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Higher derivatives in "PDEPE"

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Jaesung Lee
Jaesung Lee el 24 de Abr. de 2017
Comentada: Bill Greene el 30 de Abr. de 2017
When I want to consider "PDEPE" command for 1D PDE like Du/Dt = \nabla^4 u = \nabla \cdot \nabla^3 u, how can express the higher derivatives in f?
That is, f = \nabla^3 u. How can I modify this term with DuDx?
Thanks in advance.

Respuestas (2)

David Ding
David Ding el 27 de Abr. de 2017
HI Jaesung,
The "pdepe" function is suited to solve the PDE of the form shown in equation (1-3) of:
In your case, "f = \nabla^3 u" is unfortunately not in this form. MATLAB does offer ways to solve these kinds of PDEs numerically, for example, via the "pdeval" function.

Jaesung Lee
Jaesung Lee el 30 de Abr. de 2017
Dear Ding,
Thanks for your answer. I have attached the simplified version on my equation. The first equation is parabolic and the second qualifies as elliptic. If you don't mind, could you explain why our equations do not fit into the required pdepe format? In the original version, u(1) is the first order derivative and u(2) is the thrid order derivative. I reshaped the original ODE.
I appreciate your comments in advance.
  1 comentario
Bill Greene
Bill Greene el 30 de Abr. de 2017
Your equation for f_1 shows a derivative wrt y for u_2. I assume that is simply a typographical error? If that is the case, these equations look to me to be in a form acceptable to pdepe. What problem did you run into when you tried to solve them?

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