Question

Change the independent variable from $x$ to $u=2\sqrt{x}$ in the Bessel equation $$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-(1-x)y=0$$ and show that the equation becomes $$u^2\frac{d^{2}y}{d{u}^2}+u\frac{dy}{du}+(u^2-4)y=0$$

Short answer

The equation simplifies to $u^2\frac{d^{2}y}{d{u}^2}+u\frac{dy}{du}+(u^2-4)y=0$.

Step-by-step solution

- Substitute the Variable

Start by substituting the new variable into the equation. Let $u=2\sqrt{x}$. Given this substitution, express $x$ in terms of $u$: $x=\frac{u^2}{4}$.

- Compute $\frac{d}{dx}$ Using Chain Rule

The chain rule provides $\frac{d}{dx}=\frac{du}{dx}\frac{d}{du}$. Hence, find $\frac{du}{dx}$:$$u=2\sqrt{x}\Rightarrow \frac{du}{dx}=\frac{1}{\sqrt{x}}\Rightarrow \frac{du}{dx}=\frac{2}{u}$$. Therefore, $$\frac{d}{dx}=\frac{2}{u}\frac{d}{du}$$.

- Compute $\frac{d^2}{d{x}^2}$

Use the chain rule again for the second derivative: $$\frac{d^2}{d{x}^2}=\frac{d}{dx}\left(\frac{d}{dx}\right)=\frac{d}{dx}\left(\frac{2}{u}\frac{d}{du}\right)$$. Expand this: $$\frac{d^2}{d{x}^2}=\frac{2}{u}\frac{d}{du}\left(\frac{2}{u}\frac{d}{du}\right)=\frac{2}{u}(\frac{2d^2}{udu}-\frac{2d}{u^{2}du})=\frac{4}{u^2}\frac{d^2}{d{u}^2}-\frac{4}{u^3}\frac{d}{du}$$.

- Substitute Into Original Bessel Equation

Replace $x$ and its derivatives with $u$ and its corresponding terms. The original Bessel equation is $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-(1-x)y=0$. Substituting, we get: $${\left(\frac{u^2}{4}\right)}^2(\frac{4}{u^2}\frac{d^2}{d{u}^2}-\frac{4}{u^3}\frac{d}{du})y+\frac{u^2}{4}\left(\frac{2}{u}\frac{dy}{du}\right)-(1-\frac{u^2}{4})y=0$$.

- Simplify the Expression

Simplify the equation from Step 4: $$(\frac{u^4}{16}\cdot \frac{4}{u^2}\frac{d^{2}y}{d{u}^2}-\frac{u^4}{16}\cdot \frac{4}{u^3}\frac{dy}{dy})+\left(\frac{u^2}{4}\frac{2}{u}\frac{dy}{du}\right)-(1-\frac{u^2}{4})y=0$$. By combining terms, it reduces to $$u^2\frac{d^{2}y}{d{u}^2}+u\frac{dy}{du}+(u^2-4)y=0$$.

Key concepts

Variable Substitution

Variable substitution is a powerful technique in solving differential equations. In this context, we start with the Bessel equation: $$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-(1-x)y=0$$ Instead of solving it directly, we simplify the problem by changing the independent variable from $x$ to $u=2\text{sqrt}(x)$. This makes the equation more manageable because it can simplify complex expressions. To do this, express $x$ in terms of $u$: $$x=\frac{u^2}{4}$$This substitution merely helps in approaching the differential equation from a different perspective, making later steps, such as differentiation, more straightforward.

Chain Rule

The chain rule is essential when dealing with derivatives after a variable substitution. It helps us relate the derivatives with respect to different variables.When we substitute $u=2\text{sqrt}(x)$ into our differential equation, we modify the derivatives accordingly. First, we find the first derivative: $$\frac{du}{dx}=\frac{2}{u}$$. This translates into $$\frac{d}{dx}=\frac{2}{u}\frac{d}{du}$$The second-order derivative requires applying the chain rule twice: $$\frac{d^2}{d{x}^2}=\frac{d}{dx}(\frac{2}{u}\frac{d}{du})=\frac{2}{u}(\frac{2}{u}\frac{d^2}{d{u}^2}-\frac{2}{u^2}\frac{d}{du})=\frac{4}{u^2}\frac{d^2}{d{u}^2}-\frac{4}{u^3}\frac{d}{du}$$Hence, applying the chain rule helps us accurately transform the differential operators, crucial to changing the Bessel equation's variable.

Bessel Differential Equations

Bessel differential equations appear in various fields like physics and engineering, especially when dealing with cylindrical or spherical symmetry. Here, our goal is to show that the original Bessel equation retains its form after the variable substitution.The original equation is $$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-(1-x)y=0$$With the substitution $u=2\text{sqrt}(x)$ and changing our derivatives using the chain rule, we replace $x$ and its derivatives. Resulting transformations give us: First derivative: $$\frac{dy}{dx}=\frac{2}{u}\frac{dy}{du}$$Second derivative: $$\frac{d^{2}y}{d{x}^2}=\frac{4}{u^2}\frac{d^{2}y}{d{u}^2}-\frac{4}{u^3}\frac{dy}{du}$$Substituting these back into the original equation, we simplify:$$(\frac{u^4}{16}\frac{4}{u^2}\frac{d^{2}y}{d{u}^2}-\frac{u^4}{16}\frac{4}{u^3}\frac{dy}{du})+(\frac{u^2}{4}\frac{2}{u}\frac{dy}{du})-(1-\frac{u^2}{4})y=0$$Simplifying it gives us: $$u^2\frac{d^{2}y}{d{u}^2}+u\frac{dy}{du}+(u^2-4)y=0$$This is effectively the same form as the initial Bessel equation, only transformed to the new variable $u$.