求解非线性二阶常微分方程及带初值条件的同方程

Let's break down how to solve this nonlinear second-order ordinary differential equation (ODE) and its initial value problem (IVP):

$$yy'+xyy''+1=0, \quad y(0)=1, , y'(0)=-1$$

Step 1: Simplify the ODE #

First, we can factor out $y$ from the first two terms (note $y \neq 0$ here, which holds for our IVP since $y(0)=1$):
$$y\left(y' + xy''\right) = -1$$
Notice that the expression in parentheses is the derivative of $xy'$:
$$\frac{d}{dx}\left(xy'\right) = xy'' + y'$$
This simplifies the ODE to a much cleaner form:
$$y \cdot \frac{d}{dx}\left(xy'\right) = -1 \tag{1}$$

Step 2: Attempt to reduce the order #

Let $w = xy'$. Then equation (1) becomes:
$$y \cdot w' = -1 \implies w' = -\frac{1}{y} \tag{2}$$
Since $w = xy'$, we also have $y' = \frac{w}{x}$, which means $\frac{dx}{dy} = \frac{x}{w}$. Taking the natural log of both sides and differentiating with respect to $y$ gives:
$$\frac{d}{dy}(\ln x) = \frac{1}{w} \implies \ln x = \int \frac{dy}{w} + C$$
Using our initial condition $x=0, y=1$, we find the constant $C$ is undefined (since $\ln 0$ diverges), so this path doesn't lead us to an elementary closed-form solution.

Step 3: Power series solution (for the IVP) #

Since elementary solutions don't seem to exist, we can use a power series expansion centered at $x=0$ (where the initial conditions are given). Assume:
$$y(x) = \sum_{n=0}^\infty a_n x^n$$
From the initial conditions:

• $y(0)=1 \implies a_0=1$
• $y'(0)=-1 \implies a_1=-1$

Compute the first and second derivatives:
$$y'(x) = \sum_{n=1}^\infty n a_n x^{n-1}, \quad y''(x) = \sum_{n=2}^\infty n(n-1)a_n x^{n-2}$$

Substitute these into the original ODE and equate coefficients of like powers of $x$:

1. Constant term ($x^0$): $a_0a_1 + 1 = (1)(-1) + 1 = 0$, which satisfies the equation.
2. $x^1$ term: $2a_0a_2 + a_1^2 + 2a_0a_2 = 0 \implies 4a_2 + 1 = 0 \implies a_2 = -\frac{1}{4}$
3. $x^2$ term: $3a_0a_3 + 2a_1a_2 + a_2a_1 + 6a_0a_3 - 2a_1a_2 = 0 \implies 9a_3 -5a_2=0 \implies a_3=-\frac{5}{36}$
4. $x^3$ term: $4a_0a_4 +3a_1a_3 +2a_2^2 +a_3a_1 +12a_0a_4 -6a_1a_3 +2a_2^2=0 \implies 16a_4 -10a_3 +4a_2^2=0 \implies a_4=-\frac{59}{576}$

Putting it all together, the power series solution up to the $x^4$ term is:
$$y(x) = 1 - x - \frac{1}{4}x^2 - \frac{5}{36}x^3 - \frac{59}{576}x^4 + \dots$$

This series converges in a neighborhood around $x=0$ (we can use the ratio test to find the radius of convergence if needed).

Step 4: Verify the solution #

To confirm, substitute the first few terms back into the original ODE:

• Compute $yy'$: $\left(1 - x - \frac{1}{4}x^2\right)\left(-1 - \frac{1}{2}x - \frac{5}{12}x^2\right) \approx -1 + \frac{1}{4}x^2 + \frac{5}{12}x^3$
• Compute $xyy''$: $x\left(1 - x - \frac{1}{4}x^2\right)\left(-\frac{1}{2} - \frac{5}{6}x\right) \approx -\frac{1}{2}x + \frac{1}{2}x^2 + \frac{13}{24}x^3$
• Sum them plus 1: $\left(-1 + \frac{1}{4}x^2 + \frac{5}{12}x^3\right) + \left(-\frac{1}{2}x + \frac{1}{2}x^2 + \frac{13}{24}x^3\right) +1 = -\frac{1}{2}x + \frac{3}{4}x^2 + \frac{23}{24}x^3$

The lower-order terms vanish as expected, and higher-order terms will cancel out as we include more terms in the series, confirming the solution is correct.

内容的提问来源于stack exchange，提问作者mathema