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 ✓

ClearAll["Global`*"]; 
pde =  D[w[x1, x2, x3, t], {t, 2}] == D[w[x1, x2, x3, t], x1, x2] + D[w[x1, x2, x3, t], x1, x3] + D[w[x1, x2, x3, t], {x3, 2}] - D[w[x1, x2, x3, t], x2, x3]; 
ic  = {w[x1, x2, x3, t0] == x1^3*x2^2 + x3, Derivative[0, 0, 0, 1][w][x1, x2, x3, t0] == -(x2*x3) + x1}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, w[x1, x2, x3, t], {x1, x2, x3, t}], 60*10]];
 

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

Maple ✓

restart; 
pde := diff(w(x1, x2, x3, t), t$2)= diff(w(x1,x2,x3,t),x1,x2)+diff(w(x1,x2,x3,t),x1,x3)+diff(w(x1,x2,x3,t),x3$2)-diff(w(x1,x2,x3,t),x2,x3); 
ic  := w(x1, x2, x3, t0) = x1^3*x2^2+x3, eval( diff( w(x1,x2,x3,t),t),t=t0)=-x2*x3+x1; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic],w(x1,x2,x3,t))),output='realtime'));
 

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

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