Question

Solve this equation
d²Bz/dr² + dBz/rdr - Bz/a² = 0
Where a=(m/u ne²)½

Answer 1

Answer on Question #62954 – Math – Differential Equations

Question

Solve this equation

$$\frac {d ^ {2} B _ {z}}{d r ^ {2}} + \frac {1}{r} \frac {d B _ {z}}{d r} - \frac {B _ {z}}{a ^ {2}} = 0,$$

where $a = (\mathrm{m / u~ne}^2) / 2$

Solution

$$\frac {d ^ {2} B _ {z}}{d r ^ {2}} + \frac {1}{r} \frac {d B _ {z}}{d r} - \frac {B _ {z}}{a ^ {2}} = 0$$

Multiply equation by $r^2$, we get

$$r ^ {2} \frac {d ^ {2} B _ {z}}{d r ^ {2}} + r \frac {d B _ {z}}{d r} - \frac {r ^ {2}}{a ^ {2}} B _ {z} = 0$$

Let $\frac{r}{a} = x$, $\frac{dB_z}{dx} = \frac{dB_z}{dr} \cdot \frac{dr}{dx} = a \frac{dB_z}{dr}$, $\frac{d^2B_z}{dx^2} = a^2 \frac{d^2B_z}{dr^2}$, then the equation will become

$$x ^ {2} \frac {d ^ {2} B _ {z}}{d x ^ {2}} + x \frac {d B _ {z}}{d x} - x ^ {2} B _ {z} = 0$$

This equation is the modified Bessel's differential equation (at $\alpha = 0$)

$$x ^ {2} \frac {d ^ {2} y}{d x ^ {2}} + x \frac {d y}{d x} - (x ^ {2} + \alpha^ {2}) y = 0$$

having particular solutions $I_{\alpha}(x)$ and $K_{\alpha}(x)$ which are the modified Bessel's functions (the Bessel's functions of a purely imaginary argument) of the first and second kind respectively. In our case $\alpha = 0$, then particular solutions of the equation are $I_0(x)$ and $K_0(x)$.

The general solution is

$$B _ {z} = C I _ {0} (x) + D K _ {0} (x),$$

where $C$ and $D$ are real constants

Replacing $x$ with $\frac{r}{a}$ we get

$$B _ {z} = C I _ {0} \left(\frac {r}{a}\right) + D K _ {0} \left(\frac {r}{a}\right)$$

Answer: $B_{z} = CI_{0}\left(\frac{r}{a}\right) + DK_{0}\left(\frac{r}{a}\right)$.

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