Euler equations

characteristic polynomial

$$\left[r(r-1) + a r + b\right] x^r = 0.$$

$$\tag{2} r(r-1) + a r + b = 0.$$

(2) 這個方程就是 Euler equations 的特徵多項式. 理論上如果 (2) 的根是 $r_1$, 那 $y = x^{r_1}$ 就是方程式的解.

A. 兩相異實根

$$\tag{3} y(x) = c_1 x^{r_1} + c_2 x^{r_2},$$ 其中 $c_1$, $c_2$ 為常數.

備註:

$$x^r = e^{r\ln x}.$$ 這樣即使 $r$ 是無理數, 我們也可以輕易地計算 $x^r$.

B. 兩複數根

\begin{align} x^{\lambda + i\mu} &= e^{(\lambda + i\mu)\ln x} \\ &= e^{\lambda \ln x}\left[\cos(\mu\ln x) + i\sin(\mu\ln x)\right] \\ &= x^r\cos(\mu\ln x) + ix^r\sin(\mu\ln x). \end{align}

$$\tag{4} y(x) = c_1 x^{\lambda}\cos(\mu\ln x) + c_2 x^{\lambda}\sin(\mu\ln x),$$ 其中 $c_1$, $c_2$ 為常數.

C. 一重根

$$\tag{5} x^2 y’’ + (1 - 2r_1) xy’ + r^2_1 y = 0.$$

$$\frac{d}{dx} y_1 = r_1 x^{r_1-1} = \frac{r_1}{x} x^{r_1} = \frac{r_1}{x} y_1.$$

$$\tag{6} \left(x\frac{d}{dx} - r_1\right) y = 0.$$

Observation 1

$$\tag{7} \left(x\frac{d}{dx} - r_1\right)\left(x\frac{d}{dx} - r_1\right) y = 0.$$

Observation 2

$$\tag{8} \left(x\frac{d}{dx} - r_1\right) y_2 = cy_1,$$ 其中 $c$ 是個常數. 那我們就可以得到

$$\left(x\frac{d}{dx} - r_1\right)\left(x\frac{d}{dx} - r_1\right) y_2 = \left(x\frac{d}{dx} - r_1\right) cy_1 = 0.$$

Observation 3

$$\tag{9} xu’ = cu.$$

$$\tag{10} y(x) = c_1 x^{r_1} + c_2 x^{r_1}\ln(x),$$ 其中 $c_1$, $c_2$ 為常數.

Te-Sheng Lin (林得勝)
Associate Professor

The focus of my research is concerned with the development of analytical and computational tools, and further to communicate with scientists from other disciplines to solve engineering problems in practice.