Math Insight

Forward Euler and linear approximations

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  1. For the following differential equation \begin{align*} \diff{ u }{ t } &= 0.9 u - 2.4\\ u(0) & = -1.9 \end{align*} calculate a Forward Euler approximation to $u(2.4)$ using a time step of $\Delta t = 0.8$. 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) &= 0.9 u{\left (t \right )} - 2.4. \end{align*}

    1. The following steps will guide you in calculating the Forward Euler approximation after one step of size $\Delta t = 0.8$, yielding an estimate of $u(0.8)$.

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

      Based on that result and the differential equation, what is the slope of $u(t)$ at time $t=0$?
      $\diff{ u }{t}(0) = 0.9 u{\left (0 \right )} - 2.4 =$

      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 $t=\Delta t = 0.8$.
      $u(0.8) \approx L(0.8) =$

    2. The following steps will guide you in calculating the Forward Euler approximation after the second step of size $\Delta t = 0.8$, yielding an estimate of $u(1.6)$.

      Given the value of $u(0.8)$ estimated above and the differential equation, what is the slope of $u(t)$ at time $t=\Delta t = 0.8$?
      $\diff{ u }{t}(0.8) = 0.9 u{\left (0.8 \right )} - 2.4 =$

      Those two numbers are enough information to make a linear approximation at $t=\Delta t = 0.8$.
      $L(t) = u(0.8) + \diff{ u }{t}(0.8) (t-0.8) =$

      We use this linear approximation to approximate the value of $u(t)$ at $t=2\Delta t = 1.6$.
      $u(1.6) \approx L(1.6) =$

    3. The following steps will guide you in calculating the Forward Euler approximation after the third step of size $\Delta t = 0.8$, yielding an estimate of $u(2.4)$.

      Given the value of $u(1.6)$ estimated above and the differential equation, what is the slope of $u(t)$ at time $t=2\Delta t = 1.6$?
      $\diff{ u }{t}(1.6) = 0.9 u{\left (1.6 \right )} - 2.4 =$

      Those two numbers are enough information to make a linear approximation at $t=2\Delta t = 1.6$.
      $L(t) = u(1.6) + \diff{ u }{t}(1.6) (t-1.6) =$

      We use this linear approximation to approximate the value of $u(t)$ at $t=3\Delta t = 2.4$.
      $u(2.4) \approx L(2.4) =$

  2. For the following logistic differential equation \begin{align*} \diff{ z }{ t } &= 0.8 z \left(- \frac{z}{130} + 1\right)\\ z(0) & = 104 \end{align*} calculate a Forward Euler approximation to $z(3.9)$ using a time step of $\Delta t = 1.3$. 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{ z }{ t }(t) &= 0.8 \left(- \frac{1}{130} z{\left (t \right )} + 1\right) z{\left (t \right )}. \end{align*}

    1. The following steps will guide you in calculating the Forward Euler approximation after one step of size $\Delta t = 1.3$, yielding an estimate of $z(1.3)$.

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

      Based on that result and the differential equation, what is the slope of $z(t)$ at time $t=0$?
      $\diff{ z }{t}(0) = 0.8 \left(- \frac{1}{130} z{\left (0 \right )} + 1\right) z{\left (0 \right )} =$

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

      We use this linear approximation to approximate the value of $z(t)$ at $t=\Delta t = 1.3$.
      $z(1.3) \approx L(1.3) =$

    2. The following steps will guide you in calculating the Forward Euler approximation after the second step of size $\Delta t = 1.3$, yielding an estimate of $z(2.6)$.

      Given the value of $z(1.3)$ estimated above and the differential equation, what is the slope of $z(t)$ at time $t=\Delta t = 1.3$?
      $\diff{ z }{t}(1.3) = 0.8 \left(- \frac{1}{130} z{\left (1.3 \right )} + 1\right) z{\left (1.3 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=\Delta t = 1.3$.
      $L(t) = z(1.3) + \diff{ z }{t}(1.3) (t-1.3) =$

      We use this linear approximation to approximate the value of $z(t)$ at $t=2\Delta t = 2.6$.
      $z(2.6) \approx L(2.6) =$

    3. The following steps will guide you in calculating the Forward Euler approximation after the third step of size $\Delta t = 1.3$, yielding an estimate of $z(3.9)$.

      Given the value of $z(2.6)$ estimated above and the differential equation, what is the slope of $z(t)$ at time $t=2\Delta t = 2.6$?
      $\diff{ z }{t}(2.6) = 0.8 \left(- \frac{1}{130} z{\left (2.6 \right )} + 1\right) z{\left (2.6 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=2\Delta t = 2.6$.
      $L(t) = z(2.6) + \diff{ z }{t}(2.6) (t-2.6) =$

      We use this linear approximation to approximate the value of $z(t)$ at $t=3\Delta t = 3.9$.
      $z(3.9) \approx L(3.9) =$

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Math 1241, Fall 2020