Solving Diff.Eq. with Boundary Conditions: y(x) = x

[jt316]

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

Thanks for any help you can give.

 

[AiRAVATA]

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]

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 boundary conditions...

 

[AiRAVATA]

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, let's 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]

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

 

[AiRAVATA]

Remember to use your definitions.

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

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]

Or better yet, why don't you write $G(x)$ as

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

and use my previous post?

 

[jt316]

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]

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]

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 don't mind. I have no idea how you did that.

 

[AiRAVATA]

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, let's 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]

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]

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]

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]

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]

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]

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!

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]

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)

 

[jt316]

to verify the boundary conditions of:

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

don't I have to int( u*y(u) du ) integrated from -1 to 1 , found in the y(x) eqn.?

 

[AiRAVATA]

I don't believe so, just evaluate and all terms will cancel out.

 

[jt316]

I don't believe so, just evaluate and all terms will cancel out.

Using:

y(x)=x + 1/2 *[int( (1-x+u) *y(u) du) + int ( (1+x-u)*y(u)du)]

how could I prove that y(1) + y(-1) = 0, without integrating u*y(u) du ?

 

[AiRAVATA]

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

and the frist integral is zero, so

$$y(-1)=-1-\frac{1}{2}\int_{-1}^1 uy(u)du.$$Now, what is $y(1)$?

 

[jt316]

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

and the frist integral is zero, so

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


Now, what is $y(1)$?

but now we have

y(-1) = -1 +1/2 * [ - int ( u*y(u))du]

what can we do with:

int ( u*y(u))du

 

[jt316]

so...

y(1) = 1+ 1/2* int( u*y(u)du)

which gives me y(1) + y(-1) = 0

but don't I need a value for y(1) and y(-1) to evaluate y'(x) for y'(1) and y'(-1)?


wait... i see.. I find y'(1) and y'(-1) the same way...

 

[jt316]

I think I have a grasp on that part now...

but given the boundary conditions that we have been working with, how would i find the exact solution to:

y''(x) +y(x)=0

with BC's of:

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

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

this has gotten a little confussing because the BC's I am used to dealing with are like:
y(0)=0
y(1)=0

 

[jt316]

and your help is greatly appreciated... but how would you evaluate my last post?

the boundary conditions make it a bit confusing

 

[AiRAVATA]

Maybe there is an easier way to do it, but you'll have to ask someone else for it. The only method I know to derive such solution is not simple (I have to construct the Green function).

Right now I don't have time. If anybody hasn't answered by tomorrow morning, I'll post my rather complicated method.

 

[jt316]

Maybe there is an easier way to do it, but you'll have to ask someone else for it. The only method I know to derive such solution is not simple (I have to construct the Green function).

Right now I don't have time. If anybody hasn't answered by tomorrow morning, I'll post my rather complicated method.

I appreciate it... I found the solution using matlab... it is:

y(x)=sin(x) / cos(1)

I do think there is a simpler way, I just don't know what it is...

Thanks again for your help

 

[AiRAVATA]

Well, yeah, sure, $\sin x/\cos 1$ is the solution. I tought you wanted the integral representation.

Gald to help.

 

[jt316]

Well, yeah, sure, $\sin x/\cos 1$ is the solution. I tought you wanted the integral representation.

Gald to help.

I don't think that the solution I am looking for has the integral expression in it.

I think that we can find a solution from only:

y''(x) +y(x)=0

with BC's of:

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

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

I just don't know exactly how to do that..

I can tell that you have a great understanding of Calculas.

Thanks for helping me understand it alittle better..
if you think of an easier way to solve this, please let me know.