anonymous
  • anonymous
Problem 5: Let B = (B1(x, y), B2(x, y), 0) be a divergence-free vector, ∇ · B = 0. Find a vector function A~ = (0, 0, A(x, y)), such that ∇ × A~= B, and ∇ · A = 0.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
this is a special case of Helmholtz decomposition. we have that $$\vec B=(B_1(x,y),B_2(x,y),0)$$ and we see its divergence is: $$\frac{\partial B_1}{\partial x}+\frac{\partial B_2}{\partial y}=0\\\frac{\partial B_1}{\partial x}=-\frac{\partial B_2}{\partial y}$$ okay, now consider that we want a vector field \(\vec A=(0,0,A(x,y))\) such that $$\nabla \times \vec A=\left(-\frac{\partial A}{\partial y},\frac{\partial A}{\partial x},0\right)=B$$ so it follows we have $$B_1=-\frac{\partial A}{\partial y}\\B_2=\frac{\partial A}{\partial x}$$ and the earlier equation tells us $$-\frac{\partial^2 A}{\partial x\,\partial y}=-\frac{\partial^2 A}{\partial y\,\partial x}$$
anonymous
  • anonymous
so we can just take: $$A(x,y)=-\int B_1\, dy+f(x)\\A(x,y)=\int B_2\, dx+g(y)$$ and combine these into a single solution of the form $$A(x,y)=-\int B_1 dy+\int B_2\, dx+C$$
anonymous
  • anonymous
we call \(A\) a vector potential of \(B\): https://en.wikipedia.org/wiki/Vector_potential

Looking for something else?

Not the answer you are looking for? Search for more explanations.