solve pde with neumann boundary conditions

Good mornirg.

I have to solve a PDE: dydt=-vdydx+Dd2ydx2+Ay. The boundary condition are y=cost whent x=0 and dy/dt=0 when x=L; whereas the initial condition y=0 when t=0.

I use the method of line to solve the problem. So I rewrite the equation as dydt=-v/h*(y(i)-y(i-1))+D/h^2(2y(i-1)-2y(i)+A(i).

How can I write the boundary condition at x=L dydx=0?

thank you

[Evelyn]

To incorporate the Neumann boundary condition of dy/dx = 0 at x=L, we can use a backward difference approximation. Specifically, we can approximate the derivative at the last grid point using the following formula:

dy/dx ≈ (y(i) - y(i-1))/h

At x=L, this approximation gives:

dy/dx ≈ (y(n) - y(n-1))/h = 0

where n is the total number of grid points. Solving for y(n-1), we get:

y(n-1) = y(n)

Therefore, we can incorporate the Neumann boundary condition into our finite difference equation by replacing the expression for dy/dx at x=L with:

(y(n) - y(n-1))/h = 0

which simplifies to:

y(n) = y(n-1)

Substituting this expression into our finite difference equation for the second derivative term at x=L, we get:

D/h^2 (2y(n-1) - 2y(n) + A(n)) = 0

Solving for y(n-1), we obtain:

y(n-1) = y(n) - A(n)h^2/2D

Therefore, we can use this expression for y(n-1) in our finite difference equation to incorporate the Neumann boundary condition into the numerical solution of the PDE.

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vampire survivors

Answer 1

[Torsten]

At the right boundary (x=x_n)

D^2y/dx^2 ~

(dy_n/dx - dy_(n-1/2)/dx) /(x_n-x_(n-1/2)) ~

(dy_n/dx - (y_n-y_(n-1))/(x_n-x_(n-1)))/(x_n-x_(n-1/2))

Thus with your boundary condition dy_n/dx = 0 you get the ODE

dy_n/dt = D*(- (y_n-y_(n-1))/(x_n-x_(n-1)))/(x_n-x_(n-1/2)) + A*y_n

with x_(n-1/2) = (x_n+x_(n-1))/2