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After visiting the Whispering wall where a whisper can be clearly transmitted between two locations on the surface of a dam wall over more than 100 metres, I was puzzled as to how this occurs. I decided to try to solve the following problem: What must be the shape of a curved wall such that any sound emitted towards the wall from one corner will be reflected directly to the opposite corner as in the following diagram (co-ordinates are $x$ and $y$)?

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From this I derived the following differential equation (derivation below) but myriad substitutions and changes of variable have still left me puzzled as how to put it into a form which can be solved:

$$(y’)^2++y’\left(\frac{a^2+y^2-x^2}{ay}\right)+\frac{x}{a}=0$$

for $-a<x<0$ with $y(\pm a)=0$ and $y’(0)=0$, but I’m not sure that this is right. Can solve this equation or else find a mistake in it and find the correct equation to solve my geometrical problem?

Derivation (note that from the answers it would appear that this is slightly incorrect): Taking the following horrible paint diagram where $D$ is the point at which the sound wave is reflected and $AB$ is the tangent to the curve at that point, where we must have that the angles $BDC$ and $ADF$ marked $\theta$ are equal since a specular reflection is assumed to occur:

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(Note that this is for negative $x$ only so $-x>0$; the vertical line is through the maximum of the curve). I then take the triangle $DBC$ and divide the lengths of the sides by $a-x$ to find that (using the relation between the tangent and the derivative):

$$\theta=\arctan{y’}+\arctan{\frac{y}{a-x}}$$

I then take the triangle $ADE$ and divide the lengths by $y$ to obtain:

$$\theta=\arctan{\frac{1}{y'}}-\arctan{\frac{a+x}{y}}$$

Setting these equal and taking $\tan$ of both sides and using the law for the tangent of a sum of angles to obtain two fractions which I cross-multiply and then collecting terms I get the differential equation shown above.

(Note that interesting facts about similar walls which might better explain the Whispering wall I originally mentioned can be found here, here and here, but I am now specifically looking for a solution to my problem. Also not that the Wikipedia page in my very first link mentioned a ‘parabolic effect’, but I could not see how that would be relevant to my problem)

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Following @Djura Marinkov's remark, the correct equation to study is \begin{equation} y'^2 + \frac{x^2 - y^2 - a^2}{x y} y' -1 = 0. \tag{1} \end{equation} Using the substitution $y(x)^2 = z(\eta(x))$ where $\eta(x) = x^2$, we obtain \begin{equation} z'^2 + \left(1-\frac{z+a^2}{\eta}\right) z' - \frac{z}{\eta} = 0, \tag{2} \end{equation} which looks marginally easier than $(1)$, but not much. However, this changes when we take the derivative of $(2)$ to $\eta$: then we obtain \begin{align} z'' =& \frac{-z(1+z') + z'(-a^2 + \eta(1+z'))}{\eta(-a^2+\eta-z+2 \eta z')}\\ =& \frac{z'^2 + \left(1-\frac{z+a^2}{\eta}\right)z'-\frac{z}{\eta}}{-a^2+\eta-z+2 \eta z'}\\ =& 0, \end{align} where we used $(2)$ to obtain the last line. Therefore, $z$ must be a linear function, i.e. \begin{equation} y^2 = \alpha x^2 + \beta \tag{3}. \end{equation} You can now substitute $(3)$ in $(1)$ to find (a relation between) $\alpha$ and $\beta$, in terms of $a$. Note that this answer reflects (yes) some useful properties of ellipses, see here, here or here.

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  • $\begingroup$ Thanks a lot for that reply. I think I had made that substitution but I had not thought of taking the derivative. That's very interesting; at first I had immediately thought that the result would be an ellipse since I had heard of the property of ellipses of reflecting between foci, but since the reflections where occurring from corner to corner I thought that this would be irrelevant since no ellipse has its foci on its boundary. I wonder how this fits in? $\endgroup$ – Anon Nov 10 '16 at 22:08
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    $\begingroup$ I would say that the conclusion is that there is no wall with the desired properties that begins at $x=-a$ and ends at $x=a$. So, standing in the corner will never get you heard in the other corner. Note that you can significantly weaken your restrictions by allowing just one (or a finite number of) `exactly reflecting' directions. Then, the whispering phenomenon only occurs if you send waves in the right direction. Alternatively, you can 'cut off' your ellipse shaped wall above $x=\pm a$ and place a straight wall segment there: for the majority of speaking directions, you'll still be heard $\endgroup$ – Frits Veerman Nov 11 '16 at 11:06
  • $\begingroup$ I think it's quite interesting that no such wall can exist. Maybe you can mention this in the answer also? $\endgroup$ – Anon Nov 19 '16 at 5:44
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$$\arctan y'-\arctan \frac{y}{x-a}=\arctan \frac{y}{x+a}- \arctan y'$$ $$2\arctan y'=\arctan \frac{y}{x-a}+\arctan \frac{y}{x+a}$$ $$\frac{2y'}{1-y'^2}=\frac{\frac{y}{x-a}+\frac{y}{x+a}}{1-\frac{y^2}{x^2-a^2}}$$ $$\frac{2y'}{1-y'^2}=\frac{\frac{2xy}{x^2-a^2}}{\frac{x^2-a^2-y^2}{x^2-a^2}}$$ $$y'(x^2-y^2-a^2)=xy(1-y'^2)$$ $$y'^2+\frac{x^2-y^2-a^2}{xy}y'-1=0$$

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  • $\begingroup$ Thank you for clearing that up, I'll have to try and figure out where I went wrong. $\endgroup$ – Anon Nov 10 '16 at 22:06

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