partial differentials: ∂∂tA(y,t)+6ΛΩ(y2−y)sin(t)=∂2∂y2A(y,t)

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Grindra

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partial differentials: ∂∂tA(y,t)+6ΛΩ(y2−y)sin(t)=∂2∂y2A(y,t)

Problem
\(\displaystyle {\frac {\partial }{\partial t}}A\left( y,t \right) +6\,\Lambda\,\Omega\, \left( {y}^{2}-y \right) \sin \left( t \right) ={\frac {\partial ^{2}}{\partial {y}^{2}}}A \left( y,t \right)\)

\(\displaystyle {\frac{\partial }{\partial y}}A \left( t,0 \right) ={\frac {\partial }{\partial y}}A \left( t,1 \right) =0\)

Boundary condition

\(\displaystyle {\frac{\partial }{\partial y}}A \left( t,0 \right) ={\frac {\partial }{\partial y}}A \left( t,1 \right) =0\)


I don't know how to get this kind of solution. this is my attempt . please help.
 
Last edited:
Grindra said:
Problem
\(\displaystyle {\frac {\partial }{\partial t}}A\left( y,t \right) +6\,\Lambda\,\Omega\, \left( {y}^{2}-y \right) \sin \left( t \right) ={\frac {\partial ^{2}}{\partial {y}^{2}}}A \left( y,t \right)\)

\(\displaystyle {\frac{\partial }{\partial y}}A \left( t,0 \right) ={\frac {\partial }{\partial y}}A \left( t,1 \right) =0\)

Boundary condition

\(\displaystyle {\frac{\partial }{\partial y}}A \left( t,0 \right) ={\frac {\partial }{\partial y}}A \left( t,1 \right) =0\)


I don't know how to get this kind of solution. this is my attempt . please help.
Click to expand...

The first place you might start is the separation of variables method, see for example https://math.stackexchange.com/ques...-non-homogenous-equation-theory-clarification
...
Assume we have an inhomogeneous Partial Differential Equation of the form [FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]C[/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]D[/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]E[/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]F[/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]w[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main](1)[/FONT]

with some initial and boundary conditions.
Let us define theauxiliary linear homogeneous equation as
[FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]v[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Math]v[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]C[/FONT][FONT=MathJax_Math]v[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]D[/FONT][FONT=MathJax_Math]v[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]E[/FONT][FONT=MathJax_Math]v[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]F[/FONT][FONT=MathJax_Math]v[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main](2)[/FONT]
with the same boundary conditions as in [FONT=MathJax_Main](1)[/FONT] Then the general solution [FONT=MathJax_Math]u[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Main]general[/FONT]
of the inhomogeneous equation [FONT=MathJax_Main](1)[/FONT] can be written as the sum of a particular solution [FONT=MathJax_Math]u[/FONT][FONT=MathJax_Main]particular[/FONT] of [FONT=MathJax_Main](1)[/FONT] and a general solution [FONT=MathJax_Math]v[/FONT][FONT=MathJax_Main]general[/FONT] of the auxiliary homogeneous equation[FONT=MathJax_Main](2)[/FONT]:
[FONT=MathJax_Math]u[/FONT][FONT=MathJax_Main]general[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]v[/FONT][FONT=MathJax_Main]general[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]u[/FONT][FONT=MathJax_Main]particular[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main].[/FONT]​
Click to expand...
 
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