Simpliying this partial differential equation

Does this equation mean that:

## x+y =z ##, and
## x = y##?

I mean ## \delta_{ij} ## terms in the LHS of the eqaution equal those at the RHS ?

with knowing that ## \partial_i \partial_j x ## term dose not vanish for ## \delta_{ij}=1 ##


Any help is appreciated!

Safinaz said:

Any help is appreciated!

I think your notation is confusing. Can you clarify:

• What range does the index ##i## run over: ##\left( 1,2,3\right)## or ##\left( 0,1,2,3\right)## or something else?
• What explicitly are the variables ##w^i## that you are differentiating with respect to, i.e., ##\partial_{i}=\partial/\partial w^{i}##?
• How are these variables related to ##x,y,z##?

renormalize said:

I think your notation is confusing. Can you clarify:

• What range does the index ##i## run over: ##\left( 1,2,3\right)## or ##\left( 0,1,2,3\right)## or something else?
• What explicitly are the variables ##w^i## that you are differentiating with respect to, i.e., ##\partial_{i}=\partial/\partial w^{i}##?

renormalize said:

How are these variables related to ##x,y,z##?

Hi, thanks for reply.

• What range does the index ##i## run over: ##\left( 1,2,3\right)## or ##\left( 0,1,2,3\right)## or something else? Ans.: ##\left( 0,1,2,3\right)##
• What explicitly are the variables ##w^i## that you are differentiating with respect to, i.e., ##\partial_{i}=\partial/\partial w^{i}##? Ans.: It is ##\partial_{i}=\partial/\partial w^{i}##?
• How are these variables related to ##x,y,z##? Ans.: consider for instance ##\partial_{i} x =\partial x /\partial w^{i} ##. So that x, y and z can be written explicitly as x(w), y(w), and z(w).

OK, thanks for clearing things up!
So we have:$$\left(x+y\right)\delta_{ij}+\frac{\partial^{2}x}{\partial w^{i}\partial w^{j}}=\frac{\partial^{2}y}{\partial w^{i}\partial w^{j}}+z\delta_{ij}\;\Rightarrow\;\left(x+y-z\right)\delta_{ij}=\frac{\partial^{2}\left(y-x\right)}{\partial w^{i}\partial w^{j}}$$or:$$U\delta_{ij}=\frac{\partial^{2}V}{\partial w^{i}\partial w^{j}}\quad\text{where}\quad U\equiv x+y-z,\;V\equiv y-x\tag{1a,b,c}$$Contracting (1a) with ##\delta^{ij}## gives ##4U=\partial^{2}V/\partial w^{i}\partial w_{i}\equiv\square V##, leading to:$$\frac{1}{4}\square V\delta_{ij}=\frac{\partial^{2}V}{\partial w^{i}\partial w^{j}},\quad U=\frac{1}{4}\square V\tag{2a,b}$$The off-diagonal (##i\neq j##) terms of (2a) are: $$0=\partial_{0}\partial_{1}V=\partial_{0}\partial_{2}V=\partial_{0}\partial_{3}V=\partial_{1}\partial_{2}V=\partial_{1}\partial_{3}V=\partial_{2}\partial_{3}V$$with the unique solution ##V=k##, where ##k## is an arbitrary constant. Thus, ##U=\frac{1}{4}\square V=0##, and from (1b,c) we get finally:$$y=x+k,\;z=2x+k\;\left( x\text{ an arbitrary function},\;k\text{ an arbitrary constant}\right)$$

Thanks so much for the answer : )