I have a vector equation:
\vec{A} \times \vec{B} = \vec{C} . \vec{A} and \vec{C} are known, and \vec{B} must be determined. However, upon trying to use Cramer's rule to solve the system of three equations, I find that the determinant we need is zero. I know now that I need to choose a...
If
\int_{-\infty}^{\infty} f(k) e^{i\mathbf{k}.\mathbf{r}} d\mathbf{k} =0
then is
\ f(k)=0 ?
Is it correct to say that this is an expansion in an orthonormal basis, \ e^{ik.r} , and so linear independence demands that f(k) be zero for all k?
I have the equations
\frac{l}{u^{2}} \frac{du}{dx}=constant
and
\frac{1}{u} \frac{dl}{dx}=constant.
By "eyeball", I can say the solution is
l \propto x^{n} \ and \ u \propto x^{n-1}.
I can't see how I could arrive at these solutions 'properly', if you know what I mean
In the x-y plane, we have the equation
\nabla^{2} \Psi = - 4\pi \delta(x- x_{0}) \delta (y- y_{0})
with \Psi = 0 at the rectangular boundaries, of size L.
A paper I'm looking at says that for
R^{2} = (x-x_{0})^{2} + (y-y_{0})^{2} << L^{2} ,
that is, for points...
I've been studying Turbulence, and there's a lot of averaging of differential equations involved. The books I've seen remark offhandedly that differentiation and averaging commute
for eg. < \frac{df}{dt} > = \frac{d<f>}{dt}
Here < > is temporal averaging. If...