Phase 2:
It is all about Nitrogen Absorption rate, and for this I have to choose direction and I choose the direction along x-axis. and how the nitrogen absorbs, and when Nitrogen pass through the plant. And with the increase in Nitrogen, the growth of plants will increase and I made the complex model (As, in my model, there is a double derivative means partial double derivative, Laplace transform and it goes with both respect to time and axis. I will find in Absorption rate that how much Nitrogen is going in the plant.
M_x (∂^2 N(x,t))/(∂x^2 )-S ∂N(x,t)/∂x-∂N/∂x=N_T (x,t) ———> 1
For solving N_T (x_1 t), we have equation
N_D (∂^2 N(x,t))/(∂x^2 )- N_c N_T (x,t)-(∂N_T (x,t))/∂x=0———>2
So, in above these 2 equations:
S is Average velocity of air
M_x is coefficient of axial dispersion that describes the manner in which the Nitrogen move relative to a point.
N is change of concentration with time and distance along a particular path bounded by a structure of plant
N_c is transfer coefficient of Nitrogen
N_D is Nitrogen absorption rate
Boundary conditions for N_T
N_T (x,0)=exp⁡(-√((N_c-a)/N_D ) x) Here N_c≥0 and N_D>0
N_T (0,t)=exp⁡(-at) Here t>0
N_T (∞,t)=0 Here t>0
Boundary Conditions for N
N(x,0)=N_(0 ) Here x≥0
N(0,t)=N_1 Here t>0
N(∞,t)=0
Here N_(0 ) and N_1 are constants
Solution of N_T
Using Laplace transform on equation 2
L{N_D (∂^2 N_T (x,t))/(∂x^2 )}-N_c L{N_T (x,t)}-L{(∂N_T (x,t))/∂x}=L{0}———>3
Solving Laplace seperately
L{N_D (∂^2 N_T (x,t))/(∂x^2 )}= {N_D (d^2 N_T (x,q))/(dx^2 )}
As N_T (x,q)= L{N_T (x,t)}
L{(∂N_T (x,t))/∂x}=qN_T (x,q)-N_T (x,0)
N_T (x,0)=exp⁡(-√((N_c-a)/N_D ) x)
Putting values in equation 3
N_D (d^2 N_T (x,q))/(dx^2 )-(N_c+q) N_T (x,q)=-exp⁡(-√((N_c-a)/N_D ) x) ———>4
Finding solution by using method of characteristic equation and indeterminate coefficient methods
N_T (x,q)= c_1 exp⁡(-√((N_c-a)/N_D ) x)+c_2 exp⁡(√((N_c-a)/N_D ) x)+(1/(a+q)) exp⁡(-√((N_c-a)/N_D ) x)
Finding value of c_1 and c_2 using boundary conditions
We can obtain c_2=0 and c_1=0
So,
N_T (x,q)= (1/(a+q)) exp⁡(-√((N_c-a)/N_D ) x)
Finally
L^(-1) {N_T (x,q)}= L^(-1) {(1/(a+q)) exp⁡(-√((N_c-a)/N_D ) x) }

N_T (x,q)=exp⁡(-√((N_c-a)/N_D ) x) L_q^(-1) {(1/(a+q))}
N_T (x,t)=exp[-(at+√((N_c-a)/N_D ) x]———>5
Now, obtaining solution for equation 1
Applying Laplace Transform on equation 1
L{(∂^2 N(x,t))/(∂x^2 )}-L{S ∂N(x,t)/∂x}-L{∂N/∂x}= (exp⁡(-√((N_c-a)/N_D ) x))/M_x L{exp⁡(-at)}
We get
N(x,q)=c_1 exp[x/2(S/M_x -√((s^2+4M_x q)/M_x ))]+c_2 [x/2(S/M_x +√((s^2+4M_x q)/M_x ))]
Finding c_1 and c_2by using boundary conditions
c_2=0
c_1=N_1/q-[(N_0 (q+a)(2-N_0 (q+a)-1)/((q-∝)(q+a)-qN_0 〖(q+a)〗^2 )]
We get
N(x,q)=[1/q+1/((q-∝)(q+a))]exp[x/2(S/M_x -√((s^2+4M_x q)/M_x ))]-1/(q+a)(q-a) exp⁡(-√((N_c-a)/N_D x))———>6
Apply inverse Laplace on equation 6, we get
N(x,t)=1/2 exp⁡(Sx/(2M_x )) [exp⁡((-Sx)/(2M_x )) erf⁡((x-St)/(2√(M_x t))) ]+exp⁡(Sx/(2M_x ))erfc((x+St)/(2√(M_x t)))
±(exp⁡(∝t))/((∝+a)) {exp (±x√(S^2+4M_x∝))/(2M_x ) erf (x-t√(S^2+4M_x∝))/(2√(M_x t))}-exp⁡(x√((N_c-a)/N_D )) 1/(∝+a) [exp⁡(∝t)-exp(-∝t)]
Here, N(x,t) is the solution.

Phase 3

In the following model ‘t‘ is time and ‘x‘ is the space along the Nitrogen bounded by an layers, not allowing Nitrogen transfer easily. ‘S‘ is speed of Nitrogen and M_{x} is the axial dispersion coefficient responsible for describing how the air move in a system with respect to a point, both are positive constants.
In order to make the model more realistic, we consider the semi constrained, enabling air transfer between different sides.
It is well known that the transfer of Nitrogen affects the absorption rate. This
factor has been taken into account here. We have drawn the absorption graphs
for the case when the concentration level of Nitrogen decreases considerably. The diffusion transport of various substrates between surrounding tissue structures has been studied here. We conclude the following important results in this study. The variation of
concentration of the Nitrogen in single direction is evident from the graphs and it decreases as we
move away from the one end. Thus the absorption rate at a large distance from the surface are not capable of getting sufficient Nitrogen.

Dependant Independent
t N x ∆N ∆x ∆N/∆x x
15 0.001 0.9894 0.1 -0.3076 -0.325097529 0.9894
15 0.101 0.6818 0.1 -0.2125 -0.470588235 0.6818
15 0.201 0.4693 0.1 -0.1466 -0.68212824 0.4693
15 0.301 0.3227 0.1 -0.1009 -0.991080278 0.3227
15 0.401 0.2218 0.1 -0.0695 -1.438848921 0.2218
15 0.501 0.1523 0.1 -0.0478 -2.092050209 0.1523
15 0.601 0.1045 0.1 -0.0328 -3.048780488 0.1045
15 0.701 0.0717 0.1 -0.0226 -4.424778761 0.0717
15 0.801 0.0491 0.1 -0.0155 -6.451612903 0.0491
15 0.901 0.0336 0.1 -0.0336 -2.976190476 0.0336

Differential Equation Model is:

M_x (∂^2 N(x,t))/(∂x^2 )-S ∂N(x,t)/∂x-∂N/∂x=N_T (x,t) ———> 1
For solving N_T (x_1 t), we have equation
N_D (∂^2 N(x,t))/(∂x^2 )- N_c N_T (x,t)-(∂N_T (x,t))/∂x=0———>2
So, in above these 2 equations:
S is Average velocity of air
M_x is coefficient of axial dispersion that describes the manner in which the Nitrogen move relative to a point.
N is change of concentration with time and distance along a particular path bounded by a structure of plant
N_c is transfer coefficient of Nitrogen
N_D is Nitrogen absorption rate

The graphs show that the level of Nitrogen decreases with x until
normalcy in the condition is maintained. The graphs exhibit the
absorption rate in the one direction for specific values of parameters. With this type of information, one may predict the effects of absorption for a certain set of parameters.

Phase 4
We need to solve a system of the following two equations in MATLAB:
M_x (∂^2 N(x,t))/(∂x^2 )-S ∂N(x,t)/∂x-∂N/∂x=N_T (x,t) ———> 1
N_D (∂^2 N(x,t))/(∂x^2 )-N_c N_T (x,t)-(∂N_T (x,t))/∂x=0———>2

We can rewrite the first equation more compactly as:
M_x (∂^2 N(x,t))/(∂x^2 )-(S+1)∂N(x,t)/∂x=N_T (x,t)
We need to convert these equations into a form that is accepted by MATLAB:

This is the format that is accepted by the MATLAB PDE solver function.
We write the equations in matrix form:
[■(0&0@0&0)] ∂/∂t (■(N_T@N))=∂/∂x [■(M_x ∂N/∂x-(S+1)N@N_D ∂N/∂x-N_T )]+[■(-N_T@-N_C N_T )]
This is in the standard form required by MATLAB.
The initial conditions are:
N_T (x,0)=exp(-√((N_c-a)/N_D ) x)HereN_c≥0∧N_D>0
N(x,0)=N(x,0)=N_0 Herex≥0
The boundary conditions for the system are:
N_T (0,t)=exp(-at),t>0
N_T (∞,t)=0 ,t>0
and
N(0,t)=N_1,t>0
N(∞,t)=0
The MATLAB PDE solver expects the boundary conditions to be of the form:

So, we write them as follows:
For x=0,
[■(N_T-e^(-at)@N-N_1 )]+[■(0@0)].[■(M_x ∂N/∂x-(S+1)N@N_D ∂N/∂x-N_T )]=0
For x=∞, the boundary condition becomes:
[■(N_T@N)]+[■(0@0)].[■(M_x ∂N/∂x-(S+1)N@N_D ∂N/∂x-N_T )]=0
For simplicity, we set all unknown coefficients equal to 1 for ease of programming in MATLAB.
The graphical solution for NT is:

The graphical solution for N is: