Difficulty to Solve the Generalized Wave Equation

Here is the analytical solution. It now matches the numerical solution.

Few things were corrected including normalization. See Hermite_polynomials for references used to find correct weight. The one used is

But to use the above, it is $e^{-x^2}$ and not $e^{\frac{-x^2}{2}}$, this is why the hand solution below makes this adjustment during finding $A_n$.

Without this important change, the analytical solution was not matching the numerical solution. Another important change is that the BC must be $\pm \infty$ and not finite domain. So can not use $L=21$ for example. This explains why DEigensystem could not solve it analytically before but solved instantly when the BC was at$\pm \infty$.

It is important to note that domain length can affect the eigenvalues! For example using $L=1$ vs. $L=\pi$ one can obtain different eigenvalues.

Notice that DSolve is not able to solve this analytically. May be in Version 14.0 it will be able to do that.

(*analytical solution *) ClearAll[x, t, n, sol]; v = 1; int[n_] := Integrate[Exp[-x^2]*HermiteH[n, x], {x, -Infinity, Infinity}] lam[n_] := 2 n + 1; sol[max_] := Sum[1/(Sqrt[Pi]*2^n*n!)*int[n]*Cos[Sqrt[v/2*lam[n]]*t]*Exp[-x^2/2]* HermiteH[n, x], {n, 0, max, 1}]; 

For example, using 5 terms, the solution is

sol[5] 

Here is side by side with numerical solution

Code

ClearAll[x, t, n, sol]; v = 1; int[n_] := Integrate[Exp[-x^2]*HermiteH[n, x], {x, -Infinity, Infinity}] lam[n_] := 2 n + 1; sol[max_] := Sum[1/(Sqrt[Pi]*2^n*n!)*int[n]*Cos[Sqrt[v/2*lam[n]]*t]*Exp[-x^2/2]* HermiteH[n, x], {n, 0, max, 1}]; L = 10; bcN = {u[-L, t] == 0, u[L, t] == 0}; ic = u[x, 0] == Exp[-x^2/2]; pde = D[u[x, t], {t, 2}] - 1/2 *v*D[u[x, t], {x, 2}] + 1/2*v*x^2*u[x, t] == 0; nsol = NDSolveValue[{pde, ic, bcN}, u, {x, -L, L}, {t, 0, 10}]; Manipulate[ Module[{theSolution = sol[10]}, Row[{ Plot[Evaluate[theSolution /. {x -> x0, t -> t0}], {x0, -10, 10}, PerformanceGoal -> "Quality", PlotRange -> {Automatic, {-1.1, 1.1}}, ImageSize -> 300, PlotLabel -> "Analytical"] , Plot[Evaluate[nsol[x0, t0]], {x0, -10, 10}, PerformanceGoal -> "Quality", PlotRange -> {Automatic, {-1.1, 1.1}}, ImageSize -> 300, PlotLabel -> "Numerical"] } ] ], {{t0, 0, "time"}, 0, 10, .1, Appearance -> "Labeled", ContinuousAction -> False}, TrackedSymbols :> {t0} ] 

Hand solution

\begin{align*} u_{tt}+\frac{v}{2}x^{2}u & =\frac{v}{2}u_{xx}\\ u\left( x,0\right) & =f\left( x\right) =e^{-\frac{x^{2}}{2}}\\ u_{t}\left( x,0\right) & =0\\ u\left( -\infty,t\right) & =0\\ u\left( +\infty,t\right) & =0 \end{align*} Let $u=X\left( x\right) T\left( t\right) $, then the pde becomes \begin{align*} T^{\prime\prime}X+\frac{v}{2}x^{2}XT & =\frac{v}{2}X^{\prime\prime}T\\ \frac{T^{\prime\prime}}{T}+\frac{v}{2}x^{2} & =\frac{v}{2}\frac {X^{\prime\prime}}{X}\\ \frac{2}{v}\frac{T^{\prime\prime}}{T} & =\frac{X^{\prime\prime}}{X}-x^{2}% \end{align*} Hence \begin{align*} \frac{2}{v}\frac{T^{\prime\prime}}{T} & =-\lambda\\ T^{\prime\prime}+\frac{v}{2}\lambda T & =0 \end{align*} The solution to the above is $$ T=A\cos\left( \sqrt{\frac{v}{2}\lambda}t\right) +B\sin\left( \sqrt{\frac {v}{2}\lambda}t\right) $$ And \begin{align*} \frac{X^{\prime\prime}}{X}-x^{2} & =-\lambda\\ -X^{\prime\prime}+x^{2}X & =\lambda X\\ X\left( -\infty\right) & =0\\ X\left( \infty\right) & =0 \end{align*} The eigenvalues are (see below) $\lambda_{n}=1+2n$ for $n=0,1,2,3,\cdots$. Hence $\lambda_{n}=\left\{ 1,3,5,7,\cdots\right\} $ and corresponding eigenfunction are \begin{align*} X_{n} & =e^{-\frac{x^{2}}{2}}H_{n}\left( x\right) \qquad n=0,1,2,3\cdots\\ \lambda_{n} & =\left\{ 1,3,5,7,\cdots\right\} \end{align*} Hence the pde solution is linear combination of the solutions $X_{n}T_{n}$ or \begin{align} u & =\sum_{n=0}^{\infty}T_{n}X_{n}\tag{1}\\ & =\sum_{n=0}^{\infty}\left[ A_{n}\cos\left( \sqrt{\frac{v}{2}\lambda_{n}% }t\right) +B_{n}\sin\left( \sqrt{\frac{v}{2}\lambda_{n}}t\right) \right] e^{-\frac{x^{2}}{2}}H_{n}\left( x\right) \nonumber \end{align} At $t=0$ the above becomes, since $u\left( x,0\right) =f\left( x\right) =e^{-\frac{x^{2}}{2}}$ \begin{align*} e^{-\frac{x^{2}}{2}} & =\sum_{n=0}^{\infty}A_{n}e^{-\frac{x^{2}}{2}}% H_{n}\left( x\right) \\ e^{-x^{2}} & =\sum_{n=0}^{\infty}A_{n}e^{-x^{2}}H_{n}\left( x\right) \\ H_{m}\left( x\right) e^{-x^{2}} & =\sum_{n=0}^{\infty}A_{n}e^{-x^{2}}% H_{n}\left( x\right) H_{m}\left( x\right) \end{align*} Integrating $$ \int_{-\infty}^{\infty}H_{m}\left( x\right) e^{-x^{2}}dx=\sum_{n=0}^{\infty }A_{n}\int_{-\infty}^{\infty}e^{-x^{2}}H_{n}\left( x\right) H_{m}\left( x\right) dx $$ By orthogonality $\int_{-\infty}^{\infty}e^{-x^{2}}H_{n}\left( x\right) H_{m}\left( x\right) dx=\sqrt{\pi}2^{m}m!\delta_{nm}$ (see Wikipedia) hence the above becomes $$ \int_{-\infty}^{\infty}H_{m}\left( x\right) e^{-x^{2}}dx=A_{m}\sqrt{\pi }2^{m}m! $$ Therefore $$ A_{n}=\frac{1}{\sqrt{\pi}2^{n}n!}\int_{-\infty}^{\infty}H_{n}\left( x\right) e^{-x^{2}}dx $$ The solution (1) becomes \begin{equation} u\left( x,t\right) =\sum_{n=0}^{\infty}\left[ \left( \frac{1}{\sqrt{\pi }2^{n}n!}\int_{-\infty}^{\infty}H_{n}\left( x\right) e^{-x^{2}}dx\right) \cos\left( \sqrt{\frac{v}{2}\lambda_{n}}t\right) +B_{n}\sin\left( \sqrt{\frac{v}{2}\lambda_{n}}t\right) \right] e^{-\frac{x^{2}}{2}}% H_{n}\left( x\right) \tag{2}% \end{equation} Taking time derivative the above becomes $$ u_{t}=\sum_{n=0}^{\infty}\left[ -\left( \frac{1}{\sqrt{\pi}2^{n}n!}% \int_{-\infty}^{\infty}H_{n}\left( x\right) e^{-x^{2}}dx\right) \sqrt {\frac{v}{2}\lambda_{n}}\sin\left( \sqrt{\frac{v}{2}\lambda_{n}}t\right) +B_{n}\sqrt{\frac{v}{2}\lambda_{n}}\cos\left( \sqrt{\frac{v}{2}\lambda_{n}% }t\right) \right] e^{-\frac{x^{2}}{2}}H_{n}\left( x\right) $$ At $t=0$ and since $u_{t}=0$ simplifies to $$ 0=\sum_{n=0}^{\infty}B_{n}\sqrt{\frac{v}{2}\lambda_{n}}e^{-\frac{x^{2}}{2}% }H_{n}\left( x\right) $$ Therefore $B_{n}=0$. The final solution from (2) becomes $$ \boxed{ u\left( x,t\right) =\sum_{n=0}^{\infty}\left( \frac{1}{\sqrt{\pi}2^{n}% n!}\int_{-\infty}^{\infty}H_{n}\left( x\right) e^{-x^{2}}dx\right) \cos\left( \sqrt{\frac{v}{2}\lambda_{n}}t\right) e^{-\frac{x^{2}}{2}}% H_{n}\left( x\right) } $$ For $\lambda_{n}=\left\{ 1,3,5,7,\cdots\right\} $

Finding eigenvalues and eigenfunctions

ode = -X''[x] + x^2*X[x] == 0 DEigensystem[{First@ode, X[-Infinity] == 0, X[Infinity] == 0}, X[x], {x, -Infinity, Infinity}, 6]