very differential equation with not known source function

[khansaheb]

thank

how to do that?

Using ALGEBRA - most of the time!!

 

[khansaheb]

thank

how to do that?

In response #38 you have calculated the general form of u(x) and you want to calculate the validity of that solution. Given bdy conditions are

u(0)=u(1)=0

So calculate u(0) and u(1) → do those match given bdy. condition u(0)=u(1)=0? .....................................(1)

If those do not match - then you have made a mistake somewhere. In that case, start over.

Now derive u'(x) and u"(x) ← use those in given original DE . Does u(x) satisfy the given DE? .............(2)

If the answers to both (1) & (2) are affirmative, then most probably your solution is correct.

 

[mario99]

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
\(\displaystyle 0 = c_1e^{2} + c_2e^{-2} - \frac{1}{4}\)
\(\displaystyle c_2 = \frac{1}{4} - c_1\)
\(\displaystyle 0 = c_1e^{2} + (\frac{1}{4} - c_1)e^{-2} - \frac{1}{4}\)
\(\displaystyle 0 = c_1e^{2} + \frac{1}{4}e^{-2} - c_1e^{-2} - \frac{1}{4}\)
\(\displaystyle -\frac{1}{4}e^{-2} + \frac{1}{4} = c_1e^{2} - c_1e^{-2}\)
\(\displaystyle \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}} = c_1\)
\(\displaystyle c_2 = \frac{1}{4} - \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}}\)
\(\displaystyle u(x) = (\frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}})e^{2x} + (\frac{1}{4} - \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}})e^{-2x} - \frac{1}{4}\)

According to Mrs. Alpha, the solution is:

[imath]\displaystyle u(x) = -\frac{-e^{2-2x}-e^{2x} + 1 + e^{2}}{4 + 4e^{2}}[/imath]

Let us compare this with your solution. According to Mrs. Alpha, your solution is:

[imath]\displaystyle u(x) = \left(\frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}}\right)e^{2x} + \left(\frac{1}{4} - \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}}\right)e^{-2x} - \frac{1}{4} = \frac{e^{2x} + e^{2(1-x)} - 1 - e^{2}}{4(1 + e^{2})}[/imath]

which is the same as the first solution. Bravo

Now, let us use our green function solution in post #31. Notice that [imath]a = 0, b = 1, \ \text{and} \ f(s) = 1[/imath]. According to Mrs. Alpha, the solution is:

[imath]\displaystyle u(x) = \int_{a}^{b} G(x,s) f(s) \ ds = \int_{0}^{1} G(x,s) \ ds[/imath]

[imath]\displaystyle = -\frac{1}{2\sinh(2[1 - 0])} \left(\int_{0}^{x} \sinh(2[1 - x])\sinh(2[s - 0]) \ ds + \int_{x}^{1} \sinh(2[x - 0])\sinh(2[1 - s]) \ ds\right)[/imath]


[imath]\displaystyle = \frac{(e^{x} - e^{2 - x}) \sinh(x)}{2 (1 + e^{2})} = \frac{e^{2-2x} + e^{2x} - 1 - e^{2}}{4(1 + e^{2})}[/imath]

You can even solve this integral by hand if you want. The conclusion is that our green function solution is correct. If you notice that our green function solution [imath]G(x,s)[/imath] is independent from the source function [imath]f(s)[/imath] which makes it so powerful to solve instantly any nonhomogeneous differential equation of this type [imath]u''(x) - k^2u(x) = f(x)[/imath], [imath]a < x < b[/imath].

It will take some time to understand what this does mean, but this is the main idea of Green Function in differential equations.

 

[logistic_guy]

Using ALGEBRA - most of the time!!

i'm very good in algebra

In response #38 you have calculated the general form of u(x) and you want to calculate the validity of that solution. Given bdy conditions are

u(0)=u(1)=0

So calculate u(0) and u(1) → do those match given bdy. condition u(0)=u(1)=0? .....................................(1)

If those do not match - then you have made a mistake somewhere. In that case, start over.

Now derive u'(x) and u"(x) ← use those in given original DE . Does u(x) satisfy the given DE? .............(2)

If the answers to both (1) & (2) are affirmative, then most probably your solution is correct.

can you please show me how you'll do it to post 31?

According to Mrs. Alpha, the solution is:

[imath]\displaystyle u(x) = -\frac{-e^{2-2x}-e^{2x} + 1 + e^{2}}{4 + 4e^{2}}[/imath]

Let us compare this with your solution. According to Mrs. Alpha, your solution is:

[imath]\displaystyle u(x) = \left(\frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}}\right)e^{2x} + \left(\frac{1}{4} - \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}}\right)e^{-2x} - \frac{1}{4} = \frac{e^{2x} + e^{2(1-x)} - 1 - e^{2}}{4(1 + e^{2})}[/imath]

which is the same as the first solution. Bravo

Now, let us use our green function solution in post #31. Notice that [imath]a = 0, b = 1, \ \text{and} \ f(s) = 1[/imath]. According to Mrs. Alpha, the solution is:

[imath]\displaystyle u(x) = \int_{a}^{b} G(x,s) f(s) \ ds = \int_{0}^{1} G(x,s) \ ds[/imath]

[imath]\displaystyle = -\frac{1}{2\sinh(2[1 - 0])} \left(\int_{0}^{x} \sinh(2[1 - x])\sinh(2[s - 0]) \ ds + \int_{x}^{1} \sinh(2[x - 0])\sinh(2[1 - s]) \ ds\right)[/imath]


[imath]\displaystyle = \frac{(e^{x} - e^{2 - x}) \sinh(x)}{2 (1 + e^{2})} = \frac{e^{2-2x} + e^{2x} - 1 - e^{2}}{4(1 + e^{2})}[/imath]

You can even solve this integral by hand if you want. The conclusion is that our green function solution is correct. If you notice that our green function solution [imath]G(x,s)[/imath] is independent from the source function [imath]f(s)[/imath] which makes it so powerful to solve instantly any nonhomogeneous differential equation of this type [imath]u''(x) - k^2u(x) = f(x)[/imath], [imath]a < x < b[/imath].

It will take some time to understand what this does mean, but this is the main idea of Green Function in differential equations.

thank very much mario99

i think i'm understanding. your solution is correct. can you please show me variation of parameters?

 

[mario99]

thank very much mario99

i think i'm understanding. your solution is correct. can you please show me variation of parameters?

Show you variation of parameters?! Do you mean that you wanna solve the OP by variation of parameters?

 

[logistic_guy]

Show you variation of parameters?! Do you mean that you wanna solve the OP by variation of parameters?

yes

 

[khansaheb]

can you please show me how you'll do it to post 31?

So calculate u(0) and u(1) → do those match given bdy. condition u(0)=u(1)=0? .....................................(1)

Did you calculate u(0) & u(1) by replacing 'x' the expression derived in post 31 by '0'

and evaluating u(0) using algebra​

​

Did you calculate u(0) & u(1) by replacing 'x' the expression derived in post 31 by '1'

and evaluating u(1) using algebra​

​

Please share "detailed" work......

 

[mario99]

yes

What do you know about variation of parameters?

 

[logistic_guy]

Did you calculate u(0) & u(1) by replacing 'x' the expression derived in post 31 by '0'

and evaluating u(0) using algebra​

​

Did you calculate u(0) & u(1) by replacing 'x' the expression derived in post 31 by '1'

and evaluating u(1) using algebra​

​

Please share "detailed" work......

i'm not understand what you're sayingcan you please show me what you mean?

What do you know about variation of parameters?

nothing

 

[mario99]

nothing

Know something, then we will have a discussion!

 

[logistic_guy]

Know something, then we will have a discussion!

i think i'll sepnd couple of days reading variation of parameters

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

 

[logistic_guy]

i think i'll sepnd couple of days reading variation of parameters

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

i'm back

i think i've some idea about variation of parameters. in post number 1 my original post i solved the homogeneous equation to get
\(\displaystyle u(x) = c_1e^{kx} + c_2e^{-kx}\)

variation of parameters don't want me to write the constants \(\displaystyle c_1\) and \(\displaystyle c_2\) it want me to write two functions

\(\displaystyle u(x) = A(x)e^{kx} + B(x)e^{-kx}\)

 

[mario99]

i'm back

i think i've some idea about variation of parameters. in post number 1 my original post i solved the homogeneous equation to get
\(\displaystyle u(x) = c_1e^{kx} + c_2e^{-kx}\)

variation of parameters don't want me to write the constants \(\displaystyle c_1\) and \(\displaystyle c_2\) it want me to write two functions

\(\displaystyle u(x) = A(x)e^{kx} + B(x)e^{-kx}\)

Yes, this is one of the main ideas, but you forgot to show how variation of parameters would solve [imath]A(x)[/imath] and [imath]B(x)[/imath]!

And for simplicity, let the solution be [imath]u(x) = A(x)\cosh kx + B(x)\sinh kx[/imath].

Don't panic! They are the same solution, just written in different form and we have already shown that in previous posts.

 

[logistic_guy]

Yes, this is one of the main ideas, but you forgot to show how variation of parameters would solve [imath]A(x)[/imath] and [imath]B(x)[/imath]!

And for simplicity, let the solution be [imath]u(x) = A(x)\cosh kx + B(x)\sinh kx[/imath].

Don't panic! They are the same solution, just written in different form and we have already shown that in previous posts.

i don't panic. i think we use deruvative and integration to find the functions

 

[mario99]

i don't panic. i think we use deruvative and integration to find the functions

Still you did not show how variation of parameters will help you solve [imath]A(x)[/imath] and [imath]B(x)[/imath]. I will not make your life difficult and I will give you a big shortcut.

Variation of parameters will help you get these two equations:

[imath]A'(x)\cosh kx + B'(x)\sinh kx = 0[/imath]

[imath]kA'(x)\sinh kx + kB'(x)\cosh kx = f(x)[/imath]

Now solve for [imath]A'(x)[/imath] and [imath]B'(x)[/imath]. Then, you can get [imath]A(x)[/imath] and [imath]B(x)[/imath].

 

[logistic_guy]

Still you did not show how variation of parameters will help you solve [imath]A(x)[/imath] and [imath]B(x)[/imath]. I will not make your life difficult and I will give you a big shortcut.

Variation of parameters will help you get these two equations:

[imath]A'(x)\cosh kx + B'(x)\sinh kx = 0[/imath]

[imath]kA'(x)\sinh kx + kB'(x)\cosh kx = f(x)[/imath]

Now solve for [imath]A'(x)[/imath] and [imath]B'(x)[/imath]. Then, you can get [imath]A(x)[/imath] and [imath]B(x)[/imath].

thank

\(\displaystyle Z = \cosh kx\)
\(\displaystyle F = \sinh kx\)
\(\displaystyle G = f(x)\)



\(\displaystyle A'(x) = \frac{FG}{F^2k - Z^2k} = \frac{\sinh kx f(x)}{\sinh^2 kx k - \cosh^2 kx k}\)

\(\displaystyle B'(x) = \frac{-GZ}{F^2k - Z^2k} = \frac{-f(x)\cosh kx}{\sinh^2 kx k - \cosh^2 kx k}\)



is this correct?

 

[mario99]

is this correct?

Yes correct. But it was not necessary to use a calculator for solving a system of two equations.

With simplification, your result will be:

[imath]\displaystyle A'(x) = -\frac{f(x)\sinh kx}{k}[/imath]

[imath]\displaystyle B'(x) = \frac{f(x)\cosh kx}{k}[/imath]

As [imath]\cosh^2 kx - \sinh^2 kx = 1[/imath]

If we take the integral of both sides to each result above, we will get:

[imath]\displaystyle A(x) = \int_{\alpha}^{x}-\frac{f(s)\sinh ks}{k} \ ds[/imath]

[imath]\displaystyle B(x) = \int_{\beta}^{x}\frac{f(s)\cosh ks}{k} \ ds[/imath]

Where [imath]\alpha \ \text{and} \ \beta[/imath] are arbitrary constants. We will determine their value from the boundary conditions.

The solution so far to the original differential equation is:

[imath]\displaystyle u(x) = -\frac{\cosh kx}{k} \int_{\alpha}^{x}f(s)\sinh ks \ ds + \frac{\sinh kx}{k} \int_{\beta}^{x}f(s)\cosh ks \ ds[/imath]

Now apply the first boundary condition. What do you get?

 

[khansaheb]

is this correct?

Apply the given DE to the proposed solution .

• Does the proposed solution satisfy the DE?

If yes then - apply Bdy. and initial conditions to the solution.

• Does the proposed solution satisfy the bdy./initial condition?

If the answer is affirmative for both the questions, then the "proposed solution" is probably correct.