Solving Simple PDEs for Beginners

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SUMMARY

The discussion focuses on solving simple partial differential equations (PDEs) using specific techniques. The equations presented are: 1. \(\frac{\partial^4y(u,v)}{\partial u^2 \partial v^2} + k^2 \frac{\partial^2y(u,v)}{\partial v^2}=0\) 2. \(\frac{\partial^2y(u,v)}{\partial u^2} + k^2 y(u,v) + b=0\). The boundary conditions are specified as \(\frac{\partial y(u,v)}{\partial v}|_{v=0}=0\) and \(\frac{\partial y(u,v)}{\partial u}|_{u=0,v=0}=0\). The solution approach involves differentiating the second equation and integrating the first equation to derive two similar second-order ordinary differential equations (ODEs).

PREREQUISITES
  • Understanding of partial differential equations (PDEs)
  • Familiarity with ordinary differential equations (ODEs)
  • Knowledge of boundary value problems
  • Basic skills in mathematical analysis
NEXT STEPS
  • Study the method of separation of variables for PDEs
  • Learn about boundary value problems in the context of PDEs
  • Explore the characteristics of second-order ordinary differential equations
  • Investigate the application of Fourier series in solving PDEs
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Mathematics students, researchers in applied mathematics, and professionals working with mathematical modeling of physical systems will benefit from this discussion.

phonic
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Dear All,

I got some trobule in solving the following simple-looking PDE's. Can anyone give a hint about how to solve it? thanks a lot! I guess the solution is of the form y(u,v)=A[\cos(k(u-f(v))-B]\cosh(v)-C. But I don't know a formal way to solve.
\frac{\partial^4y(u,v)}{\partial u^2 \partial v^2} +k^2 \frac{\partial^2y(u,v)}{\partial v^2}=0

\frac{\partial^2y(u,v)}{\partial u^2} +k^2 y(u,v) +b=0

Boundary condition:

\frac{\partial y(u,v)}{\partial v}|_{v=0}=0
\frac{\partial y(u,v)}{\partial u}|_{u=0,v=0}=0
 
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If you differentiate the second equation twice with respect to v, you will see that it is basically the same equation as the first equation. Just integrate the first equation twice to get rid of the partial derivatives wrt v and you end up with two almost identical second order ODE's.

And in the second equation, the fact that y depends on v is not important, so to simplify things, you could start by solving y" + k^2y+b=0
When you have found the solution, it is trivial to extend it to the case where y depends on v.
 

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