0
$\begingroup$

Looking for some explanation to the use of boundary conditions


I have the Laplace's Equation $$u_{xx}+u_{yy}=0, \quad 0<x<L, 0<y<H$$ with boundary conditions $$u(0,y)=0, \ u(L,y)=0, \ u(x,0)=f(x), \ u(x,H)=0$$

I am analyzing the solution from the textbook (http://people.uncw.edu/hermanr/pde1/PDEbook/PDE_Main.pdf p.49) and have a question to the second ODE $$Y''-\lambda Y=0$$ The solution to the resulting quadratic equation depends on $\lambda$ so there will be three cases. Using the two conditions $Y(0)=0$ and $Y(H)=0$ I understand that the case 2 ($\lambda=0$) and case 3 ($\lambda<0$) give the trivial solution. I assume that the Y(0)=0 condition was used as first in analysis for the two cases.


Question:
Applying the Y(0)=0 condition to the first case should get the trivial solution either, shouldn't it? why don't we get the trivial solution for the case 1 either ($\lambda>0$)?


I mean $$Y(y)=c_1 e^{-\sqrt{\lambda} y} + c_2 e^{\sqrt{\lambda} y}$$ $$Y(0)=c_1 + c_2=0$$ $$c2=-c1$$ $$Y(H)=c_1 e^{-\sqrt{\lambda} H} - c_1 e^{\sqrt{\lambda} H} =0$$ $$0=c_1(e^{-\sqrt{\lambda} H} - e^{\sqrt{\lambda} H})$$ as $H>0$ then $(e^{-\sqrt{\lambda} H} - e^{\sqrt{\lambda} H})<0$ therefore $\quad c_1 = 0 \quad$ thus $\quad c_2 = 0 \quad$ trivial solution again




UPDATE

QUESTION:

I don't understand why thy $Y(0)=0$ can not be used in the stated problem as opposed to the below example where the condition can be used. Could anybody help to understand this difference please.
$$Y''+ \lambda Y=0$$ $$0<x<L, \ 0<y<H$$ $$\ u(x,0)=u(x,H)=0, \ u(L,y)=0, \ u(0,y)=v$$

$\endgroup$
  • 1
    $\begingroup$ You don't have the boundary condition $Y(0)=0$, you have the condition $u(x,0)=f(x)$, which you account for on the next page. $\endgroup$ – Aweygan Aug 3 '16 at 2:09
  • $\begingroup$ @ Awegyan are the $u(0,y)=0$ and $u(x,0)=0$ conditions not implying that $u(0,0)=0$ therefore $Y(0)=0$? $\endgroup$ – Michal Aug 3 '16 at 13:59
  • 2
    $\begingroup$ You don't have the condition $u(x,0)=0$, you have $u(x,0)=f(x)$. If you did have $u(x,0)=0$ then you are right, you will get the trivial solution. $\endgroup$ – Aweygan Aug 3 '16 at 14:15
  • $\begingroup$ @ Awegyan I had a look at similar problem with slightly different conditions $$Y''+ \lambda Y=0$$ $$0<x<L, \ 0<y<H$$ $$\ u(x,0)=u(x,H)=0, \ u(L,y)=0, \ u(0,y)=v$$ and there was the condition $Y(0)=Y(H)=0$ used. I am confused, why $Y(0)=0$ would be ok here? $\endgroup$ – Michal Aug 3 '16 at 23:01
  • 1
    $\begingroup$ That's because $u(x,0)=0$. $\endgroup$ – Aweygan Aug 3 '16 at 23:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.