Show that there's no solution to $u''e^{-u'}=2u$ satisfying given boundary conditions

Question

Let $u\in C^2([0,1])$ satisfies
$$\left\{ \begin{aligned} -u''e^{-u'}+2u&=0\ \text{in}\ [0,1] \\ u(0)&=0\\u(1)&=1 \end{aligned} \right.$$ Show that there is no such $u$ satisfying the equation or prove otherwise

I have figured out that both $u,u'$ and $u''$ are strictly increasing and $u>0\ \text{in}\ (0,1)$.

All the best, thanks.

Answer 1

Multiply the ODE with $u'$ and integrate to obtain the first integral $$ C=(1+v)e^{-v}+u^2\approx 1-\frac{v^2}2+u^2. $$ The stationary solution is $u=0$, close to the point $(u,v)=(0,0)$ the first integral looks like $1-\frac12v^2+u^2$ which indicates a hyperbolic or saddle point. The level corresponding to that point is $C=1$ and the curves $$ u=\pm\sqrt{1-(1+v)e^{-v}}=\pm\sqrt{\tfrac12v^2-\tfrac12v^3-(1+v)(e^{-v}-1+v-\tfrac12v^2)} $$ divide the plane in four parts,

$(u,v)$ phase portrait - solutions move from red to blue, in green the curves starting or ending at the saddle point

One can see and then confirm that it is impossible to move from $u=0$ to $u=1$, as the ray $u=1,~ v=0$ is separated from the positive $v$ axis.

Moreover, the points reached after time one are even closer to zero, they never reach $u=0.3$. diagram of $u(1)$ in dependence of $u'(0)=v(0)$