How do I find the steady state solution $\bar{c}(x)$ for the following PDE? $$\frac{∂c}{∂t} = D\frac{∂}{∂x}\left[\frac{∂\Phi}{∂x}c(x, t) + \frac{∂c(x,t)}{∂x}\right]$$.
Here's my attempt:
$$\frac{∂c}{∂t} = 0=D\frac{∂}{∂x}\left[\frac{∂\Phi}{∂x}c(x, t) + \frac{∂c(x,t)}{∂x} \right]$$ $$0=\frac{∂}{∂x}\left[\frac{∂\Phi}{∂x}c(x, t) + \frac{∂c(x,t)}{∂x}\right]$$
$$\frac{∂\Phi}{∂x}c(x, t) + \frac{∂c(x,t)}{∂x}=\text{constant}$$
$$\int\frac{∂\Phi}{∂x}c(x, t)\,dx + \int\frac{∂c(x,t)}{∂x}dx=\int a\space dx \tag 1$$
Do integration by parts on $$\int\frac{∂\Phi}{∂x}c(x, t)dx$$ with $u=c(x,t)$, $dv=\frac{∂\Phi}{∂x}\,dx$. So $du=dc$ and $v=\Phi$, and $$\int\frac{∂\Phi}{∂x}c(x, t)\,dx=c(x,t)\Phi-\int \Phi dc=c(x,t)\Phi - \Phi c(x,t)=0$$
Now eqn (1) becoemes $$\int\frac{∂c(x,t)}{∂x} \, dx=\int a \, dx \Longrightarrow c(x,t)=ax+b$$ for some constants $a$ and $b$.
Does this look right?
No, it is not correct. When you integrate by parts, you do it as if $\Phi$ were constant. We are looking for solutions $c(x)$ that do not depend on $t$. We arrive at the differential equation $$ \frac{\partial\Phi}{\partial x}\,c+c'=a,\quad a\text{ constant.} $$ Multiplying by $e^\Phi$ we get $$ \frac{\partial}{\partial x}(e^{\Phi}\,c)=a\,e^\Phi\implies c=b\,e^{-\Phi}+a\,e^{-\Phi}\int e^\Phi\,dx. $$ It is clear that for this to work $\Phi$ must depend only on $x$.