Undergrad Quasilinear Equation but with non-zero initial condition?

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SUMMARY

This discussion focuses on solving quasi-linear partial differential equations (PDEs) with non-zero initial conditions, specifically using the characteristic method. The participants explore the feasibility of applying alternative initial conditions, such as ##u(x,y=c) = f(x)##, instead of the traditional ##u(x,0) = f(x)##. It is established that one can redefine the independent variables to facilitate the solution, using transformations like (x, v = y - c) or (x, c - y) to adapt to different initial conditions.

PREREQUISITES
  • Understanding of quasi-linear partial differential equations (PDEs)
  • Familiarity with the characteristic method for solving PDEs
  • Knowledge of variable transformations in mathematical analysis
  • Basic concepts of initial value problems in differential equations
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  • Research the application of the characteristic method in quasi-linear PDEs
  • Explore variable transformations and their impact on PDE solutions
  • Study alternative initial conditions in initial value problems
  • Learn about backward equations in the context of PDEs
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Mathematicians, physicists, and engineers working with partial differential equations, particularly those interested in advanced methods for solving initial value problems.

LieToMe
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The way I was taught to solve many quasi-linear PDEs was by harnessing the initial condition in the characteristic method at ##u(x,0) = f(x)##. What if however I need use alternative initial conditions such as ##u(x,y=c) = f(x)## for some constant ##c##? Can the solution be propagated the same way?
 
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What do you think of shifting the ##y##-axis and solving two different initial value problems (one backwards equation) ?
 
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BvU said:
What do you think of shifting the ##y##-axis and solving two different initial value problems (one backwards equation) ?
I think that sounds reasonable but I want to make sure.
 
LieToMe said:
The way I was taught to solve many quasi-linear PDEs was by harnessing the initial condition in the characteristic method at ##u(x,0) = f(x)##. What if however I need use alternative initial conditions such as ##u(x,y=c) = f(x)## for some constant ##c##? Can the solution be propagated the same way?
Yes - you can use (x, v = y - c) as your independent variable instead of (x,y). (Or use (x, c - y) if you need to work back to y = 0.)
 

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