Here is the question:
The function $u(x,t)$ satisfies the partial differential equation $$\frac {\partial u}{\partial t}\ =\ \frac {\partial^2 u}{\partial x^2}$$ with $(0<x<L, \ \ t>0)$.
The boundary condition are given by: $$u(0,t) = a \qquad u(L,t)=b \qquad u(x,0)=f(x)$$ where $a$ and $b$ are constants and $L$ is a positive constant. Solve the following problems.
($1$) Find the solution $u_0(x)$ of the above equation when $\frac {\partial u}{\partial t} = 0$.
($2$) Derive the PDE and the boundary conditions for $$v(x,t) = u(x,t) - u_0(x).$$
Now here goes my confusion:
I can't understand what the u-subscripted zero ($u_0$(x)) means in this case.
My best guess is that $u_0(x)$ is an initial condition at $t = 0$.
But if I try to solve ($1$), I will get $u(x) = Ax + B$, which contains no $t$. Yeah, I know when $t=0$, the equation won't contain any $t$, but is this correct?
EDITED:
How can I also solve ($2$)? Why suddenly there is $v(x,t)$? As far as I know know $u_0(x)$ and $u(x,t)$ now has the same value since $\frac {\partial u}{\partial t} = 0$. Then $v(x,t)$ is zero?
Please help me, somebody. Thank you. :)
If $\frac {\partial u}{\partial t} = 0$, then the function $u$ is independent of the variable $t$, hence $u$ is a function only of $x$:
$$u(x,t)=u_0(x).$$
In this case $\frac {\partial u}{\partial t}\ =\ \frac {\partial^2 u}{\partial x^2}$ means
$$u_0''(x)=0.$$
Thus $u_0$ has the form $u_0(x)=ax+b.$