Solving a DE for modelling the spread of an infectious disease

[greswd]

This DE is related to a mathematical model of the spread of an infectious disease: $$f'\left(x\right)=a\cdot\left[f\left(x\right)-f\left(x-b\right)\right]$$

where a and b are positive constants.

I would like some pointers as to how I should begin attempting to solve it.

 

[fresh_42]

Most disturbing is the difference between the function values on the RHS. I would therefore search for a logarithmic function $f$.

 

[Infrared]

Just some thoughts: The set of solutions is a vector space containing the constant functions. Also if $ab=1$, then degree $1$ polynomials $y=px+q$ are solutions. [Edited to avoid re-using the letters $a$ and $b$]

You can also try an exponential solution $y=e^{\lambda x}$. Substituting this,

$$\lambda e^{\lambda x}=a(e^{\lambda x}-e^{\lambda (x-b)})=ae^{\lambda x}(1-e^{-\lambda b}),$$

so we want to find $\lambda$ satisfying the equation $\lambda=a(1-e^{-\lambda b}).$ Now $\lambda=0$ is always a solution (giving the constant solutions), but others will usually exist.

In fact, as long as the graphs of $\lambda$ and $a(1-e^{-\lambda b})$ are not tangent at $\lambda=0$, there will be another solution by IVT arguments. In order for them to be tangent, we need $\frac{d}{d\lambda}\big\vert_{\lambda=0}\lambda=1$ to be equal to $\frac{d}{d\lambda}\big\vert_{\lambda=0}a(1-e^{-\lambda b})=ab$, that is $ab=1$.

So, we have linear solutions when $ab=1$, and exponential solutions when $ab\neq 1$.

I've assumed $a$ and $b$ are positive, please say if this is not justified in your model.
I also haven't though about uniqueness; I don't know if there are other solutions.

 

[S.G. Janssens]

This DE is related to a mathematical model of the spread of an infectious disease: $$f'\left(x\right)=a\cdot\left[f\left(x\right)-f\left(x-b\right)\right]$$

where a and b are positive constants.

I would like some pointers as to how I should begin attempting to solve it.

This type of equation is known as a "delay differential equation" (DDE) or a "(retarded) functional differential equation" (RFDE). It occurs frequently in mathematical epidemiology, among other fields, and it has a well-established theory. Here are some pointers.

1. The initial-value problem requires the prescription of a function segment (called a "history") on the interval $[-b,0]$, so it is intrinsically infinite dimensional. Time is usually scaled such that $b = 1$ can be chosen, and as the state space one can then work with $C[-1,0]$ (with the maximum-norm), but other choices such as $L^p(-1,0)$ are also possible.

2. The characteristic equation in post #3 is analyzed in detail in Chapter XI of this text using elementary complex analysis, also see the references there to earlier literature, as well as the book by Bellman and Cooke. In general, the stability problem for linear DDEs leads to the analysis of an exponential polynomial.

3. The question of completeness of the set of exponential solutions is not trivial, precisely due to the infinite-dimensional nature of DDEs, in contrast with finite-dimensional ODEs. If you are interested in this aspect, you could consult Chapter V of the first text mentioned above.

Depending on how serious you are about pursuing this, feel free to ask me further questions.

 

[greswd]

Just some thoughts: The set of solutions is a vector space containing the constant functions. Also if $ab=1$, then degree $1$ polynomials $y=px+q$ are solutions. [Edited to avoid re-using the letters $a$ and $b$]

You can also try an exponential solution $y=e^{\lambda x}$. Substituting this,

$$\lambda e^{\lambda x}=a(e^{\lambda x}-e^{\lambda (x-b)})=ae^{\lambda x}(1-e^{-\lambda b}),$$

so we want to find $\lambda$ satisfying the equation $\lambda=a(1-e^{-\lambda b}).$ Now $\lambda=0$ is always a solution (giving the constant solutions), but others will usually exist.

In fact, as long as the graphs of $\lambda$ and $a(1-e^{-\lambda b})$ are not tangent at $\lambda=0$, there will be another solution by IVT arguments. In order for them to be tangent, we need $\frac{d}{d\lambda}\big\vert_{\lambda=0}\lambda=1$ to be equal to $\frac{d}{d\lambda}\big\vert_{\lambda=0}a(1-e^{-\lambda b})=ab$, that is $ab=1$.

So, we have linear solutions when $ab=1$, and exponential solutions when $ab\neq 1$.

I've assumed $a$ and $b$ are positive, please say if this is not justified in your model.
I also haven't though about uniqueness; I don't know if there are other solutions.

thank you, yes, $a$ and $b$ are positive, and your model has re-derived all of the infection mechanisms.

what do you think the general solution is, something that accommodates both the linear and exponential solutions?

 

[Chestermiller]

If you are going to model this even crudely, there is another key factor that you have so far neglected to include in your equations. The is the effect of interactions between infected contagious individuals and individuals that have not yet been infected. It is the product of the population density of contagious individuals with the population density of uninfected individuals (i.e., characterizing the interactions between these populations) that determines the rate of new infections. More importantly still, and not even suggested by the media, is that it is the depletion of the uninfected population density that is responsible for the characteristic of reaching a peak in the infection rate and then declining. This is a a feature of all the models currently in use. If you don't include this in your model, you will not be able to model the peak.

Current estimates are that about 30% of the New York State population has been infected, most undetected and unconfirmed. And about 10% of the US population has already been infected. This is why we are seeing peaks in the curves, even with the overall lower infection rates from social distancing.

Once you have included this missing factor in your model (as I have outlined in several of my posts in the other thread), your equations will no longer be linear, and they will not be analytically solvable. You will have to resort to numerical integration to solve the model. In the solutions I have presented in earlier posts in the other thread), I carried out the integration using Forward Euler on an Excel spreadsheet.