25 Beam PDE

 25.1 Beam PDE with zero initial velocity

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25.1 Beam PDE with zero initial velocity

problem number 159

Added January 20, 2018.

Solve \[ \frac{\partial ^2 u}{\partial t^2} + \frac{\partial ^4 u}{\partial x^4} =0 \]

With boundary conditions

\begin{align*} u(0,t) &= -12 t^2\\ f(1,t) &=1-12 t^2\\ \frac{\partial ^2 u}{\partial x^2}u(0,t) &=0 \\ \frac{\partial ^2 u}{\partial x^2}u(1,t) &=12 \end{align*}

And initial conditions

\begin{align*} u(x,0) &= x^4\\ \frac{\partial u}{\partial t}u(x,0) &=0 \end{align*}

Mathematica

\[ \left \{\left \{u(x,t)\to x^4-12 t^2\right \}\right \} \]

Maple

\[ u \left ( x,t \right ) ={x}^{4}-12\,{t}^{2} \]