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Numerical solution of fourth order differential equation

Posted 10 years ago

Could you tell me to solve numerically the equation with the next boundary conditions

\[Pi]^4 Y[y] - b^4 N0 (Y^\[Prime]\[Prime])[y] - 
  2 b^2 \[Pi]^2 (Y^\[Prime]\[Prime])[y] + 
  b^3 N0 y \[Alpha] (Y^\[Prime]\[Prime])[y] + b^4 
\!\(\*SuperscriptBox[\(Y\), 
TagBox[
RowBox[{"(", "4", ")"}],
Derivative],
MultilineFunction->None]\)[y] == 0
Y[0] == 0;
Y[1] == 0;
Y''[0] == 0;
Y''[1] == 0;

equation

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6 Replies

Thank you very much. I did not know about "AccuracyGoal". I'm solving an equation of elastic buckling of rectangular plate with linearly varying in-plane load and I expected the solution similar to Y[y] = Sin[[Pi]*y/b]. This is the solution of my equation if [Alpha] = 0 and load is constant. I expected solution similar to the last picture.

If you modify the boundary conditions, the solver gives the reasonable solution. I guess in your case the solution is very likely Y[x] == 0.

code

POSTED BY: Shenghui Yang

You are right. These deviations are just either numerical error or internal values.

POSTED BY: Shenghui Yang

But all the deviations from zero that you see, are only numerical artefacts, aren't they?

sols = ParametricNDSolve[{\[Pi]^4 Y[y] - 
     b^4 N0 (Y^\[Prime]\[Prime])[y] - 
     2 b^2 \[Pi]^2 (Y^\[Prime]\[Prime])[y] + 
     b^3 N0 y \[Alpha] (Y^\[Prime]\[Prime])[y] + 
     b^4 D[Y[y], {y, 4}] == 0,
   Y[0] == 0, Y[1] == 0, Y''[0] == 0, Y''[1] == 0}, 
  Y, {y, 0, 1}, {N0, b, \[Alpha]}, AccuracyGoal -> 100]

gives

enter image description here

I suppose that the existence and uniqueness theorem would suggest that there is only one solution, which is the "trivial" solution?

Cheers,

M.

PS: Another clue that the solutions that deviate from zero cannot be accurate is that they do not fulfil one of the boundary conditions, i.e. Y[1]==0.

POSTED BY: Marco Thiel

Here is what you can try. I believe this time Mathematica hasn't found a non-trivial solution yet. You may try give this equation a reasonable choice of parameters:

enter image description here

Attachments:
POSTED BY: Shenghui Yang

Hi,

there are quite some issues with that. First of all there are many free parameters such $b, N0, \alpha$. To integrate numerically you need to substitute values for them (or use ParametricNDSolve). Syntax that produces an output is:

sols = NDSolve[{\[Pi]^4 Y[y] - b^4 N0 (Y^\[Prime]\[Prime])[y] - 
      2 b^2 \[Pi]^2 (Y^\[Prime]\[Prime])[y] + 
      b^3 N0 y \[Alpha] (Y^\[Prime]\[Prime])[y] + 
      b^4 D[Y[y], {y, 4}] == 0 /. {b -> 1., N0 -> 1, \[Alpha] -> 1},
   Y[0] == 0, Y[1] == 0, Y''[0] == 0, Y''[1] == 0}, Y[y], {y, 0, 1}]

There might be some numerical issues though because

Plot[Y[y] /. sols, {y, 0, 1}]

gives

enter image description here

which is numerically zero, where it should be plainly zero. You can fix this by setting a higher Accuracy Goal:

sols = NDSolve[{\[Pi]^4 Y[y] - b^4 N0 (Y^\[Prime]\[Prime])[y] - 
      2 b^2 \[Pi]^2 (Y^\[Prime]\[Prime])[y] + 
      b^3 N0 y \[Alpha] (Y^\[Prime]\[Prime])[y] + 
      b^4 D[Y[y], {y, 4}] == 0 /. {b -> 1., N0 -> 1, \[Alpha] -> 1},
   Y[0] == 0, Y[1] == 0, Y''[0] == 0, Y''[1] == 0}, Y[y], {y, 0, 1}, 
  AccuracyGoal -> 100]

That gives a flatline zero.

Cheers,

Marco

POSTED BY: Marco Thiel
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