very differential equation with not known source function

[logistic_guy]

here is the question

Solve the classical differential equation $\displaystyle u'' - k^2u = f(x), \ \ \ \ a < x < b, \ \ \ \ u(a) = u(b) = 0$.

Hint: $\displaystyle u(x) = \int_{a}^{b} G(x,s) f(s) \ ds$,

where the function $\displaystyle G(x,s)$ is known as Green's function for the problem.

my attemp
before and on paper i find $\displaystyle u(x) = c_1e^{kx} + c_2e^{-kx}$

how to find particular solution for $\displaystyle f(x)$?

 

[mario99]

 

[logistic_guy]

?

i'm solving the differential equation. anyone?

 

[mario99]

?

i'm solving the differential equation. anyone?

Where is my last post? I have given you a great hint there!

 

[khansaheb]

?

i'm solving the differential equation. anyone?

Why are you posting this problem in "arithmetic" forum?

 

[logistic_guy]

Where is my last post? I have given you a great hint there!

i can't find it

Why are you posting this problem in "arithmetic" forum?

can you help?

 

[mario99]

i can't find it


can you help?

Anyway, my hint was to use the Dirac delta function which means to solve this differential equation:

$\displaystyle u'' - k^2u = \delta(x-s)$

The solution to this differential equation is $\displaystyle G(x,s)$.

Another way is to solve by using variation of parameters.

Do you know how to use these methods? Have you tried anyone of them?

 

[logistic_guy]

here is the question

Solve the classical differential equation $\displaystyle u'' - k^2u = f(x), \ \ \ \ a < x < b, \ \ \ \ u(a) = u(b) = 0$.

Hint: $\displaystyle u(x) = \int_{a}^{b} G(x,s) f(s) \ ds$,

where the function $\displaystyle G(x,s)$ is known as Green's function for the problem.

my attemp
before and on paper i find $\displaystyle u(x) = c_1e^{kx} + c_2e^{-kx}$

how to find particular solution for $\displaystyle f(x)$?

i insist to solve this differential. if i can't i'll quit engineering

Do you know how to use these methods? Have you tried anyone of them?

no

i know how to solve
$\displaystyle u'' - k^2u = 0$
$\displaystyle u'' - k^2u = 1$
$\displaystyle u'' - k^2u = x$
$\displaystyle u'' - k^2u = x^2 + e^x$
$\displaystyle u'' - k^2u = \tan x$
$\displaystyle u'' - k^2u = x\sin x$

 

[mario99]

i insist to solve this differential. if i can't i'll quit engineering

Then, solve it! What are you waiting for?

i know how to solve
$\displaystyle u'' - k^2u = 0$
$\displaystyle u'' - k^2u = 1$
$\displaystyle u'' - k^2u = x$
$\displaystyle u'' - k^2u = x^2 + e^x$
$\displaystyle u'' - k^2u = \tan x$
$\displaystyle u'' - k^2u = x\sin x$

I am better than you. I know how to solve:

$\displaystyle u'' - k^2u = 0$
$\displaystyle u'' - k^2u = 1$
$\displaystyle u'' - k^2u = x$
$\displaystyle u'' - k^2u = x^2 + e^x$
$\displaystyle u'' - k^2u = \tan x$
$\displaystyle u'' - k^2u = x\sin x$
$\displaystyle u'' - k^2u = f(x)$

 

[logistic_guy]

Then, solve it! What are you waiting for?

i can't i don't know how to solve it

I am better than you. I know how to solve:

$\displaystyle u'' - k^2u = 0$
$\displaystyle u'' - k^2u = 1$
$\displaystyle u'' - k^2u = x$
$\displaystyle u'' - k^2u = x^2 + e^x$
$\displaystyle u'' - k^2u = \tan x$
$\displaystyle u'' - k^2u = x\sin x$
$\displaystyle u'' - k^2u = f(x)$

i need the solution of the last one

 

[mario99]

i can't i don't know how to solve it


i need the solution of the last one

Did you read about my suggestion in post #7? If no, go and read about how to solve differential equations by Dirac delta function or by variation of parameters. If you don't know how to use either of these methods, it will be pointless (And waste of time) to explain the solution of the op differential equation because eventually you will give up as what happened in the complex analysis problem.

Have you at least looked at the solution in W|A?

 

[logistic_guy]

Did you read about my suggestion in post #7? If no, go and read about how to solve differential equations by Dirac delta function or by variation of parameters. If you don't know how to use either of these methods, it will be pointless (And waste of time) to explain the solution of the op differential equation because eventually you will give up as what happened in the complex analysis problem.

thank.

i think i'll spend couple of hours reading this. do you have the textbook or the website to teach this method?

Have you at least looked at the solution in W|A?

what is W|A?

 

[khansaheb]

what is W|A?

WolframAlfa.com

 

[mario99]

thank.

i think i'll spend couple of hours reading this. do you have the textbook or the website to teach this method?

Google it.

Or

Read your differential equations' book. I think that any book in differential equations will explain these two methods. (You have one, aren't you?)

This is the best video in the internet that explains the concept of Dirac delta function in differential equations. And it tells you exactly that you are just solving a green function. This video solves your OP perfectly, it just uses different boundary conditions.

what is W|A?

To know what is that, you will have to read post #13 given by professor Khan.

 

[logistic_guy]

WolframAlfa.com

thank

so it's the website

Google it.

Or

Read your differential equations' book. I think that any book in differential equations will explain these two methods. (You have one, aren't you?)

i've a textbook but i don't remember this explanation

This is the best video in the internet that explains the concept of Dirac delta function in differential equations. And it tells you exactly that you are just solving a green function. This video solves your OP perfectly, it just uses different boundary conditions.

thank

if this video solve my question i'll watch it right now

 

[logistic_guy]

i wached the video 3 times but i don't understand anything

the website W|A give me this

 

[mario99]

i wached the video 3 times but i don't understand anything

How?!



The solution of WolframAlpha proves that it will be very difficult to find the solution of the original problem. But believe it or not, with a little manipulation and by using the properties of the hyperbolic function, all this complexity in the picture can be combined (simplified) into one integral with only two functions, $f(\zeta)$ and $G(x,\zeta)$. (The same as the hint in the OP.)

We are not concerned to simplify this monster expression now, but rather we will try to find it or to find an easier version of it.

The video actually has two main ideas. The first idea is that it is telling you (at time: 1:22)

this differential equation:

$u'' - k^2 u = 0$

is almost similar to this differential equation:

$u'' - k^2 u = \delta(x - s)$

except at $s$

All of this only means that the solution that you have found in the OP will have four constants instead of two, and it must be written as a piecewise function like this:

$\displaystyle u(x) =\begin{cases}Ae^{kx} + Be^{-kx} & \ \ \text{if} \ x < s\\Ce^{kx} + De^{-kx} & \ \ \text{if} \ x > s\end{cases}$

Now, we have a problem. We have four unknown constants, but we only have two boundary conditions! How are we going to find the other two unknown constants? Here comes the other idea of the video! Can you find it?

Note: The piecewise solution is not the solution to the original differential equation, but rather it is the solution to the green function $G(x,s)$.

I should have written it like this:

$G'' - k^2G = \delta(x -s)$
$G(a) = G(b) = 0$

Then

$\displaystyle G(x) =\begin{cases}Ae^{kx} + Be^{-kx} & \ \ \text{if} \ x < s\\Ce^{kx} + De^{-kx} & \ \ \text{if} \ x > s\end{cases}$

And when we find the constants, it will be $\displaystyle G(x,s)$.

 

[logistic_guy]

this is too much information. give me some time to read it and understand it

hold on mario99, i'll get back to you soon

 

[khansaheb]

....this is too much information.

That is exactly why serious students attend classes in universities.......

 

[logistic_guy]

this is too much information. give me some time to read it and understand it

hold on mario99, i'll get back to you soon

i'm back

i think i'm understanding. it give 4 euqations in the video to find the 4 constants. i just don't understand what he did in the 3rd and 4th equatons

That is exactly why serious students attend classes in universities.......

i'm serious student. do you think i'm not?