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

[jjr]

Homework Statement

[itex]\textbf{(a)}[/itex] This is an exercise from a course on numerical analysis.

Write the system of differential equations

[itex] u''' = x^2uu'' - uv' [/itex]

[itex] v'' = xvv' + 4u' [/itex]

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

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

Homework Equations

Form of Jacobian matrix:

[itex]
J =
\begin{pmatrix}
\frac{\partial{F_1}}{\partial{x_1}} & \cdots & \frac{\partial{F_1}}{\partial{x_n}} \\
\vdots & \ddots & \vdots \\
\frac{\partial{F_m}}{\partial{x_1}} & \cdots & \frac{\partial{F_m}}{\partial{x_n}}
\end{pmatrix}
[/itex]

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 [itex] y_1 = u; y_2 = u'; y_3 = u''; y_4 = v; y_5 = v' [/itex], and thus

[itex] \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[/itex]

Could this be the correct way to transform these equations?

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

When I take the derivative of say [itex] y_2 [/itex] with respect to [itex] y_3 [/itex] is it zero or is there some implicit dependence on [itex] y_3 [/itex] in [itex] y_2 [/itex]? 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

[itex]\textbf{(a)}[/itex] This is an exercise from a course on numerical analysis.

Write the system of differential equations

[itex] u''' = x^2uu'' - uv' [/itex]

[itex] v'' = xvv' + 4u' [/itex]

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

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

Homework Equations

Form of Jacobian matrix:

[itex]
J =
\begin{pmatrix}
\frac{\partial{F_1}}{\partial{x_1}} & \cdots & \frac{\partial{F_1}}{\partial{x_n}} \\
\vdots & \ddots & \vdots \\
\frac{\partial{F_m}}{\partial{x_1}} & \cdots & \frac{\partial{F_m}}{\partial{x_n}}
\end{pmatrix}
[/itex]

This is wrong. There is only a single variable "x", not "[itex]x_1[/itex]", "[itex]x_2[/itex]", etc. the derivatives should be with respect to [itex]y_1[/itex], [itex]y_2[/itex], 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 [itex] y_1 = u; y_2 = u'; y_3 = u''; y_4 = v; y_5 = v' [/itex], and thus

[itex] \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[/itex]

Could this be the correct way to transform these equations?

Yes, that is correct.

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

When I take the derivative of say [itex] y_2 [/itex] with respect to [itex] y_3 [/itex] is it zero or is there some implicit dependence on [itex] y_3 [/itex] in [itex] y_2 [/itex]? 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 [itex]y_2[/itex] with respect to [itex]y_3[/itex] is 0 while the derivative of [itex]x^2y_1y_2- y_1y_5[/itex] with respect to [itex]y_1[/itex] is [itex]x^2y_2- y_5[/itex].

 

[jjr]

That's excellent, thank you!