# How do I solve: (p(t)f(t))''=Af(t)

1. Mar 14, 2009

### simplex

How do I solve the equation: (p(t)g(t))''=Ag(t)

I have the equation (p(t)g(t))''=Ag(t),
where:
p(t) is a known given function, for instance p(t)=sin(2*pi*f*t) but not necessarily periodical in the general case.
A is a known real constant, for instance A=A1 or A=A2, etc.
g(t) is the unknown function which I want to determine.
initial conditions:
g(0)=known value=g0
possible g'(0) known value=g0'

What I need is to display (using MATLAB) the time evolution of g=g(t), so I need an expression for this g(t).

For some A I must get an oscillating g(t) (at least for p(t) periodical) for others g(t) should die (converge) quickly to zero.

Is there an explicit formula (even with integrals that have to be numerically evaluated) for g(t)?

2. Mar 14, 2009

### arildno

It looks a bit like a Sturm-Liouville problem, but not quite..

3. Mar 16, 2009

### coomast

Hello simplex,

You can rewrite the equation as:

$$g''(t)+\left(\frac{2p'(t)}{p(t)}\right)g'(t)+\left(\frac{p''(t)-A}{p(t)} \right)g(t)=0$$

Which I might be able to solve if you have the function p(t), in case a general solution is required I do not immediately know a way. I will think about it and come back by the end of the week.

coomast

4. Mar 16, 2009

### simplex

Thank you.

Maybe you are able to get a solution for this equation:

y''(t)-q(t)y(t)=0 where q(t) is a strictely positive function

--------------------------------------
Starting with (p(t)f(t))''=Af(t), I put y(t)=not=p(t)f(t) so f(t)=y(t)/p(t) (p(t)<=>0)
Therefore, y''(t)-[A/p(t)]y(t)=0.

5. Mar 21, 2009

### coomast

Hello Simplex,

Sorry for this late reply but it has been a busy week for me. I looked in several of my books and couldn't find a definite solution to your equation. It is possible to transform it into a different type of equation, but this one is not solvable in general form and therefore I'm afraid no explicit solution exists. I started from your adapted form:

$$y''(x)-q(x) \cdot y(x)=0$$

The equation can be transformed into a Riccati equation by substituting the following:

$$u=x$$
$$v=\frac{y'}{y}$$
$$\frac{dv}{du}=\frac{y\cdot y'' - y' \cdot y'}{y^2}$$

This gives:

$$\frac{dv}{du}+v^2-q(u)=0$$

Which is a form of the Riccati equation. It can only be solved in a limited number of $q(u)$ cases.

"Introduction to nonlinear differential and integral equations" by Harold T. Davis, 1962
"Ordinary differential equations - An elementary text-book with an introduction to Lie's theory of the group of one parameter" by James Morris Page, 1897
"Examples of differential equations with rules for their solution" by George A. Osborne, 1899
"An introduction to the Lie theory of one-parameter groups with applications to the solution of differential equations" by Abraham Cohen, 1911
"An elementary treatise on differential equations" by Abraham Cohen, 1906

I know some of them have a considerable age however don't underestimate the level...

best regards,

Coomast

6. Mar 21, 2009

### simplex

Thank you.

I will see how can I solve that Ricatti equation. It seems to be nonlinear.
Even if q(u)=0, I will have to solve: $$\frac{dv}{du}+v^2=0$$ which does not seem so simple while if I make q(x)=0 in the original equation I get the trivial eq. $$y''(x)=0$$

7. Mar 22, 2009

### coomast

Hello simplex,

If $$q(u)=0[/itex], you have an ODE which can be solved by separation of the variables as: [tex]\frac{dv}{v^2}=-du$$
from which:
$$-\frac{1}{v}=-u-A$$
or:
$$v=\frac{1}{u+A}$$
Changing back to the original variables:
$$\frac{y'}{y}=\frac{1}{x+A}$$
which has as solution:
$$ln(y)=ln(x+A)+ln(B)$$
or:
$$y=C_1x+C_2$$
Now this is the same solution as the one you get from integrating $y''(x)=0$. Hope this helps.

coomast

8. Mar 22, 2009

### simplex

OK.

If q(x)=sin(x), can you solve the equation?

9. Mar 28, 2009

### coomast

Hello simplex,

By setting q(x)=sin(x) the equation becomes very difficult. I tried finding a solution for a few hours, but to no good, I keep on getting after some transformations the original equation, which is nothing more than running around in circles :-( However I did find some info in a book and this gives directly a view on the fact that it is a difficult equation. I'm not going to type everything over from the book, but will give you a few pictures on how to solve it. The equation was:

$$\frac{d^2y}{dx^2}-sin(x)\cdot y=0$$

First step, transform the variables as:

$$u=x$$
$$v=\frac{y'}{y}$$
$$\frac{dv}{du}=\frac{y''}{y}-\left(\frac{y'}{y}\right)^2$$

This gives:

$$\frac{dv}{du}=-v^2+sin(u)$$

Looking at eq_1.png you see that it is that equation with:

$$\alpha=-1$$
$$\beta=0$$
$$\gamma=1$$
$$\lambda=1$$

Applying the transformation as in eq_1.png gives:

$$\frac{dv}{dt}=-v^2+cos(t)$$

This equation is given in figure eq_2.png. Applying the transformation given there:

$$\frac{d^2z}{dt^2}-4cos(2t)\cdot z=0$$

This is the equation given in figure eq_3.png. The solution is further described in this picture. As you can see, not easy. I don't know much about Mathieu functions and can't help you on this. Hope that this brings you a step further.

best regards,

coomast

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10. Mar 29, 2009