Rudin Ch 5, #29: Uniqueness Theorem for Systems of ODEs

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Homework Statement



Specialize exercise 28 by considering the system

\y&#039;= y_{j+1} j=(1,...,k-1)<br /> y&#039;_{k}= f(x)-\sum g_{j}(x)y_{j} where the summation runs from j=1 to j=k, and g_{j} and f are continuous real functions on [a,b], and derive a uniqueness theorem for solutions of the equation<br /> <br /> y^{k}+g_{k}(x)y^{k-1}+...+g_{2}y&#039;+g_{1}(x)y = f(x)<br /> <br /> subject to initial conditions<br /> <br /> y(a)=c_{1}, y&#039;(a)= c_{2}, y^{k-1}(a) = c_{k}.[\tex]<br /> <br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br />
 
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Hi Quantumpencil! :smile:

(you use the wrong slash: :wink:)
Quantumpencil said:
Specialize exercise 28 by considering the system

y&#039;= y_{j+1}, j=(1,...,k-1),\ y&#039;_{k}= f(x)-\sum g_{j}(x)y_{j} where the summation runs from j=1 to j=k, and g_{j} and f are continuous real functions on [a,b], and derive a uniqueness theorem for solutions of the equation

y^{k}+g_{k}(x)y^{k-1}+...+g_{2}y&#039;+g_{1}(x)y = f(x)

subject to initial conditions

y(a)=c_{1}, y&#039;(a)= c_{2}, y^{k-1}(a) = c_{k}

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
yeah, so I didn't realize I made this thread; it's incomplete. Could I get it locked?

The actual thread I need help on is further-down and contains the good tech + information about how I think the solution will work out.

https://www.physicsforums.com/showthread.php?t=297047

This is the link to the actual topic.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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