Higher order D.E. to linear system of 1st order D.E.'s

[jjr]

Homework Statement

$\textbf{(a)}$ This is an exercise from a course on numerical analysis.

Write the system of differential equations

$u''' = x^2uu'' - uv'$

$v'' = xvv' + 4u'$

as a first order system of differential equations, $\textbf{y'} = \textbf{y}(x,\textbf{y})$.

$\textbf{(b)}$ Determine the Jacobian matrix $\textbf{f}_\textbf{y}(x,\textbf{y})$ for the system in (a).

Homework Equations

Form of Jacobian matrix:

$J =<br /> \begin{pmatrix}<br /> \frac{\partial{F_1}}{\partial{x_1}} & \cdots & \frac{\partial{F_1}}{\partial{x_n}} \\<br /> \vdots & \ddots & \vdots \\<br /> \frac{\partial{F_m}}{\partial{x_1}} & \cdots & \frac{\partial{F_m}}{\partial{x_n}}<br /> \end{pmatrix}$

The Attempt at a Solution

I might have solved part (a) (I'm not quite sure to be honest), but I have a few problems in part b) that I need help with.

I know the canonical way of transforming a single d'th order D.E. into a system of first order D.E.'s, but was a bit confused when there were two equations. Here is what I did:

I let $y_1 = u; y_2 = u'; y_3 = u''; y_4 = v; y_5 = v'$, and thus

$\frac{dy_1}{dx} = y_2; \frac{dy_2}{dx} = y_3; \frac{dy_3}{dx} = x^2y_1y_3 - y_1y_5; \frac{y_4}{dx} = y_5; \frac{dy_5}{dx} = xy_4y_5+4y_2$

Could this be the correct way to transform these equations?

As for b), referring to the matrix above, I reckon $F_1 = \frac{dy_1}{dx} = y_2; F_2 = \frac{dy_2}{dx} = y_3$ and so on, whereas $x_1 = y_1; x_2 = y_2$ and so on. Am I right here?

When I take the derivative of say $y_2$ with respect to $y_3$ is it zero or is there some implicit dependence on $y_3$ in $y_2$? If there is not then it should be zero, and in fact most of the entries in the Jacobian should be zero?

Thanks,
J

 

[HallsofIvy]

Homework Statement

$\textbf{(a)}$ This is an exercise from a course on numerical analysis.

Write the system of differential equations

$u''' = x^2uu'' - uv'$

$v'' = xvv' + 4u'$

as a first order system of differential equations, $\textbf{y'} = \textbf{y}(x,\textbf{y})$.

$\textbf{(b)}$ Determine the Jacobian matrix $\textbf{f}_\textbf{y}(x,\textbf{y})$ for the system in (a).

Homework Equations

Form of Jacobian matrix:

$J =<br /> \begin{pmatrix}<br /> \frac{\partial{F_1}}{\partial{x_1}} & \cdots & \frac{\partial{F_1}}{\partial{x_n}} \\<br /> \vdots & \ddots & \vdots \\<br /> \frac{\partial{F_m}}{\partial{x_1}} & \cdots & \frac{\partial{F_m}}{\partial{x_n}}<br /> \end{pmatrix}$

This is wrong. There is only a single variable "x", not "$x_1$", "$x_2$", etc. the derivatives should be with respect to $y_1$, $y_2$, etc.

The Attempt at a Solution

I might have solved part (a) (I'm not quite sure to be honest), but I have a few problems in part b) that I need help with.

I know the canonical way of transforming a single d'th order D.E. into a system of first order D.E.'s, but was a bit confused when there were two equations. Here is what I did:

I let $y_1 = u; y_2 = u'; y_3 = u''; y_4 = v; y_5 = v'$, and thus

$\frac{dy_1}{dx} = y_2; \frac{dy_2}{dx} = y_3; \frac{dy_3}{dx} = x^2y_1y_3 - y_1y_5; \frac{y_4}{dx} = y_5; \frac{dy_5}{dx} = xy_4y_5+4y_2$

Could this be the correct way to transform these equations?

Yes, that is correct.

As for b), referring to the matrix above, I reckon $F_1 = \frac{dy_1}{dx} = y_2; F_2 = \frac{dy_2}{dx} = y_3$ and so on, whereas $x_1 = y_1; x_2 = y_2$ and so on. Am I right here?

When I take the derivative of say $y_2$ with respect to $y_3$ is it zero or is there some implicit dependence on $y_3$ in $y_2$? If there is not then it should be zero, and in fact most of the entries in the Jacobian should be zero?

Thanks,
J

For the purposes of the Jacobian, treat all variables, x and all of the various "y"s, as independent. That is, the derivative of $y_2$ with respect to $y_3$ is 0 while the derivative of $x^2y_1y_2- y_1y_5$ with respect to $y_1$ is $x^2y_2- y_5$.

 

[jjr]

That's excellent, thank you!