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Change of variables in a differential equation

  1. Jan 4, 2017 #1
    1. The problem statement, all variables and given/known data

    Transform the equation:

    x2 * d2y/dx2 + 2 * x * dy/dx + (a2/x2)*y = 0

    Using:

    x=1/t

    2. Relevant equations

    The differential of a function of several variables, and the common rules of differentiation.

    https://en.wikipedia.org/wiki/Derivative

    3. The attempt at a solution

    As I have x as function of t I can calculate dx as function of t and dt. So:

    dx = (-1/t2)* dt , equation 1

    d2x = (2/t3)*dt2, equation 2 (I considered d2t=0 because it is the independent variable)

    To calculate dy/dx I symply change dx by its value at equation 1 , so I get:

    dy/dx= dy/ (-1/t2)*dt = -t2 * (dy/dt)

    (According to the book this is correct)

    Now the problem is d2y/dx2

    1.- First question, Why is it a mistake to simply substitute dx2 for its value calculated in equation 2?

    2.- According to the book I have to differenciate dy/dx which was equal to:

    -t2 dy/dt

    I do it simply calculating the differential of a product:

    (-2 * t * dt )* dy/dt -t2 * d2y/dt

    (I didn't differenciate dt, I assume d2t =0)

    Simply dividing this value by dt I get what I expected to be the right result:

    (-2*t) * dy/dt -t2 * d2y/dt2

    But according to the book, this is wrong, I am missing a term arising from dx/dt.

    This is what the book does:

    d2y/dx2 = step 1 = d/dx (dy/dx) = step 2 = (d/dt ( dy/dx))/ ( dx/dt ) to get this result:


    - ( 2t dy/dt +t2 * d2y/dt2)(-t2)

    I understand the step 1 it simply says that the second derivative is the derivative of the first derivative, but the step 2 is a mistery to me and it is the step that creates the term I am missing that is the dx/dt in the denominator. I try to differenciate thinking in terms of differentials, so I can manipulate them in the expressions like they were algebraic quantities, maybe I don't understand the operator version of the derivative.

    The last term -t2 is what I didn't get in my calculation.

    2.- Question two, in my previous calculation using differentials, Where I missed the dx/dt term?
     
  2. jcsd
  3. Jan 4, 2017 #2

    BvU

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    Differentiate to x, not to t !
     
  4. Jan 4, 2017 #3
    I don't get it, if I differenciate dy/dx to x I get d2y/dx2 right? But I am looking for d2y/dt2
     
  5. Jan 4, 2017 #4

    Ray Vickson

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    Chain rule!
    $$ \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt},$$
    etc.
     
  6. Jan 4, 2017 #5

    vela

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    It's probably best to avoid using differentials here.

    For a test function ##f##, the chain rule tells us that
    $$\frac{df}{dx} = \frac{dt}{dx} \frac{df}{dt}.$$ In terms of operators, we'd say that
    $$\frac{d}{dx} = \frac{dt}{dx} \frac{d}{dt}.$$
    To calculate ##dy/dx## then, we'd have
    $$\frac{d}{dx} y = \left(\frac{dt}{dx} \frac{d}{dt}\right) y = \frac{dt}{dx} \frac{dy}{dt},$$
    and for the second derivative, we'd get
    \begin{align*}
    \frac{d}{dx} \left(\frac{d}{dx} y\right) &= \left(\frac{dt}{dx} \frac{d}{dt}\right) \left( \frac{dt}{dx} \frac{d}{dt} \right) y \\
    &= \left(\frac{dt}{dx} \frac{d}{dt}\right) \left( \frac{dt}{dx} \frac{dy}{dt} \right) \\
    &= \frac{dt}{dx} \left[\left(\frac{d}{dt}\frac{dt}{dx}\right) \frac{dy}{dt} + \frac{dt}{dx} \frac{d}{dt} \left( \frac{dy}{dt} \right) \right]
    \end{align*}
     
  7. Jan 4, 2017 #6

    vela

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    You calculated ##\frac{d}{dt}\left(\frac{dy}{dx}\right)##, but you appear to want ##\frac{d^2 y}{dt^2} = \frac{d}{dt}\left(\frac{dy}{dt}\right).## What you actually want is ##d^2 y/dx^2## in terms of ##y##, ##dy/dt##, and ##d^2y/dt^2## to substitute into the original differential equation.
     
  8. Jan 4, 2017 #7
    Thanks folks, now I see it.
     
  9. Jan 5, 2017 #8
    Well, I have tried it hard but I don't get the right result.

    d. will mean "the differential of"

    So we have:

    dy/dt = dy/dx * dx/dt

    And I want to differenciate it, an infinitesimal change on t will change dy and dx, so let's calculate it:

    d. (dy/dt) = d. (dy/dx * dx/dt)

    I regard dt2 as equal to 0 since it is the independent variable.

    d2y/dt = d.(dy/dx) * dx/dt + d. (dx/dt) * dy/dx

    I apply the quotient rule for dy/dx so I get:

    ((dx * d2y - d2x * dy )/dx2)

    So we have this expression:

    d2y/dt= ((dx * d2y - d2x * dy )/dx2) * dx/dt +d2x/dt * dy/dx

    d2y/dt= (d2y/dx2) * (dx/dt) * dx - (dx/dt)* (d2x/dx2) *dy + (dy/dx) * (d2x/dt)

    Now I calculate d2x/dx2, which is a second order differential divided by the square of the first order differential. As we know that x=1/t I get for this quotient 2t.

    Now I divide the whole equation by dt, so finally I get:

    d2y/dt2= (d2y/dx2) * (dx/dt) * (dx/dt) - (dx/dt)* (2t) *(dy/dt) + (dy/dx) * (d2x/dt^2)

    Now I need to substitute the derivatives of x over t, eliminate dy/dx which was known from the start, and get d2y/dx2 as function of derivatives respect to t.

    I use:

    dx/dt= -1/t2

    d2x/dt2 = 2/t^3

    dy/dx = -t2 * dy/dt

    d2y/dt2= (d2y/dx2) * (1/t4) - (-1/t2)* (2t) *(dy/dt) + (-t2 * dy/dt) * (2/t3)

    The last two terms cancel out and I only get this result:

    d2y/dt2= (d2y/dx2) * (1/t4)

    Which is partially correct, but I failed to get an extra term since according to the book the right result is:

    d2y/dx2 = 2t3 * dy/dt + t4 * d2y/dt2

    So folks, Where is my mistake? Is there a fundamental conceptual problem differentiating? Or is it a small mistake somewhere?

    I apologize for my lack of understanding.
     
  10. Jan 5, 2017 #9

    BvU

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    You can't. You also cannot ignore dx2 in the original differential equation.

    Did you read Vela's #5 and #6 ? What is the reason you ignore those ?
     
  11. Jan 5, 2017 #10

    BvU

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    dy/dx is not a quotient. It is a differential quotient.
    You should get ##\displaystyle {d^2y\over dx^2} dx ## because ##\displaystyle{d\over dx}\left (\displaystyle{dy\over dx} \right ) = \displaystyle{d^2 y\over dx^2}##
     
  12. Jan 5, 2017 #11
    I didn't ignore, its just I read Euler Calculi differentialis and is the way I know to differenciate. I wanted to check if I got the right result but apparently I got it totally wrong and this was a disaster.
     
  13. Jan 5, 2017 #12

    Ray Vickson

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    Perhaps it would be better to eliminate the "differential" signs, and write the original DE as
    $$x^2 y''(x) + 2 x y'(x) +\frac{a^2}{x^2} y(x) = 0$$.
    You want to change variables to ##Y(t) = \left. y(x) \right|_{x=1/t} = y(1/t)##.

    The chain rule gives you
    $$Y'(t) = \left. y'(x) \right|_{x=1/t} \: \cdot \frac{d}{dt} \frac{1}{t} = -\frac{1}{t^2} y'(1/t)$$.
    In a similar way you can get ##Y''(t) = (d/dt) Y'(t)## in terms of ##y'(x)## and ##y''(x)##, evaluated at ##x = 1/t##.
     
  14. Jan 5, 2017 #13

    BvU

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    Not totally wrong, no disaster: the dx * d2y /dx2 is there alright !

    So far, the step to get ##\displaystyle {dy\over dx} ## in terms of ##t## has been successful. Next step is to get ##\displaystyle {d^2 y\over dx^2 }## using a similar procedure. As Ray says (and Vela and ..). Give it a try. ##\displaystyle {d^2 y\over dt^2 }## will pop up, don't worry.
     
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