for instance let's consider the following I/BVP :
$$\begin{align} u_t = u_{xx} \\ u|_{x=0} = U_0 \\ u|_{x=L} = U_L \\ u(x,t=0) = \phi(x) \end{align} $$
I know how to solve this type of problem I just don't get it why we assume that the solution is of the form $u(x,t) = u_{ss}(x) + v(x,t)$
especially why is $u_{ss}$ time-independent ?
if someone can shed some light on this it'd be great.
thanks !
$$\begin{align} u_t = u_{xx} \\ u|_{x=0} = U_0 \\ u|_{x=L} = U_L \\ u(x,t=0) = \phi(x) \end{align} $$
''why we assume that the solution is of the form $u(x,t) = u_{ss}(x) + v(x,t).$'' We do this to reduce the inhomogeneous boundary conditions to homogeneous BCs to apply the method of seperation of variables. Since $u_{ss}$ satisfies $u_{xx}=0$ with the BCs $u_{ss}(0)=U_0$ and $u_{ss}(L)=U_L$, solving this we get $u_{ss}(x)=\frac{U_L-U_0}{L}x+U_0$. Using this and $u(x,t) = u_{ss}(x) + v(x,t)$ we have a BIVP for $v(x, t)$ with homogeneous boundary conditions: $$\begin{align} v_t = v_{xx} \\ v|_{x=0} = 0 \\ v|_{x=L} = 0 \\ v(x,t=0) = \phi(x)-\frac{U_L-U_0}{L}x-U_0\qquad\text{(initial condition for}\, v) \end{align} $$ Here $U_0$ and $U_L$ are assumed as constants.