Solving second order nonlinear equation

Question

\begin{align*} y''(x)=\lambda \frac{y}{1+y} \quad \textrm{with boundary Condition: }\quad y(1)=1,y'(0)=0 \end{align*}

My idea: By multiplying $y'$ at each side, I try to take integral. However, I stuck this method...

I could not find the direction how to solve this equation..

Any help is appreciated..

Thank you!

Answer 1

First, there is a trivial solution: $y(x) = 1$ for $\lambda = 0$. Continue for $\lambda\ne 0$

Multiply through by $2y'$

$$ 2y''y' = 2\lambda \left[1-\frac{1}{1+y}\right] $$

We can integrate the above as

$$ (y')^2 = 2\lambda\big[y - \ln(1+y) - c\big] $$

where $c$ is some arbitrary constant. We can integrate using the condition $y(1)=1$ to get the inverse function

$$ x(y) = 1 \pm \int_1^y \frac{1}{\sqrt{2\lambda\big[t - \ln(1+t) - c\big]}} dt $$

Now, if $y(0) = y_0$, then $y_0$ must satisfy

$$ y_0 - \ln(1 + y_0) = c \tag{1} $$

$$ 1 \pm \int_1^{y_0} \frac{1}{\sqrt{2\lambda\big[t - \ln(1+t) - c\big]}} dt = 0 \tag{2} $$

From here, numerical methods must be used to determine $y_0$

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Discussion: Let's break this down further

$$ f(y) = y - \ln(1+y) - c $$

Then $f'(0) = 0$ and $f''(0) > 0$, which means $y=0$ is a local minimum, and $f(y) \ge -c, \forall y > -1$. This means $f(y)$ has

• No roots if $c < 0$
• One root at $y = 0$ if $c=0$
• Two distinct roots if $c > 0$: One positive, one negative.

If $c=0$, then the integral in $(2)$ will not converge at $y_0=0$, so we restrict $c > 0$. Let's denote the two roots as $y_0^+ > 0$ and $y_0^- < 0$, respectively. Then we have

• $y - \ln(1+y) - c < 0$ for $y \in (y_0^-,y_0^+)$
• $y - \ln(1+y) - c > 0$ otherwise

Note that the equation for $y'(x)$ actually has two square roots. Whether we take one, or both depends on the value of $\lambda$. There are two cases to consider

Case 1: $\lambda > 0$: The domain of convergence is either $(-\infty,y_0^-)$ or $(y_0^+, \infty)$. Since the domain needs to include $y=1$, we discard the first one, and require $y_0^+ < 1$, or $c < 1-\ln 2$.

With this restriction, the negative square root returns no roots, since

$$ 1 - \int_1^{y_0^+} \frac{1}{\sqrt{2\lambda\big[t - \ln(1+t) - c\big]}} dt = 1 + \int_{y_0}^1 \frac{1}{\sqrt{2\lambda\big[t - \ln(1+t) - c\big]}} dt > 0 $$

Thus we can solve

$$ 1 - \int_{y_0^+}^1 \frac{1}{\sqrt{2\lambda\big[t - \ln(1+t) - c\big]}} dt = 0 $$

and taking the positive square root in the solution.

Here's an example plot for $\lambda = 2$, with $y(0) = 0.602$ and $c = 0.131$. Note that taking the positive square root means $y' \ge 0$ and the solution grows without bounds.

Case 2: $\lambda < 0$: The domain of convergence is $y \in (y_0^-, y_0^+)$. Therefore, $y_0^+ > 1$ and $c > 1 - \ln 2$. This also means that $y'=0$ at both endpoints. Hence, both square roots will return a (different) solution.

• If we want $y(0) = y_0^+$, then we must take the negative square root

$$ 1 - \int_1^{y_0^+} \frac{1}{\sqrt{2\lambda\big[t - \ln(1+t) - c\big]}} dt = 0 $$

• If we want $y(0) = y_0^-$, then we must take the positive square root

$$ 1 + \int_1^{y_0^-} \frac{1}{\sqrt{2\lambda\big[t - \ln(1+t) - c\big]}} dt = 0 $$

Here's an example plot for $\lambda = -2$. The positive square root (or increasing solution) has $y(0) = -0.874$ and $c = 1.197$. The negative square root (or decreasing solution) has $y' \le 0$, has $y(0) = 1.602$ and $c = 0.643$

Note that periodic extension can be applied to this solution, like so.

But you can ignore this if you only care about $x \in [0,1]$

Answer 2

$y'= p(y) =p$ , $y''=p' (y) y' =p'p$ hence $$p'p =\lambda\frac{y}{1+y}$$ $$pdp =\lambda\frac{y}{1+y}dy$$ $$\frac{p^2}{2} =\lambda (y-\ln |1+y |) +c$$ hence $$\int_{y(x_0 )}^{y(x) }\frac{dy}{\sqrt{2\lambda (y-\ln |1+y |) +2c} }=x-x_0$$