MHB What issues can arise when solving ODEs numerically?

[Theia]

Hi
I am trying to understand numerical analysis on my freetime and today I studyed how to solve $$y' = \frac{x^2}{1 + y\sin (y^2)}$$, with initial value $$y(0) = 0$$. I asked myself two simple questions: What is $$y(1.5)$$ and what is $$y(2.5)$$?

As for to check the answers, I solved the ODE. In implicit form one can write $$2x^3 - 3 = 6y - 3\cos (y^2)$$. Now, when $$x = 1.5$$, one can solve the $$y \approx 0.941265$$ with no problems. However, when $$x = 2.5$$, there are three solutions for $$y$$, namely $$4.553574\ ;\ 4.892841\ ;\ 5.084485$$.

One knows that denominator of the ODE has two roots (in the range I'm now interested): $$1.92073\ ;\ 2.420171$$ that will ruin something. But my question for you is, what these roots will ruin? Is the solution $$y(x)$$ a continuous function even though the derivative of it exists? Because... If the $$y(x)$$ is not continuous, then I could compute for example sheeps and cows with what ever method to obtain silly answers. If the function is continuous then one just needs to find a suitable method to obtain numbers. Any thoughts? Thank you!

 

[Prove It]

Hi
I am trying to understand numerical analysis on my freetime and today I studyed how to solve $$y' = \frac{x^2}{1 + y\sin (y^2)}$$, with initial value $$y(0) = 0$$. I asked myself two simple questions: What is $$y(1.5)$$ and what is $$y(2.5)$$?

As for to check the answers, I solved the ODE. In implicit form one can write $$2x^3 - 3 = 6y - 3\cos (y^2)$$. Now, when $$x = 1.5$$, one can solve the $$y \approx 0.941265$$ with no problems. However, when $$x = 2.5$$, there are three solutions for $$y$$, namely $$4.553574\ ;\ 4.892841\ ;\ 5.084485$$.

One knows that denominator of the ODE has two roots (in the range I'm now interested): $$1.92073\ ;\ 2.420171$$ that will ruin something. But my question for you is, what these roots will ruin? Is the solution $$y(x)$$ a continuous function even though the derivative of it exists? Because... If the $$y(x)$$ is not continuous, then I could compute for example sheeps and cows with what ever method to obtain silly answers. If the function is continuous then one just needs to find a suitable method to obtain numbers. Any thoughts? Thank you!

Why solve it numerically when it can be solved exactly through separation of variables?

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{x^2}{1 + y\sin{ \left( y^2 \right) } } \\ \left[ 1 + y\sin{ \left( y^2 \right) } \right]\,\frac{\mathrm{d}y}{\mathrm{d}x} &= x^2 \\ \int{ \left[ 1 + y\sin{ \left( y^2 \right) } \right] \,\frac{\mathrm{d}y}{\mathrm{d}x} \,\mathrm{d}x} &= \int{ x^2\,\mathrm{d}x} \\ \int{ \left[ 1 + y\sin{ \left( y^2 \right) } \right] \, \mathrm{d}y } &= \frac{1}{3}\,x^3 + C_1 \\ y + \frac{1}{2} \int{ 2\,y\sin{ \left( y^2 \right) } \,\mathrm{d}y } &= \frac{1}{3}\,x^3 + C_1 \\ y + \frac{1}{2} \int{ \sin{ \left( u \right) } \,\mathrm{d}u } &= \frac{1}{3}\,x^3 + C_1 \textrm{ after substituting } u = y^2 \implies \mathrm{d}u = 2\,y\,\mathrm{d}y \\ y + \frac{1}{2}\,\left[ -\cos{\left( u \right) } \right] + C_2 &= \frac{1}{3}\,x^3 + C_1 \\ y - \frac{1}{2}\cos{ \left( y^2 \right) } + C_2 &= \frac{1}{3}\,x^3 + C_1 \\ 6\,y - 3\cos{ \left( y^2 \right) } &= 2\,x^3 + C \textrm{ where } C = 6\,\left( C_1 - C_2 \right) \end{align*}$

Now with the initial condition $\displaystyle \begin{align*} y\,\left( 0 \right) = 0 \end{align*}$ we have

$\displaystyle \begin{align*} 6\,\left( 0 \right) - 3 \cos{ \left( 0^2 \right) } &= 2\,\left( 0 \right) ^3 + C \\ -3 &= C \end{align*}$

Thus the solution is $\displaystyle \begin{align*} 6\, y - 3\cos{ \left( y^2 \right) } = 2\,x^3 - 3 \end{align*}$.

 

[Theia]

Why solve it numerically when it can be solved exactly through separation of variables?

The answer to this question is in the beginning of my first post: "I am trying to understand numerical analysis".

In addition, the solution you derived, was also mentioned in my post; thank you for verifying it. So yes, I know the implicit solution exists and I use it determine how well I can compute the numerical value.

But again, my main question was, is the solution $$y(x)$$ continuous function in the sense that a numerical method can be used? As you can see from the implicit solution, there are three different values $$y(x)$$ can get at that point. With one initial value I can compute just one value for $$y(x)$$, not three, and this is the reason I'm posting this thread. I want to understand what is happening when I compute the numerical solution in this case.

(Clearly, I should have chosen the ODE such that no one can solve it analytically to avoid exact functions completely, but this would have make it unnecessarily difficult to compare different solutions. I'm sorry, I didn't notice the rules of the forum saying anything about 'use only exact functions in the forum content, don't be interested in numerics' or 'make sure to explain why you wish to understand a certain area in mathematics better'. I must admit, I'm a bit disappointed...)

 

[I like Serena]

Hi Theia! Welcome to MHB! (Smile)

The solution is not a function of x. That's why we have multiple values for the same x.
More specifically, even the exact solution that we found is not an actual solution.
That is because the original ODE (y') is not defined everywhere on the curve.

To investigate the behavior, we should look at the values where y' is not defined, or alternatively where we have critical points.

 

[Theia]

Thank you, I like Serena.

Hmm... So because there are points where the derivative $$y'$$ is not defined, implies that numerical solution won't work because it tries to use values that don't necessarily exist. Cool!