# How to model a checking account with continuous-time compounding?

Say you have a bank account in which your invested money yields 3% every year, continuously compounded. Also, you have estimated that you spend $1000 every month to pay your bills, that are withdrawn from this account. Create a differential model for that, find its equilibriums and determine its stability. My problem here is that the \$1000 withdrawal is not continuous on time, it's discrete. The best I could achieve is, if $$S(t)$$ is the current balance: $$\dot S (t) = 0,0025S(t) - 1000$$. I'm using $$0,0025$$ as the interest rate because it yields 3% every year, so it should yield 0,25% every month. But I'm pretty confident that it's wrong. Any help would be highly appreciated! Thanks!

• Perhaps they meant to model the expenditures as continuous as well? If the expenditures are time varying then you do not have enough information to model the problem. May 27 '16 at 19:07
• Yes, I think so, but I don't know how to do that. :/ May 27 '16 at 19:07
• How is money expended? On a daily basis? Once per month? First day? Last day? May 27 '16 at 19:17
• The problem set just says "you every-day purchases sum around $1000 per month". I'm pretty confident that the question wants me to model everything continuously but I'm not sure how to do it. May 27 '16 at 19:24 • It seems to me that if$t$is to be measured in years, then$\dot S(t)=.03S-1000 \lfloor {12t}\rfloor$. May 27 '16 at 20:11 ## 3 Answers Let$x (t)$be the amount of money in the account at time$t$(years). Hence, if no money is spent, $$\dot x = r x$$ where$r = \ln (1.03)$. If$\$1000$ is spent continuously every month, then we have the ODE

$$\dot x = r x - 12000$$

We have an equilibrium point when we have

$$\bar{x} := \frac{12000}{r} \approx \406,000$$

in the account, as the interest earned per year then equals the amount of money expended per year. If we have more than $\bar{x}$ in the bank, then our wealth is growing. If we have less than $\bar{x}$ in the bank, then our wealth is decaying. Let us verify. Integrating the non-homogeneous ODE above, we obtain

$$x (t) = \bar{x} + (x_0 - \bar{x}) \, \mathrm{e}^{r t}$$

If $x_0 > \bar{x}$, our wealth is growing. If $x_0 < \bar{x}$, our wealth is decaying. If $x_0 = \bar{x}$, our wealth is stationary. Note that $\bar{x}$ is an unstable equilibrium point.

• Thanks a lot, got it! In your solution, the time is measured in years. If we represent time in months, the correct interest rate should be $r=\frac{\ln (1.03)}{12}$, right? May 28 '16 at 14:19
• @RaulGuarini Yes. And the ODE becomes $\dot x = r x - 1000$. May 28 '16 at 15:01
• @RaulGuarini Why doesn't $r = .03$ or $\frac{.03}{12}$? Jul 3 '19 at 3:05
• @user10478 If $r = \ln (1.03)$, then $$e^{rt} = \left( \exp (r) \right)^t = \left( \exp(\ln( 1.03 )) \right)^t = (1 + 0.03)^t$$ Jul 6 '19 at 10:35
• @RodrigodeAzevedo Hmm, is this because the problem says "yields $3\%$ every year," whereas $r$ would be $.03$ if it said "annual interest rate of $3\%$"? Jul 6 '19 at 11:10

The ballance must be sufficient to generate $1,000 / month in cash flow$B(e^{0.0025}-1) = 1000\\ B = \dfrac{1000}{e^{0.0025}-1} = \$399,500$

The money flow consists of two contributions $$\dot{S} = \dot{S}_y + \dot{S}_m$$ with the continous contribution $$\dot{S}_y = a S \quad (*)$$ where $a$ must be adjusted to give the yearly interest rate such that $$S_y(1\text{y}) = (1 + p) S_y(0\text{y})$$ for $p = 3\% = 3/100$ and the monthly part $$\dot{S}_m = -M \sum_{k=1}^\infty \delta(t - k\cdot 1\text{m})$$ with $M = 1000 \$$. Determining a from p: Equation (*) means$$ S_y(t) = S_y(0\text{y}) e^{at} $$and$$ S_y(1\text{y}) = S_y(0\text{y}) e^{a \cdot 1 \text{y}} = (1 + p) S_y(0\text{y}) $$so a = \ln(1+p)/1\text{y}. In summary:$$ \dot{S} = S(0\text{y}) \, (1+p)^{t/1\text{y}} - M \sum_{k=1}^\infty \delta(t - k\cdot 1\text{m})$\$