How to write equation as shown in picture using Latex?

Question

how we write this equation in latex in given form

Answer 1

With a few shortcut macros it's much easier:

plain TeX version:

\let~\catcode~`86~`j0~`X13~`C1~`D2~`M3jdefX81C~`8113jdefDXZZ81C~
`8113DXYY81CZ81jletDYLLjletYFFjfiYNNjdefYPPjpartialZIZAZBZj{Zj}Y
EEjexpandafterYOOjelseYKKjifxNI818283CL}81NAC82DNBC83Djfuturelet
jTjHDNjHCKjT}L{AOL{BF{DXUUCI)jpCjp)CDDDY VVjoverNjp)818283CKX81X
CP82VP83DOCP^C81D82VjbC81DC83DDFDYRRjrelaxNjb8182Cjd8182RDNjd81%
8283CK83RP82^C81DOP82P83FDY!!uY@@vY##wY$$xYj%%yY&&zXvv81C81^2DMM
!U!$+@U!%+#U!&+jlambda_1jleft(v!U)2! $+v@U)2!%+v#U)2!&jatop+2!@U
)2!C$%D+2@#U)2!C%&D+2!#U)2!C$&Djright)=jnuU)2!&jeqno(1.25)MMjbye

LaTeX version:

\documentclass{article}\begin{document}
\let~\catcode~`86~`j0~`X13~`C1~`D2~`M3jdefX81C~`8113jdefDXZZ81C~
`8113DXYY81CZ81jletDYLLjletYFFjfiYNNjdefYPPjpartialZIZAZBZj{Zj}Y
EEjexpandafterYOOjelseYKKjifxNI818283CL}81NAC82DNBC83Djfuturelet
jTjHDNjHCKjT}L{AOL{BF{DXUUCI)jpCjp)CDDDY VVjoverNjp)818283CKX81X
CP82VP83DOCP^C81D82VjbC81DC83DDFDYRRjrelaxNjb8182Cjd8182RDNjd81%
8283CK83RP82^C81DOP82P83FDY!!uY@@vY##wY$$xYj%%yY&&zXvv81C81^2DMM
!U!$+@U!%+#U!&+jlambda_1jleft(v!U)2!$+v@U)2!%+v#U)2!&jatop+2!@U)
2!C$%D+2@#U)2!C%&D+2!#U)2!C$&Djright)=jnuU)2!&jeqno(1.25)MMjstop

output:

---

Inspired by David Carlisle's xii :)

Answer 2

A variation on the theme, but with a greater emphasis on simplifying the input:

\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}

\newcommand{\pdiff}{\mathop{}\!\partial}

\ExplSyntaxOn
\NewDocumentCommand{\pder}{omm}
 {
  \frac{\pdiff\IfValueT{#1}{^{#1}}#2}{\faisal_pder_vars:n { #3 }}
 }
\cs_new_protected:Nn \faisal_pder_vars:n
 {
  \clist_map_inline:nn { #1 } { \pdiff##1 }
 }
\ExplSyntaxOff


\begin{document}

\begin{equation}
u\pder{u}{x}+v\pder{u}{y}+w\pder{u}{z}+
  \lambda_1
  \left(\begin{gathered}
    u^2\pder[2]{u}{x^2}+v\pder[2]{u}{y^2}+w^2\pder[2]{u}{z^2}\\
    \mspace{-\medmuskip}{}
    +2uv\pder[2]{u}{x,y}+2vw\pder[2]{u}{y,z}+2uw\pder[2]{u}{x,z}
    \end{gathered}\right)
= \nu\pder[2]{u}{z^2},
\end{equation}

\begin{equation}
\begin{split}
u\pder{u}{x}&+v\pder{u}{y}+w\pder{u}{z}
+\lambda_1\biggl(
    u^2\pder[2]{u}{x^2}+v\pder[2]{u}{y^2}+w^2\pder[2]{u}{z^2}
\\
&+2uv\pder[2]{u}{x,y}+2vw\pder[2]{u}{y,z}+2uw\pder[2]{u}{x,z}\biggr)
= \nu\pder[2]{u}{z^2},
\end{split}
\end{equation}

\end{document}

The syntax for \pder is

1. optional argument for the order of the derivative (if greater than 1)
2. function to differentiate
3. list of the variables the derivative is taken with respect to, comma separated

In the first case I used \mspace{-\medmuskip}{}+ in order to have the plus sign correctly spaced. I used gathered instead of pmatrix because it is semantically sounder.

Answer 3

Here you go:

\documentclass[10pt,a4paper]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\begin{document}
    $u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}+\lambda_1\begin{pmatrix}
    u^2\frac{\partial^2 u}{\partial x^2}+v\frac{\partial^2 u}{\partial y^2}+w^2\frac{\partial^2 u}{\partial z^2}\\+2uv\frac{\partial^2 u}{\partial x\partial y}+2vw\frac{\partial^2 u}{\partial y\partial z}+2uw\frac{\partial^2 u}{\partial x\partial z}
    \end{pmatrix} = v\frac{\partial^2 u}{\partial z^2},$
\end{document}

this will give you:

PS: Welcome to TeX.se, from next time, please provide a MWE.

Answer 4

Here is a start:

\documentclass[]{article}
\usepackage{amsmath}

\begin{document}
\begin{equation}
    u\frac{\partial u}{\partial x}+...
    +\lambda_1 
    \begin{pmatrix}
    u^2 \frac{\partial^2 u}{\partial x^2}+...\\
    + 2 uv \frac{\partial^2 u}{\partial x \partial y}+...
    \end{pmatrix} = v \frac{\partial^2 u}{\partial z^2}
\end{equation}
\end{document}

Answer 5

Borrow a little from all answer and make a little change (such as change v to \nu in the right-hand side)

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
    u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}+\lambda_1\begin{pmatrix}
    u^2\frac{\partial^2 u}{\partial x^2}+v\frac{\partial^2 u}{\partial y^2}+w^2\frac{\partial^2 u}{\partial z^2}\\+2uv\frac{\partial^2 u}{\partial x\partial y}+2vw\frac{\partial^2 u}{\partial y\partial z}+2uw\frac{\partial^2 u}{\partial x\partial z}
    \end{pmatrix} = \nu\frac{\partial^2 u}{\partial z^2},
    \tag{1.25}
\end{equation}
\end{document}

Answer 6

 \documentclass{article}

\begin{document}
 \[ u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}+ w \frac{\partial u}{\partial z} + \lambda_1 
  \left( \begin{tabular}{c}
 $\:\:\:\:\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} +  \frac{\partial^2 u}{\partial z^2}$\\ 
 $+ 2 u v \frac{\partial^2 u}{\partial x \partial y}+ 2vw  \frac{\partial^2 u}{\partial y\partial z}+2wu  \frac{\partial^2 u}{\partial z \partial u}$\\\end{tabular} \right) = \nu \frac{\partial^2 u}{\partial z^2}\] 
\end{document}

This gives desired output.

Answer 7

$$u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}+w \frac{\partial u}{\partial z}+\lambda_{1}\left(\begin{array}{c}
u^{2} \frac{\partial^{2} u}{\partial x^{2}}+v^{2} \frac{\partial^{2} u}{\partial y^{2}}+w^{2} \frac{\partial^{2} u}{\partial z^{2}} \\
+2 u v \frac{\partial^{2} u}{\partial x \partial y}+2 v w \frac{\partial^{2} u}{\partial \partial \partial z}+2 u w \frac{\partial^{2} u}{\partial x \partial z}
\end{array}\right)=\nu \frac{\partial^{2} u}{\partial z^{2}}$$