Laplace Initial Value Problem

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  • #26
Hurkyl
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Ah, that makes me happy! Two free variables, and the typical solution isn't twice differentiable at the origin.
 
  • #27
saltydog
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Interesting you note that Hurkly.

In the interest of being complete, I've calculated the initial-value problem:

[tex]ty^{''}+(t-1)y^{'}+y=0;\quad y(0.1)=3,\quad y'(0.1)=-1[/tex]

using the series I calculated above (just take the derivative of y(x)) and obtained the system to solve for A and B:

[tex]3=Ay_1(0.1)+By_2(0.1)[/tex]

[tex]-1=Ay_1^{'}(0.1)+By_2^{'}(0.1)[/tex]

Doing this, I obtain the particular solution:

[tex]y(x)\approx 24.48y_1(x)+2.9y_2(x)[/tex]

I next ploted this particular solution and compared it against Mathematica's NDSolve for this initial-value problem. The superposition of both results are below; they are identical. This along with the results above, gives me confidence the series calculated are correct.
 

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  • #28
saltydog
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So now that the general solution has been determined for the homogeneous equation, one seeks to determine the general solution for the non-homogeneous case:

[tex]xy^{''}+(x-1)y^{'}+y=x^2[/tex]

That's easy. We need only a particular solution. But we calculated an infinite number of them up there. One is:

[tex]x^2e^{-x}+\frac{1}{3}x^2[/tex]

And so the general solution is:

[tex]y(x)=Ay_1(x)+By_2(x)+x^2e^{-x}+\frac{1}{2}x^2[/tex]

With [itex]y_1(x)[/itex] and [itex]y_2(x)[/tex] defined above.

This is interesting in that I can think of no other way of solving the non-homogeneous equation analytically for the general solution. The Laplace Transform in conjunction with the power-series approach are thus shown to be effective in accomplishing this task. Note that the general solution is not defined for x=0. I am thus further led to suspect that uniqueness is lost at any point where the general solution ceases to exist.
 
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  • #29
saltydog
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Hello guys. I wish to further investigate this problem from the perspective of the following theorem (in Rainville and Bedient but no proof):

Consider the equation:

[tex]y^{''}=f(x,y,y^{'})[/tex]

If f and its partial derivatives with respect to y and [itex]y^{'}[/itex] are continuous functions in a region T defined by:

[tex]|x-x_0|\leq a,\quad |y-y_0|\leq b, \quad |y^{'}-y_0^{'}|\leq c,[/tex]

then there exists an interval [itex]|x-x_0|\leq h[/itex] AND a unique function [itex]\phi(x)[/itex] such that [itex]\phi(x)[/itex] is a solution of the differential equation for all x in the interval [itex]|x-x_0|\leq h[/itex] such that:

[tex] \phi(x_0)=y_0\quad\text{and}\quad \phi^{'}(x_0)=y_0^{'}[/tex]

Thus for:

[tex]y^{''}=t-\frac{t-1}{t}y^{'}-\frac{1}{t}y=f(t,y,y^{'})[/tex]

we have:

[tex]\frac{\partial f}{\partial y}=-\frac{1}{t}[/tex]

[tex]\frac{\partial f}{\partial y^{'}}=-\frac{t-1}{t}[/tex]

Obviously f and the partials aren't continuous at t=0 and is consistent with Arildno's analysis above.

However, that's not good enough. The theorem makes no mention of "if and only if". Are there some equations with discontinuities which still allow unique solutions? I really need to determine for myself at what specific point in the proof does the absence of continuity affects uniqueness. I'll first work on it myself and if I have problems, E.L. Ince, "Ordinary Differential Equations" has a complete proof. Or I can ask you guys. :smile:
 
  • #30
arildno
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Great work, saltydog!

I did, however say, that the non-existence of the partial derivative MIGHT indicate the lack of uniqueness..

I do not know the precise necessary and sufficient conditions for uniqueness, I'm only familiar (or rather, was, several years ago) with one particular uniqueness proof (in Marsden's "Real Analysis") which did rely upon the existence of the partial derivative.
 
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