# Nonlinear DE and Numerical System

I'm trying to investigate nonlinear system numerical methods. For the nonlinear DE x' = 2t(1+x^2). Use the value tan(1) = 1.557407724654....

a) how to find the explicit solution $x(t)$ satisfying $x(0) = 1$?

b) how to use Euler's method to approx the value of $x(1) = e$ using $\Delta t = 0.1$. I.e., recursively determine $t_k$ and $x_k$ for $k = 1,...10$ with $\Delta t = 0$, starting with $t_0 = 0$ and $x_0 = 1$.

c) Repeat using $\Delta t = 0.05$

d) Again using Euler's method but reduce step size by a factor of 5, so that delta $t = 0.01$ to approx x(1)

e) Repeat parts b, c, and d with Improved euler's method using the same step sizes

f) Repeat using Runge-Kutta

g) Calculate the error in each case, since we now have 9 different approx for the value of $x(1) = e$, three for each method.

h) Calculate how the error changes as we change the step size from 0.01 to 0.05 and then from 0.05 to 0.01

Can someone also write in the format that implicit EUler and RK 4 takes in this example?

• This exact set of questions was done with a different DEQ. Are you saying that you couldn't do a single part of this variant with that answer? Regards Apr 21, 2013 at 5:45
• @Amzoti, I'm stuck on how to get a computer calculator software to generate the values for delta t = 0.1, 0.05, and 0.01. Can you just help out with Euler's for part b,c,d and paste the output here?
– mary
Apr 21, 2013 at 19:34
• Okay, I added an example of the improved Euler. The rest is up to you! Regards Apr 23, 2013 at 0:44
• @Amzoti, thank you . However, when I use the software for Implicit Euler with h = 0.01, there seems to be an error. It works well with step sizes of h =0.05, and 0.10 but not 0.01, can you please explain why? If you get an answer, can you tell me what software you use? Thanks
– mary
Apr 23, 2013 at 1:51
• You can try this solver and see what is happening near 1.0 from the graphic I included. I suspect that your method is getting to those values and not handling them well. See: and try your problem (you have to change variable names, but it has Euler and Improved), see math-cs.gordon.edu/~senning/desolver and look at some of the values near 1.0 and above. Maybe your SW is blowing up because of this. Regards Apr 23, 2013 at 2:00

a. We are given the DEQ: $f(t, x) = x' = 2 t (1 + x^2), x(0) = \alpha = 1$.

We find the closed form solution as $\displaystyle x(t) = \tan\left(t^2 + \frac{\pi}{4}\right)$

b. To set up the Euler iteration, we have:

$h = \frac{b - a}{N} = 0.1$

$t_0 = a = 0$

$w_0 = \alpha = 1$

For $i = 1, 2, \ldots, N$:

$$w_i = w_{i-1} + h f(t, w) = w_{i-1} + 0.1\left(2~ t_{i-1} (1 + w_{i-1}^2)\right)$$

Generating the iterates yields:

$t_i ~~~~|~~ x_i$

$0.0 ~~|~~ 1.$

$0.1 ~~|~~ 1.$

$0.2 ~~|~~ 1.04$

$0.3 ~~|~~ 1.12326$

$0.4 ~~|~~ 1.25897$

$0.5 ~~|~~ 1.46577$

$0.6 ~~|~~ 1.78061$

$0.7 ~~|~~ 2.28109$

$0.8 ~~|~~ 3.14955$

$0.9 ~~|~~ 4.8967$

$1.0 ~~|~~ 9.39269$

Part c is just changing the step size for $t$.

Part d is just changing the step size for $t$.

Part e:

We have:

$x'(t) = f(t, x) = 2 t (x(t)^2+1), x(0) = 1$

$k_1 = h f(t_n, x_n)$

$k_2 = h f(t_n + (2 h)/3, x_n + (2 k_1)/3)$

$x_{n + 1} = x_n + k_1/4+(3 k_2)/4$

The iterates from this are:

$t_i ~~~~~|~~ x_i$

$0.0 ~~|~~ 1.0$

$0.1 ~~|~~ 1.02$

$0.2 ~~|~~ 1.08262$

$0.3 ~~|~~ 1.19637$

$0.4 ~~|~~ 1.37988$

$0.5 ~~|~~ 1.67285$

$0.6 ~~|~~ 2.16814$

$0.7 ~~|~~ 3.12781$

$0.8 ~~|~~ 5.5858$

$0.9 ~~|~~ 17.5821$

$1.0 ~~|~~ 467.086$

If you look at these results, things are looking shaky near the end. That is the point of this exercise.

If we look at a figure of actual versus numerical, we have: • +1 Nice step-by-step modeling...parts c and d should be doable! ;-) Apr 22, 2013 at 0:05
• @amWhy: I hope she gets it!!! Thx Apr 22, 2013 at 0:06
• @Amzoti, I got that x1 = 1.1, x2 = 1.241, and x3 = 1.4350081 with this form of Euler's method: xn = xn-1 + 0.2tn-1 + 0.1xn-1^2, which matches your format. However, I got a different answer
– mary
Apr 22, 2013 at 1:38
• Could you also provide the values for Implicit Euler and RK4 so I can check over my work on it? A setup/some values would be nice. Thanks
– mary
Apr 22, 2013 at 1:39
• @mary: You wrote xn = xn-1 + 0.2tn-1 + 0.1xn-1^2, but I think you left out a product: $x_n = x_{n-1} + 0.1(2 t_{n-1}(1 + x_{n-1}^2))$. See the issue with your result? Apr 22, 2013 at 1:41

a)$$\frac{dx}{dt}=2t(1+x^2)\Rightarrow\frac{dx}{1+x^2}=2\,t\,dt$$ $$\int\frac{dx}{1+x^2}=\int2\,t\,dt\Rightarrow\arctan(x)=t^2+C\Rightarrow x=\tan(t^2+C)$$ by initial condition $x(0)=1\Rightarrow C=\frac\pi4$