I have the following problem which I basically understand, but I cannot understand how my professor did a substitution almost at the end of the problem. Thanks a lot in advance!
Question: Find the solution $u(x,y)$ to Laplace's equation in the rectangle $0<x<a$, $0<y<b$ satisfying the boundary conditions
$$u(0,y)=0,\;\;\; u(a,y)=0\;\;\; 0<y<b$$
$$u(x,0)=h(x),\;\;\;u(x,b)=0\;\;\; 0\le x\le a.$$
Solution:
Using the method of separation of variables $u(x,y)=X(x)Y(y)$, we derive
$$X''+\lambda X=0,\;\;\;Y''-\lambda Y=0,$$
where $\lambda$ is the separation constant, and subject to the boundary conditions $X(0)=X(a)=0=Y(b)$.
The boundary problem satisfied by $X(x)$ is solved by
$$X_n(x)=\sin{\frac{n\pi x}{a}},\;\;\;n=1,2,…\;\;\;\lambda=\left(\frac{n\pi}{a}\right)^2$$
Once we know the sign of $\lambda$, we can immediately solve for $Y(y)$ to be
$$Y_n(y)=c_1 \cosh{\frac{n\pi y}{a}}+c_2 \sinh{\frac{n\pi y}{a}}.$$
Using $Y(b)=0$, we derive the constraint
$$c_1 \cosh{\frac{n\pi b}{a}}+c_2 \sinh{\frac{n\pi b}{a}}=0\implies c_2=-c_1\frac{\cosh{\frac{n\pi b}{a}}}{\sinh{\frac{n\pi b}{a}}}$$
And this is the substitution I do not understand:
Substituting back into $Y_n(y)$ and redefining the overall constant coefficient, we can write the fundamental solutions as
$$u_n(x,y)=X_n(x)Y_n(y)=\left(\sin{\frac{n\pi x}{a}}\right)\left(\sinh{\frac{n\pi(b-y)}{a}}\right).$$
How is it that $Y_n(y)=\sinh{\frac{n\pi(b-y)}{a}}$??
And we then have the general solution as
$$u(x,y)=\sum\limits_{n=1}^{\infty}c_n u_n(x,y)=\sum\limits_{n=1}^{\infty}c_n \left(\sin{\frac{n\pi x}{a}}\right)\left(\sinh{\frac{n\pi(b-y)}{a}}\right)$$
Best Answer
$$Y_{n}(y)=c_{1}\cosh\frac{n\pi y}{a}-c_{1}\frac{\cosh\frac{n\pi b}{a}}{\sinh\frac{n\pi b}{a}}\sinh\frac{n\pi y}{a} \\ =\frac{c_{1}}{\sinh\frac{n\pi b}{a}}\left(\sinh\frac{\pi n b}{a}\cosh\frac{n\pi y}{a}-\cosh\frac{n\pi b}{a}\sinh\frac{n\pi y}{a}\right) \\ =C_{n}\sinh\frac{n\pi (b-y)}{a}$$ Since $\sinh\frac{n\pi b}{a}$ is just a constant depending on $n$, and can be absorbed into the general coefficient.