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If $c=f(a+e^b)+g(a-e^b)$ where $f$ and $g$ are functions of $a+b^2$ and $a-b^2$ respectively, find $c$ such that when $b=0$, you find that $c=0$ and $\frac{\partial c}{\partial b}=1+a$.

Jay Christian
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  • Sorry for being unclear @T.Bongers, I meant that $c=0$ when evaluated at $b=0$. $c$ is a function of $a$ and $b$. I would like to find the function – Jay Christian Sep 05 '13 at 19:46
  • Wow, that's a drastic change, from $a \pm b^2$ to $a \pm e^b$... Not nice. – Kaster Sep 05 '13 at 20:03
  • Yeah sorry, I got a little confused... but it is right now, although I still can't solve it! – Jay Christian Sep 05 '13 at 20:09
  • Where did you get that problem? I might be wrong, but your restrictions seem inconsistent. – Kaster Sep 05 '13 at 20:12
  • I found it on a list of problems in an old folder from my university days. I am revisiting my notes after 15 years! Maybe they are inconsistent? Can you find solutions if you relax one restriction, say the structure of $c$, and have $c(a,b)$ instead of $c(a+e^b,a-e^b)$? Do you think its a typo and has no solution? – Jay Christian Sep 05 '13 at 20:32
  • sorry, my bad, overlooked sign somehow. I posted my answer. – Kaster Sep 05 '13 at 21:02

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You say, that $$ c(a,b) = f(a+e^b) + g(a-e^b) $$ where $f, g : \mathbb R \to \mathbb R$, and you need to find some particular form of $c(a,b)$ so that \begin{align} c(a,0) &= 0\\ \left . \frac {\partial c}{\partial b} \right |_{b = 0} &= 1 + a \end{align} Use first restriction. $$ c(a,0) = f(a+1) + g(a-1) = 0 \tag 1 $$ Now, let's find the form of of $c_b(a,b)$ $$ c_b(a,b) = f'(a+e^b)e^b - g'(a-e^b)e^b = e^b \left ( f'(a+e^b) - g'(a+e^b)\right) \tag 2 $$ so $$ c_b(a,0) = f'(a+1)-g'(a-1) = 1 + a \tag 3 $$ Now, differentiate $(1)$ $$ f'(a+1) + g'(a-1) = 0 \tag 4 $$ and therefore $$ g'(a-1) = -f'(a+1) \tag 5 $$ substitute $(5)$ to $(3)$ $$ 2f'(a+1) = a+1 $$ or $$ f'(x) = \frac x2 $$ which has obvious solution $$f(x) = \frac {x^2}4 + C$$ Now, find $g(x)$ $$ g(a-1) = -f(a+1) = -\frac {(a+1)^2}4 - C $$ or, after some trivial manipulations $$ g(x) = -\frac {(x+2)^2}4 - C $$ So, final answer is $$ c(a,b) = f(a+e^b) + g(a-e^b) = \frac {\left ( a+e^b\right )^2}4 - \frac {\left ( a-e^b+2\right )^2}4 $$

Kaster
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  • great! I had lost a negative sign halfway through and so I had $g(a-1)=f(a+1)$ and things didn't work out in the carnage and then I thought my method was just wrong. Thank you very much. – Jay Christian Sep 05 '13 at 23:09