Differential equation conceptual question.

[td21]

Homework Statement

see the picture
why the absolute sign of the logarithm can be removed? Although it is true for initial value, it is not true for x smaller than 1, right?

Homework Equations

The Attempt at a Solution

I think the initial value can conclude that the x smaller than 1 is not defined?? But i still doubt this because the initial value cannot tell the whole solution.

 

[td21]

sorry to be unclear, the above is the solution. While the original Q. is the initial value is y(0)=1 and the ODE is "y^(0.5)dx+(1+x)dy=0".

 

[vela]

A solution will only be good on one side of the singularity at x=-1 or the other. The initial condition y(0)=1 tells you which region you care about. Since x=0 is in the x>-1 region, your solution is for x>-1. In this region, |x+1|=x+1, so you don't need to keep the absolute value.

 

[td21]

A solution will only be good on one side of the singularity at x=-1 or the other. The initial condition y(0)=1 tells you which region you care about. Since x=0 is in the x>-1 region, your solution is for x>-1. In this region, |x+1|=x+1, so you don't need to keep the absolute value.

But why can't the solution includes both region by using the absolute value sign? Is it not valid to use absolute value sign in the answer?

 

[vela]

Because you can't cross the singularity. The differential equation expresses how y(x) behaves up to the singularity, but at x=-1, it breaks down. You don't have any information about what y(x) does in the region x<1.

 

[uart]

Hi td21. Your DE can be written as :

y&#039; = -\frac{\sqrt{y}}{1+x}

Take a look at the "Picard existence theorem" http://en.wikipedia.org/wiki/Picard–Lindelöf_theorem

Basically is says that in order for a DE of the form y&#039; = f(x,y) to have a unique solution you require that f(x,y) is continuous in "x" and Lipschitz continuous in "y".

BTW, Lipschitz continuous means continuous with bounded derivative.

 

[td21]

Hi td21. Your DE can be written as :

y&#039; = -\frac{\sqrt{y}}{1+x}

Take a look at the "Picard existence theorem" http://en.wikipedia.org/wiki/Picard–Lindelöf_theorem

Basically is says that in order for a DE of the form y&#039; = f(x,y) to have a unique solution you require that f(x,y) is continuous in "x" and Lipschitz continuous in "y".

BTW, Lipschitz continuous means continuous with bounded derivative.

So it means using absolute value is not allowed as it makes the solution not continuous!right? Thanks!

 

[uart]

So it means using absolute value is not allowed as it makes the solution not continuous!right? Thanks!

Well kind of, but it would be better to think of it as the absolute value not being needed rather than not allowed.

What I'm saying is, yes the solution starting from those initial conditions can't be extended back to x<=0, but it's not only whether or not the solution expression can be evaluated (over reals) that determines where the solution is valid. Sometimes you have an expression for the solution that can still be evaluated beyond the discontinuity but the solution is invalid in that region just the same.