I have a differential equation which contains a constant $a$ and a function $q(x)$. I'm interested how the second factor affects the result of the function $y(x)$ if it would be cancelled out, ie given

$$ y(x)' = \sqrt{a-q(x)} - \dfrac{q'(x)}{4(a-q(x))} \sin(2y(x)) $$

with some boundary conditions, and the same differential equation without the fraction

$$ y'(x) = \sqrt{a-q(x)} \, \, \,.$$

The constant $a$ is usually a large number. Are there some useful thereoms that can give us some conditions on the $q(x)$ function such that the difference between the solution of the two differential equations become smaller if $a$ grows?


For instance, let $q(x) = \dfrac{1}{4x^2}$ where $x \in ]0,1]$ and $a = 100886$. In this example, the difference of the (numerical) solution of the two differential equations keeps growing as $a$ becomes larger. Vice versa, $q(x)=x^2$ with $x \in [-10,10]$ gives asymptotically good results.

  • $\begingroup$ it seems too general. can you make a particular case of it. i.e. pick some numbers, functions. $\endgroup$ – abel Feb 9 '15 at 15:04

Given the differential equations $$ y' = f + g \quad (1) $$ and $$ y' = f \quad (2) $$ and introducing $$ y' = g \quad (3) $$ Using the linearity of the derivative a solution $y_2$ of $(2)$ and a solution $y_3$ of $(3)$ will obviously lead to a solution $$ y_1 = y_2 + y_3 \Rightarrow y_1' = (y_2 + y_3)' = y_2' + y_3' = f + g $$ of $(1)$ if the boundary conditions of $(1)$ and $(2)$ are compatible and transferred to $(3)$, e.g. $$ 0 = B y_1 = B (y_2 + y_3) = B y_2 + B y_3 $$

$y_3$ is the difference in trajectories $$ y_1 - y_2 = y_3 $$ So go solve $(3)$.


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