Solve second order ode with Green's functions

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SUMMARY

This discussion focuses on solving second-order ordinary differential equations (ODEs) using Green's functions, specifically the equations -u''(z) + α²u(z) = f(z) and -u''(z) + α²u'(z) = f(z) with boundary conditions u(0) = g(z) and u(z) = 0 as z approaches infinity. The key steps involve finding the Green's function G(z; z') for the operator and integrating it against the forcing function f(z') or the initial condition g(z'). The variable z is treated as complex, raising questions about the dependency of the initial condition on z.

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Mathematicians, physicists, and engineers dealing with boundary value problems in differential equations, particularly those interested in the application of Green's functions in complex variable contexts.

the king
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-u''(z)+α2u(z)=f(z), u(0)=g(z), u(z)=0 as z→∞

-u''(z)+α2u'(z)=f(z), u(0)=g(z), u(z)=0 as z→∞

I am interested to solve these two boundary problems using Green's functions. It is noticed that z is complex variable. Can someone help me to do this?
 
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Well, first things first, find your green's function for the operator given.

In other words, solve -G''(z;z') + [a^2]G(z;z') = diracdelta(z - z')

Then simply integrate the product of G(z;z') with f(z') w.r.t z' and you're done :)
 
Last edited:
Marioeden said:
Well, first things first, find your green's function for the operator given.

In other words, solve -G''(z;z') + [a^2]G(z;z') = diracdelta(z - z')

Then simply integrate the product of G(z;z') with g(z') w.r.t z' and you're done :)

g(z) is the initial condition; integrating G(z;z') against f(z') should give the solution in this case, although...

the king said:
-u''(z)+α2u(z)=f(z), u(0)=g(z), u(z)=0 as z→∞

-u''(z)+α2u'(z)=f(z), u(0)=g(z), u(z)=0 as z→∞

I am interested to solve these two boundary problems using Green's functions. It is noticed that z is complex variable. Can someone help me to do this?

Is z the only variable in this DE? How can the initial condition u(0) = g(z) depend on z?
 

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