Tengo que resolver el siguiente problema de valor límite:
$$ u_{xx}+u_{yy}=0, 0<x<a, 0<y<b \tag1$$ $$u(0,y)=u(a,y)=0, 0<y<b $$ $$u(x,0)-u_y(x,0)=0, u(x,b)=f(x), 0<x<a$$
Con el método de separación de variables he encontrado lo siguiente:
La solución es de la forma $ u(x,y)=X(x)Y(y)$
$$(1) \Rightarrow X''Y+XY''=0 \Rightarrow \frac{X''}{X}+\frac{Y''}{Y}=0 \Rightarrow \frac{X''}{X}=- \frac{Y''}{Y}=- \lambda$$
Así que tenemos los siguientes problemas:
$$\left.\begin{matrix} X''+ \lambda X=0, 0<x<a\\ X(0)=X(a)=0 \end{matrix}\right\}\tag{*}$$
$$\left.\begin{matrix} Y''- \lambda Y=0, 0<y<b\\ Y(0)-Y'(0)=0 \end{matrix}\right\}\tag{**}$$
$$u(x,b)=X(x)Y(b)=f(x)$$
$(*) \Rightarrow $ Los valores propios son $\lambda_n =\left(\frac{n \pi}{a}\right)^2$ y la función propia son $ X_n(x)= \sin\left(\frac{n \pi x}{a}\right)$ .
$$(**) \Rightarrow Y(y)=A_n \sinh\left(\frac{n \pi y}{a}\right)+B_n \cosh\left(\frac{n \pi y}{a}\right)$$
$$Y'(y)=\frac{n \pi}{a} A_n \cosh\left(\frac{n \pi y}{a}\right)+\frac{n \pi}{a}B_n \sinh\left(\frac{n \pi y}{a}\right)$$
$$Y(0)-Y'(0)=0 \Rightarrow B_n=\frac{n \pi }{a} A_n$$
Así que $Y_n(y)=A_n \sinh\left(\frac{n \pi y}{a}\right)+\frac{n \pi }{a} A_n \cosh\left(\frac{n \pi y}{a}\right)$
$$u(x,y)=\sum_{n=1}^{\infty}{X_n(x)Y_n(y)}$$
$$ u(x,y)=\sum_{n=1}^{\infty}{A_n\left[ \sinh\left(\frac{n \pi y}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi y }{a}\right)\right] \sin\left(\frac{n \pi x}{a}\right)}$$
$$u(x,b)=f(x) \Rightarrow $$
$$\sum_{n=1}^{\infty}{A_n\left[ \sinh\left(\frac{n \pi b}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi b }{a}\right)\right] \sin\left(\frac{n \pi x}{a}\right)}=f(x) \Rightarrow $$
$$A_n\left[ \sinh\left(\frac{n \pi b}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi b }{a}\right)\right] =\frac{2}{a} \int_0^a{f(x) \sin\left(\frac{n \pi x}{a}\right)}dx $$
$$ \Rightarrow A_n=\frac{2 \int_0^a{f(x) \sin\left(\frac{n \pi x}{a}\right)}dx}{a \sinh\left(\frac{n \pi b}{a}\right)+ n \pi \cosh\left(\frac{n \pi b}{a}\right)}$$
¿Es correcta mi solución?
