Hi everyone I am curious how you would apply newtons method
Hi everyone,
I am curious how you would apply newton\'s method to solve this differential equation after you have a backwards euler approximation. If you could detail your steps that would be very helpful.
A chemical compound decays over time when exposed to air, at a rate proportional to its concentration to the power of 3/2. At the same time, the compound is produced by another process. The differential equation for its instantaneous concentration is
dn(t) dt = 0.8n 3/2 + 10n1[1 exp(3t)],
where n(t) is the instantaneous concentration, n1 = 2000 is the initial concentration at t = 0. Solve the differential equation to find the concentration as a function of time from t = 0 until t = 0.5 s, using the Backward Euler method. Use a step size of h = 0.002 s and plot n(t) versus t. First clearly write down your finite difference approximation. Note that, you will obtain an ‘implicit’ algebraic equation which you will have to solve iteratively at each time step. Use the Newton’s method for iterative solution. Provide your computer program and clearly indicate your algorithm with sufficient comments on the program. How will you verify that your predicted answer is a reasonable estimate using this particular numerical scheme? Accordingly, verify your answer and show proof.
Solution
n(0.05) = 580.784
| h | t | n(t) | n\'(t) |
| 0.002 | 0 | 2000 | -71554.17528 |
| 0.002 | 0.002 | 1856.891649 | -63893.60258 |
| 0.002 | 0.004 | 1729.104444 | -57281.825 |
| 0.002 | 0.006 | 1614.540794 | -51542.76214 |
| 0.002 | 0.008 | 1511.45527 | -46534.92235 |
| 0.002 | 0.01 | 1418.385425 | -42143.68127 |
| 0.002 | 0.012 | 1334.098063 | -38275.47908 |
| 0.002 | 0.014 | 1257.547105 | -34853.41403 |
| 0.002 | 0.016 | 1187.840277 | -31813.8637 |
| 0.002 | 0.018 | 1124.212549 | -29103.87109 |
| 0.002 | 0.02 | 1066.004807 | -26679.10516 |
| 0.002 | 0.022 | 1012.646597 | -24502.25727 |
| 0.002 | 0.024 | 963.6420821 | -22541.77089 |
| 0.002 | 0.026 | 918.5585403 | -20770.82843 |
| 0.002 | 0.028 | 877.0168834 | -19166.53786 |
| 0.002 | 0.03 | 838.6838077 | -17709.27574 |
| 0.002 | 0.032 | 803.2652562 | -16382.1534 |
| 0.002 | 0.034 | 770.5009494 | -15170.58098 |
| 0.002 | 0.036 | 740.1597875 | -14061.90932 |
| 0.002 | 0.038 | 712.0359689 | -13045.13456 |
| 0.002 | 0.04 | 685.9456997 | -12110.65313 |
| 0.002 | 0.042 | 661.7243935 | -11250.05781 |
| 0.002 | 0.044 | 639.2242779 | -10455.96707 |
| 0.002 | 0.046 | 618.3123437 | -9721.881804 |
| 0.002 | 0.048 | 598.8685801 | -9042.064593 |
| 0.002 | 0.05 | 580.7844509 |
