Partial Differential Eq w Separation of variables y''+2/x*y'+A*y=dx/dt

[vodkatomic]

Diffusion within a spherical catalyst with first order reaction A->B

y''+2/x*y'+A^2*y=dy/dt

equivalent to

y(x,t) = d^2y/dx^2+2/x*(dy/dy)+A*y=dy/dt EQ-1 where A is constante -- need to use separation of variables to solve the unsteady steate!


initial conditions
y(x,0) = 0

boundary conditions

y(1,t) = f(t)
(dy/dx|x=0) = 0

with the form
Y (x, t) = YH (x, t) + YSS(x)

where YH (x, t) is the transient, homogeneous problem that takes the homogenous form of the boundary conditions,
and where YSS(x) is the steady-state problem that takes the nonhomogeneous boundary condition.

The PDE of equation is nonhomogeneous; hence we must seek superposition
of the form Y(x, t) = YH (x, t) + YSS(x), YH (x, t) takes the homogeneous
form of the PDE and boundary conditions, YSS(x) is a nonhomogeneous ODE.


The solution for the steady steady state is YH=(sinh⁡(∅*x))/(x*sinh⁡(x)) where ∅ is a constant

The question is how can I separate EQ-1 using the principle of separation of variables
is there any trick?

 

[HallsofIvy]

Need to solve the following equation

y''+2/x*y'+A^2*y=dx/dt

equivalent to

y(x,t) = d^2y/dx^2+2/x*(dy/dy)+A*y=dy/dt

No, it's not for several reasons. First, you had "dx/dt" before but I think you meant dy/dt. Second, there is no "y(x,t)" equal to that expression.

initial conditions
y(x,0) = 0

boundary conditions

y(1,t) = f(t)
(dy/dx|x=0) = 0

with the form
T (x, t) = TH (x, t) + TSS(x)

where TH (x, t) is the transient, homogeneous problem that takes the homogenous form of the boundary conditions, and where TSS(x) is the steady-state problem that takes the nonhomogeneous boundary condition.

You titled this "with separation of variables". Do you know what that means? And do you know what the "homogeneous form" and "nonhomogeneous form" mean? And do your y'' and y' mean the partial derivatives with respect to x?

Using separation of variables means that we begin by seeking a solution of the form y(x, t)= X(x)T(t). That is, where y(x,t) is a product of a function of x only and a function of t only.
With y(x, t)= X(x)T(t), \(\displaystyle \frac{\partial^2 y}{\partial x^2}= X''T\), \(\displaystyle \frac{\partial y}{\partial x}= X'T\), and \(\displaystyle \frac{\partial y}{\partial t}= XT'\) so the equation becomes
\(\displaystyle X''T+ \frac{2}{x}X'T+ XT= XT'\). Divide through by XT to get \(\displaystyle \frac{X"+ \frac{2}{x}X'+ X}{X}= \frac{T'}{T}\)

The left side is a function of x only, the right side a function of t only. In order to be equal for all x and t, each must be a constant. That is, there must be a number, \(\displaystyle \alpha\), such that \(\displaystyle \frac{X"+ \frac{2}{x}X'+ X}{X}= \alpha\), so that \(\displaystyle X"+ \frac{2}{x}X'+ X= \alpha X\), and \(\displaystyle \frac{T'}{T}= \alpha\), so that \(\displaystyle T'= \alpha T\).

For the x-equation, If you multiply both sides by \(\displaystyle x^2\) you get \(\displaystyle x^2X''+ \frac{2}{x}X'+ (1- \alpha)X= 0\) which should remind you of "Bessel's equation".

 

[vodkatomic]

Transient First order reaction within a spherical sphere where A->B
Separation of Variables in the Spherical Coordinate System
Solution of Transient Problems



1/x^2*d(x^2*dy/dx)/dx-A^2*y=dy/dt
(1/x^2)*( d^2(y(x))/dx^2+2*x*dy(x)/dx)-A^2*y=dy(t)/dt
d^2(y(x))/dx^2+(2/x)* dy(x)/dx+A^2*y=dy/dt

The PDE equations is nonhomogeneous; hence we must seek superposition
of the form T (x, t)= TH (x, t) + TSS(x), where TH (x, t) takes the homogeneous
form of the PDE and boundary conditions, while TSS(x) is a nonhomogeneous ODE.