How to Solve a 4th Order ODE with Given Boundary Conditions?

[jimmygriffey]

Hi:

I try to solve this PDE as below:

y''''-ky''+1=0

With boundary conditions of x=0.y=y''=0 and x=-h,y=y'=0.
k and h are constant.
The solution look like from the Fourier transforms.
There is one term in the solution which is :

beta_n is the positive roots of the equation of beta_n=tan(beta_n). n is from 1 to infinity.

I have no idea how this term come from. If someone can help me, I really appreciate it!

Kevin

 

[kosovtsov]

First of all your ODE has the following form

z''-kz+1=0, where z=y''

The general solution to z is obtained by the wery standard method which lead to ODE

y''= exp(k^(1/2)*x)*C1+exp(-k^(1/2)*x)*C2+1/k

and after double integration

y = C1/k*exp(k^(1/2)*x)+C2/k*exp(-k^(1/2)*x)+1/2/k*x^2+C3*x+C4

what is the general solution to your ODE. Substitution of your boundary conditions gives you 4 algebraic equations for unknowns C1,C2,C3 and C4.

 

[HallsofIvy]

Almost the same thing: the characteristic equation for the differential equation y""- ky"+ 1= 0, which is the same as y""- ky"= -1, is $r^4- kr^2= r^2(r^2- k)= r^2(r- \sqrt{k})(r+ \sqrt{k})$ which has 0 as a double root and roots $\sqrt{k}$ and $-\sqrt{k}$.

The general solution to the associated homogeneous equation is $z= C_1x+ C_2+ C_3e^{x\sqrt{k}}+ C_4e^{-x\sqrt{k}}$. Since the right hand side is a constant, normally we would try a constant solution but since a constant and x already satisfy the homogeneous solution, we try $z= Ax^2$. Then $z'= 2Ax$, $z"= 2A$ and $z"'= z""= 0$. The equation becomes 0+ k(2A)= -1 so A= -1/(2k). The general soution to that equation is
$$z(x)= C_1x+ C_2+ C_3e^{x\sqrt{k}}+ C_4e^{-x\sqrt{k}}-\frac{1}{2k}x^2$$

 

[jimmygriffey]

Hi:

Thank you for response. If I want to use Fourier transform to solve this one, how can I find the boundary conditons for y''(-h)? I need to use inverse Laplace transform to convert the y function. Therefore, it will become much easier to use the Fourier transforms.

Kevin