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I have problems finding out whether this initial value problem has an explicit form solution or if it is possible to grind out a term-by-term representation of this solution using power series expansions.

\begin{equation}f^{\prime\prime}(x)-\frac{f(x)-a}{b}f^{\prime}(x)^2=0,\quad x\in(0,1),\qquad f(1/2)=a,\,f^{\prime}(1/2)=\sqrt{2\pi b}.\end{equation} where $a\in\mathbb{R}$ and $b>0$ are constants.

Attempt: Find a solution for the easier problem, \begin{equation}f^{\prime\prime}(x)-\frac{f(x)-a}{b}f^{\prime}(x)=0,\quad x\in(0,1),\qquad f(1/2)=a,\,f^{\prime}(1/2)=\sqrt{2\pi b}.\end{equation} where $a\in\mathbb{R}$ and $b>0$ are constants. The solution for the easy problem is\begin{equation}f(x)=a+\tan\left[\frac{\sqrt{2b\sqrt{2\pi b}}(x-1/2)}{2b}\right]\sqrt{2b\sqrt{2\pi b}} \end{equation} Now, solve the original system with the $f^{\prime}(x)^2$ replacing the easier $f^{\prime}(x)$. However, I can't make sense of substituting the squared term into my calculations...

Any help or hint is greatly appreciated.

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  • $\begingroup$ Divide the equation by $f'(x)$ and then try to represent the left side as the derivative of something (it will not be difficult). $\endgroup$ – Start wearing purple Aug 23 '13 at 10:12
  • $\begingroup$ Thank you. This will give me the equation: \begin{equation}(\log \circ f^{\prime})^{\prime}(x)-\frac{f(x)-a}{b}f^{\prime}(x)=0,\qquad x\in(0,1)\end{equation} with the same initial conditions. I think I need more help on where to go from here. $\endgroup$ – semicolon Aug 23 '13 at 10:16
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    $\begingroup$ Right. And $(f(x)-a)f'(x)=(f(x)^2/2-a f(x))'$. Hence you obtain $(\mathrm{something})'=0$. $\endgroup$ – Start wearing purple Aug 23 '13 at 10:20
  • $\begingroup$ I get that something-equation to look like this:\begin{equation}\left[(\log\circ f^{\prime})(x)-\frac{1}{2b}f(x)^2-\frac{a}{b}f(x)\right]^{\prime}=0.\end{equation} $\endgroup$ – semicolon Aug 23 '13 at 10:33
  • $\begingroup$ I get that something-equation to look like this:\begin{equation}[(\log\circ f^{\prime})(x)-\frac{1}{2b}f(x)^2-\frac{a}{b}f(x)]^{\prime}=0.\end{equation} Finding an antiderivative on both sides of the equality gives us \begin{equation} (\log\circ f^{\prime})(x)-\frac{1}{2b}f(x)^2-\frac{a}{b}f(x)-C=0. \end{equation} Using the initial conditions gives $C=\log\sqrt{2\pi b}-\frac{3}{2}\frac{a^2}{b}$, which renders the following equation: \begin{equation} \log(f^\prime(x))-\frac{1}{2b}f(x)^2-\frac{a}{b}f(x)-\log\sqrt{2\pi b}+\frac{3}{2}\frac{a^2}{b}=0 \end{equation} $\endgroup$ – semicolon Aug 23 '13 at 10:40
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Let us make the change of variables: $$g(x)=\frac{f(x)-a}{\sqrt{b}},$$ which implies $g'(x)=f'(x)/\sqrt{b}$, $g''(x)=f''(x)/\sqrt{b}$. The equation for $g(x)$ is $$g''(x)-g(x)g'(x)^2=0.$$

  • Dividing this equation by $g'(x)$, we find $$\frac{g''(x)}{g'(x)}-g(x)g'(x)=\left(\ln g'(x)-\frac12 g(x)^2\right)'=0.$$ Hence $$\ln g'(x)-\frac12 g(x)^2=\mathrm{const}.\tag{1}$$

  • The constant can be fixed using the initial conditions $g(1/2)=0$, $g'(1/2)=\sqrt{2\pi}$. Substituting these values in (1), one finds that $$\ln g'(x)-\frac12 g(x)^2=\ln\sqrt{2\pi},$$ or, in another form $$g'(x)=\sqrt{2\pi}\,e^{g(x)^2/2}.\tag{2}$$

  • The equation (2) is separable. In particular, it implies that $$\frac{1}{\sqrt{2\pi}}\int_{g(1/2)}^{g(x)}e^{-h^2/2}dh=\int_{1/2}^xdt.$$ Applying the initial conditions once more, one obtains $$\frac{1}{\sqrt{2\pi}}\int_{0}^{g(x)}e^{-h^2/2}dh=x-\frac12.$$

  • The integral on the left is expressed in terms of the error function: $$\frac12\operatorname{erf}\frac{g(x)}{\sqrt{2}}=x-\frac12.$$ Hence $g(x)=\sqrt{2}\operatorname{erf}^{-1}(2x-1)$, and finally $$f(x)=a+\sqrt{2b}\operatorname{erf}^{-1}(2x-1). \tag{3}$$

The answer (3) is given in terms of special function $\operatorname{erf}^{-1}$, but one can, for example, write it in the form of a series (around $x=1/2$) with recursively determined coefficients, see e.g. here.

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