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# Introduction to Numerical Methods for Differential Equations

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### Introduction to Numerical Methods for Differential Equations

1. 1. Introduction to Numerical Methods for Diﬀerential Equations The Euler Method Matthew Henderson matthew.james.henderson@gmail.com 21 December, 2011
2. 2. Overview1 Initial Value Problems2 Approximate Solutions of Initial Value Problems3 The Euler Method Matthew Henderson () Numerical Methods 21 December, 2011 2 / 15
3. 3. An Initial Value ProblemDeﬁnitionAn initial value problem (IVP) consists of an ordinary diﬀerentialequation along with an initial condition. Matthew Henderson () Numerical Methods 21 December, 2011 3 / 15
4. 4. An Initial Value ProblemDeﬁnitionAn initial value problem (IVP) consists of an ordinary diﬀerentialequation along with an initial condition.Example y = 2(x + 3) − y, y(−1) = 3 (1) Matthew Henderson () Numerical Methods 21 December, 2011 3 / 15
5. 5. An Initial Value ProblemDeﬁnitionAn initial value problem (IVP) consists of an ordinary diﬀerentialequation along with an initial condition.Example y = 2(x + 3) − y, y(−1) = 3 (1)DeﬁnitionA solution is a function y(x) which satisﬁes both the ODE and theinitial condition. Matthew Henderson () Numerical Methods 21 December, 2011 3 / 15
6. 6. The SolutionThe IVP (1) has a unique solution: y = 2(x + 2) + e−x−1 (2)Function y satisﬁes the ODE: y = 2 − e−x−1 2(x + 3) − y = 2(x + 3) − 2(x + 2) + e−x−1 = 2x + 6 − 2x − 4 − e−x−1 = 2 − e−x−1 Matthew Henderson () Numerical Methods 21 December, 2011 4 / 15
7. 7. The SolutionThe IVP (1) has a unique solution: y = 2(x + 2) + e−x−1 (2)Function y also satisﬁes the initial condition: y(−1) = 2(−1 + 2) + e−(−1)−1 = 2(1) + e1−1 = 2 + e0 = 3 Matthew Henderson () Numerical Methods 21 December, 2011 4 / 15
8. 8. A Plot of the Solution y 10 y =2(x +2) + e−x−1 8 6 4 2 x -3 -2 -1 1 2 3 Matthew Henderson () Numerical Methods 21 December, 2011 5 / 15
9. 9. A Plot of the Solution? y 10 8 6 4 2 x -3 -2 -1 1 2 3 Matthew Henderson () Numerical Methods 21 December, 2011 6 / 15
10. 10. The Euler MethodDeﬁnitionGiven an IVP of the form: y = f(x, y), y(a) = c Matthew Henderson () Numerical Methods 21 December, 2011 7 / 15
11. 11. The Euler MethodDeﬁnitionGiven an IVP of the form: y = f(x, y), y(a) = cTo ﬁnd the approximate value of y(x + h) for some small value of h: y(x + h) = y(x) + hf(x, y) (3) Matthew Henderson () Numerical Methods 21 December, 2011 7 / 15
12. 12. The Euler MethodExampleGoing back to our example from before: y = 2(x + 3) − y, y(−1) = 3 Matthew Henderson () Numerical Methods 21 December, 2011 8 / 15
13. 13. The Euler MethodExampleGoing back to our example from before: y = 2(x + 3) − y, y(−1) = 3In this case f(x, y) = 2(x + 3) − y. Matthew Henderson () Numerical Methods 21 December, 2011 8 / 15
14. 14. The Euler MethodExampleGoing back to our example from before: y = 2(x + 3) − y, y(−1) = 3In this case f(x, y) = 2(x + 3) − y.Put h = 0.4, x = −1.0 and y = 3.0 into (3): Matthew Henderson () Numerical Methods 21 December, 2011 8 / 15
15. 15. The Euler MethodExampleGoing back to our example from before: y = 2(x + 3) − y, y(−1) = 3In this case f(x, y) = 2(x + 3) − y.Put h = 0.4, x = −1.0 and y = 3.0 into (3): y(−1.0 + 0.4) = y(−1.0) + 0.4f(−1.0, 3.0) Matthew Henderson () Numerical Methods 21 December, 2011 8 / 15
16. 16. The Euler MethodExampleGoing back to our example from before: y = 2(x + 3) − y, y(−1) = 3In this case f(x, y) = 2(x + 3) − y.Put h = 0.4, x = −1.0 and y = 3.0 into (3): y(−1.0 + 0.4) = y(−1.0) + 0.4f(−1.0, 3.0) y(−0.6) = 3.0 + 0.4(2(−1.0 + 3.0) − 3.0) Matthew Henderson () Numerical Methods 21 December, 2011 8 / 15
17. 17. The Euler MethodExampleGoing back to our example from before: y = 2(x + 3) − y, y(−1) = 3In this case f(x, y) = 2(x + 3) − y.Put h = 0.4, x = −1.0 and y = 3.0 into (3): y(−1.0 + 0.4) = y(−1.0) + 0.4f(−1.0, 3.0) y(−0.6) = 3.0 + 0.4(2(−1.0 + 3.0) − 3.0) = 3.0 + 0.4(4 − 3.0) Matthew Henderson () Numerical Methods 21 December, 2011 8 / 15
18. 18. The Euler MethodExampleGoing back to our example from before: y = 2(x + 3) − y, y(−1) = 3In this case f(x, y) = 2(x + 3) − y.Put h = 0.4, x = −1.0 and y = 3.0 into (3): y(−1.0 + 0.4) = y(−1.0) + 0.4f(−1.0, 3.0) y(−0.6) = 3.0 + 0.4(2(−1.0 + 3.0) − 3.0) = 3.0 + 0.4(4 − 3.0) = 3.4 Matthew Henderson () Numerical Methods 21 December, 2011 8 / 15
19. 19. The Euler MethodWe can compute more values in the same way . . .x = −0.6 and y = 3.4: y(−0.6 + 0.4) = 3.4 + 0.4(2(−0.6 + 3) − 3.4) y(−0.2) = 3.4 + 0.4(4.8 − 3.4) = 3.96 Matthew Henderson () Numerical Methods 21 December, 2011 9 / 15
20. 20. The Euler MethodWe can compute more values in the same way . . .x = −0.6 and y = 3.4: y(−0.6 + 0.4) = 3.4 + 0.4(2(−0.6 + 3) − 3.4) y(−0.2) = 3.4 + 0.4(4.8 − 3.4) = 3.96 x = −0.2 and y = 3.96: y(−0.2 + 0.4) = 3.96 + 0.4(2(−0.2 + 3) − 3.96) y(0.2) = bluey + 0.4(5.6 − 3.96) = 4.616 Matthew Henderson () Numerical Methods 21 December, 2011 9 / 15
21. 21. The Euler MethodWe continue to construct a table row-by-row: x y hf(x, y) -1.0 3.000000 0.400000 -0.6 0.560000 -0.2 3.960000 0.656000 0.2 4.616000 0.713600 0.6 5.329600 0.748160 1.0 6.077760 0.768896 1.4 6.846656 0.781338 1.8 7.627994 0.788803 2.2 8.416796 0.793281 2.6 9.210078 0.795969 3.0 10.00605 0.797581 Matthew Henderson () Numerical Methods 21 December, 2011 10 / 15
22. 22. The Euler MethodWe continue to construct a table row-by-row: x y hf(x, y) -1.0 3.000000 0.400000 -0.6 3.400000 0.560000 -0.2 3.960000 0.656000 0.2 4.616000 0.713600 0.6 5.329600 0.748160 1.0 6.077760 0.768896 1.4 6.846656 0.781338 1.8 7.627994 0.788803 2.2 8.416796 0.793281 2.6 9.210078 0.795969 3.0 10.00605 0.797581 Matthew Henderson () Numerical Methods 21 December, 2011 10 / 15
23. 23. The Euler Method: ComparisonNow when we plot those values we get something that begins to looklike our exact solution, albeit only on the interval [−1, 3]. y 10 y =2(x +2) + e−x−1 8 6 4 2 x -3 -2 -1 1 2 3 Matthew Henderson () Numerical Methods 21 December, 2011 11 / 15
24. 24. The Euler Method: ComparisonThe numbers which we computed for y are not exactly correct but theerrors involved are quite small: x y (approx) y (exact) Error -1.0 3.000000 3.000000 0.000000 -0.6 3.400000 3.470320 0.070320 -0.2 3.960000 4.049329 0.089329 0.2 4.616000 4.701194 0.085194 0.6 5.329600 5.401896 0.072296 1.0 6.077760 6.135335 0.057575 1.4 6.846656 6.890718 0.044062 1.8 7.627994 7.660810 0.032816 2.2 8.416796 8.440762 0.023966 2.6 9.210078 9.227324 0.017246 3.0 10.00605 10.018316 0.012269 Matthew Henderson () Numerical Methods 21 December, 2011 12 / 15
25. 25. The Euler Method: Where does it come from?Given an initial value problem of the form: y = f(x, y), y(a) = cwe want to ﬁnd the approximate value of y(b) for some b > a. Fromthe deﬁnition of derivative: y(x + h) − y(x) y (x) ≈ hfor h > 0 given and small. So, y(x + h) − y(x) f(x, y) ≈ hwhich gives y(x + h) ≈ y(x) + hf(x, y) Matthew Henderson () Numerical Methods 21 December, 2011 13 / 15
26. 26. The Euler Method: Why does it work? y =2(x +2) + e−x−1 10 8 6 4 2 -3 -2 -1 1 2 3 Matthew Henderson () Numerical Methods 21 December, 2011 14 / 15
27. 27. The EndMerry Christmas! Matthew Henderson () Numerical Methods 21 December, 2011 15 / 15