# Solution of differential equation

1. May 3, 2009

### ssky

Hi, I want to ask how to solve this equation in this way?

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2. May 3, 2009

### HallsofIvy

Staff Emeritus
It's hard to respond without know what you already know. Do you know how to get the "characteristic equation" for a "linear, homogeneous, differential equation with constant coefficients" such as this?

For this differential equation, the characteristic equation is $r^2= -\lambda_i$ and has characteristic roots $r= \pm\lambda_i i$ (that second "i" is the imaginary unit). That, in turn means that the general solution to the differential equation is $\psi_i(x)= Ae^{i\lambda_i x}+ Ce^{-i\lambda_i x}= Ccos(\lambda_i x)+ D sin(\lambda_i x)$ where A, B, C, and D are constants. Do you understand how to convert from the complex exponential to sine and cosine? It is based on $e^{ix}= cos(x)+ i sin(x)$.

Now look at the boundary conditions. From $\psi_i(x)= Ccos(\lambda_i x)+ Dsin(\lambda_i x)$, we have $d\psi_i/dx= -C\lambda_i sin(\lambda_i x)+ D\lambda_i cos(\lambda_i x)$ so $d\psi_i/dx(0)= D\lambda_i = 0$.

If $\lambda_i= 0$ the differential equation would be just $d^2\psi_i/dx= 0$ which has general solution $\psi_i(x)= Ax+ B$ (just integrate twice) and then $d\psi/dx= A= 0$ in order to satisfy $d\psi/dx(0)= 0$. Then \psi_i(1)= B= 0[/itex] from the other boundary condition. That gives $\psi_i(x)$ identically 0. That is a solution but not a very interesting or useful one! That is called the "trivial" solution. If $\lambda_i\ne 0$, we must have D= 0 so that $\psi_i(x)= C cos(\lambda_i x)$.

Now $\psi_i(1)= C cos(\lambda_i)= 0$. That is satisfied by C= 0- but again, that would mean, with both C and D equal to 0, that $\psi_i(x)$ is identically 0. In order to have a "non trivial" solution, we must have $cos(\lambda_i)= 0$. cos(x)= 0 for x an "odd multiple of $\pi/2$": $\pi/2$, $3\pi/2$, etc. That means we must have $\lambda_i= (2i- 1)\pi/2$. That is the reason for the subscript "i" on the functions. We now have $\psi(x)= Ccos(((2i-1)\pi/2)x)$.

The intial $\sqrt{2}$ is the "normalizing" part. That makes the integral of the square of the function, from 0 to 1, equal to 1.

Last edited: May 8, 2009
3. May 3, 2009

### ssky

HallsofIvy, i do not know how to thank you.
thank you very very very ... very much.
but, i don't understand :shy: the last part

can you help me pleas

The intial is the "normalizing" part. That makes the integral of the square of the function, from 0 to 1, equal to 1

Last edited: May 3, 2009
4. May 4, 2009

### Cyosis

For a function f(x) to be normalised on an interval [a,b] it has to satisfy $$\int_a^b |f(x)|^2 dx=1$$. Solve this for C.

5. May 4, 2009

### ssky

thank you very very much ----> Cyosis
I understood ... but when I solved another problem I do not get the same solution.
look at the first and the second parts of the solution why we wrote cosh and sinh

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6. May 4, 2009

### Cyosis

The difference between the previous differential equation and this one is that the first one had a second derivative in it and this one a fourth derivative. So the characteristic equation changes from $r^2=-\lambda_i^2$ to $r^4=\mu_i^4$. This has the solutions $r=\{-\mu_i,\mu_i,-i \mu_i, i\mu_i \}$ as a result the general solution will become $\psi(x)=A e^{-\mu x}+B e^{\mu x}+C e^{- i \mu x}+D e^{i \mu x}$. The exponents with i in their powers can be written as cosine and sine again and the real exponents can be written as cosh and sinh.

7. May 4, 2009

### HallsofIvy

Staff Emeritus
Notice an important difference between your two problems:
The first was $d^2\psi/dx^2= -\lambda_i^2 \psi$ which has characteristic equation $r^2= -\lambda_i^2$ with roots $\pm i\lambda_i$ while the second was $d^4\psi/dx^4= /mu^4\psi$ which has characteristic equation $r^4= \mu^4$. We can solve that by first taking the square root: $r^2= \pm \mu^2$. Taking the positive root, $r^2= \mu^2$, and taking the square root again gives $r= \pm \mu$ while taking the negative root, $r^2= -\mu^2$ and taking the square root gives $r= \pm i\mu$. The four roots of the equation are $\mu$, $-\mu$, $i\mu$, and $-i\mu$.

The imaginary roots, $i\mu$ and $-i\mu$, give $sin(\mu x)$ and $cos(\mu x)$ solutions as before. The solutions corresponding to the real roots, $\mu$, and $-\mu$, can be written $e^{\mu x}$ and $e^{-\mu x}$, but because
$$cosh(\mu x)= \frac{e^{\mu x}+ e^{-\mu x}}{2}$$
and
$$sinh(\mu x)= \frac{e^{\mu x}- e^{-\mu x}}{2}$$
they can be written in terms of sinh and cosh also. Since cosh(0)= 1 and sinh(0)= 0, those functions are better suited for initial value problems where we are given values of the function and its derivative at x= 0.

8. May 5, 2009

### ssky

thank you ... but :surprised if we write in the solution $e^{\mu x}$ and $e^{-\mu x}$ is it right ? or we have to write cosh and sinh in the solution

9. May 5, 2009

### Cyosis

You don't have to write it in cosh, sinh form. Halls gave a good reason why it is nice to do so however.

10. May 5, 2009

### ssky

I am so grateful for your interesting in my problem and I will be more grateful if you tell to found the constants c1, c2 , c3 and c4 because I tried but I failed.

11. May 5, 2009

### Cyosis

Show us all the steps you've done so far so we can see where you failed.

12. May 6, 2009

### ssky

sorry, i can not write it by latex .

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13. May 6, 2009

### Cyosis

That is all correct so far. You can now solve equation (5) for C3 and then plug it in to solve for C4. After that you can solve C1 and C2. It doesn't look like it's going to be pretty though, I'll have a look at it later to see if you can simplify it.

14. May 6, 2009

### ssky

thank you but I do not understand what you mean.

15. May 6, 2009

### Cyosis

From the picture you linked you get the following two equations:

\begin{align} & -C_3 \cosh \mu_i-C_4 \sinh \mu_i+C_3 \cos \mu_i+C_4 \sin \mu_i=0 \\ & -C_3 \sinh \mu_i-C_4 \cosh \mu_i - C_3 \sin \mu_i+C_4 \cos \mu_i=0 \end{align}

Solving equation (1) for C3 yields:

$$C_3=-C_4 \frac{\sin \mu_i -\sinh \mu_i}{\cos \mu_i-\cosh \mu_i}$$

Plug this value for C3 into equation (2) and solve for C4.

Edit: This is a pretty weird problem, because the only function that seems to meet all requirements is 0.

Last edited: May 6, 2009
16. May 6, 2009

### ssky

I did as you say but i can not solve it because it is very difficult. please, look to the picture

why $$cosh(\mu) cos(\mu)=1$$ ?

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17. May 8, 2009

### ssky

pleas pleas pleas help me

18. May 8, 2009

### HallsofIvy

Staff Emeritus
What "boundary conditions" is that talking about? Is it the same problem as you posted in response #5?

19. May 8, 2009

### ssky

yes, it is the same problem as i posted in response #5 and #16

20. May 8, 2009

### HallsofIvy

Staff Emeritus
Okay, the general solution is
$$\psi(x)= C_1cosh(\mu x)+ C_2sinh(\mu x)+ C_3cos(\mu x)+ C_4sin(\mu x)$$
so
$$\psi'(x)= \mu C_1sinh(\mu x)+ \mu C_2 cosh(\mu x)- \mu C_3sin(\mu x)+ \mu C_4cos(\mu x)$$

and the conditions are
$$\psi(0)= d\psi(0)/dx= 0$$
$$\psi(1)= d\psi(1)/dx= 0$$

$$\psi(0)= C_1+ C_3= 0$$
so
$$C_3= -C_1$$

$$\psi'(0)= \mu C_2+ \mu C_4= 0$$
so
$$C_4= -C_2$$.

That tells us that we can write
$$\psi(x)= C_1cosh(\mu x)+ C_2sinh(\mu x)- C_1cos(\mu x)+ C_2sin(\mu x)$$

Now we use
$$\psi(1)= C_1cosh(\mu)+ C_2sinh(\mu)- C_1cos(\mu)+ C_2sin(\mu)= 0$$
and
$$\psi'(1)= \mu C_1sinh(\mu)+ \mu C_2 cosh(\mu)+ \mu C_1sin(\mu)- \mu C_2cos(\mu)= 0$$

We can write those as
$$C_1(cosh(\mu)- cos(\mu))+ C_2(sinh(\mu)- sin(\mu))= 0$$
and
$$C_1(sinh(\mu)+ sin(\mu)+ C_2(cosh(\mu)- cos(\mu))= 0$$

In order to solve for, say, $C_1$, we would have to eliminate $C_2$.
We could do that by multiplying the first equation by $cosh(\mu)- cos(\mu)$, the second equation by $sinh(\mu)- sin(\mu)$, and subtracting the second from the first.

That gives
$$(cosh(\mu)- cos(\mu))(cosh(\mu)- cos(\mu))C_1- (sinh(\mu)- sin(\mu))(sinh(\mu)+ sin(\mu)C_1= 0$$
$$C_1(cosh^2(\mu)- 2sinh(\mu)sin(\mu)+ cos^2(\mu)- sinh^2(\mu)+ sin^2(\mu))= 0$$.

But $cosh^2(\mu)- sinh^2(\mu)= 1$ and $cos^2(\mu)+ sin^2(\mu)= 1$ so this becomes
$$(2- 2cosh(\mu)cos(\mu))C_1= 0$$

Now, one obvious solution is $C_1= 0$ but that leads to $C_2= 0$ also which means $\psi(x)$ is identically 0- the "trivial" solution. In order to have a "non-trivial" solution, we must have $C_1\ne 0$ which means we must have the coefficient
$$2- 2cosh^2(\mu)cos(\mu)= 0$$
which leads immediately to $2cosh(\mu)cos(\mu)= 2$ or
[tex]cosh(\mu)cos(\mu)= 1[/itex].

Last edited: May 8, 2009