I am working on understanding single-shooting methods. I couldn't find a similar problem to mine unfortunatly.
So, given is the boundary value problem
$$ \begin{cases} u''(x)= (u'(x))^2 , \forall x \in (0, \frac\pi 2 )\\ u(0)= 0 \\ u( \frac\pi 2 ) = 1 \end{cases}$$
I need to solve the exact solution $ v= v( . ; s) $ and determine the existance range.
So I generate an IVP first
$$ \begin{cases}v''(x)= (v'(x))^2, \forall x \in (o, \frac\pi 2) \\ v(0)=0 \\ v'(0)= s \end{cases} $$
I think i get the idea..but it fails with the conversion.
I do appreciate any help :-)
There is a defect in the problem specification. Since the differential equation for $u$ only requires $x \in (0,\pi/2)$, the solution to the differential equation is independent of the constraints on the endpoints. For instance, what forces $u$ continuous at $0$ or $\pi/2$? But let's assume we actually meant to have the differential equation hold for $x$ in an open set containing $[0,\pi/2]$ so that $u$ is defined, continuous, differentiable, and twice differentiable up to (and past) the specified endpoints. ("Twice differentiable" implies the other three.) The problem's request for an existence range suggests that this is intended, in fact, we are to find the largest such open set on which the differential equation can hold, meet the boundary conditions, and include the interval $[0,\pi/2]$.
The general solution to $v'' = (v')^2$ is $v(x) = c_2 - \ln(x+ c_1)$. (I suppress the absolute values here. I've never seen anyone recognize there are three cases here when the absolute values were explicit.) Matching the initial conditions, $v(0) = 0$ forces $c_2 = \ln(c_1)$ and then $v'(0) = s$ forces $c_1 = \frac{-1}{s}$. Then $$ v(x) = \ln \left( \frac{-1}{s} \right) - \ln\left( x+\frac{-1}{s} \right) = \ln\left( \frac{1}{1-sx}\right) \tag{*} $$ as long as $\frac{-1}{s} > 0$ and $x + \frac{-1}{s} > 0$ (... since we must land in the domain of both logarithms. (Case I)) These inequalities simplify to $s < 0, x > \frac{1}{s}$, which automatically includes $x \in [0,\pi/2]$.
If $v(\pi/2) = 1$, then $s = \frac{2\mathrm{e}-2}{\mathrm{e} \pi} > 0$, so we have a problem. Look at the middle of (*). It says: at $x = 0$, start at height $0$, then as $x$ increases, descend. There's no way this can ever get back up to height $1$.
So what do we do? The other way for the equality in (*) to hold is for both arguments to the logarithm to be negative so their ratio is positive. (Recall that we are really working with $\ln(|\dots|)$, so we choose the other sign under both the absolute values. (Case II) If we only choose to reverse one sign, then the resulting form doesn't even satisfy the differential equation. (Case III)) These new inequalities simplify to $s > 0$ and $x < \frac{1}{s}$. To allow $x \in [0,\pi/2]$, we must have $s < \frac{2}{\pi}$.
It happens that $s = \frac{2\mathrm{e}-2}{\mathrm{e} \pi} = 0.4024\dots < 0.636\dots = \frac{2}{\pi}$, so we have a valid solution on $[0,\pi/2]$, $$ u(x) = \ln \left( \frac{1}{1-\frac{2\mathrm{e}-2}{\mathrm{e}\pi}x} \right) \text{.} $$
In fact, the domain of this solution is $x \in \left(-\infty, \frac{\mathrm{e}\pi}{2\mathrm{e}-2} \right)$.