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I tried to bilinearize the two equations:

eq = {Laplacian[psi[t, x, y], {x, y}] + 
     6 u[t, x, y] psi[t, x, y] + (x^2 + y^2) u[t, x, y] == 
    I D[u[t, x, y], 
      t], -(I /Sqrt[2] (D[u[t, x, y], x] + D[u[t, x, y], y]) + 
       psi[t, x, y]) == 1/10000 D[psi[t, x, y], t]};

and I have looked at some posts here on stackexchange, but found no posts that considered the matter of bilinearization. Does anyone have a tip of where to start?

nssm2 = NonlinearStateSpaceModel[eq, psi[t,x,y], u[t,x,y], t]

but to no avail. I read also this:

https://library.wolfram.com/infocenter/Articles/2831/

But I couldn't find a hint there either.

Thanks

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  • $\begingroup$ What means "bilinearization" ? u & psi small from same order? $\endgroup$ Jun 11, 2021 at 11:07
  • $\begingroup$ Yes, it means to get a linear form of the two PDEs where the two , u and psi, are part of the linearized resulting PDE. $\endgroup$ Jun 11, 2021 at 11:50

4 Answers 4

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Try

eq1=Normal[Series[eq /. {u -> Function[{t, x, y}, eps u[t, x, y]] ,psi -> Function[{t, x, y}, eps psi[t, x, y]]}, {eps,0,1}]] /.eps -> 1

to get the linearized two odes !

enter image description here

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  • $\begingroup$ Thanks a lot, this is a great start. $\endgroup$ Jun 11, 2021 at 12:03
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    $\begingroup$ You're welcome, hope it helps solving your problem! $\endgroup$ Jun 11, 2021 at 12:04
  • $\begingroup$ Yes, that last correction in your posts gave the right answer. It is in fact the same as Daniel Hubers. Thanks $\endgroup$ Jun 11, 2021 at 12:16
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I assume that by "bilinearize" you mean linearization relative to u[t, x, y] and psi[t, x, y]. This can be achieved using Series. Note, Series relative to more than one variable does does first linearization relative to the first variable. Then, the result is linearized relative to the second variable, e.t.c. That means, we get cross terms, that need to be eliminated. E.g.

eq = {Laplacian[psi[t, x, y], {x, y}] + 
     6 u[t, x, y] psi[t, x, y] + (x^2 + y^2) u[t, x, y] == 
    I D[u[t, x, y], 
      t], -(I /Sqrt[2] (D[u[t, x, y], x] + D[u[t, x, y], y]) + 
       psi[t, x, y]) == 1/10000 D[psi[t, x, y], t]};

To linearize, we do:

ser = Series[eq, {u[t, x, y], 0, 1}, {psi[t, x, y], 0, 1}];

This is a SeriesData object. For further work we need to get rid of the O[..] terms:

ser= ser // Normal;

To eliminate the cross terms, we first need to expand the expression and then set the cross terms to zero:

Expand[ser] /. u[t, x, y] psi[t, x, y] -> 0

enter image description here

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  • $\begingroup$ Thanks for this. If I linearize it with respect to the second variable, do I get 4 ODEs, which are all relevant? $\endgroup$ Jun 11, 2021 at 12:04
  • $\begingroup$ ' Series[eq, {psi[t, x, y], 0, 1}] // Normal' gives 2PDE's $\endgroup$ Jun 11, 2021 at 13:13
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Look what @Jens has done:

TaylorSeries[expr_, vars_?ListQ, start_?ListQ, order_?IntegerQ] := 
 Normal[Series[expr /. Thread[vars -> t*(vars - start) + start], {t, 0, order}]] /. t -> 1

eq = {Laplacian[psi[t, x, y], {x, y}] + 6 u[t, x, y] psi[t, x, y] + (x^2 + y^2) u[t, x, y] == 
I D[u[t, x, y], t], -(I/Sqrt[2] (D[u[t, x, y], x] + D[u[t, x, y], y]) + psi[t, x, y]) == 1/10000 D[psi[t, x, y], t]};

TaylorSeries[eq, {u[t, x, y], psi[t, x, y]}, {0, 0}, 1]

enter image description here

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Hint.

Assuming you have an approximation $(u_k,\psi_k)$ then you can proceed linearly as

$$ \cases{ \mathcal{D}_1[u_{k+1}]+6(u_k \psi_k +u_k(\psi_{k+1}-\psi_k)+\psi_k(u_{k+1}-u_k))+(x^2+y^2)u_{k+1}=i \mathcal{D}_2[u_{k+1}]\\ -\frac{i}{\sqrt{2}}\mathcal{D}_3[u_{k+1}]+\psi_{k+1}=10^{-4}\mathcal{D}_4[\psi_{k+1}] } $$

solving for $(u_{k+1},\psi_{k+1})$ to obtain the next approximation. Here $\mathcal{D}_j[\cdot]$ are differential operators.

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