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# Numerical solution of fourth order differential equation

Posted 9 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; Y == 0; Y'' == 0; Y'' == 0;  Attachments:
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Posted 9 years ago
 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.
Posted 9 years ago
 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. Posted 9 years ago
 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, Y == 0, Y'' == 0, Y'' == 0}, Y, {y, 0, 1}, {N0, b, \[Alpha]}, AccuracyGoal -> 100] gives 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==0.
Posted 9 years ago
 You are right. These deviations are just either numerical error or internal values.
Posted 9 years ago
 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:  Attachments:
Posted 9 years ago
 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, Y == 0, Y'' == 0, Y'' == 0}, Y[y], {y, 0, 1}] There might be some numerical issues though because Plot[Y[y] /. sols, {y, 0, 1}] gives 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, Y == 0, Y'' == 0, Y'' == 0}, Y[y], {y, 0, 1}, AccuracyGoal -> 100] That gives a flatline zero.Cheers,Marco