Question

someone knows how to solve this issue?

Answer 1

What issue? I don't see anything.

Answer 2

r

Answer 3

woah...english

Answer 4

English would be nice.

Answer 5

I'll translate just a minute

Answer 6

Essentially, "show that the function satisfies the Laplace equation..."

Answer 7

Take the appropriate partial derivatives and plug them into the differential equation.

Answer 8

1) Show that the function F (X, Y) = ln (sqrt (x ^ 2 + y ^ 2)) satisfies the Laplace equation in two-dimensional

((& PartialD,) ^ (2) F) / (& PartialD, x ^ (2)) + ((& PartialD,) ^ (2) F) / (& PartialD; ^ y (2)) = 0

Answer 9

@fabiomartins do you know how to take partial derivatives?

Answer 10

Already tried all ways, but I can not, I wish someone would make it complete and then explain to me if possible ...

Answer 11

\[f(x,y)=\ln\sqrt{x^2+y^2}\]
\[\begin{align*}\frac{\partial f}{\partial x}&=\frac{\dfrac{\partial}{\partial x}\left[(x^2+y^2)^{1/2}\right]}{\sqrt{x^2+y^2}}\\\\
&=\frac{\dfrac{1}{2}(x^2+y^2)^{-1/2}(2x)}{\sqrt{x^2+y^2}}\\\\
&=\frac{2x}{2\sqrt{x^2+y^2}\sqrt{x^2+y^2}}\\\\
&=\frac{x}{x^2+y^2}\end{align*}\]
\[\begin{align*}\frac{\partial f}{\partial y}&=\frac{\dfrac{\partial}{\partial y}\left[(x^2+y^2)^{1/2}\right]}{\sqrt{x^2+y^2}}\\\\
&=\frac{\dfrac{1}{2}(x^2+y^2)^{-1/2}(2y)}{\sqrt{x^2+y^2}}\\\\
&=\frac{2y}{2\sqrt{x^2+y^2}\sqrt{x^2+y^2}}\\\\
&=\frac{y}{x^2+y^2}\end{align*}\]
These are your first order partial derivatives. Agreed?

Answer 12

And these are your second-order partial derivatives:
\[\begin{align*}\frac{\partial^2 f}{\partial x^2}&=\frac{\partial}{\partial x}\left[\frac{\partial f}{\partial x}\right]\\\\
&=\frac{\partial }{\partial x}\left[\frac{x}{x^2+y^2}\right]\\\\
&=\frac{(x^2+y^2)\dfrac{\partial}{\partial x}[x]-x\dfrac{\partial }{\partial x}[x^2+y^2]}{(x^2+y^2)^2}\\\\
&=\frac{(x^2+y^2)(1)-x(2x)}{(x^2+y^2)^2}\\\\
&=\frac{x^2+y^2-2x^2}{(x^2+y^2)^2}\\\\
&=\frac{y^2-x^2}{(x^2+y^2)^2}\end{align*}\]
\[\begin{align*}\frac{\partial^2 f}{\partial y^2}&=\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial y}\right]\\\\
&=\frac{\partial }{\partial y}\left[\frac{y}{x^2+y^2}\right]\\\\
&=\frac{(x^2+y^2)\dfrac{\partial}{\partial y}[y]-y\dfrac{\partial }{\partial y}[x^2+y^2]}{(x^2+y^2)^2}\\\\
&=\frac{(x^2+y^2)(1)-y(2y)}{(x^2+y^2)^2}\\\\
&=\frac{x^2+y^2-2y^2}{(x^2+y^2)^2}\\\\
&=\frac{x^2-y^2}{(x^2+y^2)^2}\end{align*}\]

Answer 13

So now you plug these back into the Laplace equation to see if it is true:
\[\begin{align*}
\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}&=\frac{y^2-x^2}{(x^2+y^2)^2}+\frac{x^2-y^2}{(x^2+y^2)^2}\\\\
&=\frac{(y^2-y^2)+(x^2-x^2)}{(x^2+y^2)^2}\\\\
&=0
\end{align*}\]
which means the equation is indeed satisfied by \(f(x,y)=\ln\sqrt{x^2+y^2}\)

Answer 14

It's over?