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frank0520
 one year ago
Suppose f(x,y) is harmonic (f has continuous second partials and fxx +fyy =0) and define the vector field
Perp(f)=<fy,fx>
a) Show that Perp(f) is a conservative vector field that is orthogonal to Grad(f).
Hint: To show that Perp(f) is conservative use one of the equivalent properties.
b) Since Perp(f) is conservative it has a potential function g(x,y) called the harmonic conjugate of f(x,y). Show that g(x,y) is in fact harmonic.
frank0520
 one year ago
Suppose f(x,y) is harmonic (f has continuous second partials and fxx +fyy =0) and define the vector field Perp(f)=<fy,fx> a) Show that Perp(f) is a conservative vector field that is orthogonal to Grad(f). Hint: To show that Perp(f) is conservative use one of the equivalent properties. b) Since Perp(f) is conservative it has a potential function g(x,y) called the harmonic conjugate of f(x,y). Show that g(x,y) is in fact harmonic.

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frank0520
 one year ago
Best ResponseYou've already chosen the best response.0Here is a picture of the question.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1For part a, use the given hint : If the vector field is conservative, then the curl is 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\text{Perp(f)} = \langle f_y,~f_x\rangle\] find the curl

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if \(\nabla g=\langle f_y,f_x\rangle\) then it follows \(g_x=f_y,g_y=f_x\) so it follows $$g_{xx}=\frac{\partial}{\partial x}(f_y)=f_{xy}\\g_{yy}=\frac{\partial}{\partial y}(f_x)=f_{yx}$$so it follows by symmetry of mixed partials \(f_{xy}=f_{yx}\) that \(g_{xx}+g_{yy}=f_{xy}+f_{yx}=0\), i.e. \(g\) is indeed harmonic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to prove \(\langle f_y,f_x\rangle=\langle f_y,f_x,0\rangle\) as a vector field on \(\mathbb{R}^3\) is conservative, we can show \(\nabla\times f=\langle0,0,\frac{\partial}{\partial x}(f_x)\frac{\partial}{\partial y}(f_y)\rangle=\langle0,0,f_{xx}+f_{yy}\rangle=0\) since we're told \(f\) is harmonic (i.e. \(f_{xx}+f_{yy}=0\))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to show it's orthogonal to \(\nabla f=\langle f_x,f_y\rangle\) you can compute their dot product: $$\langle f_y,f_x\rangle\cdot\langle f_x,f_y\rangle=f_yf_x+f_xf_y=0$$
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