2nd Order Liner PDE’s Part 1

**bugatti79** · March 24th 2011, 03:26 AM

a) Determine whether the given eqn is hyperbolic, parabolic or elliptic
b) Determine the subset of R^2
c) Determine the characteristics curves

$ u_{xx}+2xu_{yx}+u_{yy}+Cos(xy)u_{y}=u(x,y) $

$ B^2-AC = x^2-1(1) = x^2-1 $

a)

Hyperbolic if $x^2-1 > 0$
ie $x^2>1$
$ x>1, x<-1 $

Elliptic if $x^2-1< 0$
ie $x^2< 1$
$ -1<x<1 $

Parabolic if $x^2-1 = 0$
ie $x^2=1$
$ x=-1, x=+1 $

b) How do you write the subset query correctly?

c)

$ \frac{dy}{dx} = \frac{B \pm \sqrt{B^2-AC}}{A} = x \pm \sqrt{x^2-1} $

My initial attempt to solve for y on this by letting u = x^2 -1 and use substitution doesnt seem to work?

Thanks

**HallsofIvy** · March 24th 2011, 08:03 AM

Originally Posted by bugatti79

a) Determine whether the given eqn is hyperbolic, parabolic or elliptic
b) Determine the subset of R^2
c) Determine the characteristics curves

$ u_{xx}+2xu_{yx}+u_{yy}+Cos(xy)u_{y}=u(x,y) $

$ B^2-AC = x^2-1(1) = x^2-1 $

a)

Hyperbolic if $x^2-1 > 0$
ie $x^2>1$
$ x>1, x<-1 $

Elliptic if $x^2-1< 0$
ie $x^2< 1$
$ -1<x<1 $

Parabolic if $x^2-1 = 0$
ie $x^2=1$
$ x=-1, x=+1 $

b) How do you write the subset query correctly?

I think you already have. "Determine the subset of R^2" by itself makes no sense- but I suspect the question was asking you to determine the subsets on which the equation is "hyperbolic", "elliptic", and "parabolic".

c)

$ \frac{dy}{dx} = \frac{B \pm \sqrt{B^2-AC}}{A} = x \pm \sqrt{x^2-1} $

My initial attempt to solve for y on this by letting u = x^2 -1 and use substitution doesnt seem to work?

Thanks

$y= \int x\pm\sqrt{x^2- 1}$

To integrate $\int \sqrt{x^2- 1}dx$ use either the trig substitution x= sec(u) or the hyperbolic substitution x= sinh(u). $sec^2(x)- 1= tan^2(u)$ and $sinh^2(u)- 1= cosh^2(u)$ so either will help you get rid of the square root.

**bugatti79** · March 24th 2011, 08:29 AM

Originally Posted by HallsofIvy

I think you already have. "Determine the subset of R^2" by itself makes no sense- but I suspect the question was asking you to determine the subsets on which the equation is "hyperbolic", "elliptic", and "parabolic".

$y= \int x\pm\sqrt{x^2- 1}$

To integrate $\int \sqrt{x^2- 1}dx$ use either the trig substitution x= sec(u) or the hyperbolic substitution x= sinh(u). $sec^2(x)- 1= tan^2(u)$ and $sinh^2(u)- 1= cosh^2(u)$ so either will help you get rid of the square root.

Thanks HallsofIvy,

I will tackle that this evening. How did you infer the correct substitution...is it based on experience?

I mean how does one know if a substitution is correct, trial and error?

Thanks

**bugatti79** · March 24th 2011, 12:14 PM

Originally Posted by HallsofIvy

I think you already have. "Determine the subset of R^2" by itself makes no sense- but I suspect the question was asking you to determine the subsets on which the equation is "hyperbolic", "elliptic", and "parabolic".

$y= \int x\pm\sqrt{x^2- 1}$

To integrate $\int \sqrt{x^2- 1}dx$ use either the trig substitution x= sec(u) or the hyperbolic substitution x= sinh(u). $sec^2(x)- 1= tan^2(u)$ and $sinh^2(u)- 1= cosh^2(u)$ so either will help you get rid of the square root.

Ok,

For the hyperbolic case $x^2 -1 >0$ I calculate the integral $\displaystyle y = \int (x \pm \sqrt{x^2-1})dx$ to yield 2 solutions

$y_1 = \displaystyle \frac{1}{2}[x^2 +x \sqrt{x^2-1} -ln(x + \sqrt{x^2-1}) +C]$
and

$y_2 = \displaystyle \frac{1}{2}[x^2 -x \sqrt{x^2-1} +ln(x + \sqrt{x^2-1}) +C]$

For the parabolic case $x^2-1=0$ the integral yields

$y = \frac{x^2}{2} +C$

For the elliptic case $x^2-1<0$ , I am not sure how to evaluate the integral since x^2 is less than 0 and hence there would be a negative under the sqrt...I know the solutions will involve complex numbers but I cant put my finger on it...

Thanks

**bugatti79** · March 26th 2011, 02:28 AM

Originally Posted by bugatti79

For the elliptic case $x^2-1<0$ , I am not sure how to evaluate the integral since x^2 is less than 0 and hence there would be a negative under the sqrt...I know the solutions will involve complex numbers but I cant put my finger on it...

Thanks

For the elliptic case my attempt is to modify the $\sqrt{x^2 -1}$ term to handle x^<1, i change it to

$\sqrt{-(-(x^2))-1} = i \sqrt{- x^2-1}$

This doesnt look right because if x=0 then we get another imaginary term

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