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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.

**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.

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