I am working through example $3.3$ in Griffiths Electrodynamics in section $3.3$ on Separation of Variables. The example involves solving the $2$-dimensional version of Laplace's Equation for the following setup:
Where we have $2$ grounded metal plates in the $xz$-plane and we would like to find the potential in the space between the plates. Obviously, this involves solving Laplace's equation $$\nabla^2V = \frac{\partial^2V}{\partial x^2} + \frac{\partial^2V}{\partial y^2} = 0$$ Since there is no charge in the space between the plates.
I'll spare most of the details of the solution itself as I'm sure anyone answering this question is more than well aware of how to perform separation of variables for a kind of situation like this. My question involves the part of the solution where we determine constants via boundary conditions anyways.
We end up with the following solution: $$V(x,y) = X(x)\cdot Y(y) = [Ae^{kx} + Be^{-kx}][C\sin(ky) + D\cos(ky)]$$ Subject to the following boundary conditions: $$V(x,0) = V(x,a) = 0$$ $$V(0,y) = V_0$$ $$V \rightarrow 0\ \text{as}\ x \rightarrow \infty$$
The example goes on to determine the coefficients via the BC's. First by noting that via the last BC we must have $A = 0$, which makes sense. But then, the example goes on to say "we are then left with $Be^{-kx}[C\sin(ky) + D\cos(ky)]$ which we absorb $B$ into the constants $C$ and $D$ and are left with $e^{-kx}[C\sin(ky) + D\cos(ky)]$. My question is:
- Why can we absorb $B$ into the other constants $C$ and $D$? Is it just the case that when multiplying through to get a more explicit form the solution we have $$V(x,y) = BCe^{-kx}\sin(ky) + BDe^{-kx}\cos(ky)$$ so that we are just saying $BC \rightarrow C$ and $BD \rightarrow D$? I think I may be starting to see it as I type up this question but I just didn't understand the reasoning as to why it was done in the order that it was. I initially thought that we would be losing information by absorbing the constant $B$, since we still had the boundary condition $V(0,y) = V_0$ which could possibly tell us something about $B$. However, I suppose it doesn't since imposing that condition we are still left with $V_0 = B[C\sin(ky) + D\cos(ky)]$. Anyway, this has become a bit more meandering than I initially intended when formulating this question due to my sudden realization, so any verification that my thinking is correct would be good to know. I'll keep the question up for posterity in case anyone else wonders the same. Thank you.
Let's start with the solution through separation of variables:
$$V(x,y)=(Ae^{kx}+Be^{-kx})(C\sin(ky)+D\cos(ky))\tag1$$
If we impose the condition on $(1)$ that $\lim_{x\to\infty}V(x,y)$ is finite, then $A$ must be $0$. At this point the solution to the $2$-D Laplace's equation with finite solution as $x\to\infty$ is
$$V(x,y) = Be^{-kx}(C\sin(ky)+D\cos(ky))\tag2$$
If we impose the condition on $(2)$ that for all $x>0$, $V(x,0)=V(x,a)=0$, then we $D$ must be $0$ and $k=n\pi/a$ for any $n\in \mathbb{N_{>0}}$.
At this point, the solution to Laplace's equation that satisfies the boundary conditions at $y=0$ and $y=a$ along with the condition as $x\to \infty$ is for any $n\in \mathbb{N}_{>0}$
$$\begin{align} V(x,y)&=BCe^{-n\pi x/a}\sin(n\pi y/a)\\\\ &=\alpha e^{-n\pi x/a}\sin(n\pi y/a)\tag3 \end{align}$$
where the constant $B\times C$ has been denoted by $\alpha$.
Now, inasmuch as $(3)$ is applicable for each $n\in \mathbb{N_{>0}}$, we can associate a different constant for each $n$. Moreover, linearity of Laplace's equation permit us to sum over all $n$ and write a general solution as
$$\begin{align} V(x,y)&=\sum_{n=1}^\infty \alpha_n e^{-n\pi x/a}\sin(n\pi y/a)\\\\ \end{align}$$
Finally, we find $\alpha_n$ by applying the last boundary condition $V(0,y)=V_0(y)$. Proceeding we have
$$\alpha_n = \frac2a\int_0^a V_0(y')\sin(n\pi y'/a)\,dy'$$
and hence for $x>0$
$$V(x,y)=\frac2a \int_0^a V_0(y')\sum_{n=1}^\infty e^{-n\pi x/a}\sin(n\pi y'/a) \sin(n\pi y/a)\,dy' $$