Solve second order ode with Green's functions

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To solve the second-order ordinary differential equations using Green's functions, the first step is to determine the Green's function for the specified operator. The equations presented are -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 solution involves integrating the product of the Green's function G(z; z') with the function f(z') over z'. There is a question regarding the dependence of the initial condition g(z) on the variable z, which raises concerns about the formulation of the problem. The discussion emphasizes the importance of correctly identifying the Green's function to proceed with the solution.
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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