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I am reading the book "Laplace transforms" by Murray Spiegel and I am not able to completely solve the one before last exercise in the book, i.e. number 69 of chapter 8. The exercise is the following: Solve the boundary-value problem $$\frac{\partial^2 Y}{\partial t^2}+b^2 \frac{\partial^4 Y}{\partial x^4}=0$$ in which $0<x<L$ and $t>0$. The boundary conditions are $Y(0,t)=0$, $Y(L,t)=0$, $Y(x,0)=0$, $Y_t(x,0)=0$, $Y_{xx}(L,t)=0$ and $EIY_{xx}(0,t)=P_0sin(\omega t)$. Not explicitly given in this exercise but mentioned earlier in the book is $b^2=\frac{EIg}{\mu}$, all constants.

The solution is given as well and is the following:

$$Y(x,y)=\frac{b P_0 sin(\omega t)}{2EI \omega}\cdot \left[\frac{sinh((L-x)\sqrt{\omega/b})}{sinh(L\sqrt{\omega/b})}-\frac{sin((L-x)\sqrt{\omega/b})}{sin(L\sqrt{\omega/b})} \right]$$ $$+\frac{2\omega P_0 b}{\pi EI} \sum_{n=1}^{\infty}\frac{sin(n\pi x/L) \cdot sin(bn^2\pi^2t/L^2)}{n(\omega^2-b^2n^4\pi^4/L^4)}$$

I managed to get to a solution in the Laplace domain but it seems to me it is incorrect because I would have expected an infinite amount of poles (looking at the resulting summation) and I just have three of them...

So, I did the following. Using the Laplace transform on the equation gives, using the initial conditions, the equation:

$$\frac{d^4y}{dx^4}+\frac{s^2}{b^2}y=0$$

The solution to this is:

$$y(x,s)=A \cdot cosh(\sqrt{\frac{s}{2b}}x) \cdot cos(\sqrt{\frac{s}{2b}}x)+B \cdot cosh(\sqrt{\frac{s}{2b}}x) \cdot sin(\sqrt{\frac{s}{2b}}x)$$ $$+C \cdot sinh(\sqrt{\frac{s}{2b}}x) \cdot cos(\sqrt{\frac{s}{2b}}x)+D \cdot sinh(\sqrt{\frac{s}{2b}}x) \cdot sin(\sqrt{\frac{s}{2b}}x)$$

I get an error that my post looks like spam. I will shorten it and hope it is OK.

Applying the boundary conditions gives as solution:

$$y(x,s)=D\cdot sinh(\sqrt{\frac{s}{2b}}x) \cdot sin(\sqrt{\frac{s}{2b}}x)$$ $$+D\cdot \frac{sin(\sqrt{\frac{s}{2b}}L) \cdot cos(\sqrt{\frac{s}{2b}}L)}{sinh^2(\sqrt{\frac{s}{2b}}L) + sin^2(\sqrt{\frac{s}{2b}}L)} \cdot sinh(\sqrt{\frac{s}{2b}}x) \cdot cos(\sqrt{\frac{s}{2b}}x)$$ $$-D\cdot \frac{sinh(\sqrt{\frac{s}{2b}}L) \cdot cosh(\sqrt{\frac{s}{2b}}L)}{sinh^2(\sqrt{\frac{s}{2b}}L) + sin^2(\sqrt{\frac{s}{2b}}L)} \cdot cosh(\sqrt{\frac{s}{2b}}x) \cdot sin(\sqrt{\frac{s}{2b}}x)$$

with:

$$D=\frac{P_0 \omega b}{EI} \cdot \frac{1}{s(s^2+\omega^2)}$$

This means now that there are only three poles, $s=0$, $s=i\omega$ and $s=-i\omega$ and I do not see how an inverse transform could give the solution from the book. Is it possible to check the result I have obtained so far before I work on the contour integration and to show how to get to the summation?

I am aware that the question is not the world's shortest one but any help is greatly appreciated.

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A quick look tells me your on the right track ... 3 poles is correct.

BTW, my background is electrical filter designs in industry. LaPlace transforms and pole/zero locations are key to my designs.

Hope my comments are of help :)

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  • $\begingroup$ Thanks @Bru for the quick response. I gave the result in the Laplace domain a new look and the poles are indeed infinite in number due to the sin and sinh functions. I overlooked this... I will now try to get the inverse transformation. It is about 25 years since I did this, let's see how much of my skills are left :-) $\endgroup$
    – Dimitri
    Jul 22, 2020 at 16:18
  • $\begingroup$ I am not able to transform the solution back to the time domain. Any help is welcome. $\endgroup$
    – Dimitri
    Jul 22, 2020 at 20:08
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    $\begingroup$ Only 25 years ago! Your a young guy on the block. I'm thinking 60 years back. Can you show me your Laplace equations and I'll see if my old memory can help. $\endgroup$
    – Bru
    Jul 23, 2020 at 14:35

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