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The database contains 327 equations (8 equations are awaiting activation).

Equation data
Category:3. Linear Partial Differential Equations
Subcategory:3.5. Higher-Order Equations
Equation(s):$\displaystyle 
\frac{\partial^2w}{\partial t^2}+a\frac{\partial^4w}{\partial x^4}=0$.
Solution(s),
Transformation(s),
Integral(s)
:
$1^\circ$. Solution:\hfill\break
$$
w(x,t)= \sum^\infty_{n=0}[C_{0n}w_{0n}(x,t)+C_{1n}w_{1n}(x,t)+C_{2n}w_{2n}(x,t)+C_{3n}w_{3n}(x,t)],$$
where $C_{0n}$, $C_{1n}$, $C_{2n}$ and $C_{3n}$ are arbitrary constants, and the
functions $w_{0n}$, $w_{1n}$, $w_{2n}$, and $w_{3n}$ are defined by the relations\\
$\displaystyle w_{0n}(x,t)=\sum^{n-2k\ge0}_{k=0}{\frac{(-1)^kn!}{(4k)!(n-2k)!a^k}x^{4k}t^{n-2k}}$,\\
$\displaystyle w_{1n}(x,t)=\sum^{n-2k\ge0}_{k=0}{\frac{(-1)^kn!}{(4k+1)!(n-2k)!a^k}x^{4k+1}t^{n-2k}}$,\\
$\displaystyle w_{2n}(x,t)=\sum^{n-2k\ge0}_{k=0}{\frac{(-1)^kn!}{(4k+2)!(n-2k)!a^k}x^{4k+2}t^{n-2k}}$,\\
$\displaystyle w_{3n}(x,t)=\sum^{n-2k\ge0}_{k=0}{\frac{(-1)^kn!}{(4k+3)!(n-2k)!a^k}x^{4k+3}t^{n-2k}}$.
\medskip

$2^\circ$. Solution:\hfill\break
$$w(x,t)= \sum^\infty_{n=0}[C_{n0}w_{n0}(x,t)+C_{n1}w_{n1}(x,t)],$$
where $C_{n0}$ and $C_{n1}$ are arbitrary constants, and
the functions $w_{n0}$ and $w_{n1}$ are defined by the relations\\
$\displaystyle w_{n0}(x,t)=\sum^{n-4k\ge0}_{k=0}{\frac{(-1)^kn!a^k}{(2k)!(n-4k)!}x^{n-4k}t^{2k}}$,\\
$\displaystyle w_{n1}(x,t)=\sum^{n-4k\ge0}_{k=0}{\frac{(-1)^kn!a^k}{(2k+1)!(n-4k)!}x^{n-4k}t^{2k+1}}$.
Remarks:This equation is encountered in studying transverse vibration of 
elastic rods.
Novelty:New solution(s) / integral(s)
Admin's Comment:For other solutions see, for example, "Handbook of Linear Partial Differential Equations for Engineers and Scientists" (Chapman & Hall/CRC Press, 2002, pp. 605-608)
by A. D. Polyanin.
Author/Contributor's Details
Last name:Stepuchev
First name:Valeriy
Middle(s) name:Germanovich
Country:Latviya
City:Sigulda
Statistic information
Submission date:Mon 10 Mar 2008 21:36
Edits by author:0

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