Fourth order differential equation I have this physics mathematical problem : (see link in comment)
$$EI \frac{∂^4u}{∂x^4}= f \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$ 
The boundary conditions are:  $\,\,\frac{∂^2u}{∂x^2} = 0\,\,$   and $\,\,\,EI\frac{∂^3u}{∂x^3}=±F$
where $E$  is Young’s modulus, $I$ is second moment of area, $f$ is force per unit
length applied to the beam and $F$ is the force applied to the edges and the $±$ applies the the left and right edges.
The upper load placed on the chip is modelled as two point forces while the
loads exerted by the pins of the chip are modelled as (localised) Hookean springs
as shown in figure [$1$](in the link). Thus the equations become:
$EI\frac{∂^4u}{∂x^4} = −F \,[δ(x − B) + δ(x + B)] \,− \,k_2\,u(0)\, δ(x)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$
with boundary conditions :
$EI\frac{∂^3u}{∂x^3} = -k_1\,u(−A)\,\,\,\,\,,\,\,\,\,\,\,\,\, x = −A \,\,\,\,\,\,\,\,\,\,\,\,(3)$
$EI\frac{∂^3u}{∂x^3} = k_1\,u(A)\,\,\,\,\,\,,\,\,\,\,\,\,\, x = A \,\,\,\,\,\,\,\,\,\,\,\,(4)$
$\frac{∂^2u}{∂x^2} = 0\,\,\,\,\,\,\,,\,\,\,\,\, x = ±A \,\,\,\,\,\,\,\,\,\,\,(5)$
Given the displacement the tensile strain $\ebselen_{xx}$ can be calculated from the equation
$\ebselen_{xx}=\frac{h}{2} \frac{∂^2u}{∂x^2} \,\,\,\,\,\,\,\,\,\,\,\, (6)$
where $h$ is the height of the chip.
Show that calculating the displacements can be reduced to the problem of
solving a set of linear equations.
could you help please

Trying to answer:
$x$ axis:$\,\,\,\,\,\,$
$-A$______$-B$_____$0$____$B$_____$A$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,I\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, II\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, III \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,IV$


*

*In area $I$ , $II$ , $III$ and $IV$ equation $(2)$ become ZERO: $EI\frac{∂^4u}{∂x^4} =0$
solving this equation we get a linear equation system:


$$
\left\{ 
\begin{array}{c}
u_I=a_1x^3+a_2x^2+a_3x+a_4 \\ 
u_{II}=b_1x^3+b_2x^2+b_3x+b_4 \\ 
u_{III}=c_1x^3+c_2x^2+c_3x+c_4 \\
u_{IV}=d_1x^3+d_2x^2+d_3x+d_4
\end{array}
\right. 
$$
which has $16$ unknowns ($a$'s,$b$'s,$c$'s and $d$'s) 
 A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}%
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
Let's $\ds{EI \equiv {1 \over \beta}}$. We understoot that $-A < -B < B < A$ where $A >0 $ and $B > 0$.
Then,
\begin{align}
\partiald[3]{{\rm u}\pars{x}}{x}
&=
-\beta k_{1}{\rm u}\pars{-A}
+
\beta\int_{-A}^{x}\braces{%
-F\bracks{\delta\pars{x' - B} + \delta\pars{x' + B}}
-
k_{2}{\rm u}\pars{0}\delta\pars{x'}}\,\dd x'
\\[3mm]&=
-\beta k_{1}{\rm u}\pars{-A}
-
\beta F\Theta\pars{x - B}
-
\beta F\Theta\pars{x + B}
-
k_{2}{\rm u}\pars{0}\Theta\pars{x}
\tag{1}
\end{align}
Since
$\ds{%
\left.
\partiald[3]{{\rm u}\pars{x}}{x}\right\vert_{x = A}
=
\beta k_{1}{\rm u}\pars{A}}$, we get the condition
$$
\beta k_{1}\bracks{{\rm u}\pars{-A} + {\rm u}\pars{A}}
=
-2\beta F - k_{2}{\rm u}\pars{0}
$$
From $\pars{1}$:
\begin{align}
\partiald[2]{{\rm u}\pars{x}}{x}
&=
-\beta k_{1}{\rm u}\pars{-A}\pars{x + A}
\\[3mm]&+
\int_{-A}^{x}\bracks{%
-
\beta F\Theta\pars{x' - B}
-
\beta F\Theta\pars{x' + B}
-
k_{2}{\rm u}\pars{0}\Theta\pars{x'}}\,\dd x'
\tag{2}
\end{align}
I hope the OP can take it from here. Just repeat the above procedures.
