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Help with difficult differential equation

by hushish
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hushish
#1
May8-12, 01:10 AM
P: 27
Hi,

I need help with a difficult differential equation. The following is the differential equation for a beam on a varying-stiffness elastic foundation. Essentially the beam sits on a cantilever, whose stiffness varies to the third power with length. Does anybody know how to solve such an equation as a closed form solution (if possible)?

E*I*y(x)''''+k*y(x)/x^3=0

where k is a constant.
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tiny-tim
#2
May8-12, 07:04 AM
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hi hushish! welcome to pf!

(try using the X2 button just above the Reply box )

have you tried y = a power of x ?
chiro
#3
May8-12, 07:13 AM
P: 4,572
Quote Quote by hushish View Post
Hi,

I need help with a difficult differential equation. The following is the differential equation for a beam on a varying-stiffness elastic foundation. Essentially the beam sits on a cantilever, whose stiffness varies to the third power with length. Does anybody know how to solve such an equation as a closed form solution (if possible)?

E*I*y(x)''''+k*y(x)/x^3=0

where k is a constant.
Hey hushish and welcome to the forums.

Are you familiar with integral transforms applied to linear differential equations? Have you ever come across the Laplace transform or the Fourier Transform?

hushish
#4
May8-12, 09:36 AM
P: 27
Help with difficult differential equation

I haven't done Laplace transforms since University. I'm open to anything that will yield a solution, but I'll need to be guided through it...
LCKurtz
#5
May8-12, 02:02 PM
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If you lump the constants together and multiply the equation by ##x^4## it is of the form$$
x^4y'''' +cxy=0$$That is an Euler - Cauchy equation. See

http://en.wikipedia.org/wiki/Cauchy%...Euler_equation

which discusses various ways to solve it. One of the most direct is to follow Tiny Tim's suggestion of looking for a solution of the form ##y = x^p##. You don't want to mess with LaPlace transforms for this one.
tiny-tim
#6
May8-12, 05:57 PM
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Quote Quote by LCKurtz View Post
You don't want to mess with LaPlace
'nuff said!
hushish
#7
May10-12, 03:15 AM
P: 27
Thanks for the advice and the link. I am now left with the following equation:

m4-6m3+11m2-6m+kx=0

Any suggestions for the roots of the equation? Some of them will be complex...a little too much maths for my engineering brain.
LCKurtz
#8
May10-12, 12:46 PM
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Quote Quote by hushish View Post
Thanks for the advice and the link. I am now left with the following equation:

m4-6m3+11m2-6m+kx=0

Any suggestions for the roots of the equation? Some of them will be complex...a little too much maths for my engineering brain.
You mean ##m^4-6m^3+11m^2-6m+k## with no ##x##. I would suggest a mathematics program such as Maple of Mathematica. Maple seems to be able to solve it explicitly. If you don't have access to such a program, tell me what ##k## is and I will give you the roots. Or are you hoping for a general formula containing ##k## as a parameter?
pasmith
#9
May10-12, 01:28 PM
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Quote Quote by LCKurtz View Post
If you lump the constants together and multiply the equation by ##x^4## it is of the form$$
x^4y'''' +cxy=0$$That is an Euler - Cauchy equation.
Except it isn't, because ##xy## is not of the form ##x^ny^{(n)}##.
This means that substituting ##y = x^m## doesn't work because it leads to:

Quote Quote by hushish View Post
m4-6m3+11m2-6m+kx=0

Any suggestions for the roots of the equation? Some of them will be complex...a little too much maths for my engineering brain.
Don't bother: If ##y = x^m## were actually a solution, you would have ended up with a fourth-order polynomial for ##m## whose coefficients are independent of ##x##. (As an aside, a pair of complex conjugate roots ##m = p \pm iq## would result in real-valued solutions of the form ##x^p \cos(q\ln(x))## and ##x^p \sin(q\ln(x))##).

I suspect the only method of solving this one (aside from doing it numerically, which would be my first choice here) is to use a variant of Frobenius' Method: pose a power series solution of the form
[tex]y(x) = x^k \sum_{n=0}^{\infty} a_nx^n[/tex]
and choose ##k## so that ##a_0## is not required to be zero. Substituting this into
the ODE then gives the condition
[tex]a_0 k(k-1)(k-2)(k-3) = 0[/tex]
Unfortunately the roots of ##k(k-1)(k-2)(k-3) = 0## differ by integers, so we only get one linearly independent solution from this; and for ##a_1## to be finite we must take ##k = 3##. We then have the recurrence relation
[tex] a_{n+1} = {{-c a_n} \over {(n+4)(n+3)(n+2)(n+1)}}
= -c a_n {{n!} \over {(n+4)!}}[/tex]
which gives
[tex] a_n = a_0 (-1)^n c^n \prod_{r=0}^{n-1} {{r!} \over {(r+4)!}}[/tex]
The resulting power series has an infinite radius of convergence: for ##n \geq 4##, we have
[tex]|a_0| |c|^n |x|^{n+3} \prod_{r=0}^{n-1} {{r!} \over {(r+4)!}} =
|a_0| |x|^3 |cx|^n {{2!3!} \over {n!(n+1)!(n+2)!(n+3)!}} <
|a_0| |x|^3 2!3! {{|cx|^n} \over {n!}} [/tex]
Thus, for ##n \geq 4##, the absolute value of the ##n##th term of the series for ##y(x)## is strictly less than a constant (##12|a_0| |x|^3##) times the ##n##th term of the series for ##e^{|cx|}##, which converges for all finite ##|cx|##. Since the behaviour of the first three terms does not affect convergence, the series for ##y(x)## also converges for all finite ##|cx|##.

That gives one solution; there are three others to find.
LCKurtz
#10
May10-12, 01:47 PM
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Quote Quote by pasmith View Post
Except it isn't, because ##xy## is not of the form ##x^ny^{(n)}##.
You're absolutely correct. Big woops on my part.
hushish
#11
May11-12, 07:39 AM
P: 27
Thanks a mil guys. That seems like the way to go. I have a few understanding questions on the maths, and if I come across any problems I'll post it here.
pasmith
#12
May11-12, 11:26 AM
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On further reflection, I think the only ways of getting all four linearly independent solutions are (1) an integral transform method (possibly Laplace) which hopefully will yield four seperate contours on the complex plane each of which will contribute a solution, or (2) solve it by fiat: give names to the following four special cases and declare that the solution is a linear combination of them: (a) ##y(1) = 1##, ##y'(1) = y''(1) = y'''(1) = 0##, (b) ##y(1) = y''(1) = y'''(1) = 0##, ##y'(1) = 1## , (c) ##y(1) = y'(1) = y'''(1) = 0##, ##y''(1) = 1##, and (d) ##y(1) = y'(1) = y''(1) = 0##, ##y'''(1) = 1##. Obviously, if you're interested in the actual behaviour or values of these functions you'll either have to use the first method or calculate them by numerical solution of the ODE. But frankly, if you're reduced to solving it numerically you may as well just solve it for the particular initial or boundary conditions you have. (Be aware that ##x = 0## is a singular point of the ODE, so one or more of these might blow up there. That's why I'm applying conditions at ##x = 1##, where all four should be well-behaved.)
hushish
#13
May14-12, 01:19 AM
P: 27
Thanks pasmith. Unfortunately the maths that you describe is beyond me. I'm just a simple engineer, and so I have never manipulated the maths in the way you describe. Is there any way that we can work together on the solution somehow-perhaps you can guide me in the steps. In return, I can source you in any documentation that comes out as a result of the research...
chiro
#14
May14-12, 04:31 AM
P: 4,572
Quote Quote by hushish View Post
Thanks pasmith. Unfortunately the maths that you describe is beyond me. I'm just a simple engineer, and so I have never manipulated the maths in the way you describe. Is there any way that we can work together on the solution somehow-perhaps you can guide me in the steps. In return, I can source you in any documentation that comes out as a result of the research...
For your equation: E*I*y(x)''''+k*y(x)/x^3=0, you can transform this to x^3*E*I*y(x)''''+k*y(x)=0, where you will have to not use x=0 (unless your initial condition y(0) = 0 holds) and like hushish said, use Laplace techniques.

You say you are an engineer, so I'm surprised you have not come across the Laplace transform technique before.

If you don't need to prove it rigorously then use something like Wolfram Alpha, but if you want to prove it an understand it, then you will really need to learn the Laplace Transform and the identities for the forward transformation and for the backward (inverse) transformation.

You can find everything you need to do for forward transformation by looking at a decent table of Laplace transformations and also by the definition of the Laplace transformation by it's integral representation and then deriving results using normal calculus techniques that usually involve integration by parts.
Dickfore
#15
May14-12, 05:05 AM
P: 3,014
Quote Quote by LCKurtz View Post
If you lump the constants together and multiply the equation by ##x^4## it is of the form$$
x^4y'''' +cxy=0$$That is an Euler - Cauchy equation. See

http://en.wikipedia.org/wiki/Cauchy%...Euler_equation

which discusses various ways to solve it. One of the most direct is to follow Tiny Tim's suggestion of looking for a solution of the form ##y = x^p##. You don't want to mess with LaPlace transforms for this one.
No, it is not. An Euler equation would be:
[tex]
x^4 y^{\mathrm{IV}} + c \, x \, y' = 0
[/tex]
or
[tex]
x^4 \, y^{\mathrm{IV}} + c \, y = 0
[/tex]
LCKurtz
#16
May14-12, 11:54 AM
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Quote Quote by Dickfore View Post
No, it is not. An Euler equation would be:
[tex]
x^4 y^{\mathrm{IV}} + c \, x \, y' = 0
[/tex]
or
[tex]
x^4 \, y^{\mathrm{IV}} + c \, y = 0
[/tex]
As was noted earlier and I have already acknowledged a couple of days ago in post #10. Is there some reason to point it out again?
pasmith
#17
May14-12, 01:00 PM
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Quote Quote by chiro View Post
For your equation: E*I*y(x)''''+k*y(x)/x^3=0, you can transform this to x^3*E*I*y(x)''''+k*y(x)=0, where you will have to not use x=0 (unless your initial condition y(0) = 0 holds) and like hushish said, use Laplace techniques.

You say you are an engineer, so I'm surprised you have not come across the Laplace transform technique before.

If you don't need to prove it rigorously then use something like Wolfram Alpha, but if you want to prove it an understand it, then you will really need to learn the Laplace Transform and the identities for the forward transformation and for the backward (inverse) transformation.

You can find everything you need to do for forward transformation by looking at a decent table of Laplace transformations and also by the definition of the Laplace transformation by it's integral representation and then deriving results using normal calculus techniques that usually involve integration by parts.
That's not really helpful here: the equation has non-constant coefficients. Indeed Wolfram Alpha can't handle the Laplace transform of ##x^3y^{(4)}##: (annoyingly I can't include links, so go to Wolfram alpha, type in "laplace transform of t^3f''''(t)" and see what you get). Nor can Wolfram alpha solve the ODE directly.

What you have to do is to let
[tex]
y(x) = \int_{\gamma} e^{-px} f(p) dp
[/tex]
(I don't know why I used e^{-px} as my kernel; e^{px} would have been the obvious choice)
where f(p) is to be determined and ##\gamma## is a contour in the complex p-plane which is to be chosen to our advantage. Substituting this, we end up with
[tex]
[ e^{-px} g(x,p)]_{\gamma} + \int_{\gamma} e^{-px} D(f) dp = 0
[/tex]
where g(x,p) is obtained by integration by parts and D is a differential operator with respect to p. The idea is then to solve ##D(f) = 0## for ##f(p)##, and then to choose a contour ##\gamma## so that ##\gamma## encloses one or more singularities of ##f## and the boundary term ##[ e^{-px} g(x,p)]_{\gamma}## vanishes (hopefully choosing different contours will give us the four linearly independent solutions of the original ODE).

For the equation in question, ##D(f)## reduces to ##p^4u'''(p) + cu(p)## where ##u(p) = p^4 f(p)## which is a slight improvement (we've reduced it to third order) and still linear but still not straightforward to solve.

Really, I think it would be better to spend time and effort on writing a program which will take initial/boundary conditions and constants and solve the ODE numerically, rather than trying to solve it analytically.
pasmith
#18
May14-12, 03:47 PM
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P: 949
I believe I have found all four solutions.

We have:
[tex]y^{(4)} + {{ky} \over {x^3}} = 0[/tex]

My first step was to substitute ##y = x^3f(x)## which gave

[tex]x^3 f^{(4)} + 12x^2 f^{(3)} + 36xf'' + 24f' + kf = 0[/tex]

I then looked for a power series solution for f of the form
[tex]f(x) = x^{\lambda}\sum_{n=0} a_n(\lambda) x^n[/tex]
where ##\lambda## will be chosen such that ##a_0 \neq 0##. Substituting the above into the ODE for f gave the following condition on ##\lambda##:

[tex]\lambda(\lambda^3 + 6\lambda^2 + 47\lambda - 30) = 0[/tex]

The solution ##\lambda = 0## was not a surprise: it corresponds to the series I found before. But the roots of the cubic factor are new, and according to Wolfram alpha are not integers: there's a real solution ##\lambda_0 ≈ 0.6## and a complex conjugate pair ##p \pm iq## with a negative non-integer real part.

The recurrence relation for ##a_n(\lambda)##, ##n \geq 1##, is
[tex]a_n(\lambda) = {{-ka_{n-1}(\lambda)} \over {(n+\lambda)((n+\lambda)^3 + 6(n+\lambda)^2 + 47(n+\lambda) - 30)}}[/tex]

For the complex roots, it's not obvious that ##a_n(\lambda)## is real if ##a_0(\lambda)## is real. However we can set ##a_n(\lambda) = b_n(\lambda) + ic_n(\lambda)## for real ##b_n(\lambda)## and ##c_n(\lambda)##. And in fact if we take ##a_0(p+iq) = A + iB## and ##a_0(p-iq)=A-iB## for arbitrary real A and B, then the coefficients obtained from ##\lambda = p-iq## will be the complex conjugates of those obtained from ##\lambda = p+iq##. So we get the same ##b_n## for each, with ##c_n## differing only by a sign, which we fix by considering ##\lambda = p+iq##. We then have ##b_0 = A##, ##c_0 = B##.

Taking ##\lambda = p+iq## gives
[tex]f(x) = x^{p+iq} \sum_{n=0}^{\infty} (b_n + ic_n)x^n
= x^p e^{iq\ln x} \sum_{n=0}^{\infty} (b_n + ic_n)x^n
= x^p (\cos(q\ln x) + i\sin(q\ln x)) \sum_{n=0}^{\infty} (b_n + ic_n)x^n[/tex]
whose real and imaginary parts (provided ##x > 0##) are
[tex]\Re(f(x)) = x^p\cos(q\ln x) \sum_{n=0}^{\infty} b_n x^n -
x^p\sin(q\ln x) \sum_{n=0}^{\infty} c_n x^n [/tex]
and
[tex]\Im(f(x)) = x^p\sin(q\ln x) \sum_{n=0}^{\infty} b_n x^n +
x^p\cos(q\ln x) \sum_{n=0}^{\infty} c_n x^n [/tex]
Taking ##\lambda = p-iq## simply gives the complex conjugate of the above.

So the four real-valued solutions are
[tex]\lambda = 0: y(x) =x^3f(x) = x^3 \sum_{n=0}^{\infty} a_n(0)x^n[/tex]
[tex]\lambda = \lambda_0: y(x) =x^3f(x) = x^{3+\lambda_0} \sum_{n=0}^{\infty} a_n(\lambda_0)x^n[/tex]
[tex]\lambda = p \pm iq: y(x) =x^3f(x) = x^{p+3}\cos(q\ln x) \sum_{n=0}^{\infty} b_n x^n -
x^{p+3}\sin(q\ln x) \sum_{n=0}^{\infty} c_n x^n[/tex]
[tex]\lambda = p \pm iq: y(x) =x^3f(x) = x^{p+3}\sin(q\ln x) \sum_{n=0}^{\infty} b_n x^n + x^{p+3}\cos(q\ln x) \sum_{n=0}^{\infty} c_n x^n[/tex]
where the last two are valid only for ##x > 0##.

I don't propose to solve the recurrence relations, but I can prove convergence of the power series by the ratio test:

[tex] \lim_{n \to \infty} \left| {{a_n(\lambda) x^n} \over {a_{n-1}(\lambda)x^{n-1}}} \right|
= \lim_{n \to \infty} \left| {{kx} \over {(n+\lambda)((n+\lambda)^3 + 6(n+\lambda)^2 + 47(n+\lambda) - 30)}}\right| = 0[/tex]

It's satisfying to have found all four solutions analytically, but I still think that, in practice, numerical solution is the way to go.


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