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I am not been able to wrire a suitable code to solve the equation for pde

1 vue (au cours des 30 derniers jours)
Tapobrata
Tapobrata le 17 Jan 2024
L (∂^4 u)/(∂z^4 )+M (∂^2 u)/(∂z^2 )-S (∂^4 u)/(∂z^2 ∂t^2 )-X (∂^3 u)/(∂z^2t)+c ∂u/∂t+m (∂^2 u)/(∂t^2 )+P (∂^3 u)/(∂t^3 )+Q (∂^4 u)/(∂t^4 )+k_f u =f(z,t)

Réponses (1)

Aman
Aman le 6 Fév 2024
Hi Tapobrata,
I understand that you want to solve the mentioned PDE using MATLAB and want to know how to do it.
The PDE mentioned can be represented in MATLAB using the below code.
syms u(z,t) L M S X c m P Q k_f f(z, t);
pde = L*diff(u, z, 4) + M*diff(u, z, 2) - S*diff(diff(u,t,2),z,2) - X*diff(diff(u,t,1),z,3) + c*diff(u,t,1) + m*diff(u,t,2) + P*diff(u,t,3) + Q*diff(u,t,4) + k_f*u - f(z,t);
pretty(pde);
4 3 2 3 4 2 d d d d d d Q --- u(z, t) + P --- u(z, t) - S --- #1 + m #1 - X --- #2 + c #2 + L --- u(z, t) + M --- u(z, t) + k_f u(z, t) - f(z, t) 4 3 2 3 4 2 dt dt dz dz dz dz where 2 d #1 == --- u(z, t) 2 dt d #2 == -- u(z, t) dt
As the PDE contains high-order derivatives and mixed spatial-temporal terms, it would be difficult to find a solution for it, as in order to solve it, we need to have information on the initial condition, boundry condition, coefficients, and the function "f(z,t)". Once we have all this information, we can try to use the "pdepe" function to find the solution, but note that the "pdepe" function is limited to parabolic and elliptic PDEs in one spatial dimension and time. Refer to the following link to learn more about the "pdepe" function.
I hope it helps!
  2 commentaires
Tapobrata
Tapobrata le 9 Fév 2024
Déplacé(e) : John D'Errico le 9 Fév 2024
I have initial/boundary conditions as space and time derivatives all to be zero ,yet pdepe is not giving any answers.
Aman
Aman le 9 Fév 2024
Déplacé(e) : John D'Errico le 9 Fév 2024
Could you please share what you have tried? Additionally, let me know what the expected output should be in that case.

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