Regarding steady state solution of $u_{t}= c^2 u_{xx}$ with $u_{x}(0,t) = c_{1}$ and $u(L,t) = c_{1}$?

Question

Suppose we have the one dimensional diffusion equation $u_{t} = c^2 u_{xx}$ with the boundary condition $u(L,t) = c_{2} $ and $u_{x}(0,t) = c_{1}$. I donot recognize which type of condition is it? badly, it seems to be the Robin condition.

Suppose I proceed by assuming $u(x,t) = X(x) T(t)$ then $XT' = c^2 X'' T$, but I am thinking how to incorporate the conditions given into the PDE or the process which can give us the steady state solution?

EDIT:

I am still thinking of the solution, the question is attached as pic. The solution $u(x,t)$ is independent of $t$ in the below solution, may be the steady state then also how to answer the 2nd part as it hints the dependence of time there:

Answer 1

Let $u(x,t) = w(x) + v(x,t)$ where $w(x)$ is the steady-state. Then we have

$$ w''(x) = 0, \quad w'(0) = w(L) = c_1 $$

which gives $w(x) = c_1(x-L) + c_1$

Now you can use separation of variables to find $v(x,t)$, which is homogeneous on the boundary (of "mixed" type).

Answer 2

$$u_{t} = c^2 u_{xx}$$ with conditions $\quad u(L,t) = c_{2}\quad \text{and}\quad u_{x}(0,t) = c_{1}$

Inspection only gives the solution : $$u(x,t)=c_1 x-c_1L+c_2$$ Proof :

$u_t=0$ and $u_{xx}=0$ satisfy $u_t=u_{xx}$

$u(L,t)=c_1 L-c_1L+c_2=c_2$ satisfies the condition $u(L,t) = c_{2}$ .

$u_x(t,x)=c_1$ satisfies the condition $u_t(0,t)=c_1$ .

Comment : Since the solution appears too simple and trivial, probably there is something missing or wrong in the wording of the question.