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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.