62 lines
2.8 KiB
Plaintext
62 lines
2.8 KiB
Plaintext
Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is an explicit method for solving initial value problems (IVPs), as described in [[wp:Euler method|the wikipedia page]].
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The ODE has to be provided in the following form:
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::: <big><math>\frac{dy(t)}{dt} = f(t,y(t))</math></big>
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with an initial value
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::: <big><math>y(t_0) = y_0</math></big>
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To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:
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::: <big><math>\frac{dy(t)}{dt} \approx \frac{y(t+h)-y(t)}{h}</math></big>
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then solve for <math>y(t+h)</math>:
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::: <big><math>y(t+h) \approx y(t) + h \, \frac{dy(t)}{dt}</math></big>
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which is the same as
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::: <big><math>y(t+h) \approx y(t) + h \, f(t,y(t))</math></big>
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The iterative solution rule is then:
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::: <big><math>y_{n+1} = y_n + h \, f(t_n, y_n)</math></big>
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where <big><math>h</math></big> is the step size, the most relevant parameter for accuracy of the solution. A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.
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'''Example: Newton's Cooling Law'''
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Newton's cooling law describes how an object of initial temperature <big><math>T(t_0) = T_0</math></big> cools down in an environment of temperature <big><math>T_R</math></big>:
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::: <big><math>\frac{dT(t)}{dt} = -k \, \Delta T</math></big>
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or
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::: <big><math>\frac{dT(t)}{dt} = -k \, (T(t) - T_R)</math></big>
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<br>
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It says that the cooling rate <big><math>\frac{dT(t)}{dt}</math></big> of the object is proportional to the current temperature difference <big><math>\Delta T = (T(t) - T_R)</math></big> to the surrounding environment.
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The analytical solution, which we will compare to the numerical approximation, is
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::: <big><math>T(t) = T_R + (T_0 - T_R) \; e^{-k t}</math></big>
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;Task:
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Implement a routine of Euler's method and then to use it to solve the given example of Newton's cooling law with it for three different step sizes of:
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:::* 2 s
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:::* 5 s and
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:::* 10 s
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and to compare with the analytical solution.
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;Initial values:
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:::* initial temperature <big><math>T_0</math></big> shall be 100 °C
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:::* room temperature <big><math>T_R</math></big> shall be 20 °C
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:::* cooling constant <big><math>k</math></big> shall be 0.07
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:::* time interval to calculate shall be from 0 s ──► 100 s
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<br>
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A reference solution ([[#Common Lisp|Common Lisp]]) can be seen below. We see that bigger step sizes lead to reduced approximation accuracy.
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[[Image:Euler_Method_Newton_Cooling.png|center|750px]]
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