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seen Dec 4 at 15:19

Nov
25
accepted solvability condition for differential operator
Nov
25
comment solvability condition for differential operator
@TZakrevskiy, from your comment now I understand what is meaning of self-adjoint differential operator. Honestly I could not understand answer given by you at all because my ineptness in this field. Anyway thanks for your answer and explanation..!
Nov
25
answered solvability condition for differential operator
Nov
24
revised solvability condition for differential operator
edited tags
Nov
24
revised solvability condition for differential operator
deleted 6 characters in body
Nov
24
asked solvability condition for differential operator
Oct
1
comment finding a parameter to match the solution asymptotically
@JuliánAguirre, If I solve the equation with boundary conditions $y'(-3) =0$ and $y'(3) = 1$ in Mathematica, fine tuning of parameter $b$ shows that, I can make $y(-3)$ as small as possible. But if I start with boundary conditions $y'(-3) =0$ and $y(-3) = 0$, Mathematica declares problem as ill conditioned.
Oct
1
comment finding a parameter to match the solution asymptotically
@JuliánAguirre, that x axis is an asymptote to y(x) as x tend to -3.
Oct
1
asked finding a parameter to match the solution asymptotically
Sep
10
revised Finding family of curve for given asymptotes
edited tags
Sep
10
asked Finding family of curve for given asymptotes
Aug
22
asked solution of $y' + y^2 = \varphi^2(x)$
Aug
19
asked Solving second order differential equation numerically with values given at intermediate points.
Aug
19
accepted Proof of definite integral is finite
Aug
19
accepted function with zero first to n'th derivative at end points
Aug
19
comment Solution of $y''(x) -k = \delta(x-x_0)y(x)$
yes, I understood this when I carefully read the boundary conditions. Thanks again for the explanation.
Aug
19
comment Solution of $y''(x) -k = \delta(x-x_0)y(x)$
@Dmoreno, After what you have derived we assume $u_y =0$(i.e. there is no flow in the direction perpendicular to gravity which follows by symmetry) and arrive at equation in the question. While solving Usual NS equations we specify value of velocity at the boundary. Here friction force is added to make sure velocity in the vicinity of wall be made gradually zero. Velocity profile can be then expected to be not parabolic nearby wall.
Aug
19
accepted Solution of $y''(x) -k = \delta(x-x_0)y(x)$
Aug
19
comment Solution of $y''(x) -k = \delta(x-x_0)y(x)$
@Winter, Thanks a lot for the answer..!!
Aug
19
comment Solution of $y''(x) -k = \delta(x-x_0)y(x)$
@Dmoreno, I have added the physical background.