How to handle a differential equation whose RHS contains the derivative of the Dirac delta function?Solution of differential equation with Dirac DeltaGetting 0 solving Schrodinger equation with Dirac delta by Fourier transformIntegral of delta dirac functionDiffusion equation involving dirac delta termFirst order differential equation with a derivative delta functionAbsolute value: First Derivative Heaviside Function + Second Derivative Dirac Delta Function DistributionDifferential Equation with Delta DiracSolving differential equation with the Dirac Delta FunctionVerifying the delta function satisfies Poisson's EquationHow to solve the differential equation with delta function on right hand side
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How to handle a differential equation whose RHS contains the derivative of the Dirac delta function?
Solution of differential equation with Dirac DeltaGetting 0 solving Schrodinger equation with Dirac delta by Fourier transformIntegral of delta dirac functionDiffusion equation involving dirac delta termFirst order differential equation with a derivative delta functionAbsolute value: First Derivative Heaviside Function + Second Derivative Dirac Delta Function DistributionDifferential Equation with Delta DiracSolving differential equation with the Dirac Delta FunctionVerifying the delta function satisfies Poisson's EquationHow to solve the differential equation with delta function on right hand side
$begingroup$
I have a differential equation of the following form.
beginalign
beginsplit
fracmathrmd^4psi(eta)mathrmdeta^4-beta^4psi(eta)=psi'(zeta),delta'(eta-zeta)
endsplit
endalign
How to handle the derivative of the Dirac delta function on the right-hand side? I am trying to find a closed form solution using Green's functions.
Suppose if a differential equation contains both Dirac delta function and its derivative, can I linearly superimpose the solution of the differential equation of just Dirac delta function plus the solution of the differential equation with derivative of Dirac delta function?
beginalign
beginsplit
fracmathrmd^4psi(eta)mathrmdeta^4-beta^4psi(eta)=psi(zeta),delta(eta-zeta)+psi'(zeta),delta'(eta-zeta)
endsplit
endalign
ordinary-differential-equations closed-form dirac-delta
$endgroup$
add a comment |
$begingroup$
I have a differential equation of the following form.
beginalign
beginsplit
fracmathrmd^4psi(eta)mathrmdeta^4-beta^4psi(eta)=psi'(zeta),delta'(eta-zeta)
endsplit
endalign
How to handle the derivative of the Dirac delta function on the right-hand side? I am trying to find a closed form solution using Green's functions.
Suppose if a differential equation contains both Dirac delta function and its derivative, can I linearly superimpose the solution of the differential equation of just Dirac delta function plus the solution of the differential equation with derivative of Dirac delta function?
beginalign
beginsplit
fracmathrmd^4psi(eta)mathrmdeta^4-beta^4psi(eta)=psi(zeta),delta(eta-zeta)+psi'(zeta),delta'(eta-zeta)
endsplit
endalign
ordinary-differential-equations closed-form dirac-delta
$endgroup$
$begingroup$
Actually ,my the goal is to extract the solution using Green's function approach
$endgroup$
– musimathics
Mar 16 at 11:31
$begingroup$
Actually I was editing the question, It may be happened accddentally. Please suggest , so that I can make changes to the question
$endgroup$
– musimathics
Mar 16 at 11:33
1
$begingroup$
Assuming zero initial conditions (all four of them), the Laplace transform is $$Psi (s) = psi'(zeta) left(fracss^4 - beta^4right)$$ From here, one can use partial fraction expansion to conclude what the form of the solution is.
$endgroup$
– Rodrigo de Azevedo
Mar 16 at 11:47
$begingroup$
This is using Laplace transforms. Is there any ways that I can fit a Green's function to this DE.
$endgroup$
– musimathics
Mar 16 at 11:49
$begingroup$
I am not familiar with Green's functions. At least, not that I know of.
$endgroup$
– Rodrigo de Azevedo
Mar 16 at 11:50
add a comment |
$begingroup$
I have a differential equation of the following form.
beginalign
beginsplit
fracmathrmd^4psi(eta)mathrmdeta^4-beta^4psi(eta)=psi'(zeta),delta'(eta-zeta)
endsplit
endalign
How to handle the derivative of the Dirac delta function on the right-hand side? I am trying to find a closed form solution using Green's functions.
Suppose if a differential equation contains both Dirac delta function and its derivative, can I linearly superimpose the solution of the differential equation of just Dirac delta function plus the solution of the differential equation with derivative of Dirac delta function?
beginalign
beginsplit
fracmathrmd^4psi(eta)mathrmdeta^4-beta^4psi(eta)=psi(zeta),delta(eta-zeta)+psi'(zeta),delta'(eta-zeta)
endsplit
endalign
ordinary-differential-equations closed-form dirac-delta
$endgroup$
I have a differential equation of the following form.
beginalign
beginsplit
fracmathrmd^4psi(eta)mathrmdeta^4-beta^4psi(eta)=psi'(zeta),delta'(eta-zeta)
endsplit
endalign
How to handle the derivative of the Dirac delta function on the right-hand side? I am trying to find a closed form solution using Green's functions.
Suppose if a differential equation contains both Dirac delta function and its derivative, can I linearly superimpose the solution of the differential equation of just Dirac delta function plus the solution of the differential equation with derivative of Dirac delta function?
beginalign
beginsplit
fracmathrmd^4psi(eta)mathrmdeta^4-beta^4psi(eta)=psi(zeta),delta(eta-zeta)+psi'(zeta),delta'(eta-zeta)
endsplit
endalign
ordinary-differential-equations closed-form dirac-delta
ordinary-differential-equations closed-form dirac-delta
edited Mar 16 at 11:43
Rodrigo de Azevedo
13.2k41960
13.2k41960
asked Mar 16 at 11:18
musimathicsmusimathics
1418
1418
$begingroup$
Actually ,my the goal is to extract the solution using Green's function approach
$endgroup$
– musimathics
Mar 16 at 11:31
$begingroup$
Actually I was editing the question, It may be happened accddentally. Please suggest , so that I can make changes to the question
$endgroup$
– musimathics
Mar 16 at 11:33
1
$begingroup$
Assuming zero initial conditions (all four of them), the Laplace transform is $$Psi (s) = psi'(zeta) left(fracss^4 - beta^4right)$$ From here, one can use partial fraction expansion to conclude what the form of the solution is.
$endgroup$
– Rodrigo de Azevedo
Mar 16 at 11:47
$begingroup$
This is using Laplace transforms. Is there any ways that I can fit a Green's function to this DE.
$endgroup$
– musimathics
Mar 16 at 11:49
$begingroup$
I am not familiar with Green's functions. At least, not that I know of.
$endgroup$
– Rodrigo de Azevedo
Mar 16 at 11:50
add a comment |
$begingroup$
Actually ,my the goal is to extract the solution using Green's function approach
$endgroup$
– musimathics
Mar 16 at 11:31
$begingroup$
Actually I was editing the question, It may be happened accddentally. Please suggest , so that I can make changes to the question
$endgroup$
– musimathics
Mar 16 at 11:33
1
$begingroup$
Assuming zero initial conditions (all four of them), the Laplace transform is $$Psi (s) = psi'(zeta) left(fracss^4 - beta^4right)$$ From here, one can use partial fraction expansion to conclude what the form of the solution is.
$endgroup$
– Rodrigo de Azevedo
Mar 16 at 11:47
$begingroup$
This is using Laplace transforms. Is there any ways that I can fit a Green's function to this DE.
$endgroup$
– musimathics
Mar 16 at 11:49
$begingroup$
I am not familiar with Green's functions. At least, not that I know of.
$endgroup$
– Rodrigo de Azevedo
Mar 16 at 11:50
$begingroup$
Actually ,my the goal is to extract the solution using Green's function approach
$endgroup$
– musimathics
Mar 16 at 11:31
$begingroup$
Actually ,my the goal is to extract the solution using Green's function approach
$endgroup$
– musimathics
Mar 16 at 11:31
$begingroup$
Actually I was editing the question, It may be happened accddentally. Please suggest , so that I can make changes to the question
$endgroup$
– musimathics
Mar 16 at 11:33
$begingroup$
Actually I was editing the question, It may be happened accddentally. Please suggest , so that I can make changes to the question
$endgroup$
– musimathics
Mar 16 at 11:33
1
1
$begingroup$
Assuming zero initial conditions (all four of them), the Laplace transform is $$Psi (s) = psi'(zeta) left(fracss^4 - beta^4right)$$ From here, one can use partial fraction expansion to conclude what the form of the solution is.
$endgroup$
– Rodrigo de Azevedo
Mar 16 at 11:47
$begingroup$
Assuming zero initial conditions (all four of them), the Laplace transform is $$Psi (s) = psi'(zeta) left(fracss^4 - beta^4right)$$ From here, one can use partial fraction expansion to conclude what the form of the solution is.
$endgroup$
– Rodrigo de Azevedo
Mar 16 at 11:47
$begingroup$
This is using Laplace transforms. Is there any ways that I can fit a Green's function to this DE.
$endgroup$
– musimathics
Mar 16 at 11:49
$begingroup$
This is using Laplace transforms. Is there any ways that I can fit a Green's function to this DE.
$endgroup$
– musimathics
Mar 16 at 11:49
$begingroup$
I am not familiar with Green's functions. At least, not that I know of.
$endgroup$
– Rodrigo de Azevedo
Mar 16 at 11:50
$begingroup$
I am not familiar with Green's functions. At least, not that I know of.
$endgroup$
– Rodrigo de Azevedo
Mar 16 at 11:50
add a comment |
0
active
oldest
votes
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$begingroup$
Actually ,my the goal is to extract the solution using Green's function approach
$endgroup$
– musimathics
Mar 16 at 11:31
$begingroup$
Actually I was editing the question, It may be happened accddentally. Please suggest , so that I can make changes to the question
$endgroup$
– musimathics
Mar 16 at 11:33
1
$begingroup$
Assuming zero initial conditions (all four of them), the Laplace transform is $$Psi (s) = psi'(zeta) left(fracss^4 - beta^4right)$$ From here, one can use partial fraction expansion to conclude what the form of the solution is.
$endgroup$
– Rodrigo de Azevedo
Mar 16 at 11:47
$begingroup$
This is using Laplace transforms. Is there any ways that I can fit a Green's function to this DE.
$endgroup$
– musimathics
Mar 16 at 11:49
$begingroup$
I am not familiar with Green's functions. At least, not that I know of.
$endgroup$
– Rodrigo de Azevedo
Mar 16 at 11:50