Solving PDE w/Fourier: Obtain All Solutions

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SUMMARY

The discussion focuses on solving the partial differential equation (PDE) given by the equation ∂²u/∂x² - ∂u/∂y = u, with solutions of the form u(x,y) = (A cos(αx) + B sin(αx))f(y). The derived solution is u = -2 Σ from n=1 to ∞ [((-1)ⁿ/n)e^((1+n)(1-y)) sin(nx)]. The method of separation of variables is employed, leading to the equation X''/X - Y'/Y = 1, which allows for the identification of the constants necessary for the solution.

PREREQUISITES
  • Understanding of partial differential equations (PDEs)
  • Familiarity with Fourier series and their applications
  • Knowledge of boundary value problems
  • Proficiency in separation of variables technique
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  • Study the method of separation of variables in detail
  • Explore Fourier series and their convergence properties
  • Learn about boundary value problems in PDEs
  • Investigate the application of eigenvalue problems in solving PDEs
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Mathematics students, physicists, and engineers who are solving partial differential equations, particularly those interested in Fourier analysis and boundary value problems.

walter9459
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Homework Statement


Obtain all solutions of the equation partial ^2 u/partial x^2 - partial u/partial y = u of the form u(x,y)=(A cos alpha x + B sin alphax)f(y) where A, B and alpha are constants. Find a solution of the equation for which u=0 when x=0; u=0 when x = pi, u=x when y=1.



Homework Equations

The solution is u = -2 summuation from n=1 to infinity (((-1)^n)/n)e^((1+n)(1-y)) sin nx.


The Attempt at a Solution


I believe the next step is to use u(x,Y) = X(x)Y(y) so the equation then becomes (1/x) partial ^2 X/partial x^2 - (1/y)partial Y/partial y = u. Then I get lost, can I get some help on how I would solve this problem?
 
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walter9459 said:

The Attempt at a Solution


I believe the next step is to use u(x,Y) = X(x)Y(y) so the equation then becomes (1/x) partial ^2 X/partial x^2 - (1/y)partial Y/partial y = u.

No, you forgot to divide the right hand side by u. You should have gotten the following:

\frac{X''}{X}-\frac{Y'}{Y}=1

Since the first term on the left side depends only on x and the second depends only on y, and since they differ by a constant (namely 1), try setting X''/X=-\alpha^2 and Y'/Y=-\alpha^2-1. That way you get the desired form of the solution and when you subtract them you get 1.
 

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