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I want to solve a PDE like this:

$\frac{\partial y}{\partial t}=a\frac{\partial ^2y}{\partial x^2}-b\frac{\partial y}{\partial x}-c y,(a,b,c\in \mathbb{R})\tag{1}$

with the boundary conditions:

$ \begin{equation}\begin{cases} y|_{x=0}=y_0+y_1 \cos (\text{$\omega $t})\\ y|_{x\rightarrow +\infty }=y_2\\ \end{cases}\ \tag{2} \end{equation}\ $

In the second BC, $y_2\neq 0$

I uncoupled the problem using superposition principle into two sub problems:

$ \begin{equation}\begin{cases} \frac{\partial y}{\partial t}=a\frac{\partial ^2y}{\partial x^2}-b\frac{\partial y}{\partial x}-c y\\ y|_{x=0}=y_0\\ y|_{x\rightarrow +\infty }=y_2\\ \end{cases} \tag{3} \end{equation} $

and

$ \begin{equation}\begin{cases} \frac{\partial y}{\partial t}=a\frac{\partial ^2y}{\partial x^2}-b\frac{\partial y}{\partial x}-c y\\ y|_{x=0}=y_1 \cos (\text{$\omega $t})\\ y|_{x\rightarrow +\infty }=0\\ \end{cases} \tag{4} \end{equation} $

I have 3 questions:

  1. Is this uncoupling correct? If not, how can I use superposition principle to uncouple this problem?

  2. If the actual system is only changed by $\cos (\text{$\omega $t})$. In other words, if there's no $\cos (\text{$\omega $t})$ term, the system is in steady state. On this occasion can I write sub problem(3) like this: $ \begin{equation}\begin{cases} 0=a\frac{\partial ^2y}{\partial x^2}-b\frac{\partial y}{\partial x}-c y\\ y|_{x=0}=y_0\\ y|_{x\rightarrow +\infty }=y_2\\ \end{cases} \tag{5} \end{equation} $

  3. The problem(5) is actually an ODE. If I solve it, I get into trouble. The general solution of the ODE is: $y=C_1 \exp \left(\frac{b+\sqrt{4 a c+b^2}}{2 b}\right)+C_2 \exp \left(\frac{b-\sqrt{4 a c+b^2}}{2 b}\right)\tag{6}$ But I cannot deal with the second boundary condition. The first term of the general solution tend to positive infinity and the second term tend to zero. No matter what the value of $C_1$ and $C_2$ is, the condition is not satisfied. Does this trouble imply that I cannot set the boundary condition like $y|_{x\rightarrow +\infty }=y_2$? I wonder if the BC here can only be set like: $y|_{x\rightarrow +\infty }=0$ or $\left.\frac{\partial y}{\partial x}\right|_{x\rightarrow +\infty }=0$?

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