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 jt316 Apr22-07 04:19 PM

diff.eqn. help

I need help figuring out the solution to this diff.eq.

y(x) = x + (1/2)*∫(from u=-1 to 1)[ ( 1-| x – u | ) y(u) du] , x є [ -1, 1]

I have to show that:
y(x) + y(x) = 0 , x є [ -1, 1]

subject to:

y(1) + y(-1) = 0

y(1) + y(-1) = 2

 AiRAVATA Apr22-07 07:50 PM

What you mean is you need to prove that

$$y(x)=x+\frac{1}{2}\int_{-1}^x (1-|x-u|)y(u)du ,\qquad x\in[-1,1]$$

is a solution of $y''(x)+y(x)=0$ with the given boundary conditions or viceversa?

If it is the frist one, you just have to remember the definition for absolute value is

$$|x|=\left\{\begin{array}{rl} x & \hbox{if }x\ge 0, \\ -x & \hbox{if }x<0\end{array}\right.$$

So, the function $y(x)$ is actually

$$y(x)=x+\frac{1}{2}\left\{\int_{-1}^x [1-(x-u)]y(u)du+\int_{x}^1 [1-(u-x)]y(u)du\right\}$$

Hence, all you have to do is use the fundamental theorem of calculus, and prove that it satisfies the boundary conditions.

 jt316 Apr22-07 09:56 PM

Quote:
 Quote by AiRAVATA (Post 1311079) What you mean is you need to prove that $$y(x)=x+\frac{1}{2}\int_{-1}^x (1-|x-u|)y(u)du ,\qquad x\in[-1,1]$$ is a solution of $y''(x)+y(x)=0$ with the given boundary conditions or viceversa? If it is the frist one, you just have to remember the definition for absolute value is $$|x|=\left\{\begin{array}{rl} x & \hbox{if }x\ge 0, \\ -x & \hbox{if }x<0\end{array}\right.$$ So, the function $y(x)$ is actually $$y(x)=x+\frac{1}{2}\left\{\int_{-1}^x [1-(x-u)]y(u)du+\int_{x}^1 [1-(u-x)]y(u)du\right\}$$ Hence, all you have to do is use the fundamental theorem of calculus, and prove that it satisfies the boundary conditions.

That is what I tried to do but for some reason I'm not coming up with the correct solution for the given boundry conditions...

 AiRAVATA Apr23-07 12:08 AM

Are you deriving correctly?

The reason I ask this is because you are not going to use a standard form of the fundamental theorem of calculus. For example, lets define the function

$$F(x)=\int_a^x f(u,x)du.$$

By definition,

$$F'(x)=\lim_{h \rightarrow 0} \frac{F(x+h)-F(x)}{h}=\lim_{h \rightarrow 0} \frac{1}{h}\left\{ \int_a^{x+h} f(u,x+h)du-\int_a^x f(u,x)du\right\}.$$

Rewrittig the right hand of the equation,

$$F'(x)=\lim_{h \rightarrow 0}\frac{1}{h}\left\{\int_x^{x+h} f(u,x+h)du +\int_a^x \left[f(u,x+h)-f(u,x)\right]du\right\}.$$

If the function $f(u,x)$ is well behaved in the domain of definition (as yours is), then we can exchange the integral and the limit, and using the fundamental theorem of calculus,

$$F'(x)=f(x,x)+\int_a^x \frac{\partial f}{\partial x}(u,x)du.$$

Now, you tell me what is the derivative of $G(x)$ when

$$G(x)=\int_x^b g(u,x)du.$$

 jt316 Apr23-07 12:20 AM

I have no idea... I thought I knew but now I don't..

 AiRAVATA Apr23-07 12:35 AM

$$G'(x)=\lim_{h \rightarrow 0} \frac{G(x+h)-G(x)}{h}=\lim_{h \rightarrow 0} \frac{1}{h} \left\{\int_{x+h}^b g(u,x+h)du-\int_x^b g(u,x)du \right\}.$$

Rewritting the right side of the equality,

$$G'(x)=\lim_{h \rightarrow 0} \frac{1}{h}\left\{\int_{x+h}^x g(u,x+h)du+\int_x^b g(u,x+h)du-\int_x^b g(u,x)du\right\}.$$

Can you take it from here?

 AiRAVATA Apr23-07 12:39 AM

Or better yet, why dont you write $G(x)$ as

$$G(x)=-\int_b^x g(u,x)du$$

and use my previous post?

 jt316 Apr23-07 01:00 AM

It's not completely done, but I got:

y(x)= x + (1/2)*y(x) - (1/2)*y(x)*x^2

does it look like I'm on the right track?

using this I can get y(x) + y(x) = 0
and y(x) = (2*x) / (x^2 + 1)
which will give me: y(1) + y(-1) = 0
but I can't get the last condition to work out.

I can't get y(1) + y(-1) = 0

????

 AiRAVATA Apr23-07 01:32 AM

Do it carefully:

$$y'(x)=1+\frac{1}{2}\left\{-\int_{-1}^x y(u)du+\int_x^1 y(u)du\right\}$$

and there you go. Calculating $y''(x)$ should be a piece of cake.

 jt316 Apr23-07 01:41 AM

how did you evaluate that and leave the integrals in? I evaluated the integrals and found an expression for y(x) and then took the deriv. and found y(x). Can I not do it that way? I would like to see the method you used to work that out, if you dont mind. I have no idea how you did that.

 AiRAVATA Apr23-07 01:45 AM

As I stated in my previous posts, you are not using the standar fundamental theorem of calculus, because the function you are trying to derive depends on $x$ on both the limits of integration, as on the integrand. Thats why there is an extra integral term (think of it as some sort of chain rule).

The Fundamental Theorem of Calculus (roughly) states that

$$\frac{d}{dx}\left(\int_a^x f(u)du\right)=f(x).$$

In your case, the integrand does not only depends on $u$, but also on $x$. Thats why there is the extra term. In a previous post, I've proven that

$$\frac{d}{dx} \left(\int_a^x f(u,x)du\right)=f(x,x)+\int_a^x \frac{\partial f}{\partial x}(u,x) du.$$

In your case, lets write $y(x)$ as

$$y(x)=x+\frac{1}{2}\left\{\int_{-1}^x f(u,x)du +\int_x^1 g(u,x)du\right\},$$

where $f(u,x)=[1-(x-u)]y(u)$ and $g(u,x)=[1-(u-x)]y(u)$.

Using my previous posts, the derivative of $y(x)$ is

$$y'(x)=1+\frac{1}{2}\left\{f(x,x)+\int_{-1}^x\frac{\partial f}{\partial x}(u,x)du-g(x,x)+\int_x^1 \frac{\partial g}{\partial x}(u,x)du\right\}.$$

Substituting $f,\,g,\,\frac{\partial f}{\partial x},\,\frac{\partial g}{\partial x}$, you should come up with the correct answer.

 jt316 Apr23-07 02:08 AM

I did the exact same thing but I didnt have the f(x,x) and g(x,x) terms. what are f(x,x) and g(x,x)?

 HallsofIvy Apr23-07 07:46 AM

They are exactly what they say: f(x,y) and g(x,y) with y replaced by x. For example if f(x,y)= (x+y)2, then f(x,x)= (x+ x)2= 4x2.

That rather complicated formula AiRAVATA is giving you is a special case of Leibniz's formula:
$$\frac{d}{dx}\left[\int_{\alpha(x)}^{\beta(x)} \phi(x,t)dt\right]= \phi(x,\beta(x))\frac{d\beta(x)}{dx}- \phi(x,\alpha(x))\frac{d\alpha(x)}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial\phi(x,t)}{\partial x}dt$$

 jt316 Apr23-07 01:13 PM

I think I'm getting close but I am stumped on int( u*y(u) du ) integrated from -1 to 1 ? The y(u) is throwing me off....

 AiRAVATA Apr23-07 02:43 PM

You don't have to integrate anything, you have to diferentiate. In fact, you can't integrate as $y(u)$ is unknown.

Lets recapitulate: if $f(u,x)=[1-(x-u)]y(u)$, then

$$f(x,x)=y(x),$$

and

$$\frac{\partial f}{\partial x}(u,x)=-y(u),$$

so

$$\frac{d}{dx}\left(\int_{-1}^x [1-(x-u)]y(u)du\right)=y(x)-\int_{-1}^x y(u)du.$$

What is the derivative of the next term?

 jt316 Apr23-07 04:10 PM

Quote:
 Quote by AiRAVATA (Post 1311813) You don't have to integrate anything, you have to diferentiate. In fact, you can't integrate as $y(u)$ is unknown. Lets recapitulate: if $f(u,x)=[1-(x-u)]y(u)$, then $$f(x,x)=y(x),$$ and $$\frac{\partial f}{\partial x}(u,x)=-y(u),$$ so $$\frac{d}{dx}\left(\int_{-1}^x [1-(x-u)]y(u)du\right)=y(x)-\int_{-1}^x y(u)du.$$ What is the derivative of the next term?
For the second half to that part I get: - y(x) + int( y(u) du) from x->1

so to find y(x), what is the partial derivative for d [y(u)] / dx = ? wouldn't it be zero?
I know I need to get: y(x)= -y(x) and d [y(u)] / dx = 0 wouldn't get me there. Do I still use Leibnitz rule to find y(x) ?

I also have another question... How do you use the symbolic notation?.. that would make this much easier.

I appreciate the help.

 AiRAVATA Apr23-07 05:15 PM

Quote:
 For the second half to that part I get: - y(x) + int( y(u) du) from x->1
That is correct!, so

$$y'(x)=1+\frac{1}{2}\left\{-\int_{-1}^x y(u)du + \int_x^1 y(u)du\right\}.$$

For the second derivative, using the fundamental theorem of calculus,

$$\frac{d}{dx} \left(\int_{-1}^x y(u)du\right)=y(x),$$

(the other term is for you to calculate) and that's it!

Quote:
 I also have another question... How do you use the symbolic notation?.. that would make this much easier.
Click on the formulae to find out :)

 jt316 Apr23-07 05:41 PM

Quote:
 Quote by AiRAVATA (Post 1311937) That is correct!, so $$y'(x)=1+\frac{1}{2}\left\{-\int_{-1}^x y(u)du + \int_x^1 y(u)du\right\}.$$ For the second derivative, using the fundamental theorem of calculus, $$\frac{d}{dx} \left(\int_{-1}^x y(u)du\right)=y(x),$$ (the other term is for you to calculate) and that's it! Click on the formulae to find out :)
I see.... I think I have it now... I'll let you know how it comes out (now that you have helped me along)

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