Math Insight

Forward Euler and linear approximations

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  1. For the following differential equation \begin{align*} \diff{ v }{ t } &= - 1.3 v - 4.3\\ v(0) & = -8 \end{align*} calculate a Forward Euler approximation to $v(1.5)$ using a time step of $\Delta t = 0.5$. In this problem, we guide your through the steps of the calculation. It may easier to understand the steps by writing the differential equation using an explicit argument of $t$ \begin{align*} \diff{ v }{ t }(t) &= - 1.3 v{\left (t \right )} - 4.3. \end{align*}

    1. The following steps will guide you in calculating the Forward Euler approximation after one step of size $\Delta t = 0.5$, yielding an estimate of $v(0.5)$.

      What is the initial condition? $v(0) = $

      Based on that result and the differential equation, what is the slope of $v(t)$ at time $t=0$?
      $\diff{ v }{t}(0) = - 1.3 v{\left (0 \right )} - 4.3 =$

      Those two numbers are enough information to make a linear approximation at $t=0$.
      $L(t) = v(0) + \diff{ v }{t}(0) (t-0) =$

      We use this linear approximation to approximate the value of $v(t)$ at $t=\Delta t = 0.5$.
      $v(0.5) \approx L(0.5) =$

    2. The following steps will guide you in calculating the Forward Euler approximation after the second step of size $\Delta t = 0.5$, yielding an estimate of $v(1.0)$.

      Given the value of $v(0.5)$ estimated above and the differential equation, what is the slope of $v(t)$ at time $t=\Delta t = 0.5$?
      $\diff{ v }{t}(0.5) = - 1.3 v{\left (0.5 \right )} - 4.3 =$

      Those two numbers are enough information to make a linear approximation at $t=\Delta t = 0.5$.
      $L(t) = v(0.5) + \diff{ v }{t}(0.5) (t-0.5) =$

      We use this linear approximation to approximate the value of $v(t)$ at $t=2\Delta t = 1.0$.
      $v(1.0) \approx L(1.0) =$

    3. The following steps will guide you in calculating the Forward Euler approximation after the third step of size $\Delta t = 0.5$, yielding an estimate of $v(1.5)$.

      Given the value of $v(1.0)$ estimated above and the differential equation, what is the slope of $v(t)$ at time $t=2\Delta t = 1.0$?
      $\diff{ v }{t}(1.0) = - 1.3 v{\left (1.0 \right )} - 4.3 =$

      Those two numbers are enough information to make a linear approximation at $t=2\Delta t = 1.0$.
      $L(t) = v(1.0) + \diff{ v }{t}(1.0) (t-1.0) =$

      We use this linear approximation to approximate the value of $v(t)$ at $t=3\Delta t = 1.5$.
      $v(1.5) \approx L(1.5) =$

  2. For the following logistic differential equation \begin{align*} \diff{ u }{ t } &= 0.3 u \left(- \frac{u}{790} + 1\right)\\ u(0) & = 1106 \end{align*} calculate a Forward Euler approximation to $u(9.9)$ using a time step of $\Delta t = 3.3$. In this problem, we guide your through the steps of the calculation. It may easier to understand the steps by writing the differential equation using an explicit argument of $t$ \begin{align*} \diff{ u }{ t }(t) &= 0.3 \left(- \frac{1}{790} u{\left (t \right )} + 1\right) u{\left (t \right )}. \end{align*}

    1. The following steps will guide you in calculating the Forward Euler approximation after one step of size $\Delta t = 3.3$, yielding an estimate of $u(3.3)$.

      What is the initial condition? $u(0) = $

      Based on that result and the differential equation, what is the slope of $u(t)$ at time $t=0$?
      $\diff{ u }{t}(0) = 0.3 \left(- \frac{1}{790} u{\left (0 \right )} + 1\right) u{\left (0 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=0$.
      $L(t) = u(0) + \diff{ u }{t}(0) (t-0) =$

      We use this linear approximation to approximate the value of $u(t)$ at $t=\Delta t = 3.3$.
      $u(3.3) \approx L(3.3) =$

    2. The following steps will guide you in calculating the Forward Euler approximation after the second step of size $\Delta t = 3.3$, yielding an estimate of $u(6.6)$.

      Given the value of $u(3.3)$ estimated above and the differential equation, what is the slope of $u(t)$ at time $t=\Delta t = 3.3$?
      $\diff{ u }{t}(3.3) = 0.3 \left(- \frac{1}{790} u{\left (3.3 \right )} + 1\right) u{\left (3.3 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=\Delta t = 3.3$.
      $L(t) = u(3.3) + \diff{ u }{t}(3.3) (t-3.3) =$

      We use this linear approximation to approximate the value of $u(t)$ at $t=2\Delta t = 6.6$.
      $u(6.6) \approx L(6.6) =$

    3. The following steps will guide you in calculating the Forward Euler approximation after the third step of size $\Delta t = 3.3$, yielding an estimate of $u(9.9)$.

      Given the value of $u(6.6)$ estimated above and the differential equation, what is the slope of $u(t)$ at time $t=2\Delta t = 6.6$?
      $\diff{ u }{t}(6.6) = 0.3 \left(- \frac{1}{790} u{\left (6.6 \right )} + 1\right) u{\left (6.6 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=2\Delta t = 6.6$.
      $L(t) = u(6.6) + \diff{ u }{t}(6.6) (t-6.6) =$

      We use this linear approximation to approximate the value of $u(t)$ at $t=3\Delta t = 9.9$.
      $u(9.9) \approx L(9.9) =$

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Math 1241, Fall 2020