1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. May 18, 2014 #1


    User Avatar

    1. The problem statement, all variables and given/known data

    [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).

    2. Relevant equations

    Form of Jacobian matrix:

    J =
    \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}}

    3. 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?

    Last edited: May 18, 2014
  2. jcsd
  3. May 18, 2014 #2


    User Avatar
    Science Advisor

    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.

    Yes, that is correct.

    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].
  4. May 18, 2014 #3


    User Avatar

    That's excellent, thank you!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted