Partial Differential Equation with variable coefficients

[NicolaiTheDane]

Homework Statement

This question relates to a very large project I have been assigned in a course on mathematical methods in structural engineering. I have to solve the following equation, in a specific way:

(17)

Now we have to assume the following solution:

(18)

It wants me insert this sum, into the equation and use ortogonality properties of the eigenfunctions of $Yn(x)$ (Earlier in the project, I have found eigenfunctions for simplified examples of the equation. I assume these are the ones it refers to. They are basically just sin functions) to express the above on the following matrixform:

(19)

where the column vector

contains the $N$ time dependent variable from the finite "row expansion" of (18). Then I have to rewrite this 2nd order ODE system to a 1st order ODE system on the standard form:

where

Now there are more to this then that, but I'll hold it here for now.

Homework Equations

Listed above where appropriate.

The Attempt at a Solution

None. I have absolute no idea where to start, as we haven't been taught this method. I considered trying to insert (18) into (17) as the assign wants, and try to use power series method, as that is an approach I'm familiar with. However this clearly won't work, because of the $\alpha(x)$ coefficients. More over it wouldn't bring me an closer to the solution that is actually requested. So I'm totally lost on where to start. Any help would be greatly appreciated!

 

[MathematicalPhysicist]

If you try separation of variables, i.e plugging in the PDE $\xi(t,\epsilon)=X(x)T(t)$, you would get:
$$X''''T-k^2(\alpha(x)^{-1}X')'T+12d^4\alpha^{-3}(x)XT+(\Omega_y \ell^2)^{-2}XT''=0$$

Where $'$ denotes differentiation with respect to the argument of the function.

Now, if $\Omega_y$ is a constant with respect to $t$ and $x$ (I denote $\Omega \ell^2=C=constant$, then after dividing by $XT$ we get:
$$X''''/X-k^2(\alpha(x)^{-1}X''-\alpha(x)^{-2} \alpha'(x)X')/X+12d^4\alpha^{-3}(x)+CT''/T=0$$
Now, isolate the part that depends only on x and the part that depends only on t.

In order to solve the equation for x you need to know the exact form of $\alpha(x)$, if they want through a power series then substitute for $\alpha(x)=\sum_{n=0}^{\infty} \alpha_n x^n$ (if the solution is defined for around x=0), and substitute $X(x)=\sum_{n=0}^{\infty} X_n x^n$.
In order to find reciprocal of $\alpha(x)$, use Cauchy series product theorem.

 

[MathematicalPhysicist]

You need to find a recursive relations for the series coefficients.

 

[NicolaiTheDane]

If you try separation of variables, i.e plugging in the PDE $\xi(t,\epsilon)=X(x)T(t)$, you would get:
$$X''''T-k^2(\alpha(x)^{-1}X')'T+12d^4\alpha^{-3}(x)XT+(\Omega_y \ell^2)^{-2}XT''=0$$

Where $'$ denotes differentiation with respect to the argument of the function.

Now, if $\Omega_y$ is a constant with respect to $t$ and $x$ (I denote $\Omega \ell^2=C=constant$, then after dividing by $XT$ we get:
$$X''''/X-k^2(\alpha(x)^{-1}X''-\alpha(x)^{-2} \alpha'(x)X')/X+12d^4\alpha^{-3}(x)+CT''/T=0$$
Now, isolate the part that depends only on x and the part that depends only on t.

In order to solve the equation for x you need to know the exact form of $\alpha(x)$, if they want through a power series then substitute for $\alpha(x)=\sum_{n=0}^{\infty} \alpha_n x^n$ (if the solution is defined for around x=0), and substitute $X(x)=\sum_{n=0}^{\infty} X_n x^n$.
In order to find reciprocal of $\alpha(x)$, use Cauchy series product theorem.

But how is the ever going to lead to the matrix they want me to use?

 

[MathematicalPhysicist]

Well,
$$
X''''/X-k^2(\alpha(x)^{-1}X''-\alpha(x)^{-2} \alpha'(x)X')/X+12d^4\alpha^{-3}(x)=-CT''/T=k$$
where k is some constant with respect to t and x.

So you get for t the equation:
$$T''+(k/C)T=0$$
Which is your ODE in (19).

I recommend that you read Arfken's textbook, it's the best for applied physicists and engineers.

 

[NicolaiTheDane]

Well,
$$
X''''/X-k^2(\alpha(x)^{-1}X''-\alpha(x)^{-2} \alpha'(x)X')/X+12d^4\alpha^{-3}(x)=-CT''/T=k$$
where k is some constant with respect to t and x.

So you get for t the equation:
$$T''+(k/C)T=0$$
Which is your ODE in (19).

I recommend that you read Arfken's textbook, it's the best for applied physicists and engineers.

Unforunately I only have another 4 days to complete this, so buying and reading a whole textbook on the subject isn't an option. I will keep it in mind though, for when I get some time off, as I find the subject very fascinating.

As for your proposed solution. I get where your going, but again I just don't see how its helpful. As I see it, it'll merely leave me with a huge mess of summations. The only math course I have had on this subject, deals with constant variables, and what little was there on varying constants, the book basically said "Beyond the scope of this course". Also all separations I have done so far, moves all the bulk of the equations onto the $T(t)$ part of the equations. Wouldn't solving this part in this manor, basically invalidate all other work I have done?

 

[NicolaiTheDane]

If you try separation of variables, i.e plugging in the PDE $\xi(t,\epsilon)=X(x)T(t)$, you would get:
$$X''''T-k^2(\alpha(x)^{-1}X')'T+12d^4\alpha^{-3}(x)XT+(\Omega_y \ell^2)^{-2}XT''=0$$

Where $'$ denotes differentiation with respect to the argument of the function.

Now, if $\Omega_y$ is a constant with respect to $t$ and $x$ (I denote $\Omega \ell^2=C=constant$, then after dividing by $XT$ we get:
$$X''''/X-k^2(\alpha(x)^{-1}X''-\alpha(x)^{-2} \alpha'(x)X')/X+12d^4\alpha^{-3}(x)+CT''/T=0$$
Now, isolate the part that depends only on x and the part that depends only on t.

In order to solve the equation for x you need to know the exact form of $\alpha(x)$, if they want through a power series then substitute for $\alpha(x)=\sum_{n=0}^{\infty} \alpha_n x^n$ (if the solution is defined for around x=0), and substitute $X(x)=\sum_{n=0}^{\infty} X_n x^n$.
In order to find reciprocal of $\alpha(x)$, use Cauchy series product theorem.

I have an exact expression for the $\alpha(x)$ btw

. Does that change anything?