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$$\frac{\partial^4 X}{\partial x^4}+\frac{2}{Y}\frac{\partial ^2 X}{\partial x^2}\frac{\partial^2 Y}{\partial y^2}+\frac{X}{Y}\frac{\partial^4 Y}{\partial y^4}=0\tag{3-70}$$ Requiring that $(\partial^2 Y/\partial y^2)/Y$ and $(\partial^4 Y/\partial y^4)/Y$ be independent of $y$, so that the variables are separable, $Y$ must be of the form $$Y=a\sin ky+b\cos ky\tag{3-71}$$

This is the part of book "Theory of Dislocations (1982)" that is focused on solving differential equations.

$X$ is a one variable function of $x$,

$Y$ is a one variable function of $y$.

Could you explain why $\dfrac{\partial^2 Y}{\partial y^2}\dfrac{1}{Y}$ and $\dfrac{\partial^4 Y}{\partial y^4}\dfrac{1}{Y}$ should be constant here?

K.defaoite
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SGJ
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  • Your question may benefit from some math formatting: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Mailbox Feb 10 '24 at 23:49
  • This would be a good question if you spent more time posing it more fully. – A rural reader Feb 23 '24 at 22:40

1 Answers1

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There is a bit of confusion with terminology. By "separable", the author means, essentially, that the equation can be reduced to an ODE in one variable. Looking at the equation, we can write it two ways. First, by focusing on the $Y$ variable:

$${\frac{\partial^4 X}{\partial x^4}}+\left(\frac{\partial^2X}{\partial x^2}\right)\color{green}{\boldsymbol{\frac{2}{Y}\frac{\partial^2Y}{\partial Y^2}}}+(X)~\color{green}{\boldsymbol{\frac{1}{Y}\frac{\partial ^4Y}{\partial y^4}}}$$ The only way that this equation can reduce to a single variable ODE in $Y$ is if all of the black terms are constant w.r.t the variable $x$. However, since this requires that $X$ be constant, this would be considered a trivial case, and so we ignore it.

On the other hand, we may write the equation focusing on the variable $X$:

$$\color{green}{\boldsymbol{\frac{\partial^4 X}{\partial x^4}}}+{\left(\frac{2}{Y}\frac{\partial^2 Y}{\partial y^2}\right)}\color{green}{\boldsymbol{\frac{\partial ^2 X}{\partial x^2}}}+{\left(\frac{1}{Y}\frac{\partial^4 Y}{\partial y^4}\right)} \color{green}{\boldsymbol X}=0$$

In this equation, it is possible to reduce it to an ODE in $X$ if both $(2/Y)Y''$ and $(1/Y)Y''''$ are both constant, which has a non-trivial solution. So, the author focuses on this case.

K.defaoite
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  • But is this method ok? Are the solutions found by this method can be thought as the only solutions? – SGJ Feb 11 '24 at 03:18
  • @SGJ I don't know. That would require detailed analysis outside the scope of your original question. – K.defaoite Feb 11 '24 at 04:34
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    The separated solutions are far from the only ones, but the hope is that any solution can be obtained as a (usually infinite) linear combination of separated solutions. – Hans Lundmark Feb 11 '24 at 12:50