# Thread: Help in 2nd order boundary value problem

1. ## Help in 2nd order boundary value problem

I want to solve:

$y(x)''-(\frac{m\pi}{a})^2y(x)=0$

With boundary condition y(0)=y(a)=0.

First part is very easy using constant coef. which give:

$y(x)=c_1 cosh(\frac{m\pi}{a}x) + c_2 sinh(\frac{m\pi}{a}x)$

$y(0)=0 \;\Rightarrow\; c_1=0 \;\Rightarrow\; y(x) = c_2 sinh(\frac{m\pi}{a}x)$

$y(a)=0 \;\Rightarrow\; y(x) = c_2 sinh(\frac{m\pi}{a}(a-x))$

Obviously the two boundary condition give two different answer. The book just gave:

$h_1(x)=sinh(\frac{m\pi}{a}(a-x)) \;\; and \;\; h_2(x)=sinh(\frac{m\pi}{a}x)$

My understanding is the general solution is a combination of the two solution that satisfy both boundary conditions. But in this case, each solution meet only one of the boundary condition and not the other. The general solution is the linear combination of the two solution which will not meet either one of the boundary condition!!!

2. Hi. This is what it looks to me: The general solution to $y''-\left(\frac{m \pi}{a}\right)^2 y=0$ is $y(x)=c_1 e^{m\pi x/a}+c_2 e^{-m\pi x/a}$. For $y(0)=0$ implies $c_1=-c_2$. For the second boundary condition, we have $c_1 e^{m\pi}=c_1 e^{-m\pi}$ or $m=\pm i$ therefore the solution is $y(x)=c_1\left(e^{\pi x i/a}-e^{-\pi x i/a}\right)=c_1 \sin(\pi x/a)$.

3. Originally Posted by shawsend
Hi. This is what it looks to me: The general solution to $y''-\left(\frac{m \pi}{a}\right)^2 y=0$ is $y(x)=c_1 e^{m\pi x/a}+c_2 e^{-m\pi x/a}$. For $y(0)=0$ implies $c_1=-c_2$. For the second boundary condition, we have $c_1 e^{m\pi}=c_1 e^{-m\pi}$ or $m=\pm i$ therefore the solution is $y(x)=c_1\left(e^{\pi x i/a}-e^{-\pi x i/a}\right)=c_1 \sin(\pi x/a)$.

This is actually part of solving Poisson equation using single series expansion instead of two series by lumping y into the coef. And solve with only x as shown:

Given . $\nabla^2u(x,y)=f(x,y) \;\;,\;\;0

We start with $\phi_m(x)= \;\;,\;\; \nabla^2\phi_m(x)=-\lambda_m\phi_m(x) \;\;,\;\;\phi_m(0) =\phi_m(a) = 0 \;\;,\;\; \lambda_m=(\frac{m\pi}{a})^2$

$u(x,y)=\sum_{m=1}^{\infty} E_m(y) \; sin(\frac{m\pi x}{a})$

$\Rightarrow \nabla^2u(x,y)= \sum_{m=1}^{\infty} [E''_m(y) \;-\; (\frac{m\pi}{a})^2 \; E_m(y)]\;\; sin(\frac{m\pi x}{a}) \;=\; f(x,y)$

$let\; b_m(y) \;=\; E''_m(y) \;-\; (\frac{m\pi}{a})^2E_m(y) \Rightarrow\; b_m(y)=\frac{2}{a}\int_0^a f(x,y) sin(\frac{m\pi x}{a}) dx$

From the original boundary condition we come up with:

$E''_m(y)- (\frac{m\pi}{a})^2E_m(y) = b_m(y)$
$E_m(0)=0 \;\;,\;\; E_m(b)=0$

Starting below is same as my original post, just substitude . $y(x)=E_m(y)$

We want to solve for $E_m(y)$ by solving the associate homogeneous equation:

$E''_m(y)- (\frac{m\pi}{a})^2E_m(y)=0$

The book then gave:

$h_1(y)=sinh((\frac{m\pi}{a})(b-y)) \;\;,\;\; h_2(y)=sinh((\frac{m\pi}{a})y)$

$W(h_1,h_2)= (\frac{m\pi}{a}) sinh( (\frac{m\pi}{a}) b)$

Where W is the Wronskian.

I just don't follow the last two steps. I copy exactly from the book.

Obviously the two boundary condition give two different answer. The book just gave:

$h_1(x)=sinh(\frac{m\pi}{a}(a-x)) \;\; and \;\; h_2(x)=sinh(\frac{m\pi}{a}x)$