# Thread: Quasilinear first order PDE-weird parametrization?!

1. ## Quasilinear first order PDE-weird parametrization?!

I am confused by the following example about solving quasilinear first order PDEs.

For the part I circled, the solution is just x^2 + y^2 = k where k is an arbitrary constant. To parametrize it in terms of t, can't we just put x = a cos(t), y = a sin(t) ? Here we only have one arbitrary constant a.
But in the example, they used a weird parametrization of a circle that includes TWO arbitary constants a and b. So my point is: why introduce another extra arbitrary constant when it is completely unnecessary to do so?

Can someone please explain why it is absolutely necessary to parametrize the circle in the way they do?

Any help is greatly appreciated!

[note: also under discussion in S.O.S. math cyberboard]

2. Originally Posted by kingwinner
I am confused by the following example about solving quasilinear first order PDEs.

For the part I circled, the solution is just x^2 + y^2 = k where k is an arbitrary constant. To parametrize it in terms of t, can't we just put x = a cos(t), y = a sin(t) ? Here we only have one arbitrary constant a.
But in the example, they used a weird parametrization of a circle that includes TWO arbitary constants a and b. So my point is: why introduce another extra arbitrary constant when it is completely unnecessary to do so?

Can someone please explain why it is absolutely necessary to parametrize the circle in the way they do?

Any help is greatly appreciated!

[note: also under discussion in S.O.S. math cyberboard]
I think I might. You have the equations

$\frac{dx}{dt} = y,\;\;\; \frac{dy}{dt} = - x$ (1)

so

$\frac{d^2 x}{dt^2} = \frac{dy}{dt} = -x$ or $\frac{d^2x}{dt^2} + x = 0$

A second order ODE with the solution $x = a \cos t + b \sin t$. From (1) $y = \frac{dx}{dt} = - a \sin t + b \cos t$

3. OK, now I see why the circled part is correct by using your method. Your way actually makes more sense to me

But why is the author trying to combine the two equations dx/dt = y, dy/dt = -x ? How is this going to help us to solve the system?
Combining these two, we get dy/dx = -x/y, the general solution is just x^2 + y^2 = k where k is an arbitrary constant. To parametrize this general solution it in terms of t, just put x = a cos(t), y = a sin(t), right? This parametrization satisfies the equation x^2 + y^2 = k, so it must be a correct parametrization. What is wrong with this approach? Can you please point out where this line of logic fails?

Thank you!

4. Originally Posted by kingwinner
OK, now I see why the circled part is correct by using your method. Your way actually makes more sense to me

But why is the author trying to combine the two equations dx/dt = y, dy/dt = -x ? How is this going to help us to solve the system?
Combining these two, we get dy/dx = -x/y, the general solution is just x^2 + y^2 = k where k is an arbitrary constant. To parametrize this general solution it in terms of t, just put x = a cos(t), y = a sin(t), right? This parametrization satisfies the equation x^2 + y^2 = k, so it must be a correct parametrization. What is wrong with this approach? Can you please point out where this line of logic fails?

Thank you!
You have two first order differential equations so your solution should have two constants of integration. You have only one constant of integration so you only have one solution not the general solution!