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

I have tried the considering the following: Rewrite the $f^{\prime}$ term in order to obtain an equation of first order. But I get stuck in the substitutions as I do not know what to make of the $f$ term.

Any help or hint is greatly appreciated.

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First, $g(x):=f(x)-a$ gives us a simplification $$g''=g'g/b,\quad g(1/2) =0,\quad g'(1/2)=\sqrt {2\pi b}.$$ Next, change of variables $y=2x-1$, so the interval becomes $(-1,1)$, initial data is taken at zero.

$$h(y):=g((y+1)/2),\quad 4h''(y) = g'' ((y+1)/2),\quad \quad 2h'(y) = g' ((y+1)/2).$$ The equation becomes $$h''(y)= \frac{h(y)h'(y)}{2b},\quad h(0) =0,\quad h'(0)=\frac{\sqrt {2\pi b}}{2}. $$

We can integrate this equation: $$h'(y) = \frac{h^2}{4b}+C,\quad h(0) =0,\quad h'(0)=\frac{\sqrt {2\pi b}}{2}.$$ We find $C$ from initial conditions: $C= h'(0)=\frac{\sqrt {2\pi b}}{2}>0$. At last, we obtain something of the first order. Once we recall that $\tan x'= \tan^2 x+1$, the further direction of reasoning is clear. We make another scaling of variables; if $$h(y) = A\tan(By),$$then$$ h' (y) = AB(\tan^2(By)+1)= \frac BA h^2(y) +AB= \frac{h^2}{4b}+ \frac{\sqrt {2\pi b}}{2}.$$

All we have to do now is to solve $$\frac BA =\frac{1}{4b},$$ $$ AB= \frac{\sqrt {2\pi b}}{2},$$ plug $A$, $B$ back to $h$ and revert all our changes of variables and translations back to $f$.

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substitute values of $f(\frac{1}{2})$ &$f'(\frac{1}{2})$ on the above equation and you get $f''(\frac{1}{2}) = 0$.approximating using Taylor expansion for $f(x)$ centered at $x = \frac{1}{2}$ is, $$f(x) = \frac{f( \frac{1}{2} )}{0!} + \frac{f'( \frac{1}{2} )}{1!}x + \frac{f''( \frac{1}{2} )}{2!}x^2 $$ or $$f(x)_{x = \frac{1}{2}} \approx a + x\sqrt{2πb}$$

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If we think of $f$ as the independent variable (this step might take some justification), we can write $f'(x)=g(f)$. Then, by the chain rule, $f''(x) =\frac{dg}{dx} = \frac{dg}{df}\frac{df}{dx}$

Using this substitution, we can rewrite the DE as

$$ \frac{dg}{df}f'(x)-\frac{f-a}{b}f'(x)=0\\ \frac{dg}{df}=\frac{f-a}{b} $$

This is a separable DE, which we can solve by integration. This gives us a solution for $g$. We can then use this to perform a substitution which will reduce the order of the original DE by 1. The resulting equation will also be separable, and can be solved by integration to find the general solution.

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  • $\begingroup$ I am not sure how to process this information. By the chain rule, $(f^{\prime}\circ f)^{\prime}=(f^{\prime\prime}\circ f)\cdot f^{\prime}$. How is this equal to $f^{\prime\prime}$ and furthermore, how does this help me reduce the equation? $\endgroup$ – semicolon Aug 22 '13 at 15:28
  • $\begingroup$ I am not even sure that the expression $f^\prime\circ f$ makes sense for all $x\in(0,1)$, since $f$ may take values outside the domain of $f^\prime$. $\endgroup$ – semicolon Aug 22 '13 at 15:37
  • $\begingroup$ I meant $f'' = \frac{df'}{dx} = \frac{df'}{df}\frac{df}{dx}$. I'll elaborate a bit more in my answer. $\endgroup$ – hasnohat Aug 22 '13 at 17:23
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The solution by TZakrevskiy is nice. I will add that after the same change of variables, he suggests, to obtain: $$ g'' = \frac{g}{2b} g'$$ with boundary conditions of: $$ g(0)=0$$ and $$ g'(0)=\frac{\sqrt{2 \pi b}}{2} $$ you can apply the general solution of that nonlinear ODE as given in the handbook by Polyanin and Zaitsev:

For a general equation of the form $g''=f(g)g'$ and $g=g(x)$ the solution is:

$$ \int \frac{dg}{F(g) + C_1}=C_2 + x $$

where $C_1$ and $C_2$ are the integration constants and $F(g)=\int f(g) dg$. Applying this will yield the same $tan$ based solution as noted above.

Cheers,

Paul Safier

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