Solving Linear & Nonlinear System w/ Reparametrization Function

In summary: Your Name]In summary, the individual is seeking help with a problem involving a model for transport of a solute and solvent across a permeable membrane. They have attempted to find a solution using a proposition, but have encountered an issue in defining the function $\rho$. They have crossposted their question on a mathematics forum but have not yet received a response. The expert suggests that the issue lies in the way the function $\rho$ has been defined and provides guidance on how to proceed with finding the correct transformation between solutions of the linear and nonlinear systems. The expert also offers further assistance if needed.
  • #1
kalish1
99
0
I have a problem on which I am stuck and would like help on how to proceed. The problem and my work is fairly lengthy, so please bear with me.

**Problem:** A model for transport of a solute (moles of salt) and solvent (volume of water) across a permeable membrane has the form $$\dot{W}=A(k-\frac{M}{W}),\dot{M}=B(k-\frac{M}{W})$$

where $k$ is a parameter representing the bulk solute concentration and $A$ and $B$ are parameters that represent the permeability of the membrane.

(a) The water volume $W$ is a positive quantity. Show that the system can be made linear by a reparametrization.

(b) Determine the transformation between solutions of the linear and nonlinear systems.

I need to make use of this proposition as well:

**Proposition:**

If $g:U \rightarrow \mathbb{R}$ is a positive, smooth function where $U$ is an open set, $J \subset \mathbb{R}$ is an open interval containing the origin and $\gamma:J \rightarrow \mathbb{R^n}$ is a solution of the differential equation $\dot{x}=f(x)$ with $\gamma(0) = x_0 \in U$, then the function $B:J \rightarrow \mathbb{R}$ given by $$B(t) = \int_0^t\frac{1}{g(\gamma(s))}ds$$ is invertible on its range $K \subseteq \mathbb{R}$. If $\rho:K \rightarrow J$ is the inverse of $B$, then the identity $$\rho'(t) = g(\gamma(\rho(t)))$$ holds for all $t\in K$, and the function $\sigma:K \rightarrow \mathbb{R^n}$ given by $\sigma(t) = \gamma(\rho(t))$ is the solution of the differential equation $\dot{x}=g(x)f(x)$ with initial condition $\sigma(0) = x_0$.

**My attempt (part a):**
$$\begin{pmatrix} W' \\ M' \\ \end{pmatrix}=\frac{1}{W}\begin{pmatrix} Ak & -A \\ Bk & -B \\ \end{pmatrix} \begin{pmatrix} W \\ M \\ \end{pmatrix}
$$

where $$g(t)=\frac{1}{W}, \delta(t) = \begin{pmatrix} Ak & -A \\ Bk & -B \\ \end{pmatrix}, f(t)=\begin{pmatrix} Ak & -A \\ Bk & -B \\ \end{pmatrix} \begin{pmatrix} W \\ M \\ \end{pmatrix}$$

The (**correct as has been verified with Mathematica and WolframAlpha**) solution of the linear system $$\dot{X}=\delta X, W(0)=w_0, M(0)=m_0,$$ where $X = \begin{pmatrix} W \\ M \\ \end{pmatrix}$ is:

$$X(t) = \frac{1}{kA-B}\left\{(Am_0-Bw_0)\begin{pmatrix} 1 \\ k \end{pmatrix} + (kw_0-m_0)\begin{pmatrix} A \\ B \end{pmatrix}e^{(kA-B)t}\right\}=\gamma(t).$$

Now, because $\rho$ in the proposition is the inverse of $B$, we have the formula $$t = \int_0^\rho\frac{1}{g(\gamma(s))}ds.$$

So, $$t=\int_0^\rho\frac{1}{\frac{1}{\gamma(s)}}ds=
\int_0^\rho\gamma(s)ds = \int_0^\rho \frac{1}{kA-B}\left\{(Am_0-Bw_0)\begin{pmatrix} 1 \\ k \end{pmatrix} + (kw_0-m_0)\begin{pmatrix} A \\ B \end{pmatrix}e^{(kA-B)t}\right\}ds.$$

**My question is: How can $t$ on the LHS be equal to the integral of a vector function on the RHS? Where have I gone wrong?**

Thanks for any help.

By the way, I have crossposted this on differential equations - Reparametrization function as transformation between linear and nonlinear systems - Mathematics Stack Exchange

but have not gotten any replies after 1 day.
 
Last edited:
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  • #2

Thank you for reaching out for help with your problem. I understand the frustration of being stuck on a problem and not being able to proceed. I will do my best to assist you in finding a solution.

After reviewing your attempt at solving the problem, I believe the issue lies in the way you have defined the function $\rho$. In the proposition, the function $\rho$ is defined as the inverse of $B$, which is a scalar function. However, in your solution, you have defined $\rho$ as the inverse of the vector function $\gamma$. This is not correct.

To find the correct transformation between solutions of the linear and nonlinear systems, you need to define $\rho$ as the inverse of the scalar function $B$. This means that $t$ on the left-hand side of the equation should be equal to the integral of a scalar function on the right-hand side.

I hope this helps you to proceed with your problem. If you have any further questions or need clarification, please do not hesitate to ask. Good luck with your work.
 

1. How do you define a linear system?

A linear system is a set of equations where each equation is a linear combination of the variables. This means that each equation can be written in the form of y = mx + b, where m and b are constants and x is the variable. In other words, the variables in a linear system have an exponent of 1 and are not multiplied together.

2. What is the purpose of reparametrization function in solving linear and nonlinear systems?

The reparametrization function is used to transform a nonlinear system into a linear one, making it easier to solve. It does this by introducing new variables and parameters that simplify the equations and eliminate nonlinear terms. This allows for a more efficient and accurate solution to the system.

3. What is the difference between a linear and nonlinear system?

A linear system has equations that can be written in a linear form, while a nonlinear system has equations that cannot be written in a linear form. In other words, the variables in a linear system have an exponent of 1 and are not multiplied together, while the variables in a nonlinear system may have exponents greater than 1 and can be multiplied together.

4. How do you solve a linear system using reparametrization function?

To solve a linear system using reparametrization function, first identify the nonlinear terms in the equations. Then, introduce new variables and parameters to eliminate these nonlinear terms and transform the system into a linear one. Finally, use traditional methods such as substitution or elimination to solve the linear system and find the values of the original variables.

5. Can all nonlinear systems be solved using reparametrization function?

No, not all nonlinear systems can be solved using reparametrization function. Some systems may have nonlinear terms that cannot be eliminated through reparametrization, making it impossible to transform the system into a linear one. In these cases, other methods such as numerical approximations may be used to find a solution.

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