Discussion Forum

Hi

try to make an intermediate new variable equal to the second order derivative and solve for all then you can get to the 4th order

Good luck
Ivar

Hi Mr Kjelberg,

thank you for your response, but I have another problem:
the PDE is

B * d'''' y / d x'''' + dy / dt = 0

and the boundary conditions are

y ( - infinite , t ) = y ( +infinite , t ) = 0

d y(0,t) / dx = m

d''' y / dx''' = M

where B,m,M are constants and the variable y (heigth) is a function of x (lenght) and t (time).
If I make an intermediate new variable equal to the second order derivative, like d'''y/dx''''=d"u/du", how
I can treat the last boundary condition (d''' y / dx''' = M), which is a third order derivative ?

Thanks and regards,

Raffaele

Hi

I have a slight problem to follow your notation:

you have coordiante x, and time t and dependent variable y=f(x,t) no ?

so what is wrong by saying you have z=g(x,t) such that z=d^2y/dx^2, then your BC remain and you have dz/dx=M.
and you have two dependent variables y,z.

Now I would change the names of the dependent vaiables to v and w to avoid any conflicts with Comsols internal names of the geometrical variables x,y,z ...

But apart from that, why cannot you just continue like that ?

Good luc
Ivar