2.16.8 second order in time, Linear PDE, initial conditions at \(t=t_0\)

problem number 147

Added December 20, 2018.

Example 27, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Solve for \(w(x_1,x_2,x_3,t)\) \[ \frac {\partial w^2}{\partial t^2} = \frac {\partial w^2}{\partial x_1 x_2} + \frac {\partial w^2}{\partial x_1 x_3} + \frac {\partial w^2}{\partial x_3^2} - \frac {\partial w^2}{\partial x_2 x_3} \] With initial condition \begin {align*} w(x_1,x_2,x_3,t_0) &= x_1^3 x_2^2 + x_3 \\ \frac {\partial w}{\partial t}(x_1,x_2,x_3,t_0) &= -x_2 x_3 + x_1 \end {align*}

Mathematica


\[\left \{\left \{w(\text {x1},\text {x2},\text {x3},t)\to t^2 \left (3 \text {t0}^2 \text {x1}-\frac {\text {t0}}{2}+3 \text {x1}^2 \text {x2}\right )+t^3 \left (\frac {1}{6}-2 \text {t0} \text {x1}\right )+\frac {t^4 \text {x1}}{2}+t \left (-2 \text {t0}^3 \text {x1}+\frac {\text {t0}^2}{2}-6 \text {t0} \text {x1}^2 \text {x2}+\text {x1}-\text {x2} \text {x3}\right )+3 \text {t0}^2 \text {x1}^2 \text {x2}+\frac {\text {t0}^4 \text {x1}}{2}-\frac {\text {t0}^3}{6}-\text {t0} \text {x1}+\text {t0} \text {x2} \text {x3}+\text {x1}^3 \text {x2}^2+\text {x3}\right \}\right \}\]

Maple


\[w \left (\mathit {x1} , \mathit {x2} , \mathit {x3} , t\right ) = 3 t^{2} \mathit {x1}^{2} \mathit {x2} +\frac {\mathit {t0}^{4} \mathit {x1}}{2}+\mathit {x1}^{3} \mathit {x2}^{2}+\frac {t^{3}}{6}-t \mathit {x2} \mathit {x3} +\frac {\left (-12 \mathit {x1} t -1\right ) \mathit {t0}^{3}}{6}+\frac {\left (18 t^{2} \mathit {x1} +18 \mathit {x1}^{2} \mathit {x2} +3 t \right ) \mathit {t0}^{2}}{6}+\frac {\left (-36 t \mathit {x1}^{2} \mathit {x2} -3 t^{2}+6 \mathit {x2} \mathit {x3} +\left (-12 t^{3}-6\right ) \mathit {x1} \right ) \mathit {t0}}{6}+\frac {\left (3 t^{4}+6 t \right ) \mathit {x1}}{6}+\mathit {x3}\]

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