Trouble solving a differential equation

[rmas]

Hi,

How can I solve this equation (finding f(x,y)) ?

$\left (\frac{\partial f}{\partial x} \right )^2+ \left( \frac{\partial f}{\partial y} \right )^2=0$

Thanks a lot.

 

[jackmell]

Can't you factor that with some i's? You know, like if I had:

$$(x^2+y^2)=0$$ and I wanted to factor it, then I'd write:

$$(x+iy)(x-iy)=0$$

Then that means:

$$(x+iy)=0$$

or:

$$x-iy=0$$

Ok, same dif then. Now do yours.

Nothing wrong with complex solutions. Try not to be intimidated by them.

 

[hunt_mat]

Both term in the equation are greater than or equal to zero, so the equation says that they must both be zero and hence:
$$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$$
So what does this say about f?

 

[rmas]

Thank you very much jackmell, that is very helpful.

Both term in the equation are greater than or equal to zero, so the equation says that they must both be zero and hence:
$$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$$
So what does this say about f?

The partial derivatives vanish, I suppose it means that f is constant ?

 

[hunt_mat]

Yep.

 

[opsb]

What about f(x,y) = exp(x)exp(i*y)

Then first derivative w.r.t. x is just f(x,y)

First derivative w.r.t. y is i*f(x,y)

Sum of their squares is zero, yet they are not constant.

f is necessarily constant under the constraint that f has first partial derivatives which are functions mapping to the real numbers. Then the two squared terms are necessarily zero or greater, so must be zero. Then f would have to be constant.