2nd order PDE using integration by parts

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SUMMARY

The discussion focuses on solving the second-order partial differential equation (PDE) given by (\zeta - \eta)^2 \frac{\partial^2 u(\zeta,\eta)}{\partial\zeta \, \partial\eta}=0. The user attempts to find the general solution by setting X = \partial u/\partial\eta and applying integration by parts. The resulting equation, (\zeta - \eta)^2X - 2\int \zeta X \, d\zeta + 2\eta \int X\, d\zeta = f(\eta), indicates a need for further clarification on the method and next steps in the solution process.

PREREQUISITES
  • Understanding of second-order partial differential equations
  • Familiarity with integration by parts technique
  • Knowledge of independent variables in PDEs
  • Basic concepts of function notation and derivatives
NEXT STEPS
  • Study the method of characteristics for solving PDEs
  • Learn advanced integration techniques applicable to PDEs
  • Explore the implications of boundary conditions on PDE solutions
  • Investigate the role of independent variables in multi-variable calculus
USEFUL FOR

Students and researchers in mathematics, particularly those focusing on differential equations, as well as educators seeking to enhance their understanding of integration techniques in the context of PDEs.

perishingtardi
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Homework Statement


Find the general solution of the equation
(\zeta - \eta)^2 \frac{\partial^2 u(\zeta,\eta)}{\partial\zeta \, \partial\eta}=0,
where ##\zeta## and ##\eta## are independent variables.

Homework Equations


The Attempt at a Solution


I set ##X = \partial u/\partial\eta## so that (\zeta - \eta)^2 \frac{\partial X}{\partial\zeta}=0. Then \int (\zeta - \eta)^2 \frac{\partial X}{\partial\zeta} \, d\zeta=f(\eta). I used integration by parts to obtain
(\zeta - \eta)^2X - 2\int \zeta X \, d\zeta + 2\eta \int X\, d\zeta = f(\eta), but I'm not sure if this is the correct method, or how to proceed.
 
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Hint: what is
\frac{ \partial \zeta} { \partial\eta}
 
dirk_mec1 said:
Hint: what is
\frac{ \partial \zeta} { \partial\eta}

its zero?? how does that help though?
 

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