MHB How to solve this boundary value problem-Method of separation of variables

[mathmari]

Hey!

I have a question.. (Wasntme)

When we have the following boundary value problem:
$$u_{xx}+u_{yy}=0, 0<x<a, 0<y<b (1)$$
$$u_x(0,y)=u_x(a,y)=0, 0<y<b$$
$$u(x,0)=x, u_y(x,b)=0, 0<x<a$$using the method of separation of variables, the solution would be of the form $u(x,y)=X(x) \cdot Y(y)$
$$(1) \Rightarrow X''Y+XY''=0 \Rightarrow \frac{X''}{X}+\frac{Y''}{Y}=0 \Rightarrow \frac{X''}{X}=- \frac{Y''}{Y}=- \lambda$$

So we have the following problems:
$$\left.\begin{matrix}
X''+\lambda X=0, 0<x<a\\
X'(0)=X'(a)=0
\end{matrix}\right\}(2)$$

and

$$\left.\begin{matrix}
Y''-\lambda Y=0, 0<y<b\\
Y'(b)=0
\end{matrix}\right\}(3)$$

$$u(x,0)=X(x)Y(0)=x$$

So solving the problem $(2)$ we get the eigenvalues and the corresponding eigenfunctions.
If we have the following boundary value problem:
$$u_{xx}+u_{yy}=0, 0<x<1, 0<y<1 (1')$$
$$u(0,y)=0, u(1,y)= \sin{(\pi y)} \cos{(\pi y)}, 0<y<1$$
$$u(x,0)=u(x,1)=0, 0<x<1$$

using the method of separation of variables, the solution would be of the form $u(x,y)=X(x) \cdot Y(y)$
$$(1') \Rightarrow X''Y+XY''=0 \Rightarrow \frac{X''}{X}+\frac{Y''}{Y}=0 \Rightarrow \frac{X''}{X}=- \frac{Y''}{Y}= \lambda$$

So we have the following problems:
$$\left.\begin{matrix}
X''-\lambda X=0, 0<x<1\\
X(0)=0
\end{matrix}\right\}(2')$$

and

$$\left.\begin{matrix}
Y''+\lambda Y=0, 0<y<1\\
Y(0)=Y(1)=0
\end{matrix}\right\}(3')$$

$$u(1,y)=X(1)Y(y)= \sin{(\pi y)} \cos{(\pi y)}$$

So do we have to solve in this case the problem $(3')$ to find the eigenvalues nd the corresponding eigenfunctions?
Or did I have to set $\frac{X''}{X}=- \frac{Y''}{Y}=- \lambda$ instead of $\frac{X''}{X}=- \frac{Y''}{Y}= \lambda$, as at the other boundary value problem? (Wondering)

 

[mathmari]

Or could we solve this problem as followed? (Wondering)

$$u(x,y)=v(x,y)+s(x)$$

$$v_{xx}+v_{yy}+s''(x)=0, 0<x<1, 0<y<1$$
$$v(0,y)+s(0)=0, 0<y<1$$
$$v(1,y)+s(1)=\sin{( \pi y)} \cos{( \pi y)}, 0<y<1$$
$$v(x,0)+s(x)=v(x,1)+s(x)=0, 0<x<1$$

So we have the following problems:

$$s''(x)=0, 0<x<1$$
$$s(0)=0, s(1)= \sin{( \pi y)} \cos{( \pi y)}$$

$$v_{xx}+v_{yy}=0, 0<x<1, 0<y<1$$
$$v(0,y)=0, v(1,y)=0, 0<y<1$$
$$v(x,0)=v(x,1)=-s(x), 0<x<1$$

 

[mathmari]

Or could we solve this problem as followed? (Wondering)

$$u(x,y)=v(x,y)+s(x)$$

$$v_{xx}+v_{yy}+s''(x)=0, 0<x<1, 0<y<1$$
$$v(0,y)+s(0)=0, 0<y<1$$
$$v(1,y)+s(1)=\sin{( \pi y)} \cos{( \pi y)}, 0<y<1$$
$$v(x,0)+s(x)=v(x,1)+s(x)=0, 0<x<1$$

So we have the following problems:

$$s''(x)=0, 0<x<1$$
$$s(0)=0, s(1)= \sin{( \pi y)} \cos{( \pi y)}$$

$$v_{xx}+v_{yy}=0, 0<x<1, 0<y<1$$
$$v(0,y)=0, v(1,y)=0, 0<y<1$$
$$v(x,0)=v(x,1)=-s(x), 0<x<1$$

The solution of the problem $$s''(x)=0, 0<x<1$$
$$s(0)=0, s(1)= \sin{( \pi y)} \cos{( \pi y)}$$
is $s(x)=\sin{( \pi y)} \cos{( \pi y)} x$, isn't it? (Wondering)To solve the problem $$v_{xx}+v_{yy}=0, 0<x<1, 0<y<1$$
$$v(0,y)=0, 0<y<1$$
$$v(1,y)=0, 0<y<1$$
$$v(x,0)=v(x,1)=-s(x), 0<x<1$$
we use the method of separation of variables, the solution is of the form $v(x,y)=X(x)Y(y)$

So we become the following two problems:

$$\left.\begin{matrix}
X''+ \lambda X=0, 0<x<1\\
X(0)=X(1)=0
\end{matrix}\right\}(*)$$

$$\left.\begin{matrix}
Y''- \lambda Y=0, 0<x<1\\
Y(0)=Y(1)=-s(x)=-\sin{( \pi y)} \cos{( \pi y)} x
\end{matrix}\right\}(**)$$

Right? (Wondering)

From the problem $(*)$ we get that the eigenvalues are $\lambda_n=n^2 \pi^2$ and the corresponding eigenfunctions are $X_n(x)=\sin{(n \pi x)}$.

From the problem $(**)$, we get $Y(y)=c_n e^{-n \pi y}+d_n e^{n \pi y}$, don't we?

$Y(0)=-s(x) \Rightarrow c_n+d_n=-\sin{( \pi y)} \cos{( \pi y)} x$
Is this relation correct? Can we set at the left part $y=0$ and the right part of the relation there is still the variable $y$? (Wondering)

 

[I like Serena]

Erm... this seems to be quite a work in progress... (Sweating)

You seem to know what you're doing! (Nod)

Or do you still have questions? (Wondering)

 

[mathmari]

Erm... this seems to be quite a work in progress... (Sweating)

You seem to know what you're doing! (Nod)

Or do you still have questions? (Wondering)

I finally used the first way I mentioned at my first post, I mean using the method of separation of variables ($u(x,y)=X(x)Y(y)$) and found that $Y_n(y)=\sin{(n \pi y)}$ and $X_n(x)=A_n \sinh{(n \pi x)}$. So the solution is $$u(x,y)=\sum_{n=1}^{\infty}{A_n \sinh{(n \pi x)}\sin{(n \pi y)}}$$
Then using the conditions the final solution is $$u(x,y)=\frac{1}{2 \sinh{(2 \pi})} \sinh{(2 \pi x)} \sin{(2 \pi y)}$$ which satisfies the conditions! (Sun)