Very difficult partial differential equation

[Mrx]

Thanks for your reply. The equation you got for X(x) is the same I tried to solve. Let's do it again:

$$\lambda^2+2VX'\lambda + (V^2-c^2)X''=0$$

has solution

$$X(x)=\exp\left(r_{\pm} x\right)$$

With $$r=\frac{V\lambda \pm c\sqrt{\lambda^2}}{V^2-c^2}$$

Now you can distinguish three cases:

• $$\lambda$$ purely real
• $$\lambda$$ purely imaginary
• $$\lambda$$ = a + bi

$$\lambda$$ purely real
Then
$$r=\frac{-V\lambda \pm c\lambda}{V^2-c^2}=\frac{-V\lambda \pm c\lambda}{(V+c)(V-c)}=\frac{-\lambda}{V \pm c}$$
So
$$X(x)=A\exp\left(\frac{-\lambda}{V+c}x\right)+B\exp\left(\frac{-\lambda}{V-c}x\right)$$
With boundary conditions: $$X(0)=X(L)=0$$ it follows that A + B = 0 and
$$X(L)=A\exp\left(\frac{-\lambda}{V+c}L\right)-A\exp\left(\frac{-\lambda}{V-c}L\right)=0\Leftrightarrow A=0$$

It doesn't matter which values lambda takes ($$\lambda>0,\lambda<0, \lambda=0$$). $$\lambda$$ cannot be purely real.

$$\lambda$$ purely imaginary
Say $$\lambda = b i$$ Then $$r=\frac{-b V i \pm c b i}{V^2-c^2}$$.
And
$$X(x)=A\exp\left(\frac{-b V i + c b i}{V^2-c^2}x \right)+B\exp\left(\frac{-b V i - c b i}{V^2-c^2}x \right)$$
$$X(0)=0$$ gives: B=-A, so
$$X(x)=A\exp\left(\frac{-bVi}{V^2-c^2}x \right) \left(\exp\left(\frac{c b i}{V^2-c^2}x \right)-\exp\left(\frac{- c b i}{V^2-c^2}x \right)\right)$$
$$X(x)=2iA\exp\left(\frac{-bVi}{V^2-c^2}x \right) \left(\sin\left( \frac{bcx}{V^2-c^2}\right))\right)$$
$$X(L)=0$$ gives $$b=\frac{n\pi \left( V^2 -c^2 \right)}{cL}$$
so $$\lambda = \frac{n\pi \left( V^2 -c^2 \right)}{cL} i$$

In the end then:

$$u(x,t)= \sum_{n=1}^{\infty} a_n \sin\left(\frac{n \pi x}{L} \right)\exp\left(\frac{n\pi V x}{cL}i \right) \exp\left(\frac{n\pi (V^2-c^2) t}{cL}i \right)$$

Did you get this as a solution?

I haven't tried the third case yet.

 

[dirk_mec1]

Thanks for your reply. The equation you got for X(x) is the same I
$$u(x,t)= \sum_{n=1}^{\infty} a_n \sin\left(\frac{n \pi x}{L} \right)\exp\left(\frac{n\pi V x}{cL}i \right) \exp\left(\frac{n\pi (V^2-c^2) t}{cL}i \right)$$

Did you get this as a solution?

Yes, that's the solution I got! It's tough but we've made it! Note that you can write this in sine and cosine and you can determine the (in this case 2) coefficients via the initial conditions.

 

[Mrx]

Okay great! thank you for your help.

Do you know what this equation ectually represents? I can't find anything about the physical meaning, or a derivation of this equation. I tried searching on conveyor belt equation but got nothing.

 

[Bendavid2]

Could anyone explain a little bit of the physical meaning of this PDE?

If V=0 than you have the wave equation. So c is the speed of the wave (in the belt) and u is the displacement in the y-direction. So what does it means that V=0, I think it means the belt is not moving. What is very strange to me because V is a constant and not time depending.

In the solution given by

$$u(x,t)= \sum_{n=1}^{\infty} a_n \sin\left(\frac{n \pi x}{L} \right)\exp\left(\frac{n\pi V x}{cL}i \right) \exp\left(\frac{n\pi (V^2-c^2) t}{cL}i \right)$$

you can bring the exp-parts together and you get a x-v't where v' is a the speed of the wave relative to the frame. (c is speed relative to the belt and V the speed of the belt relative to the frame).

$$2 V u_x_t + V^2 u_x_x$$ is the "source therm" I think, first of all since you have just a wave equations and the other therms are source therms. Second reason is that the units of $$u_x_t$$ are Hertz and mabye it has to do with the rotations of one of the wheels.

The 2V has something to do with the speed between upper belt and lower (returning) belt.

This is all what I could thought about.
Maybe somebody can make a 3dplot in maple?

Greetz Bendavid2