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I have the function $$f(x) = \int\limits_{k=0}^x e^{f(k)} dk$$

How can I solve for an explicit formula of $f$? In my current solution, it becomes undefined past a certain value of $x$

Atempts: Taking a derivative, we have $$f'(x) = e^f(x)$$ $$\Longleftrightarrow$$ $$\ln f'(x) = f(x)$$ Taking a second derivative, $$\frac{f''(x)}{f'(x)} = f'(x)$$ $$\Longleftrightarrow$$ $$f'(x)^2 = f''(x)$$ Similarly, $$f^{(n)}(x) = (n-1)!f'(x)^n$$

Now, by the Taylor expansion (and substituting $g(x)=f'(x)$): $$g(x) = g(0)+xg'(0)+\frac12x^2g''(0)+\cdots$$ $$=g(0)+xg(0)^2+x^2g(0)^3+\cdots$$ $$=g(0)[1+xg(0)+x^2g(0)^2+\cdots]$$ $$=\frac{g(0)}{1-xg(0)}$$ Therefore, $$f(x)=\int g(x) = \int \frac{f'(0)}{1-xf'(0)}$$ $$\Longleftrightarrow$$

$$f(x) = f(0)-ln(1-xf'(0))$$

So, it seems I have a solution. But what happens when $x>\frac{1}{f'(0)}?$

PS: Original problem is from ocw, #2 on this link: https://ocw.mit.edu/courses/physics/8-01sc-classical-mechanics-fall-2016/assignments/MIT8_01F16_pset4.pdf

In this case, $f'(0) < 0$, and $x\ge 0$, hence $f(x)$ is always defined.

Lutz Lehmann
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programjames
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1 Answers1

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You can solve this much more simply. First you read off the original integral equation also the initial condition $f(0)=0$.

Then by the method of separation of variables $$ f'(x)=e^{f(x)}\implies e^{-f(x)}f'(x)=1\\ \implies e^{-f(x)}-e^{-f(0)}=-x,\\f(x)=-\ln(1-x). $$

Lutz Lehmann
  • 126,666
  • I don't see how you got the middle equation. – programjames Aug 20 '19 at 18:43
  • @programjames, The second implication follows from applying the fundamental theorem of calculus to $(e^{-f})' = -f' e^{-f}$, or equivalently $\int f' e^{-f} , \mathrm{d}x = -e^{-f} + \mathsf{const}$. Also, this answer tells that there is no way we can 'naturally' extend $f$ beyond $x < 1$. This kind of blow-up phenomenon is quite common in ODE business. – Sangchul Lee Aug 20 '19 at 18:46