Consider $u''(x)+u'(x)=f(x)$ and $u'(0)=u(0)=\frac{1}{2}[u'(l)+u(l)]$. Does a solution necessarily exist, or is there a condition that f(x) must satisfy for existence?
I proved that the solution is not unique, but I am not sure the strategy in tackling this problem. Any help is appreciated for me to begin.
We first solve the IVP $$u''+u'=f(x),\qquad u(0)=u'(0)=0\ .$$ The auxiliary function $v(x):=e^x u'(x)$ satisfies $$v'(x)=e^x f(x),\qquad v(0)=0\ .$$ It follows that $$v(x)=\int_0^x e^\tau f(\tau)\>d\tau$$ and then $$u(x)=\int_0^x e^{-t} v(t)\>dt=\int_0^x \left( e^{-t} \int_0^t e^\tau f(\tau)\>d\tau\right)\>dt$$ This particular solution $u_0$ can after some manipulation of integrals be rewritten as $$u_0(x)=\int_0^x\bigl(1-e^{\tau-x}\>f(\tau)\bigr)\>d\tau\ .$$ Therefore we obtain the following general solution of the original ODE: $$u(x)=c_0+ c_1e^{-x}+u_0(x)\qquad(0\leq x\leq \ell)\ .$$ Now we have to take care of the boundary conditions $$c_0+c_1=-c_1={1\over2}\bigl(c_0 +C\bigr)\ ,\tag{1}$$ where $$C=u_0'(\ell)+u_0(\ell)=e^{-\ell}v(\ell)+u_0(\ell)=\int_0^\ell f(\tau)\>d\tau\ .$$ The first condition $(1)$ enforces $c_1=-{c_0\over 2}$, and then the second condition can only be satisfied if $C=0$, whereby $c_0$ may be arbitrary. Therefore we have the following alternative:
If $C:=\int_0^\ell f(\tau)\>d\tau\ne0$ the given boundary value problem has no solution. If $C=0$ the problem has infinitely many solutions given by $$u(x)=c_0\left(1-{1\over2}e^{-x}\right)+u_0(x),\qquad c_0\in{\mathbb R}\ .$$