Given $(\beta u')' = f(x)$, $u(-1) = 0, u(1) = 1$ where $\beta = 1+ H(x)$ and $H$ is the Heaviside function. For $$f= 0, \quad f=H, \quad f=\delta,\quad f=\delta'$$ I would like to find the jump condition at $x=0$ and the solution $u$.
This is from an old exam, and I have not seen this type of problem before. I am not sure what is the jump condition, so I tried by solving with integration and here is my attempt.
For $f=0$, then we know $\beta u' =c$ thus $$u'=c/\beta = \begin{cases} c \quad & x\leq 0 \\ c/2 \quad & x>0 \end{cases}$$ we just simply integrate $$u(x) = \begin{cases} cx+d \quad & x\leq 0 \\ cx/2+d \quad & x>0 \end{cases}$$ we want $u(-1) = 0$, thus $-c+d = 0$, and $u(1) = 1$, thus $c/2+d = 1$.
For $f=H$, we know $$\beta u' =\begin{cases} c \quad & x\leq 0\\ c+x \quad & x >0 \end{cases}$$ which implies $$u' =\begin{cases} c \quad & x\leq 0\\ \frac{c+x}{2} \quad & x >0 \end{cases}$$ and again we simply integrate and find the integration constant $c$ and $d$.
$f= \delta$, we know $\beta u' = H + c$ so $$u' =\begin{cases} c \quad & x\leq 0\\ \frac{c+1}{2} \quad & x >0 \end{cases}$$ and again we find integration constant $c,d$.
$f= \delta'$, we know $\beta u' = \delta + c$, so $$u' =\begin{cases} \delta + c \quad & x\leq 0\\ \frac{\delta + c}{2} \quad & x >0 \end{cases}$$
$$u =\begin{cases} H + cx + d \quad & x\leq 0\\ \frac{H + cx}{2} + d \quad & x >0 \end{cases}$$ and we find $c,d$ again with boundary conditions.
Are these correct? Thanks!