general solution to infinitely differential equations of a certain form... :D

Question

general solution to infinitely differential equations of a certain form... >:D

kainui · Answer

Here's a fun thing I came up with, so I'll ask it to anyone who wants to try. :D

What's the general solution to this differential equation: $$\sum_{m=0}^{M} a_m \frac{\partial^n u}{\partial x_m^n}=0$$
where $a_m$ are arbitrary constants, and n is an integer.

---

For example a specific case when M=2 and n=2 we have:

$$a_0 \frac{\partial^2 u}{\partial x_0^2}+a_1 \frac{\partial^2 u}{\partial x_1^2}+a_2 \frac{\partial^2 u}{\partial x_2^2}=0$$

If we relabel the variables $x_0=t$, $x_1=x$ and $x_2=y$ and relabel the constants $a_0=-1$, $a_1=a_2=c^2$ we have the wave equation in 2D.

divinesolar · Answer

If only I was smart enough for this ;p. Have not learned all of calculus yet, I only took calculus in middle school haha.

thomas5267 · Answer

This is a really hard question.  I think even in the M=2 and n=2 case, you have very different solutions depending on the coefficients.  You have the heat equation, the wave equation, and the Poisson's equation.  I could not even fathom how bad it will be as M and n increases.

kainui · Answer

You're right about the wave equation, but it turns out this is less general than it seems. Poisson's equation is the nonhomegeneous version of Laplace's equation, and this only encompasses Laplace's equation because it's homogeneous! Similarly, the heat equation has derivatives of different orders, but n is a constant so the first time derivative and the Laplacian's second derivatives end up not mixing together here!

So really, this thing is less interesting and not as general as we'd like it to be, but at least this has a general solution haha

thomas5267 · Answer

This has a general solutiom????????  Impossiburu!

thomas5267 · Answer

For n=2, it seems like this is some sort of wave equation in higher dimensions, although I am unsure what effect the signs of the coefficients will have on the solutions.  For other n, I have no idea and I need to sleep.

kainui · Answer

Haha I'll type up the answer I guess then, unless you want to try to figure it out :P

thomas5267 · Answer

If m=1, you can simply have equal and opposite partial derivatives?  For other m,n, I have no idea.  Probably have something to do with the lack of mixed derivatives?  Something related to eigen-something?

thomas5267 · Answer

Just type up the solution.  Unless there are some exquisite tricks I do not see how I can solve this using year 1 maths.

kainui · Answer

Let's start from here:

$$\sum_{m=0}^M a_m \frac{\partial^n u}{\partial x_m^n} = 0$$

Now what looks similar to this?

$$\sum_{m=0}^M e^{i 2 \pi m / (M+1)}=0$$

Now it'd be nice if we could just say:

$$e^{i 2 \pi m / (M+1)}=a_m \frac{\partial^n u}{\partial x_m^n}$$

But as it is, we can't do that. However, for any arbitrary function $u$ we can write:

$$u(y)$$

where the $y$ is a linear function of the $x_m$ variables. 
$$y=\sum_{m=0}^M b_mx_m$$ 
So when we take its derivative, by the chain rule

$$ \frac{\partial^n u}{\partial x_m^n} = \frac{d^nu}{dy^n} \left(\frac{\partial  y}{\partial x_m}\right)^n=\frac{d^nu}{dy^n}b_m^n $$

Plugging this back into our equation gets us:


$$\sum_{m=0}^M a_m \frac{\partial^n u}{\partial x_m^n} = 0$$$$\sum_{m=0}^M a_m \frac{d^nu}{dy^n}b_m^n=\frac{d^nu}{dy^n}\sum_{m=0}^M a_m b_m^n = 0$$$$\sum_{m=0}^M a_mb_m^n = 0$$

Now we can do what we wanted earlier:
$$e^{i 2 \pi m / (M+1)}=a_mb_m^n $$
$$b_m = \sqrt[n]{\frac{e^{i 2 \pi m / (M+1)}}{a_m}}$$
$$y= \sum_{m=0}^M \sqrt[n]{\frac{e^{i 2 \pi m / (M+1)}}{a_m}} x_m$$ 

So quick recap, ALL possible functions you choose for $u $ work, for example $u=\sin y$ or $u=-e^{-y^3}$, you can more naturally choose what u should be to solve your problem based on your initial and boundary conditions though.

kainui · Answer

Check this out, it's exactly how d'Alembert found his general solution to the wave equation, I merely extended it to more cases: http://mathworld.wolfram.com/dAlembertsSolution.html

thomas5267 · Answer

I am surprised that any differentiable function works for u.  Since there are so many differentiable functions out there, I could not think how can you construct a basis to the solution using arbitrary u.  Probably some Gram-Schimdt but the solution will be ugly.  Furthermore, there might be some issue posed by the complex number under the radicals.  I don't know whether the "pricinple part" will suffices in this case.

kainui · Answer

Ah, I almost included that extra bit but I decided that was probably enough.

Basically two facts are that since the DE is linear we can combine any sum of functions to make a new one that works this way. The other thing is if u has a power series expansion (probably not necessary actually), then we can immediately reconstruct u with pieces of it from linear combinations of the complex versions of it.

The thing is messy so I'll give an example of how the argument goes, let's use: $\omega=e^{i 2 \pi/3}$ and throw it into some arbitrary function f(x) with a power series.

$$f(x) = a_0+a_1x+a_2x^2+a_3x^3+\cdots$$$$f(\omega x) = a_0+a_1x\omega +a_2x^2\omega ^2+a_3x^3+\cdots$$$$f(\omega^2 x) = a_0+a_1x\omega^2 +a_2x^2\omega +a_3x^3+\cdots$$
So adding all of these together in different ways we get:
$$\frac{f(x)+f(\omega x)+f(\omega^2 x)}{3} = a_0+a_3x^3+\cdots$$
$$\frac{f(x)+\omega f(\omega x)+\omega^2 f(\omega^2 x)}{3} =a_2x^2+a_5x^5+\cdots$$
$$\frac{f(x)+\omega f(\omega x)+\omega f(\omega^2 x)}{3} = a_1x +a_4x^4+\cdots$$

So the point is the functions on the left are all functions of real variables multiplied by complex numbers and when you combine them in this way you end up getting entirely real functions pop out. So through a lot of messing around like this, you can make your function real. phew