Can we ignore terms of differential equation if coefficients go to zero at a point?

Question

Given a differential equation with variable coefficients, if at a given point some coefficients go to zero, can I take them to be zero and assume that the answer is correct for that point?

I mean, let me give a concrete example: considering the following differential equation: $$ a(x)\frac{dy}{dx}+b(x)\left(\frac{dy}{dx}\right)^{-1/2}+c(x)\left(\frac{dy}{dx}\right)^{1/2}+d(x)\left(\frac{dy}{dx}\right)^{-1}=0~. $$ if at a given point, say $x=0$, the coefficients $c(x)$ and $d(x)$ are zero, can I solve the simplified equation: $$ a(x)\frac{dy}{dx}+b(x)\left(\frac{dy}{dx}\right)^{-1/2}=0~, $$ and get a valid result for $y(0)$? Moreover, does the validity of the result depends of, by solving the simplified equation, we get $\frac{dy}{dx}(0)=0$, $\frac{dy}{dx}(0)=\infty$ or $\frac{dy}{dx}(0)=\text{const.}$?

Answer 1

I assume that all functions are differentiable and smooth wherever required.

The equation of interest has form

$$f(y'(x), a(x), b(x), c(x), d(x)) = 0,$$

with $c(0) = d(0) = 0$ and the subject of interest are solutions near $x = 0.$

Assuming that $\left.\frac{\partial f}{\partial y'}\right|_{x = 0} \neq 0,$ by implicit function theorem there is a smooth function of four variables $g(a, b, c, d)$ so that $$y'(x) = g(a(x), b(x), c(x), d(x))$$ in some neighborhood of $x= 0.$

Using continuity, it is possible to express $y'(x)$ in the simplified equation as $$y'(x) = g(a(x), b(x), c(0) + o(1), d(0) + o(1) = g(a(x), b(x), c(0), d(0)) + o(1)$$

and integrate the resulting equation as $$y(x) = y(0) + \int\limits_0^x g\left(a(\xi), b(\xi), c(0), d(0)\right) d\xi + o(x) \equiv \tilde{y}(x) + o(x).$$

So the solution of the simplified equation has error $o(x)$ (see more about little-o notation).

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However, it is interesting to get more precise estimation of the error term than just $o(x).$

Using Taylor expansion, it is possible to obtain that $$y'(x) = g(a(x), b(x), c(0) + c'(0)x + o(x), d(0) + d'(0)x + o(x)) = \\ g(a(x), b(x), c(0), d(0)) + \left. \frac {\partial g(a(x), b(x), c(x), d(0))} {\partial c(x)} \right | _{c(x) = c(0)}\cdot c'(0)x + \left. \frac {\partial g(a(x), b(x), c(0), d(x))} {\partial d(x)} \right | _{d(x) = d(0)}\cdot d'(0)x + o(x). $$

Using smoothness of $g(a, b, c, d),$ $a(x),$ and $b(x),$ it is possible to approximate the derivatives of $g$ by their values at $a(x) = a(0)$ and $b(x) = b(x)$ keeping the same total error term $o(x):$

$$y'(x) = g(a(x), b(x), c(0), d(0)) + \left. \frac {\partial g(a(0), b(0), c(x), d(0))} {\partial c(x)} \right | _{c(x) = c(0)}\cdot c'(0)x + \left. \frac {\partial g(a(0), b(0), c(0), d(x))} {\partial d(x)} \right | _{d(x) = d(0)}\cdot d'(0)x + o(x).$$

Getting back to the implicit function theorem, it is possible to express derivates of $g(a, b, c, d)$ using derivatives of $f(y', a, b, c, d)$ as

$$\frac{\partial g}{\partial c} = -\frac{\partial f}{\partial c} \left / \frac{\partial f}{\partial y'}\right.,\ \frac{\partial g}{\partial d} = -\frac{\partial f}{\partial d} \left / \frac{\partial f}{\partial y'}\right..$$ in some neighbourhood of $x = 0.$

So $$y'(x) = g(a(x), b(x), c(0), d(0)) - \frac{ \left.\frac{\partial f}{\partial c}\right|_{x = 0}c'(0) + \left.\frac{\partial f}{\partial d}\right|_{x = 0}d'(0)} {\left.\frac {\partial f}{\partial y'}\right|_{x = 0}}x + o(x).$$

Integrating it by $x$ it is easy to obtain that $$y(x) = \tilde{y}(x) - \frac{ \left.\frac{\partial f}{\partial c}\right|_{x = 0}c'(0) + \left.\frac{\partial f}{\partial d}\right|_{x = 0}d'(0)} {2 \left.\frac {\partial f}{\partial y'}\right|_{x = 0}}x^2 + o(x^2)$$ where $\tilde{y}(x)$ is the solution of the simplified equation.

So the solution of the simplified equation is close to the solution of the original one only when $\left|x\right| \ll {\sqrt{\left|\alpha\right|}},$ where $$\alpha = \left.\frac {2\frac{\partial f}{\partial y'}}{\frac{\partial f}{\partial c}c'(x)+\frac{\partial f}{\partial d}d'(x)}\right|_{x = 0}.$$