How does the line integral method solve ODEs?

[saltydog]

Though you guys might find this interesting. It's something I've not seen before. Consider the ODE:

$$\frac{dy}{dx}=f(x,y)$$

and the attached plot below.

The blue represents the surface for the function f(x,y) (some particular example for the plot) in the first quadrant.

The vertical lines represents the line integral along the curve y(x) that is a (particular) solution for the ODE.

That is, we can say:

$$y(x)=y_0+\int_{y(x)} f(u,v) ds$$

And thus solving the differential equation amounts to finding a curve in the x-y plane such that the value of the line integral starting from the point $(x_0,y_0)$, along the curve y(x), to the point (x,y(x)), plus $y_0$ is the same as the value of y(x).

Why do I find that amazing?

 

[saltydog]

Though you guys might find this interesting. It's something I've not seen before. Consider the ODE:

$$\frac{dy}{dx}=f(x,y)$$

and the attached plot below.

The blue represents the surface for the function f(x,y) (some particular example for the plot) in the first quadrant.

The vertical lines represents the line integral along the curve y(x) that is a (particular) solution for the ODE.

That is, we can say:

$$y(x)=y_0+\int_{y(x)} f(u,v) ds$$

And thus solving the differential equation amounts to finding a curve in the x-y plane such that the value of the line integral starting from the point $(x_0,y_0)$, along the curve y(x), to the point (x,y(x)), plus $y_0$ is the same as the value of y(x).

Why do I find that amazing?

For the record I was in error above.

$$y(x) \ne y_0+\int_{y(x)} f(u,v) ds$$

but rather simply:

$$y(x)=y_0+\int_a^b f(t,y(t)) dt$$

Thank God I'm not a Mathematician and I apologize if anyone was led astray by this.

 

[saltydog]

I believe I've resolved my mis-interpretation of the integral equation derived from the first order ODE:

$$\frac{dy}{dx}=f(x,y)$$

Upon integration of both sides, we obtain:

$$y(x)=y_0+\int_{x_0}^x f(t,y(t))dt$$


Erroneously above, I interpreted this to mean integration around the path that is y(x) when actually, I'm beginning to realize it's more like integration of the projection of the line integral onto a flat surface behind the x-y-plane. I hereby cristen this operation a "projection integral" and wish to define it in general:

Take any continuous function f(x,y) in the first quadrant and mark a trajectory upon it's surface that is one-to-one from the points $(x_0,y_0,z_0)$ to $(x_1,y_1,z_1)$. Project the resulting curve onto a flat surface orthogonal to the x-y plane (and behind it). Integrate this curve from $x_0$ to $x_1$. The resulting integral is:

$$\int_{x_0}^{x_1} P(t) dt$$

where P is the projection function. Now, if the trajectory so marked on f(x,y), results from a path taken along the x-y plane in terms of some function g(x) which satisfies the following equation:

$$g(x)=y_0+\int_{x_0}^x P(t)dt$$

Then g(x) satisfies the differential equation:

$$\frac{dg}{dt}=f(x,y)$$

I believe this is correct and will pursue verifying it.

 

[saltydog]

Consider the equation:

$$\frac{dy}{dx}=y;\quad y(0)=1$$

Solving we obtain:

$$y(x)=e^x$$

We know that this function satisfies the integral equation:

$$y(x)=1+\int_0^x y(t)dt$$

The first plot shows the equation f(x,y)=y in blue. Underneath this plot, shown in vertical lines, is the path that the solution $y(x)=e^x$ takes connecting it with the points immediately above it on the surface f(x,y). I was incorrect above to suggest that the corresponding line-integral along this path was directly related to the integral or differential equation. However, it's becomming clear to me that it's projection to a wall orthogonal to the x-y plane is related. The second plot is the projection of those lines to such a wall. They form a function I call C(x). The thick line is the curve [itex]y(x)=e^x[/tex] which fit it nicely. $y(x)=e^x$ of course is the solution to the differential equation.

What I find interesting is that this is the case for more complicated expressions for the function f(x,y). In fact, I suggest our goal of solving an ODE is to determine the projection function C(x) such that:

$$g(x)=y_0+\int_{x_0}^x C(t)dt$$

where g(x) satisfies the differential equation. I know this may be obvious but I still find it interesting to look at the solution of an ODE from this perspective.


Edit:Oh yea, I suspect an analogous interpretation can be made for higher-ordered ODEs. That is, is there and if so, what is the projection function for a second-order ODE? Is it a surface, the volume underneath of which as a function of x and y, satisfy the differential equation?

 

[Integral]

Just a general comment.

You are on a pretty good track, of course a good fundamental course in ODE's would cover much of what you are playing with.

Hint: You are being very caviler with your limits of integration, take better care of them, understand what they mean in terms of your integral.

Hopefully Halls of Ivy will drop in and add his $.02.

 

[saltydog]

Just a general comment.

You are on a pretty good track, of course a good fundamental course in ODE's would cover much of what you are playing with.

Hint: You are being very caviler with your limits of integration, take better care of them, understand what they mean in terms of your integral.

Hopefully Halls of Ivy will drop in and add his $.02.

Hello Integral. Yes, a fundamental course. I don't believe I had that with Rainville and Bedient. And Devaney's book, which I've used a lot, is mostly qualitative. Might you kindly tell me specifically where I'm not being rigorous with the limits of integration?

 

[saltydog]

Now, if the trajectory so marked on f(x,y), results from a path taken along the x-y plane in terms of some function g(x) which satisfies the following equation:

$$g(x)=y_0+\int_{x_0}^x P(t)dt$$

Then g(x) satisfies the differential equation:

$$\frac{dg}{dt}=f(x,y)$$

I believe this is correct and will pursue verifying it.

Suppose I need to make another correction: g(x) above satisfies the differential equation only if:

$$P(t)=f(t,g(t))$$

with:

$$\frac{dg}{dx}=f(x,g)$$

Suppose that's kind of a circular arguement. The geometry is still interesting though. Take for example:

$$\frac{dy}{dx}=\frac{1}{y};\quad y(0)=0$$

Ignoring for the moment the Lipschitz condition is not met and we obtain two solutions, consider for the moment the positive solution:

The first plot shows $f(x,y)=\frac{1}{y}$ in blue along with the path of the solution underneath it.

The second plot is the solution.

The third plot is the projection function C(t) with y(x) described in terms of this function as:

$$y(x)=y_0+\int_{x_0}^x C(t)dt$$

 

[Integral]

I looked up the references you gave the first appears to be a first "how to solve" course the second appears to an applications course. By fundamental I meant a ODE theory course I took mine from Ordinary Differential Equations in Rⁿ by Piccinini,Stampacchia, & Vidossich. This book starts from page 1 with constructions similar to your approach. Essentially each solution to a ODE is one of a family of "curves". The "curve" followed is defined by the initial conditions.

I may have been off the mark in my initial assement. Let me spend some more time looking at what you are doing.

 

[saltydog]

I looked up the references you gave the first appears to be a first "how to solve" course the second appears to an applications course. By fundamental I meant a ODE theory course I took mine from Ordinary Differential Equations in Rⁿ by Piccinini,Stampacchia, & Vidossich. This book starts from page 1 with constructions similar to your approach. Essentially each solution to a ODE is one of a family of "curves". The "curve" followed is defined by the initial conditions.

I may have been off the mark in my initial assement. Let me spend some more time looking at what you are doing.

You know Integral, I'm really interested in uniqueness and what determines the number of solutions at a point which fails the Lipschitz condition: some ODEs have 2 solutions, some 3, some an infinite number. But I understand you're retiring so that's ok if I don't here from you. Look at the bright side: You could be going to Louisiana. It's in the mid-90's now. The gumbo's good though.:tongue2: