Taken from Mathematica DSolve help pages
Initial value problem with Dirichlet boundary conditions. 1D, zero potential.
Solve for \(f(x,t)\) \[ I f_t = - 2 f_{xx} \] With boundary conditions \begin {align*} f(5,t) &= 0\\ f(10,t) &=0 \\ \end {align*}
And initial conditions \(f(x,2)=f(x)\) where \(f(x)=-350 + 155 x - 22 x^2 + x^3\)
Mathematica ✓
ClearAll["Global`*"]; pde = I*D[f[x, t], t] == -2*D[f[x, t], {x, 2}]; g[x_] := -350 + 155*x - 22*x^2 + x^3; ic = f[x, 2] == g[x]; bc = {f[5, t] == 0, f[10, t] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, f[x, t], {x, t}], 60*10]]; sol = sol /. K[1] -> n;
\[\left \{\left \{f(x,t)\to \sum _{n=1}^{\infty }\frac {100 \left (7+8 (-1)^n\right ) e^{-\frac {2}{25} i n^2 \pi ^2 (t-2)} \sin \left (\frac {1}{5} n \pi (x-5)\right )}{n^3 \pi ^3}\right \}\right \}\]
Maple ✓
restart; pde :=I*diff(f(x,t),t)=-2*diff(f(x,t),x$2); bc:=f(5,t)=0,f(10,t)=0; g:=x->-350+155*x-22*x^2+x^3; ic:=f(x,2)=g(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],f(x,t))),output='realtime'));
\[f \left ( x,t \right ) =\sum _{n=1}^{\infty }{\frac { \left ( 800\, \left ( -1 \right ) ^{n}+700 \right ) \sin \left ( 1/5\,n\pi \, \left ( x-5 \right ) \right ) {{\rm e}^{-{\frac {2}{25}}\,i{\pi }^{2}{n}^{2} \left ( t-2 \right ) }}}{{n}^{3}{\pi }^{3}}}\]
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