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

Question

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 Also, you have estimated that you spend $1000 every month to pay your bills, that are withdrawn from this account.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!

Answer 1

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.

Answer 2

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: $$ \dot{S} = S(0\text{y}) \, (1+p)^{t/1\text{y}} - M \sum_{k=1}^\infty \delta(t - k\cdot 1\text{m}) $$

Answer 3

The ballance must be sufficient to generate $1,000 / month in cash flow

$B(e^{0.0025}-1) = 1000\\ B = \dfrac{1000}{e^{0.0025}-1} = \$399,500$