I have a two-point boundary value problem
$$\left(\tau u_{x}\right)_{x}=f(x), \quad 0<x<1, \quad u(0)=0, \quad u(1)=0$$
where $\tau$ is constant, is the surface tension coefficient of a string. We assume $f(x)$ is a unit point force at some point $\alpha$ where $0<\alpha < 1$, so that
$$\int_{0}^{1} f(x) \phi(x) d x=\lim _{\epsilon \rightarrow 0} \int_{0}^{1} \delta_{\epsilon}(x-\alpha) \phi(x) d x=\int_{0}^{1} \delta(x-\alpha) \phi(x) d x=\phi(\alpha)$$
$\phi(x) \in C^{1}[0,1] \text { vanishing at } x=0 \text { and } x=1, \text { where } \delta_{\epsilon}(x)$ is a continuous nonnegative function with compact support such that
$$\int_{-\infty}^{\infty} \delta_{\epsilon}(x) dx=1$$
Integrating,
$$\int_{\alpha^{-}}^{\alpha^{+}}\left(\tau u_{x}\right)_{x} d x=\int_{\alpha^{-}}^{\alpha^{+}} \delta(x-\alpha) d x=1$$
This leads to $$\tau u_{x}\left(\alpha^{+}\right)=\tau u_{x}\left(\alpha^{-}\right)+1$$.
My question is, why is the exact solution then given by the following?
$$u(x)=\left\{\begin{array}{ll}{-\frac{(1-\alpha) x}{\tau}} & {\text { if } 0 \leq x \leq \alpha} \\ {-\frac{\alpha(1-x)}{\tau}} & {\text { if } \alpha<x \leq 1}\end{array}\right.$$ ?
In particular, I don't see where the solution comes from? It seems we > are just using ftc on both sides of the singular point $\alpha$; however i > don't see where $\alpha$ term on the rhs of $u$ comes from..
You can just integrate in this case, using $Θ$ to denote the Heaviside function (instead of $u$ for the unit ramp function) $$ (τu_x)_x=δ(x-α)\implies τu_x(x)=C+\Theta(x-α))\implies τu(x)=Cx+D+\max(0,x-α). $$ Now adjust the constants to satisfy the boundary conditions, $D=u(0)=0$, $C+(1-α)=u(1)=0$ to find $$ u(x)=\frac1τ(-(1-α)x+\max(0,x-α))=\frac1τ\max(-(1-α)x,-α(1-x)). $$