Hi guys! I need your help! How can i compute the div(grad(cos(wt)))??

2. ## Re: Computation of div(grad(f)))

I'm not sure if the problem makes any sense, how would you compute the gradient of cos(wt) ? What are the variables?

3. ## Re: Computation of div(grad(f)))

Normally "grad" is applied to a function of 2 or 3 variables: if f is a function of x, y, and z, f(x,y,z) then $grad f= \frac{\partial f}{\partial x}i+ \frac{\partial f}{\partial y}j+ \frac{\partial f}\partial z}k$. The div of a vector function $f_xi+ f_yj+ f_zk$ is the scalar function $\frac{\partial f_x}{\partial x}+ \frac{\partial f_y}{\partial y}+ \frac{\partial f_z}{\partial z}$. So div(grad(f)) would be $\frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2}+ \frac{\partial^2 f}{\partial z^2}$.

With a single variable like t, one way to interpret this is with that single variable replacing one of the x, y, z variables, the others 0. In that case, we would have "div(grad(f))" equal to $\frac{d^2f}{dt^2}$, just the second derivative. Another way to interpret this is as a one component of a two or three component vector. But then I don't know how you would apply "grad" to a vector valued function. Perhaps it would help if you posted the entire problem.