# Velocity of the solution of a diff. eq.

1. Nov 25, 2004

### da_willem

Suppose you have a scalar field $\psi(x,t)$ subjected to a certain differential equation. Is there an easy way to find at wich (phase) speed dx/dt this field propagates without actually solving the differential equation.

E.g. it is well known the differential equation

$$\frac{\partial^2 \psi}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2}$$

has solution with phase speed c

or that

$$\frac{\partial^2 \psi}{\partial x^2} = 0$$

has solutions with an infinite phase speed.

2. Nov 25, 2004

### dextercioby

My answer is "yes and no".
It is well known that the first equation describes waves propagating with speen "c"
in both directions of the real axis Ox.The so-called progressive and regressive waves.One way to show that is to solve the equation.Alternatively,you find that "c" has dimensions of speed/velocity by analyzing the dimentionality of the equation.But,you cannot be sure that "c" is the velocity of the waves,unless solving the equation and finding the proper phases.The velocity could be a constant multiplied by "c",the only way to make sure it is "c" is solving the wave equation.For the second equation it suffices to make the limit c->infinity in the original wave equation,but only if you know for sure that "c" is the velocity in question.

3. Nov 28, 2004

### da_willem

Can I deduce from your post there is no way to find the 'velocity of the solutions' but to solve the differential equation? E.g.

$$\frac{\partial^2 \psi}{\partial x^2} = c \frac{\partial \psi}{\partial t}$$

I believe this one has solution with an infinite velocity. Can you see that from the equation or do you have to solve it?

Can you say linear differential equations with equal even derivatives to space and time always yield harmonic solutions and thus have a finite speed associated with the oscillatory pattern of the harmonics?

Last edited: Nov 28, 2004
4. Nov 28, 2004

### ReyChiquito

I dont understand quite well what you are saying, but i know that the solution for the wave equation doesnt have to be a harmonic solution, thats for sure.

$$U_{xx}-c^2U_{tt}=0$$
$$U(x,0)=f(x)$$
$$U_{x}(x,0)=g(x)$$

then

$$U(x,t)=\frac{f(x+ct)+f(x-ct)}{2}+\frac{1}{2c}\int_{x-ct}^{x+ct}g(\tau)d\tau$$

5. Nov 28, 2004

### da_willem

Ok, you're right, not harmonic solutions but functions of x+/-ct. But doesn't the general solution you gave imply solutions of the wave equation always have a phase velocity c? Can this result be extended to other differential equations like

$$\frac{\partial^4 \psi}{\partial x^4} = \frac{1}{c^4} \frac{\partial^4 \psi}{\partial t^4}$$

6. Nov 29, 2004

### dextercioby

The heat equation you posted states that heat propagates along the x axis.The trick is that c is not a velocity.U can see that very clear,by checking the dimentionality of the equation.Solve the equation and see that it is a wave equation with finite speed,as it describes waves just like a normal wave equation encountered in optics/acustic/classical field theory.The speed for the heat waves is +k/c,where "c"is the constant from your equation,and "k" is the modulus of the wave vector,which appears in the solution in forms of $$exp(\pm ikx)$$.
Not necessary hamonic solutions,as it's proved above.

7. Nov 29, 2004

### da_willem

I learned the solution to the heat equation does not have a finite speed and definitely does not describe waves as the wave equation does. The solution to the equation involves an error function. This error function does spread as a function of time, but its value can change at infinity in an infinitesimaly short time period!

8. Dec 1, 2004

### ReyChiquito

This is what i would do

$$\frac{\partial^4 \psi}{\partial x^4} - \frac{1}{c^4} \frac{\partial^4 \psi}{\partial t^4}=\left(\frac{\partial^2 }{\partial x^2} + \frac{1}{c^2} \frac{\partial^2 }{\partial t^2}\right)\left(\frac{\partial^2 }{\partial x^2} - \frac{1}{c^2} \frac{\partial^2 }{\partial t^2}\right)\psi$$

Solve the inhomogeneous wave equation

$$W=\left(\frac{\partial^2 }{\partial x^2} - \frac{1}{c^2} \frac{\partial^2 }{\partial t^2}\right)\psi$$

and then

$$\left(\frac{\partial^2 }{\partial x^2} + \frac{1}{c^2} \frac{\partial^2 }{\partial t^2}\right)W=0$$

hey, nobody said it was easy :P