# Methods to linearize terms of PDE

• robby991
In summary, the conversation discusses the coding of a diffusion equation with a nonlinear term added to it. The method being considered is using Newton's Method to linearize the nonlinear term, but there is confusion as to whether it is necessary and if it can be integrated into the diffusion equation using Euler's Method. The code provided shows the central difference iteration method being used to solve the equation, but the addition, subtraction, or removal of the nonlinear term does not affect the solution plot.

#### robby991

HI, I have solved the diffusion equation using the central difference scheme. Next, I would like to code this diffusion equation with a nonlinear term added to the equation. The full equation is as follows:

Code:
dS/dt = Ds * d^2S/dx^2 - aS/b+S

Since aS/b+S is a nonlinear term, I need to linearize it. I was thinking using Newton's Method to perform this, where:

Code:
aS/b+S = f(S) = So +f'(So)(S-So)

Would this be the best way to do this? This will then have to be subtracted from the diffusion term, evaluated using central difference.

I initially thought I did not have to do this, however there is no change at all in the plot when I add, subtract, or remove this function from the central difference interation in my code. This iteration is as follows:

Code:
for j=1:numt-1

%2nd Derivative Central Difference Iteration%

for i=2:numx-1
S(j+1,i) = S(j,i) + (dt/dx^2)*Ds*(S(j,i+1) - 2*S(j,i) + S(j,i-1[B]))-((Vmax*dt*S(j,i))/(Km+S(j,i))); [/B]

end

end

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Strictly speaking, "Newton's method" is a method for numerically solving an equation- of course, it involves replacing the function by its tangent approximation at some point so that may be what you mean. Yes, at $(x_0, f(x_0))$ the best linear approximation, its tangent approximation, is $y= f'(x_0)(x- x_0)+ f(x_0)$.

Of course, you have to choose $$\displaystyle x_0$$ carefully- and perhaps change it as x changes in your solution algorithm.

Thank you for the reply. I would like to know if this needs to be linearized by this method, or if it can be integrated into the diffusion equation via Eulers Method. The full equation is as follows

Code:
dS/dt = Ds * d^2S/dx^2 - VmaxS/Km+S

Where dS/dt = VmaxS/Km+S is known as the Michaelis Menten equation. My confusion arrises due to the fact that alone, this can be solved (I think) using the Eulers Method

Code:
S(ndt) = S((n-1)dt) - [dt*Vmax*S(n-1)dt/Km + S((n-1)dt)]

However, when coupled with the second order diffusion equation (evaluated with the central difference method), does the proposed solution still hold? Or does it have to be linearized?

Like I mentioned in my first post, the addition, subtraction, or removal of this term does not change the solution plot at all. My full code is as follows

Code:
clear all;

numx = 10;                      %number of grid points in space
numt = 1000;                    %number of time steps to be iterated over
tmax = .0045;
Length = 1E-6;                  %length of grid
Ds = .019E-9;                      %requirement Ds(dt)/dx^2 < .5
Vmax = 275E-6;
Km = 3E-3;

x = linspace(0,Length,numx);   %vector of x values, to be used for plotting
t = linspace(0,tmax,numt)';      %vector of t values, to be used for plotting
S = zeros(numt,numx);           %initialize everything to zero

dx = x(2)-x(1);                 %Define grid spacing in time
dt = t(2)-t(1);                 %Define grid spacing in time

%specify initial conditions%

t(1) = 0;      %1st t position = 0

S(1,:) = sin((pi/2)+(x*pi/2));
S(:,numx) = 0;   %dirichlet boundry

%iterate central difference equation%

for j=1:numt-1

%2nd Derivative Central Difference Iteration%

for i=2:numx-1
S(j+1,i) = S(j,i) + (dt/dx^2)*Ds*(S(j,i+1) - 2*S(j,i) + S(j,i-1))-((Vmax*dt*S(j,i)*dt)/(Km+S(j,i)*dt));
end

S(j+1,1)=S(j,1)+dt*Ds*2*(S(j,2)-S(j,1))./dx.^2;     %Neumann Boundary Condition

end

plot(x,S(numt,:));

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## 1. What is the purpose of linearizing terms in a PDE?

The purpose of linearizing terms in a PDE (partial differential equation) is to simplify and solve the equation by breaking it down into smaller, more manageable parts. This is especially useful for non-linear PDEs, which are difficult to solve directly.

## 2. How do you linearize a PDE?

There are several methods for linearizing a PDE, including the method of characteristics, the method of separation of variables, and the method of similarity solutions. Each method has its own advantages and is suitable for different types of PDEs.

## 3. What are the benefits of linearizing a PDE?

Linearizing a PDE can make the equation easier to understand and solve, as well as provide insights into the behavior of the system described by the PDE. It can also help in the development of numerical methods for solving PDEs.

## 4. Are there any limitations to linearizing a PDE?

Linearizing a PDE may not always be possible, especially for highly non-linear equations. Additionally, the linearized equation may not accurately reflect the behavior of the original non-linear equation in certain regions of the solution space.

## 5. Can linearization be applied to all types of PDEs?

No, linearization is not applicable to all types of PDEs. It is most commonly used for first-order and second-order PDEs, but may also be used for higher-order PDEs in certain cases. It is important to carefully consider the properties of the PDE before attempting to linearize it.