# Math Help - 2nd Order Liner PDE’s Part 1

1. ## 2nd Order Liner PDE’s Part 1

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$

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

2. 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$

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.

3. 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

4. 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

5. 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