Analyzing solutions of y'= r-ky using W lambert function?

  • Thread starter kochibacha
  • Start date
  • Tags
    Function
In summary: This is an IV infusion problem where you are given the amount of a drug at time t, the infusion rate, and the elimination rate constant. You are also given the initial amount of the drug. You are interested in finding the time to reach steady-state. If y0 = 0, you can find the time to reach steady-state using any knowledge or technology from mathematics. However, if y0 is not zero, you cannot find the time to reach steady-state using any knowledge or technology from mathematics.
  • #1
kochibacha
14
0
consider this IVP
y'=r-ky , y(0)=y0
y= (y0)e^(-kt) + (r/k)(1-e^(-kt))
if y,y0,r,t are provided, we should be able to solve for k and that's the problem but what I'm really interested is analyzing this problem

if we let y=0.99 (r/k) find t in terms of all other variables

Of courses, if y0 = 0 we can see that t= -(ln 0.01)/k

i wonder if y0 is not zero is it possible to analyze variable t using any knowledge or technology from mathematics?

this problems is derived from real application of pharmacokinetics IV infusion where

y= amount of drug at time t
y0 = initial amount of drug at t = 0
r = infusion rate
k = elimination rate constant
(r/k) = amount of drug at steady-state (as t --> infinity the amount of drug will approach this value and 99% of (r/k) is a good approximation of amount of drug at steady-state)

so what I really ask is how the initial amount of drug reflects the time to reach steady-state ( for example. how much we increase y0 in order to halve the time to reach steady-state )
 
Physics news on Phys.org
  • #2
I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
 
  • #3
Let's see what I can whip up...
[itex]y' + ky = r[/itex] is a first-order, linear, non-homogenous ODE with [itex]k \neq 0[/itex] and [itex]r \neq 0[/itex], both of which I assume to be constant.
The general solution is (by partitioning into homogenous and particular solutions) [itex]C_0 e^{-kt} + \frac{r}{k}[/itex]. Given an initial value, [itex]y(0) = y_0[/itex], the unique solution is [itex](y_0 - \frac{r}{k})e^{-kt} + \frac{r}{k}[/itex]. (Same as yours, but rewritten.)

You then ask what I assume to be is the time at which this solution is equal to [itex]0.99\frac{r}{k}[/itex].

[tex](y_0 - \frac{r}{k})e^{-kt} + \frac{r}{k} = 0.99\frac{r}{k}[/tex]
[tex](y_0 - \frac{r}{k})e^{-kt} + 0.01\frac{r}{k} = 0[/tex]
[tex](y_0 - \frac{r}{k})e^{-kt} = -0.01\frac{r}{k}[/tex]
[tex]e^{-kt} = -0.01\frac{\frac{r}{k}}{y_0 - \frac{r}{k}}[/tex]
[tex]e^{-kt} = -0.01\frac{r}{k y_0 - r}[/tex]
Immediately, we see that [itex]\frac{r}{k y_0 - r} < 0[/itex] for there to be a real-valued, finite solution.

It's worth noting that if [itex]k = 0[/itex], we get a linear polynomial as a solution to the ODE, for which there is no steady state unless [itex]r = 0[/itex] as well, in which case it's [itex]y_0[/itex] as the solution is just a constant. If [itex]k \neq 0[/itex] and [itex]r = 0[/itex], the only steady state would be 0, for which your approximation fails to give a finite solution to the problem. If [itex]k y_0 - r = 0[/itex], the solution to the ODE would be [itex]\frac{r}{k}[/itex], which has a steady-state but the approximation fails as well.

In any case, assuming [itex]\frac{r}{k y_0 - r} < 0[/itex], the solution is:
[tex]e^{-kt} = -0.01\frac{r}{k y_0 - r}[/tex]
[tex]-kt = \ln{\left(-0.01\frac{r}{k y_0 - r}\right)}[/tex]
[tex]t = -\frac{\ln{\left(-0.01\frac{r}{k y_0 - r}\right)}}{k} = -\frac{\ln{\left(0.01\frac{r}{r - k y_0}\right)}}{k}[/tex]
 

1. What is the W Lambert function?

The W Lambert function, also known as the Omega function, is a special mathematical function that is defined as the inverse of the function y=xe^x. It is denoted by W(x) and is commonly used in solving equations involving exponential and logarithmic functions.

2. How is the W Lambert function used in analyzing solutions of y'= r-ky?

The W Lambert function is used in analyzing solutions of the differential equation y'= r-ky by providing a closed-form solution. It allows us to express the solution in terms of the Lambert function, which can be easily evaluated using numerical methods.

3. What is the significance of the parameter k in the differential equation y'= r-ky?

The parameter k represents the rate at which the solution y decreases over time. It is often referred to as the decay rate and plays a crucial role in determining the behavior of the solution.

4. Can the W Lambert function be used for all values of r and k?

No, the W Lambert function can only be used for specific values of r and k. In particular, it can only be used when the value of k is greater than or equal to 1/r. If this condition is not met, other methods such as numerical approximations may be necessary to analyze the solution of y'= r-ky.

5. Are there any real-life applications of analyzing solutions of y'= r-ky using the W Lambert function?

Yes, the W Lambert function has various applications in fields such as physics, chemistry, and engineering. It is commonly used in modeling population growth and radioactive decay, as well as in analyzing heat transfer and electrical circuits.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Calculus
Replies
3
Views
944
  • Differential Equations
Replies
1
Views
676
Replies
1
Views
1K
  • MATLAB, Maple, Mathematica, LaTeX
Replies
1
Views
1K
Replies
131
Views
4K
  • Calculus and Beyond Homework Help
Replies
3
Views
429
  • Differential Equations
Replies
5
Views
1K
Replies
26
Views
1K
  • Special and General Relativity
Replies
9
Views
965
Back
Top