserting \(u\)'s expression into \(v\)'s equation, we have \(x_2 = \frac{v - h(u)}{\sin u}\). The vector field is rectified in the coordinates \((u, v)\). For the third vector field:

Identify the given vector field

The given vector field is \(x_{1} \partial / \partial x_{1}+\left(1-x_{1}^{2}\right) \partial / \partial x_{2}\) for \(x_{1}^{2}<1\).

Find the differential equations for the coordinates

Again, equate the given vector components with the constant field: \(u\partial / \partial x_1 + v \partial / \partial x_2\). So, \(\frac{\partial u}{\partial x_1} = x_1\), \(\frac{\partial v}{\partial x_2} = 1-x_1^2\).

Solve the differential equations

First, \(\int \frac{\partial u}{\partial x_1} dx_1 = \int x_1 dx_1 \implies u = \frac{1}{2}x_1^2 + g(v).\) Next, \(\int \frac{\partial v}{\partial x_2} dx_2 = \int (1-x_1^2) dx_2 \implies v = (1-x_1^2)x_2 + h(u).\)

Combine the equations to find the new coordinates

Now, we create a system of equations: \(u = \frac{1}{2}x_1^2 + g(v)\), \(v = (1-x_1^2)x_2 + h(u)\). Rearranging and inserting \(v\)'s expression into \(u\)'s equation, we have \(x_1 = \sqrt{2(u - g((1-x_1^2)x_2 + h(u))}\). Similarly, rearranging and inserting \(u\)'s expression into \(v\)'s equation, we have \(x_2 = \frac{v - h(\frac{1}{2}x_1^2 + g(v))}{1-x_1^2}\). The vector field is rectified in the coordinates \((u, v)\).