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No such function

Prove that there exists no function f:[0, 2] \rightarrow \mathbb{R} such that f(0)=0 and

(1)   \begin{equation*}f'(x) - f^4(x) =1 \quad \text{forall} \;\; x \in [0, 2] \end{equation*}

Solution

Since f'(x)>1 forall x\in [0, 2] this means that the function g(x)=f(x)-x \;, \; x \in [0, 2] is strictly increasing in [0, 2]. Hence g(x)>g(0) forall x \in [0, 2] which in turn implies f(x)>x forall x \in [0, 2]. Thus, f(1)>1. On the other hand , since f is strictly increasing we get

(2)   \begin{equation*} 1< f(1) < f(2) \end{equation*}

From the first equation we have successively:

    \begin{align*} f'(x) - f^4(x) =1 &\Rightarrow f'(x) = 1 + f^4(x) \\ &\Rightarrow \frac{f'(x)}{1+f^4(x)} = 1 \\ &\Rightarrow \int_{0}^{x} \frac{f'(t)}{1+f^4(x)} \, \mathrm{d}t = \int_0^x \, \mathrm{d}t\\ &\Rightarrow \int_{0}^{f(x)} \frac{\mathrm{d}t}{1+t^4} = x \end{align*}

Thus,

(3)   \begin{equation*} \int_{0}^{f(x)} \frac{\mathrm{d}t}{1+t^4} = x \end{equation*}

Using equation (3) we conclude that

(4)   \begin{equation*} \int_{f(1)}^{f(2)} \frac{\mathrm{d}t}{t^4+1} = 1 \end{equation*}

Using equation (2) we get that

    \begin{align*} 1 &=\int_{f(1)}^{f(2)} \frac{\mathrm{d}t}{t^4+1} \\ &\leq \int_{f(1)}^{f(2)} \frac{\mathrm{d}t}{t^2+1} \\ &= \arctan f(2) - \arctan f(1) \end{align*}

Finally,

    \begin{align*} 1 \leq \arctan f(2) - \arctan f(1) & \Rightarrow 1 \leq \arctan \left ( \frac{f(2)-f(1)}{1+f(2) f(1)} \right ) \\ &\Rightarrow \arctan \left ( \tan 1 \right ) \leq \arctan \left ( \frac{f(2)-f(1)}{1+f(2) f(1)} \right ) \\ &\Rightarrow 1 \leq \frac{f(2)-f(1)}{1+f(1)f(2)} \\ &\Rightarrow 1+f(1)f(2)-f(2)+f(1) \leq 0\\ &\Rightarrow 1+f(1)+f(2) \left ( f(1)-1 \right ) \leq 0 \end{align*}

which is an obscurity. Hence , there is no such function!

 

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