very differential equation with not known source function

[mario99]

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

Correct. He just used the properties of Green function.

Property I:
[imath]G'(s^{+}) - G'(s^{-}) = 1[/imath]

This just means that:

[imath]\displaystyle \lim_{x \rightarrow s^{+}} G'(x) - \lim_{x \rightarrow s^{-}} G'(x) = 1[/imath]

Property II:
[imath]G(s^{-}) = G(s^{+})[/imath]

This just means that:

[imath]\displaystyle \lim_{x \rightarrow s^{-}} G(x) = \lim_{x \rightarrow s^{+}} G(x)[/imath]

Note: The first property can be proved easily, but the second property requires variation of parameters which I think is beyond your math level.

If you are interested to know the proofs, tell me. All remains now is just to apply the four conditions. Watch the video again (time 9:30) and apply them one by one. I will correct you if you go wrong.

 

[logistic_guy]

Correct. He just used the properties of Green function.

Property I:
[imath]G'(s^{+}) - G'(s^{-}) = 1[/imath]

This just means that:

[imath]\displaystyle \lim_{x \rightarrow s^{+}} G'(x) - \lim_{x \rightarrow s^{-}} G'(x) = 1[/imath]

Property II:
[imath]G(s^{-}) = G(s^{+})[/imath]

This just means that:

[imath]\displaystyle \lim_{x \rightarrow s^{-}} G(x) = \lim_{x \rightarrow s^{+}} G(x)[/imath]

Note: The first property can be proved easily, but the second property requires variation of parameters which I think is beyond your math level.

If you are interested to know the proofs, tell me. All remains now is just to apply the four conditions. Watch the video again (time 9:30) and apply them one by one. I will correct you if you go wrong.

i'll try

\(\displaystyle Ae^{ka} + Be^{-ka} = 0\)
\(\displaystyle Ce^{kb} + De^{-kb} = 0\)
\(\displaystyle kCe^{ks^{+}} - kDe^{-ks^{+}} - kAe^{ks^{-}} - kBe^{-ks^{-}} = 1\)
\(\displaystyle Ce^{ks^{+}} + De^{-ks^{+}} - Ae^{ks^{-}} + Be^{-ks^{-}} = 0\)

 

[mario99]

i'll try

\(\displaystyle Ae^{ka} + Be^{-ka} = 0\)
\(\displaystyle Ce^{kb} + De^{-kb} = 0\)
\(\displaystyle kCe^{ks^{+}} - kDe^{-ks^{+}} - kAe^{ks^{-}} - kBe^{-ks^{-}} = 1\)
\(\displaystyle Ce^{ks^{+}} + De^{-ks^{+}} - Ae^{ks^{-}} + Be^{-ks^{-}} = 0\)

Almost correct. The video put brackets to not lose track of the signs, but you did not put. Also, don't substitute [imath]s^{+}[/imath] and [imath]s^{-}[/imath] when you do the limit. Just substitute the letter [imath]s[/imath] without any signs.

You should write it like this:

[imath]Ae^{ka} + Be^{-ka} = 0[/imath]
[imath]Ce^{kb} + De^{-kb} = 0[/imath]
[imath]kCe^{ks} - kDe^{-ks} - (kAe^{ks} - kBe^{-ks}) = 1[/imath]
[imath]Ce^{ks} + De^{-ks} - (Ae^{ks} + Be^{-ks}) = 0[/imath]

After simplifying, we will get:

[imath]Ae^{ka} + Be^{-ka} = 0[/imath]
[imath]Ce^{kb} + De^{-kb} = 0[/imath]
[imath]kCe^{ks} - kDe^{-ks} - kAe^{ks} + kBe^{-ks} = 1[/imath]
[imath]Ce^{ks} + De^{-ks} - Ae^{ks} - Be^{-ks} = 0[/imath]

Now, you have 4 equations with 4 unknowns. It will be tedious to solve the equations by hand, so it is better to use a calculator. Before we solve these four equations, I will show a trick that can solve green function instantly. (almost instantly.)

I will go back to our green function solution:

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

This solution can be written with hyperbolic functions like this:

[imath]\displaystyle G(x) =\begin{cases}A\sinh kx + B\cosh kx & \ \ \text{if} \ x < s\\C\sinh kx + D\cosh kx & \ \ \text{if} \ x > s\end{cases} [/imath]

I am lazy to use different constants, so I will use the same constants.

The trick will allow us to get rid of two constants which will make it very easy to find the remaining two constants.

Let us assume that [imath]A = 0[/imath] and let us apply the first boundary condition.

[imath]B\cosh ka = 0[/imath]

It is either [imath]B = 0[/imath] or [imath]\cosh ka = 0[/imath].

If you are familiar with the graph of the function [imath]\cosh x[/imath], you will know that it is impossible [imath]\cosh ka = 0[/imath], so it means that [imath]B = 0[/imath] which gives us the trivial solution [imath]G(x) = 0[/imath]. Our assumption was wrong.

Now let us assume that [imath]B = 0[/imath] and let us apply the first boundary condition again.

[imath]A\sinh ka = 0[/imath]

It is either [imath]A = 0[/imath] or [imath]\sinh ka = 0[/imath]. If [imath]A = 0[/imath], we will get the trivial solution [imath]G(x) = 0[/imath], so it must be [imath]\sinh ka = 0[/imath]. [imath]\sinh ka = 0[/imath] only when [imath]ka = 0[/imath]. We know [imath]k \neq 0[/imath], so it must be [imath]a = 0[/imath]. But this means that our interval will always start at zero. We want our interval to start at any number, not only zero. What should we do?

The solution is to introduce two new arbitrary constants. Let us write our solution so far.

[imath]\displaystyle G(x) =\begin{cases}A\sinh kx & \ \ \text{if} \ x < s\\C\sinh kx & \ \ \text{if} \ x > s\end{cases} [/imath]

We can write [imath]A\sinh kx[/imath] as:

[imath]\displaystyle \frac{A}{2}(e^{kx} - e^{-kx})[/imath]

Let us introduce our two new arbitrary constants, [imath]m[/imath] and [imath]n[/imath].

[imath]\displaystyle \frac{A}{2}(me^{kx} - ne^{-kx})[/imath]

Let us define [imath]\displaystyle m = e^{-ka}[/imath] and [imath]\displaystyle n = e^{ka}[/imath].

[imath]\displaystyle \frac{A}{2}(e^{-ka}e^{kx} - e^{ka}e^{-kx}) = \frac{A}{2}(e^{k[x-a]} - e^{-k[x - a]}) = A\sinh(x - a)[/imath].

A similar process will be done to [imath]C\sinh kx[/imath].
We are allowed to introduce any arbitrary constants and define them with our choice as long as the main conditions are satisfied.

Our new and enhanced solution is now:

[imath]\displaystyle G(x) =\begin{cases}A\sinh(k[x-a]) & \ \ \text{if} \ x < s\\C\sinh(k[b - x]) & \ \ \text{if} \ x > s\end{cases} [/imath]

Now the boundary conditions are satisfied and we were able to get rid of two constants. Also we are free to choose any interval.

Let us apply green function properties:

[imath]-Ck\cosh(k[b - s]) - Ak\cosh(k[s - a]) = 1[/imath]

[imath]A\sinh(k[s - a]) = C\sinh(k[b - s])[/imath]


Can you find the constants [imath]A[/imath] and [imath]C[/imath]?

 

[logistic_guy]

Almost correct. The video put brackets to not lose track of the signs, but you did not put. Also, don't substitute [imath]s^{+}[/imath] and [imath]s^{-}[/imath] when you do the limit. Just substitute the letter [imath]s[/imath] without any signs.

You should write it like this:

[imath]Ae^{ka} + Be^{-ka} = 0[/imath]
[imath]Ce^{kb} + De^{-kb} = 0[/imath]
[imath]kCe^{ks} - kDe^{-ks} - (kAe^{ks} - kBe^{-ks}) = 1[/imath]
[imath]Ce^{ks} + De^{-ks} - (Ae^{ks} + Be^{-ks}) = 0[/imath]

After simplifying, we will get:

[imath]Ae^{ka} + Be^{-ka} = 0[/imath]
[imath]Ce^{kb} + De^{-kb} = 0[/imath]
[imath]kCe^{ks} - kDe^{-ks} - kAe^{ks} + kBe^{-ks} = 1[/imath]
[imath]Ce^{ks} + De^{-ks} - Ae^{ks} - Be^{-ks} = 0[/imath]

Now, you have 4 equations with 4 unknowns. It will be tedious to solve the equations by hand, so it is better to use a calculator. Before we solve these four equations, I will show a trick that can solve green function instantly. (almost instantly.)

I will go back to our green function solution:

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

This solution can be written with hyperbolic functions like this:

[imath]\displaystyle G(x) =\begin{cases}A\sinh kx + B\cosh kx & \ \ \text{if} \ x < s\\C\sinh kx + D\cosh kx & \ \ \text{if} \ x > s\end{cases} [/imath]

I am lazy to use different constants, so I will use the same constants.

The trick will allow us to get rid of two constants which will make it very easy to find the remaining two constants.

Let us assume that [imath]A = 0[/imath] and let us apply the first boundary condition.

[imath]B\cosh ka = 0[/imath]

It is either [imath]B = 0[/imath] or [imath]\cosh ka = 0[/imath].

If you are familiar with the graph of the function [imath]\cosh x[/imath], you will know that it is impossible [imath]\cosh ka = 0[/imath], so it means that [imath]B = 0[/imath] which gives us the trivial solution [imath]G(x) = 0[/imath]. Our assumption was wrong.

Now let us assume that [imath]B = 0[/imath] and let us apply the first boundary condition again.

[imath]A\sinh ka = 0[/imath]

It is either [imath]A = 0[/imath] or [imath]\sinh ka = 0[/imath]. If [imath]A = 0[/imath], we will get the trivial solution [imath]G(x) = 0[/imath], so it must be [imath]\sinh ka = 0[/imath]. [imath]\sinh ka = 0[/imath] only when [imath]ka = 0[/imath]. We know [imath]k \neq 0[/imath], so it must be [imath]a = 0[/imath]. But this means that our interval will always start at zero. We want our interval to start at any number, not only zero. What should we do?

The solution is to introduce two new arbitrary constants. Let us write our solution so far.

[imath]\displaystyle G(x) =\begin{cases}A\sinh kx & \ \ \text{if} \ x < s\\C\sinh kx & \ \ \text{if} \ x > s\end{cases} [/imath]

We can write [imath]A\sinh kx[/imath] as:

[imath]\displaystyle \frac{A}{2}(e^{kx} - e^{-kx})[/imath]

Let us introduce our two new arbitrary constants, [imath]m[/imath] and [imath]n[/imath].

[imath]\displaystyle \frac{A}{2}(me^{kx} - ne^{-kx})[/imath]

Let us define [imath]\displaystyle m = e^{-ka}[/imath] and [imath]\displaystyle n = e^{ka}[/imath].

[imath]\displaystyle \frac{A}{2}(e^{-ka}e^{kx} - e^{ka}e^{-kx}) = \frac{A}{2}(e^{k[x-a]} - e^{-k[x - a]}) = A\sinh(x - a)[/imath].

A similar process will be done to [imath]C\sinh kx[/imath].
We are allowed to introduce any arbitrary constants and define them with our choice as long as the main conditions are satisfied.

Our new and enhanced solution is now:

[imath]\displaystyle G(x) =\begin{cases}A\sinh(k[x-a]) & \ \ \text{if} \ x < s\\C\sinh(k[b - x]) & \ \ \text{if} \ x > s\end{cases} [/imath]

Now the boundary conditions are satisfied and we were able to get rid of two constants. Also we are free to choose any interval.

Let us apply green function properties:

[imath]-Ck\cosh(k[b - s]) - Ak\cosh(k[s - a]) = 1[/imath]

[imath]A\sinh(k[s - a]) = C\sinh(k[b - s])[/imath]


Can you find the constants [imath]A[/imath] and [imath]C[/imath]?

thank

this is too much information

can we first solve the 4 equations then i'll read your trick? i don't want to lose concentration

i appreciate your efforts

 

[mario99]

thank

this is too much information

can we first solve the 4 equations then i'll read your trick? i don't want to lose concentration

i appreciate your efforts

Yeah, sure.

You can use this website to solve the system:

Matrix calculator

Matrix addition, multiplication, inversion, determinant and rank calculation, transposing, bringing to diagonal, row echelon form, exponentiation, LU Decomposition, QR-decomposition, Singular Value Decomposition (SVD), solving of systems of linear equations with solution steps

matrixcalc.org

On the left, you will see System of equations calculator, click on it. Then, change [imath]x_1, x_2, x_3, \text{and} \ x_4 \ \text{to} \ A, B, C, \text{and} \ D[/imath], respectively. The calculator will not accept entries like [imath]e^{ka}[/imath], so you will have to replace it with a letter. Don't use the letter [imath]E[/imath], it has a special meaning in that calculator.

I prefer to choose these letters:

[imath]Z = e^{ka}[/imath]
[imath]F = e^{-ka}[/imath]
[imath]G = e^{kb}[/imath]
[imath]H = e^{-kb}[/imath]
[imath]M = e^{ks}[/imath]
[imath]N = e^{-ks}[/imath]

Solve the system, and let me see what you will get.

 

[logistic_guy]

is this correct?

\(\displaystyle A = \frac{FHM - FGN}{2FGMNk - 2HMNZk}\)

\(\displaystyle B = \frac{-HMZ + GNZ}{2FGMNk - 2HMNZk}\)

\(\displaystyle C = \frac{FHM - HNZ}{2FGMNk - 2HMNZk}\)

\(\displaystyle D = \frac{-FGM + GNZ}{2FGMNk - 2HMNZk}\)

 

[mario99]

is this correct?

From the set up of the matrix, I guess yes. If there is any error, we will know about it during simplification. (Errors will make our life difficult to simplify!)

Let us substitute back the values of [imath]A[/imath] from post #25.


[imath]\displaystyle A = \frac{FHM - FGN}{2FGMNk - 2HMNZk} = \frac{e^{-ka}e^{-kb}e^{ks} - e^{-ka}e^{kb}e^{-ks}}{2e^{-ka}e^{kb}e^{ks}e^{-ks}k - 2e^{-kb}e^{ks}e^{-ks}e^{ka}k}[/imath]


Let us try to simplify this expression. If you look at the denominator, you will see:

[imath]MN = e^{ks}e^{-ks} = e^{ks - ks} = e^{0} = 1[/imath]. The expression will be now:


[imath]\displaystyle A = \frac{e^{-ka}e^{-kb}e^{ks} - e^{-ka}e^{kb}e^{-ks}}{2e^{-ka}e^{kb}k - 2e^{-kb}e^{ka}k}[/imath]


We know [imath]\sinh x[/imath] has this form:

[imath]\displaystyle \sinh x = \frac{1}{2}(e^{x} - e^{-x})[/imath]

Or

[imath]\displaystyle 2\sinh x = e^{x} - e^{-x}[/imath]


Look at the denominator again, you will see this expression:

[imath]\displaystyle 2e^{-ka}e^{kb}k - 2e^{-kb}e^{ka}k = 2k(e^{k[b - a]} - e^{-k[b - a]}) = 4k\sinh(k[b - a])[/imath]


[imath]\displaystyle A = \frac{e^{-ka}e^{-kb}e^{ks} - e^{-ka}e^{kb}e^{-ks}}{4k\sinh(k[b - a])}[/imath]


Now we will try to simplify the numerator. We have this expression in the numerator:

[imath]\displaystyle e^{-ka}e^{-kb}e^{ks} - e^{-ka}e^{kb}e^{-ks} = e^{-ka}(e^{k[s - b]} - e^{-k[s - b]}) = 2e^{-ka}\sinh(k[s - b])[/imath]


[imath]\displaystyle A = \frac{2e^{-ka}\sinh(k[s - b])}{4k\sinh(k[b - a])} = \frac{e^{-ka}\sinh(k[s - b])}{2k\sinh(k[b - a])}[/imath]


We have simplified the constant [imath]A[/imath] successfuly. This means that you have set up the matrix correctly. I will simplify [imath]B[/imath] and you simplify [imath]C[/imath] and [imath]D[/imath].

Don't panic, it is not difficult to simplify. Once you see the structure of the [imath]\sinh x[/imath] function, you are done! Just follow what I have done.

 

[logistic_guy]

From the set up of the matrix, I guess yes. If there is any error, we will know about it during simplification. (Errors will make our life difficult to simplify!)

Let us substitute back the values of [imath]A[/imath] from post #25.


[imath]\displaystyle A = \frac{FHM - FGN}{2FGMNk - 2HMNZk} = \frac{e^{-ka}e^{-kb}e^{ks} - e^{-ka}e^{kb}e^{-ks}}{2e^{-ka}e^{kb}e^{ks}e^{-ks}k - 2e^{-kb}e^{ks}e^{-ks}e^{ka}k}[/imath]


Let us try to simplify this expression. If you look at the denominator, you will see:

[imath]MN = e^{ks}e^{-ks} = e^{ks - ks} = e^{0} = 1[/imath]. The expression will be now:


[imath]\displaystyle A = \frac{e^{-ka}e^{-kb}e^{ks} - e^{-ka}e^{kb}e^{-ks}}{2e^{-ka}e^{kb}k - 2e^{-kb}e^{ka}k}[/imath]


We know [imath]\sinh x[/imath] has this form:

[imath]\displaystyle \sinh x = \frac{1}{2}(e^{x} - e^{-x})[/imath]

Or

[imath]\displaystyle 2\sinh x = e^{x} - e^{-x}[/imath]


Look at the denominator again, you will see this expression:

[imath]\displaystyle 2e^{-ka}e^{kb}k - 2e^{-kb}e^{ka}k = 2k(e^{k[b - a]} - e^{-k[b - a]}) = 4k\sinh(k[b - a])[/imath]


[imath]\displaystyle A = \frac{e^{-ka}e^{-kb}e^{ks} - e^{-ka}e^{kb}e^{-ks}}{4k\sinh(k[b - a])}[/imath]


Now we will try to simplify the numerator. We have this expression in the numerator:

[imath]\displaystyle e^{-ka}e^{-kb}e^{ks} - e^{-ka}e^{kb}e^{-ks} = e^{-ka}(e^{k[s - b]} - e^{-k[s - b]}) = 2e^{-ka}\sinh(k[s - b])[/imath]


[imath]\displaystyle A = \frac{2e^{-ka}\sinh(k[s - b])}{4k\sinh(k[b - a])} = \frac{e^{-ka}\sinh(k[s - b])}{2k\sinh(k[b - a])}[/imath]


We have simplified the constant [imath]A[/imath] successfuly. This means that you have set up the matrix correctly. I will simplify [imath]B[/imath] and you simplify [imath]C[/imath] and [imath]D[/imath].

Don't panic, it is not difficult to simplify. Once you see the structure of the [imath]\sinh x[/imath] function, you are done! Just follow what I have done.

thank

i'm very excited. give me some time to read this

thank again mario99

 

[mario99]

thank

i'm very excited. give me some time to read this

thank again mario99

You are welcome.

We have simplified the constant [imath]A[/imath] successfuly. This means that you have set up the matrix correctly. I will simplify [imath]B[/imath] and you simplify [imath]C[/imath] and [imath]D[/imath].

I will do my homework now.

[imath]\displaystyle B = \frac{GNZ - HMZ}{2FGMNk - 2HMNZk} = \frac{e^{kb}e^{-ks}e^{ka} - e^{-kb}e^{ks}e^{ka}}{2e^{-ka}e^{kb}e^{ks}e^{-ks}k - 2e^{-kb}e^{ks}e^{-ks}e^{ka}k}[/imath]


All of the 4 constants have the same denominator, so we have this result from post #27:

[imath]\displaystyle B = \frac{e^{kb}e^{-ks}e^{ka} - e^{-kb}e^{ks}e^{ka}}{4k\sinh(k[b - a])}[/imath]


The expression in the numerator is:

[imath]\displaystyle e^{kb}e^{-ks}e^{ka} - e^{-kb}e^{ks}e^{ka} = e^{ka}(e^{k[ b - s]} - e^{-k[b - s]}) = 2e^{ka}\sinh(k[b - s])[/imath]


[imath]\displaystyle B = \frac{2e^{ka}\sinh(k[b - s])}{4k\sinh(k[b - a])} = \frac{e^{ka}\sinh(k[b - s])}{2k\sinh(k[b - a])}[/imath]


It is your turn now. I will be waiting to see your simplification.

 

[logistic_guy]

i'm back

\(\displaystyle C = \frac{FHM - HNZ}{2FGMNk - 2HMNZk} = \frac{e^{-ka}e^{-kb}e^{ks} - e^{-kb}e^{-ks}e^{ka}}{4k\sinh(k[b - a])}\)


\(\displaystyle e^{-ka}e^{-kb}e^{ks} - e^{-kb}e^{-ks}e^{ka} = e^{-kb}(e^{k[s - a]} - e^{-k[s - a]}) = 2e^{-kb}\sinh(k[s - a])\)


\(\displaystyle C = \frac{2e^{-kb}\sinh(k[s - a])}{4k\sinh(k[b - a])} = \frac{e^{-kb}\sinh(k[s - a])}{2k\sinh(k[b - a])}\)


\(\displaystyle D = \frac{-FGM + GNZ}{2FGMNk - 2HMNZk} = \frac{-e^{-ka}e^{kb}e^{ks} + e^{kb}e^{-ks}e^{ka}}{4k\sinh(k[b - a])}\)

i'll reverse the numator to change the negative sign

\(\displaystyle D = \frac{e^{kb}e^{-ks}e^{ka} - e^{-ka}e^{kb}e^{ks}}{4k\sinh(k[b - a])}\)


\(\displaystyle e^{kb}e^{-ks}e^{ka} - e^{-ka}e^{kb}e^{ks} = e^{kb}(e^{k[a - s]} - e^{-k[a - s]}) = 2e^{kb}\sinh(k[a - s])\)


\(\displaystyle D = \frac{2e^{kb}\sinh(k[a - s])}{4k\sinh(k[b - a])} = \frac{e^{kb}\sinh(k[a - s])}{2k\sinh(k[b - a])}\)

i'm sorry to cheat i copy and paste your method exactlyi'm spend too much time in the simplification so i think i'm correct

 

[mario99]

i'm sorry to cheat i copy and paste your method exactly

Don't be sorry. This is the main idea, you copy, understand, and learn.

i'm spend too much time in the simplification so i think i'm correct

It looks good. If there is an error, we will not be able to simplify!

Let us go back to the piecewise solution of Green function.

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

We will replace the constants with what we got.

[imath]\displaystyle \Large G(x,s) =\begin{cases}\left(\frac{e^{-ka}\sinh(k[s - b])}{2k\sinh(k[b - a])}\right)e^{kx} + \left(\frac{e^{ka}\sinh(k[b - s])}{2k\sinh(k[b - a])}\right)e^{-kx} & \ \ \text{if} \ x < s\\[10pt]\left(\frac{e^{-kb}\sinh(k[s - a])}{2k\sinh(k[b - a])}\right)e^{kx} + \left(\frac{e^{kb}\sinh(k[a - s])}{2k\sinh(k[b - a])}\right)e^{-kx} & \ \ \text{if} \ x > s\end{cases} [/imath]


Now we will use the property [imath]\sinh(-x) = -\sinh(x)[/imath] to simplify the expression further. Let us work on the first row of the solution.

[imath]\displaystyle \left(\frac{e^{-ka}\sinh(k[s - b])}{2k\sinh(k[b - a])}\right)e^{kx} + \left(\frac{e^{ka}\sinh(k[b - s])}{2k\sinh(k[b - a])}\right)e^{-kx}[/imath]


[imath]\displaystyle =\left(\frac{\sinh(k[s - b])}{2k\sinh(k[b - a])}\right)e^{-ka}e^{kx} - \left(\frac{\sinh(k[s - b])}{2k\sinh(k[b - a])}\right)e^{ka}e^{-kx}[/imath]


[imath]\displaystyle =\left(\frac{\sinh(k[s - b])}{2k\sinh(k[b - a])}\right)\left(e^{-ka}e^{kx} - e^{ka}e^{-kx}\right)[/imath]


[imath]\displaystyle =\left(\frac{\sinh(k[s - b])}{2k\sinh(k[b - a])}\right)\left(e^{k[x - a]} - e^{-k[x - a]}\right)[/imath]


[imath]\displaystyle =\left(\frac{\sinh(k[s - b])}{2k\sinh(k[b - a])}\right)2\sinh(k[x - a])[/imath]


[imath]\displaystyle =\frac{\sinh(k[x - a])\sinh(k[s - b])}{k\sinh(k[b - a])}[/imath]


We will do the same process to the second row, and our green function will be:

[imath]\displaystyle \Large G(x,s) =\begin{cases}\frac{\sinh(k[x - a])\sinh(k[s - b])}{k\sinh(k[b - a])} & \ \ \text{if} \ x < s\\[10pt]\frac{\sinh(k[x - b])\sinh(k[s - a])}{k\sinh(k[b - a])} & \ \ \text{if} \ x > s\end{cases} [/imath]


Since [imath]s - b[/imath] in the first row produces negative values, for convenience, we prefer to write it as [imath]-(b - s)[/imath]. Also in the second row for [imath]x - b = -(b - x)[/imath]. With these changes as well as writing the interval in its full version, our final answer for green function is:

[imath]\displaystyle \Large G(x,s) =\begin{cases}-\frac{\sinh(k[x - a])\sinh(k[b - s])}{k\sinh(k[b - a])} & \ \ \text{if} \ a \leq x < s\\[10pt]-\frac{\sinh(k[b - x])\sinh(k[s - a])}{k\sinh(k[b - a])} & \ \ \text{if} \ s < x \leq b \end{cases} [/imath]


And we have solved the OP problem successfully

[imath]\displaystyle \Large u(x) = \int_{a}^{b} G(x,s)f(s) \ ds[/imath], where [imath]\displaystyle \Large G(x,s)[/imath] is the green function above.

 

[logistic_guy]

thank very much mario99

this is too much information

give me time to read it

i appreciate it

 

[mario99]

thank very much mario99

this is too much information

give me time to read it

i appreciate it

And don't forget to read my trick in post #24.

 

[logistic_guy]

i'm back

i think i'm understanding your simplification. i also write it on paper myself to practice

you give me very nice solution

thank very much mario99

And don't forget to read my trick in post #24.

\(\displaystyle -Ck\cosh(k[b - s]) - Ak\cosh(k[s - a]) = 1\)

\(\displaystyle A\sinh(k[s - a]) = C\sinh(k[b - s])\)

i'll try to find the letters

\(\displaystyle A = \frac{C\sinh(k[b - s])}{\sinh(k[s - a])}\)

\(\displaystyle -Ck\cosh(k[b - s]) - \frac{C\sinh(k[b - s])}{\sinh(k[s - a])}k\cosh(k[s - a]) = 1\)

\(\displaystyle \sinh(k[s - a])Ck\cosh(k[b - s]) + C\sinh(k[b - s])k\cosh(k[s - a]) = -\sinh(k[s - a])\)

\(\displaystyle C = \frac{-\sinh(k[s - a])}{\sinh(k[s - a])k\cosh(k[b - s]) + \sinh(k[b - s])k\cosh(k[s - a])}\)

it don't match your answer post 31i'm sure my simplification correct

 

[mario99]

it don't match your answer post 31i'm sure my simplification correct

You need to simplify it further.

Let us take [imath]C[/imath] again.

\(\displaystyle C = \frac{-\sinh(k[s - a])}{\sinh(k[s - a])k\cosh(k[b - s]) + \sinh(k[b - s])k\cosh(k[s - a])}\)

It is tedious to simplify the denominator by hand, so I will ask Mrs. W|A.

According to this woman:

[imath]\sinh(k[s - a])k\cosh(k[b - s]) + \sinh(k[b - s])k\cosh(k[s - a]) = k\sinh(k[s - a] + k[b - s])[/imath]

which is just [imath]k\sinh(k[b - a])[/imath], therefore:

[imath]\displaystyle C = \frac{-\sinh(k[s - a])}{k\sinh(k[b - a])}[/imath]

Then

[imath]\displaystyle A = \frac{C\sinh(k[b - s])}{\sinh(k[s - a])} = \frac{\frac{-\sinh(k[s - a])}{k\sinh(k[b - a])}\sinh(k[b - s])}{\sinh(k[s - a])} = \frac{-\sinh(k[b - s])}{k\sinh(k[b - a])}[/imath]

which is the same answer that we have found by solving for the four constants, [imath]A, B, C, \ \text{and} \ D[/imath]. (But shorter!)



Believe it or not, the two methods that we have used were the boring way to solve the problem. The only real method is to solve the differential equation by variation of parameters. Why?

Because..
1. It is the standard way to solve such problems.
2. The two methods that we have used were derived from it.
3. It is a powerful method.
4. You don't need to know more about it than what you have already known!

If you don't know when to prefer one method over another, I will give you a scenario. Imagine that you have three friends, Tom, Jack, and George.

You use the trick method when you are with Tom because he does not like mathematics.
You use the Dirac delta method when you are with Jack because he likes to solve simultaneous equations.
You use variation of parameters when you are with George because he is fancy.

 

[logistic_guy]

thank

You need to simplify it further.

Let us take [imath]C[/imath] again.

It is tedious to simplify the denominator by hand, so I will ask Mrs. W|A.

According to this woman:

woman?

[imath]\sinh(k[s - a])k\cosh(k[b - s]) + \sinh(k[b - s])k\cosh(k[s - a]) = k\sinh(k[s - a] + k[b - s])[/imath]

which is just [imath]k\sinh(k[b - a])[/imath], therefore:

[imath]\displaystyle C = \frac{-\sinh(k[s - a])}{k\sinh(k[b - a])}[/imath]

Then

[imath]\displaystyle A = \frac{C\sinh(k[b - s])}{\sinh(k[s - a])} = \frac{\frac{-\sinh(k[s - a])}{k\sinh(k[b - a])}\sinh(k[b - s])}{\sinh(k[s - a])} = \frac{-\sinh(k[b - s])}{k\sinh(k[b - a])}[/imath]

i can't simplify but thank for showing that

Believe it or not, the two methods that we have used were the boring way to solve the problem. The only real method is to solve the differential equation by variation of parameters. Why?

Because..
1. It is the standard way to solve such problems.
2. The two methods that we have used were derived from it.
3. It is a powerful method.
4. You don't need to know more about it than what you have already known!

what is variation of parameters?

If you don't know when to prefer one method over another, I will give you a scenario. Imagine that you have three friends, Tom, Jack, and George.

You use the trick method when you are with Tom because he does not like mathematics.
You use the Dirac delta method when you are with Jack because he likes to solve simultaneous equations.
You use variation of parameters when you are with George because he is fancy.

you solve the question in 2 methods. i'v a question how to check if the solution is correct or wrong?

 

[mario99]

woman?

Yes, Mrs. Alpha.

what is variation of parameters?

One of the main methods for solving differential equations.

you solve the question in 2 methods. i'v a question how to check if the solution is correct or wrong?

If you were given this problem:

[imath]u''(x) - 4u(x) = 1[/imath]
[imath]0 < x < 1[/imath]
[imath]u(0) = u(1) = 0[/imath]

How would you solve it?

 

[logistic_guy]

If you were given this problem:

[imath]u''(x) - 4u(x) = 1[/imath]
[imath]0 < x < 1[/imath]
[imath]u(0) = u(1) = 0[/imath]

How would you solve it?

homegenous solution
\(\displaystyle u(x) = c_1e^{2x} + c_2e^{-2x}\)
particular solution
\(\displaystyle u(x) = a_0\)
\(\displaystyle 0 - 4u_0 = 1\)
\(\displaystyle u_0 = \frac{-1}{4}\)
general solution
\(\displaystyle u(x) = c_1e^{2x} + c_2e^{-2x} - \frac{1}{4}\)
first condition
\(\displaystyle 0 = c_1e^{0} + c_2e^{0} - \frac{1}{4}\)
\(\displaystyle c_1 + c_2 = \frac{1}{4}\)
second condition
\(\displaystyle 0 = c_1e^{2} + c_2e^{-2} - \frac{1}{4}\)
\(\displaystyle c_2 = \frac{1}{4} - c_1\)
\(\displaystyle 0 = c_1e^{2} + (\frac{1}{4} - c_1)e^{-2} - \frac{1}{4}\)
\(\displaystyle 0 = c_1e^{2} + \frac{1}{4}e^{-2} - c_1e^{-2} - \frac{1}{4}\)
\(\displaystyle -\frac{1}{4}e^{-2} + \frac{1}{4} = c_1e^{2} - c_1e^{-2}\)
\(\displaystyle \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}} = c_1\)
\(\displaystyle c_2 = \frac{1}{4} - \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}}\)
\(\displaystyle u(x) = (\frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}})e^{2x} + (\frac{1}{4} - \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}})e^{-2x} - \frac{1}{4}\)

 

[khansaheb]

if the solution is correct or wrong

Check whether the given bdyConditions and the initial conditions are satisfied and whether the given DE/s are satisfied with "your" derived expression.

 

[logistic_guy]

Check whether the given bdyConditions and the initial conditions are satisfied and whether the given DE/s are satisfied with "your" derived expression.

thank

how to do that?