distributional derivative of two variable function Can any one help me to show that $ u(x,y) = \log |(x+y)/(x-y)| $ is locally integrable on $R^{2}$ ? I guess yes because it only can problem near $y= x$. Further how to find its distributional derivative $u_{xy}$.
 A: If we make a linear change of variable $z=x+y$, $x-y$, then our function $u$ becomes $\log|z|-\log|w|$. Note that $\log|z|$ is locally integrable on $\mathbb{R}$, thus it's also locally integrable on $\mathbb{R}^2$. Similarly with $\log|w|$. Since a linear change of variable does not change local integrability, $u$ is locally integrable in the original variables $x,y$. 
Now we begin to compute $u_{xy}$. Note that for $\varphi\in C_c^\infty(\mathbb{R}^2)$, $\varphi=\varphi(z,w)$, we have the following identity:
$$\frac{\partial^2}{\partial x\partial y}\varphi(x+y,x-y)=
\varphi_{zz}(x+y,x-y)-\varphi_{ww}(x+y,x-y).$$
This identity extends to distributions by duality. Now write
$$u(x,y)=f(x+y,x-y)-g(x+y,x-y)=\log|x+y|-\log|x-y|$$ for 
$$f(z,w)=\log|z|$$ and $$g(z,w)=\log|w|,$$ then we have 
$$u_{xy}(x+y,x-y)=f_{zz}(x+y,x-y)+g_{ww}(x+y,x-y).$$
Let's now compute $f_{zz}$. Note that $\log|z|$ defines a distribution on $\mathbb{R}$, and let's denote this distribution by $a$. Then $f=a\otimes \text{Id}$ where $\text{Id}$ is the identity distribution on the $w$ variable. Thus $f_{zz}=a''\otimes\text{Id}$. Similarly, $g_{ww}=\text{Id}\otimes a''$, and thus $u_{xy}=a''\otimes\text{Id}+\text{Id}\otimes a''$.
Now for $a(z)=\log|z|$, $\varphi\in C_c^\infty(\mathbb{R})$, write 
$$(a,\varphi')=\int \log|z| \varphi(z)\ dz
=\lim_{\varepsilon\to 0+}\int_{|z|>\varepsilon}\log|z|\varphi'(z)\ dz$$
and perform integration by part to get 
$$(a',\varphi)=\lim_{\epsilon\to 0+}\int_{|z|>\epsilon}\frac{\varphi(z)}{z}\ dz=\int_{|z|>1}\frac{\varphi(z)}{z}\ dz+\int_{|z|\leq 1}\frac{\varphi(z)-\varphi(0)}{z}\ dz.$$
Then 
$$(a'',\varphi)
=-(a',\varphi')=-\int_{|z|>1}\frac{\varphi'(z)}{z}\ dz-\lim_{\varepsilon\to 0+}\int_{\varepsilon<|z|\leq 1}\frac{\varphi'(z)-\varphi'(0)}{z}\ dz.$$
By integration by parts again, we get 
$$(a'',\varphi)=-\int_{|z|>1}\frac{\varphi(z)}{z^2}\ dz-
\lim_{\varepsilon\to 0+}\int_{\varepsilon<|z|\leq 1}
\frac{\varphi(z)-\varphi'(0)z-\varphi(0)}{z^2}\ dz+2\varphi'(0).$$
So in short, $u_{xy}=a''\otimes\text{Id}+\text{Id}\otimes a''$. Explicitly, 
for $\varphi\in C_c^\infty(\mathbb{R}^2)$, 
$$(u_{xy},\varphi)=(a''\otimes\text{Id},\varphi)+(\text{Id}\otimes a'',\varphi)$$
where 
$$(a''\otimes\text{Id},\varphi)
=-\int_{|x+y|>1}\frac{\varphi(x,y)}{(x+y)^2}\ dxdy-
\lim_{\varepsilon\to 0+}\int_{\varepsilon<|x+y|\leq 1}
\frac{\varphi(x,y)-\frac{1}{2}(\varphi_x(\frac{x-y}{2},\frac{y-x}{2})+\varphi_y(\frac{x-y}{2},\frac{y-x}{2}))(x+y)-\varphi(\frac{x-y}{2},\frac{y-x}{2})}{(x+y)^2}\ dxdy+\varphi_x(\frac{x-y}{2},\frac{y-x}{2})+\varphi_y(\frac{x-y}{2},\frac{y-x}{2})$$
and 
$$(\text{Id}\otimes a'',\varphi)
=-\int_{|x-y|>1}\frac{\varphi(x,y)}{(x-y)^2}\ dxdy-
\lim_{\varepsilon\to 0+}\int_{\varepsilon<|x-y|\leq 1}
\frac{\varphi(x,y)-\frac{1}{2}(\varphi_x(\frac{x+y}{2},\frac{x+y}{2})-\varphi_y(\frac{x+y}{2},\frac{x+y}{2}))(x-y)-\varphi(\frac{x+y}{2},\frac{x+y}{2})}{(x-y)^2}\ dxdy+\varphi_x(\frac{x+y}{2},\frac{x+y}{2})-\varphi_y(\frac{x+y}{2},\frac{x+y}{2}).$$
