Solving Generalized Heat Equation by separation of variables

Questions
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1

Given the following problem

u_{t}-\frac{v}{2}\cdot u_{xx}+\frac{v}{2}\cdot x^2 \cdot u(x)=0
u(x,0)=f(x)

Where f(x)=y_1(x)+0.2y_4(x)+0.01y_6 (x)

y_n are the Eigenfunctions that defined as y_n(x)=\exp\left(\displaystyle{\frac{-x^2}{2}}\right)H_n(x)

By separation we assume

u(x,t)=X(x)T(x)

I have found that the general solution of T is
T(x)=c_1e^{-\lambda v t}
and the general solution of X is
X_n(x)=\exp\left(\displaystyle{\frac{-x^2}{2}}\right)H_n(x)

So the general solution to our initial problem is
u_n(x,t)=\sum_{n=1}^{\infty}A_n \exp\left(\displaystyle{\frac{-x^2}{2}}\right)H_n(x) c_1e^{-\lambda v t}

Using the I.C we got

u(x,0)=\sum_{n=1}^{\infty}A_n \exp\left(\displaystyle{\frac{-x^2}{2}}\right)H_n(x)=f(x)

I have the following two questions

  1. How many terms does the expansion of the solution have? Use the Manipulate command, in conjunction with Plot, to construct animated graphics for the evolution of the solution. For the first case
  2. With appropriate snapshots at 3 different moments of your choice (or even using Manipulate), compare the analytical and numerical solutions. Plot using Plot3D the solution u( x, t ).

I have tried this code but it does not seem right to me.
Any suggestions?

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2

Just a question: You are using the LegendreP expansion together with the Gaussian measure, but the eigenfunctions on the infinite interval are Parabolic cylinder functions with Hermite polynomials as the special case of normalizable functions on all of R.

The LegendreP is of no use outside (-1,1), What is the intention of their use here?

Regards Roland

3

I used it for the orthogonality of Hermit polynomials. what would you advise me to do; put the Hermite command in place of LegendreP?

4

Hi Athanasios, seems I dont quite understand..

LegrendreP[n,x] is orthogonal on (-1,1) with measure (1-x^2),
HermiteH[n,x ] is orthogonal on -inf, inf with measure e^(-x^2).
e^(-x^2) H_n(x) is orthogonal on -inf, inf with measure 1
e^(-x^2) H_n(x) are eigenfunctions of -d_xx +x^2 with eigenvalue 2n+1

It follows without any approximation that

Sum[a_n  e^(-(2n+1)t  HermiteH[n,x]) ,{ n, 0, inf}]

solves

d_t == d_xx +  x^2

Yor start distribution is of that form for some a_n

Of course you may approximate e^-x2/2 HermiteH[n,x] by LegendreP[m,x] on (-1.1), But only using the measure (1-x^2) dx and normailzation to an orthonormal basis will yield the correect approximation.

Regards Roland

5

Could you show me that in this particular problem? Because I am a bit confused

Thank you in advance

6

In[1]:=
DOp[v_] = ((D[#, t] - v/2 D[#, {x, 2}] + v/2 x^2 #) &);

In[3]:= Assuming[n \[Element] PositiveIntegers , 
 DOp[v][E^(-(n + 1/2) v t - x^2/2)  HermiteH[n, x]] // FullSimplify]

Out[3]= 0

In[5]:=  
DOp[v][f0x[t_, x_, v_] = 
   E^(-(1 + 1/2) v t - x^2/2)  HermiteH[1, x] + 
    2/10 E^(-(4 + 1/2) v t - x^2/2)  HermiteH[4, x] +
    1/100  E^(-(6 + 1/2) v t - x^2/2)  HermiteH[6, x] ] // FullSimplify

Out[5]= 0

In[6]:= nds[p_, {t_, x_}, v_, {t_, 0, T_}, {x_, a_, b_}] :=
 NDSolve[   {DOp[ p[t,x] ==0,   
   p[0, x] ==  2 E^(-(x^2/2)) x + 1/5 E^(-(x^2/2)) (12 - 48 x^2 + 16 x^4) + 
     1/100 E^(-(x^2/2)) (-120 + 720 x^2 - 480 x^4 + 64 x^6)       
   Derivative[0, 1][p][t, a] == 0, Derivative[0, 1][p][t, b] == 0}, 
  p[t, x], {t, 0, T}, {x, a, b}]

In[7]:= 
q[t_, x_] = p[t, x] /. {nds[p, {t, x}, 1, {t, 0, 1}, {x, -4, 4}][[1]]}

During evaluation of In[7]:= NDSolve::ibcinc: Warning: boundary and initial conditions are inconsistent.

Out[7]= {InterpolatingFunction[{{0., 1.}, {-4., 
   4.}}, {5, 5, 1, {76, 75}, {4, 6}, 0, 0, 0, 0, Automatic, {}, {}, 
    False}, {CompressedData[s=
"]}, {Automatic, Automatic}][t, x]}

Manipulate[
 Plot[{q[t, x], f0x[t, x, 1]}, {x, -4, 4}, 
  PlotStyle -> {{Red, Thickness[0.02]}, {Black}}, 
  PlotRange -> {-6, 6}], {t, 0, 1}]
7

I am trying this part of your code but it does not work

nds[p_, {t_, x_}, v_, {t_, 0, T_}, {x_, a_, b_}] :=
NDSolve[   {DOp[ p[t,x] ==0,   
p[0, x] ==  2 E^(-(x^2/2)) x + 1/5 E^(-(x^2/2)) (12 - 48 x^2 + 16 x^4) + 
1/100 E^(-(x^2/2)) (-120 + 720 x^2 - 480 x^4 + 64 x^6)       
Derivative[0, 1][p][t, a] == 0, Derivative[0, 1][p][t, b] == 0}, 
p[t, x], {t, 0, T}, {x, a, b}]
8

You are right:.

DOp[v][f0x[t_, x_, v_] = 
   E^(-(1 + 1/2) v t - x^2/2) HermiteH[1, x] + 
    2/10 E^(-(4 + 1/2) v t - x^2/2) HermiteH[4, x] + 
    1/100 E^(-(6 + 1/2) v t - x^2/2) HermiteH[6, x]] // FullSimplify

After making v a parameter to  DOp  and defining the start distribution f0x,  forgot to  update the definition

  nds[p_, {t_, x_}, v_, {t_, 0, T_}, {x_, a_, b_}] :=
 NDSolve[
  {DOp[v][p[t, x]] == 0,
   p[0, x] == f0x[0, x, v],
   Derivative[0, 1][p][t, a] == 0, Derivative[0, 1][p][t, b] == 0},
  p[t, x], {t, 0, T}, {x, a, b}]

nds[p, {t, x}, 1, {t, 0, 1}, {x, -4, 4}]


q[t_, x_] = 
 p[t, x] /. {nds[p, {t, x}, 1, {t, 0, 1}, {x, -4, 4}][[1]]} // Quiet


Manipulate[
  Plot[{q[t, x], f0x[t, x, 1]}, {x, -4, 4}, 
    PlotStyle -> {{Red, Thickness[0.02]}, {Black}}, 
    PlotRange -> {-6, 6}], {t, 0, 1}]
9

I figured out how to solve the problem numerically. Could it be solved analytically if we apply the solution representation formula to find the solution to the problem with the initial condition f(x)?