# Solving a simple second order ODE with initial condition

Okay i've been at it for far too long now. It comes from a bigger question from working with a PDE. I did seperation of variables and now I am stuck near the end of the problem. Here is the ODE in question $$\Phi''(y)= \lambda^2\cdot\Phi(y)$$ with the following initial condition $$\Phi'(H)=0$$ where $H$ is a positive number. Also I know $\displaystyle \lambda = \frac{n\pi}{L} >0$. There is another condition but I dont think it can help $$\Phi(0) = \begin{cases} 0 & x > L/2 \\ 1 & x < L/2 \end{cases}$$ sorry, i dont know how to do cases in latex and yes, that is an $x$ in the initial condition. Like i said this is a bigger problem that has both x and y.

The solution should be $$\Phi(y) = B \cdot \cosh{(\lambda(H-y))}$$

I've tried going through the following general solutions $$\Phi(y) = Ae^{\lambda y} + B e^{-\lambda y}$$ and $$\Phi(y) = A \sinh{\lambda y} + B \cosh{\lambda y}$$ but no luck that way :(

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It is probably easiest just to remember as a general fact that $$f(x)=\begin{cases}A\cosh(\lambda(x-H)) & A, H\in\mathbb R\\ A\sinh(\lambda(x-H)) & A, H\in\mathbb R\\ e^{\lambda(x-H)} & H\in\mathbb R\\ e^{-\lambda(x-H)} & H\in\mathbb R\end{cases}$$ are a complete set of solutions to $f''=\lambda^2f$ in general. To remember that they are solutions, just note that since the ODE does not mention the independent variable we can slide any known solution to the left and right. Proving them complete is slightly more involved -- one strategy would be to show that together they cover all possible pairs of $f(0), f'(0)$, by case analysis on their logarithmic derivatives.

Then, since you're already given a $H$ such that $f'(H)=0$ you can immediately see that it must be of the form $A\cosh(x-H)$ for some $A$, since the other forms have no stationary points. Finding $A$ such that your condition for $\Phi(0)$ holds is then a simple matter of scaling.

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What does it mean by " since the ODE does not mention the independent variable we can slide any known solution to the left and right." Also, what does it mean by saying the other forms have no stationary points – Tyler Hilton Nov 26 '11 at 22:36
The ODE is $\frac{d^2y}{dx^2}=\lambda^2 y$, and it doesn't mention $x$, only differences in $x$. So if you add or subtract a constant from all $x$s (thereby moving its graph to the left or right), it cannot make a solution into a nonsolution or vice versa. -- A stationary point of a function $f$ is an $x$ such that $f'(x)=0$ There are clearly no such points for $\sinh$ or the pure exponential forms. – Henning Makholm Nov 26 '11 at 22:45
Thankyou. To be honest I still dont understand the "differences" part but I'll do some more research on my own now that I know what it is. Also would $x=0$ be considered a stationary point or are zeros not considered a stationary point. – Tyler Hilton Nov 26 '11 at 22:51
The differences point is simply that $x\mapsto f(x)$ is a solution if and only of $x\mapsto f(x+c)$ is a solution. $x=0$ is a stationary point if and only if $f'(0)=0$. It's not a deep concept -- I could just as well have said "... since $f'(x)$ is always nonzero for the other forms". – Henning Makholm Nov 26 '11 at 22:54
Also what does it mean by having $A, H \in \mathbb{R}$ – Tyler Hilton Nov 27 '11 at 2:36

The forms $Ae^{\lambda y} + Be^{-\lambda y}$ and $D \cosh(\lambda(C - y))$ are really different ways of writing the same thing. For $$Ae^{\lambda y} + Be^{-\lambda y} = \sqrt{AB} ( \sqrt{A \over B} e^{\lambda y} + \sqrt{B \over A} e^{-\lambda y})$$ $$= \sqrt{AB} ({ e^{\lambda y + {1 \over 2}\ln({A \over B})} + e^{-\lambda y - {1 \over 2}\ln({A \over B})}})$$ Letting $C = -{1 \over 2\lambda}\ln{A \over B}$, this becomes $$\sqrt{AB} ( e^{\lambda y - \lambda C}+ e^{-\lambda y + \lambda C})$$ $$=2\sqrt{AB} \cosh(\lambda(y - C)) = 2\sqrt{AB} \cosh(\lambda(C - y))$$ Thus letting $D = 2\sqrt{AB}$ you get the other form. Plug in your initial condition to show $C = H$.

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thankyou. is there another way to do this? – Tyler Hilton Nov 26 '11 at 21:59
you could just keep the solution in the form $Ae^{\lambda y} + Be^{-\lambda y}$. Your initial condition will give $A$ in terms of $B$ so your solutions will be $B \times$ function, which will also be correct although written a different way. – Zarrax Nov 26 '11 at 22:04
Do you have a A in terms of B so I can compare my answer? I got $A = B e^{H\lambda}$ – Tyler Hilton Nov 26 '11 at 22:32
I think it's $A = Be^{-2H\lambda}$. – Zarrax Nov 27 '11 at 1:31