IVP Unique Does y' = x^5 - y^5 + 2xe^y Have 1?

[Benny]

Q. Determine whether the initial value problem y' = x^5 - y^5 + 2xe^y ,y\left( 3 \right) = \pi has a unique solution.

I've seen things like Lipschitz bound and conditions on the partial derivative of f(x,y) with respect to y where y' = f(x,y) but such things mean very little to me. The reason is because I've never seen any examples where something like a Lipschitz bound is applied. I've also yet to see the df/dy(partial) continuity conditions applied. For this example if I take the RHS to be f(x,y) then obviously the partial derivative of f wrty is continuous everywhere so the additional conditions which come with df/dy don't really apply.

I'm just wondering if the IVP has a unique solution. My limited knowledge of DEs is inclined to suggest to me that there is a unique solution to the DE because the partial derivative of the RHS with respect to y is continuous everywhere. Can someone offer some input? Any help appreciated.

 

[saltydog]

Q. Determine whether the initial value problem y' = x^5 - y^5 + 2xe^y ,y\left( 3 \right) = \pi has a unique solution.

I've seen things like Lipschitz bound and conditions on the partial derivative of f(x,y) with respect to y where y' = f(x,y) but such things mean very little to me. The reason is because I've never seen any examples where something like a Lipschitz bound is applied. I've also yet to see the df/dy(partial) continuity conditions applied. For this example if I take the RHS to be f(x,y) then obviously the partial derivative of f wrty is continuous everywhere so the additional conditions which come with df/dy don't really apply.

I'm just wondering if the IVP has a unique solution. My limited knowledge of DEs is inclined to suggest to me that there is a unique solution to the DE because the partial derivative of the RHS with respect to y is continuous everywhere. Can someone offer some input? Any help appreciated.

Benny, dude, I bet a PF dollar that IVP has a unique solution. Yep, a whole dollar. That's because I trust the Existence and Uniqueness Theorems for this type of differential equation. That is, The RHS is continuous everywhere and so is the partial with respect to y.

I tell you what, if you would carefully review the Existence and Uniqueness theorem for first-order ODEs, every single small detail, then I bet it time you'd come away with confidence too to look at an ODE like that one and know if it has a unique solution.

 

[Brad Barker]

Benny, dude, I bet a PF dollar that IVP has a unique solution. Yep, a whole dollar. That's because I trust the Existence and Uniqueness Theorems for this type of differential equation. That is, The RHS is continuous everywhere and so is the partial with respect to y.

I tell you what, if you would carefully review the Existence and Uniqueness theorem for first-order ODEs, every single small detail, then I bet it time you'd come away with confidence too to look at an ODE like that one and know if it has a unique solution.

i'd be willing to bet more than a dollar on this one!

 

[HallsofIvy]

For this example if I take the RHS to be f(x,y) then obviously the partial derivative of f wrty is continuous everywhere so the additional conditions which come with df/dy don't really apply.

I don't understand this at all! The conditions for the "Existance and Uniqueness" Theorem for initial value problems are simply that the right hand side be continuous and "Lipschitz in y" in some region around (x₀,y₀). "Lipschitz" may be difficult to apply but it's easy to show that \frac{\partial f}{\partial y} is sufficient (though not necessary). You say "obviously the partial derivative of f wrty is continuous" so that certainly does apply!

 

[Benny]

Sorry for the confusion.

The notes I have don't specifically refer to an existence and uniqueness theorem, although I'm pretty sure that's the message being communicated. I just went over the material that I have again and I have 2 parts:

(1) y' = f(x,y), y(x_0) = y_0 has a unique solution if it satisfies:

(i) has a solution in some interval -d < x - x_0 < d if f(x,y) is continuous in an open set containing (x_0,y_0).

(ii) has a unique solution in the interval if, in addition to continuity, we have a "Lipschitz bound" |f(x,y) - f(x_0,y_0)| <= K|y_1 - y_2| (where K is a constant) for all (x,y) in the neighbourhood of (x_0,y_0).

Probably irrelevant but the above looks similar to the the epsilon delta definition for functions of two variables.

Also, a statement at the end of the theorem says condition (ii) asks for less than continuity of df/dy(partial). So I guess in this case since df/dy is continuous everywhere then the IVP has a unique solution.

There is also another theorem.

(2) Let f(x,y) and df/dy(the partial derivative) be continuous in the rectangle x_0 <= x <= a, |y-y_0| <= b. The IVP y' = f(x,y) with y(x_0) = y_0 has precisely one solution in the interval x_0 <= x <= alpha, where:

<br /> \alpha = \min \left\{ {a,\frac{b}{M}} \right\},M = \mathop {\max }\limits_{x_0 \le x \le a,\left| {y - y_0 } \right|} \left| {f\left( {x,y} \right)} \right|<br />

I can kind of understand the bit about rectangles but I don't really get why a minimum for alpha is taken.

 

[saltydog]

The Lipchitz condition is based on the Mean Value Theorem applied to f(x,y) as a function of y. Thus there exists a number y_m between y_1 and y_2 such that:

f(x,y_1)-f(x,y_2)=\frac{\partial f}{\partial y}(x,y_m)(y_1-y_2)

Thus if the partial is continuous then it is bounded, that is:

\left|\frac{\partial f}{\partial y}\right|\leq K

for some K. There you go, the Lipschitz Condition:

|f(x,y_1)-f(x,y_2)|\leq K|y_1-y_2|

This crucial inequality is then used in both the Existence and Uniqueness Theorems to show that a particular sequence of functions is bounded by a convergent series and thus converges also. Later in the theorem, it is shown that this sequence of functions converges to the solution of the IVP in at least some small neighborhood around (x_0,y_0) .

 

[HallsofIvy]

Sorry for the confusion.

The notes I have don't specifically refer to an existence and uniqueness theorem, although I'm pretty sure that's the message being communicated. I just went over the material that I have again and I have 2 parts:

(1) y' = f(x,y), y(x_0) = y_0 has a unique solution if it satisfies:

(i) has a solution in some interval -d < x - x_0 < d if f(x,y) is continuous in an open set containing (x_0,y_0).

(ii) has a unique solution in the interval if, in addition to continuity, we have a "Lipschitz bound" |f(x,y) - f(x_0,y_0)| <= K|y_1 - y_2| (where K is a constant) for all (x,y) in the neighbourhood of (x_0,y_0).

Basically, that is the "existance and uniqueness" theorem for initial value problems.

Probably irrelevant but the above looks similar to the the epsilon delta definition for functions of two variables.

Also, a statement at the end of the theorem says condition (ii) asks for less than continuity of df/dy(partial). So I guess in this case since df/dy is continuous everywhere then the IVP has a unique solution.

Yes, exactly.

There is also another theorem.

(2) Let f(x,y) and df/dy(the partial derivative) be continuous in the rectangle x_0 <= x <= a, |y-y_0| <= b. The IVP y' = f(x,y) with y(x_0) = y_0 has precisely one solution in the interval x_0 <= x <= alpha, where:

<br /> \alpha = \min \left\{ {a,\frac{b}{M}} \right\},M = \mathop {\max }\limits_{x_0 \le x \le a,\left| {y - y_0 } \right|} \left| {f\left( {x,y} \right)} \right|<br />

I can kind of understand the bit about rectangles but I don't really get why a minimum for alpha is taken.

This second one has a bit more detail. The first theorem just says that there is a unique solution is some region around x₀ but doesn't say how small that region is. The second is trying to give you some idea of the size of the region in which you can expect that solution to be defined. The α and M are really taken from the proof of the first theorem.

The basic idea is that you convert from the differential equation
\frac{dy}{dx}= f(x,y) with y(x_0)= y_0
to the corresponding "integral equation"
y(x)= \int_{x_0}^x f(t,y(t))dt+ y_0
then convert that to an operator
F(y)= \int_{x_0}^x f(t,y(t))dt+ y_0

y satisfies the integral equation and the initial value problem if and only if
F(y)= y- if y is a "fixed point" of F.

The "Banach Fixed Point" theorem says that F has a unique fixed point as long as F is a "contraction function" from some set M to itself. "Contraction function" means |F(y_1)-F(y_0)|\leq c|y_1- y_0| where c is strictly between 0 and 1.

In order to apply that to the F above, we take M to be the set of all functions y such that y is continuous on some interval around x₀ such that y(x) remains in the set where f(x,y) is continuous and Lipschitz (and |y| is lub y(x) on that set). Of course, we have to make sure that if y is in that set, so is F(y). If we just pick the interval around x₀ so that it stays in the region where f(x,y) is continuous, then in general that is not true- it depends on the interval. But we can "shrink" the interval so that is true. The first condition in theorem (2) above is precisely that "shrink".

We also need to be sure it is a contraction map. You may notice that the "contraction map" definition looks a lot like the "Lipschitz" definition! That is used but notice also that the "C" for Lipschitz does not have to be less than 1. Again, it turns out that F is not necessarily a "contraction"- it depends on the x interval. Again, we can "shrink" that interval so that it is. That second condition in theorem (2) above is precisely that "shrink". We need both conditions to "shrink" the interval enough to be sure that F does map our set of function M into itself and the F is a contraction.

Of course, you don't need to know that or even know theorem (2) to answer the origina question:
"Does the initial value problem y&#039;= x^5- y^5+ 2xe^y with y(3)= \pi have a unique solution?

Here f(x)= x⁵-y⁵+ 2xe^y is continuous and differentiable for all (x,y) so it is continuous and Lipschitz in y any neighborhood of (3,\pi) and so has a unique solution defined in some neighborhood of x= 3. You don't have to know theorem (2) because the problem does not ask how large that neighborhood is.