Existence and Uniqueness Theorems for ODEs

[twoflower]

Hi all,
I'd be happy if someone could clarify these two things to me:

1. While solving linear first-order ODE, I first solve homogenous equation (with the right side equal to 0) and eventually I get to the point (just an example):

$$\log |y| = \log C(e^{x} - 1)$$

Now, is it ok to compute $C$ for given non-homogenous equation and then write there two solutions

$$y = C(e^{x} - 1)$$

$$y = -C(e^{x} - 1)\mbox{ ?}$$

Because, you know, both satisfy

$$\log |y| = \log C(e^{x} - 1)$$

Anyway, sometimes it gave me correct results (both $C(e^{x} - 1)$ and $-C(e^{x} - 1)$ were solutions) while other times only plus-signed solution was ok.
2. Solving Bernoulli's ODE, let's say

$$y' = 2y + 2x\sqrt{y}$$

and substituting

$$z = \sqrt{y}$$

$$y = z^2$$

$$y' = 2zz'$$

We get

$$2zz' = 2z^2 + 2xz$$

and first thing I do is dividing with $2z$ and so getting the condition $z \neq 0$
After some computing, I get the result

$$z = Ce^{x} - x - 1$$

and thus

$$y = (Ce^{x} - x - 1)^2$$

a) first question
I'd suppose this expression must not get zeroed because at the beginning we divided the equation with $2z$. Anyway, even if $C=1$ and $x=0$ (and thus $z = 0 = y$), the original equation holds true. How it comes?b) second question
Having the substitution above in mind, I know that $z$ itself must be $\geq 0$. Anyway, if we take $C \in (0, 1)$, then the equation

$$Ce^{x} -x - 1 = 0$$

has two roots, $x_0$ and $x_1$. Then we know that
for $x \in (x_0, x_1)$

$$Ce^{x} -x - 1 < 0$$

Ok, still remembering that this is $z$ and it must be $\geq 0$, we shouldn't accept this interval for $x$. However, what's the problem with this solution on this interval? When you put it in the original equation, it's ok I think.

I can't understand why the interval $(x_0,x_1)$ is excluded from the solution (according to our professor).Thank you for any suggestions.

 

[saltydog]

b) second question
Having the substitution above in mind, I know that $z$ itself must be $\geq 0$. Anyway, if we take $C \in (0, 1)$, then the equation

$$Ce^{x} -x - 1 = 0$$

has two roots, $x_0$ and $x_1$. Then we know that
for $x \in (x_0, x_1)$

$$Ce^{x} -x - 1 < 0$$

Ok, still remembering that this is $z$ and it must be $\geq 0$, we shouldn't accept this interval for $x$. However, what's the problem with this solution on this interval? When you put it in the original equation, it's ok I think.

I can't understand why the interval $(x_0,x_1)$ is excluded from the solution (according to our professor).


Thank you for any suggestions.

Well, I would've lost money on that one Twoflower. Apparently when you assign:

$$z=y^{1/2}$$

You implicitly define z$\geq 0$.

Note that the solution:

$$y(t)=(ce^x-(1+x))^2$$

does NOT satisfiy the ODE in the interval where z is less than zero. For example, take the plot below for c=0.5 and:

$$y(t)=(ce^x-(1+x))^2$$

For (approx):

$$x\in (-0.77,1.68)$$

the curve does not satisfy the ODE. New for me . Thanks!

 

[saltydog]

I wish to clear up something with this problem from the perspective of existence and uniqueness:

Consider:

$$y^{'}=f(x,y)=2y+2x\sqrt{y},\quad y(0)=0.25$$

Now,since f(x,y) and the partial of f with respect to y exists in a bounded region about the point (0,0.25), we can expect a unique solution passing through this point.

The solution above was determined to be:

$$y(x)=(ce^x-1-x)^2\;\text{with}\;ce^x-1-x\geq 0$$

Solving for c, I obtain:

$$c=0.5\;\text{and}\;c=1.5$$

Now, initially this would suggest two solutions satisfy the equation. However, the value of c=0.5 would not satisify the restiction [itex]ce^x-1-x\geq 0[/tex] and so must be discarded. I'm left then with the "unique" solution:<br /> <br /> $$y(x)=(1.5e^x-1-x)^2$$[/itex]

 

[twoflower]

Thank you Saltydog, now I'm clear about that, it helped me a lot.

Now,since f(x,y) and the partial of f with respect to y exists in a bounded region about the point (0,0.25), we can expect a unique solution passing through this point.

What does this follow from? Is there some theorem saying this?

Could you clarify those two other questions to me too? Thank you very much.

 

[saltydog]

What does this follow from? Is there some theorem saying this?

This is based on Existence and Uniqueness Theorems for ordinary differential equations:

Consider:

$$y^{'}=f(x,y)$$

Let T denote the rectangular region defined by:

$$|x-x_0|\leq a\quad |y-y_0|\leq b$$

with the point $(x_0,y_0)$ at its center. Let f and $\frac{\partial f}{\partial y}$ be continuous at each point in T, then there exists an interval$|x-x_0|\leq h$ and a function y(x) such that:

1. y(x) is a solution to the ODE on the interval $|x-x_0|\leq h$.

2. y(x) is unique in the sense that if another equation h(x) satisfies the ODE, then y(x)=h(x).

Note that the interval $|x-x_0|\leq h$ can be very small and nothing is said about uniqueness and existence if f and the partial do not exist at the point x0.

Thus for a particular IVP, I know if the above is satisfied, I can be guaranteed a unique solution. However, I'm not sure about the case of y=0. Note that for the IVP:

$$y^{'}=2y+2x\sqrt{y};\quad y(0)=0$$

The partial is indeterminant at the initial condition so uniqueness and existence cannot be guaranteed.

Could you clarify those two other questions to me too? Thank you very much.

Not sure I can do better than the above. Perhaps someone in here more knowledgeable than I could elaborate on the matter.