Solving a PDE with Characteristic Curves and Initial Conditions

[gtfitzpatrick]

Homework Statement

sin(y)[itex]\frac{ \partial u}{ \partial x} + \frac{ \partial u}{ \partial y} = (xcos(y)-sin^2(y))u[/itex]

where ln(u(x,[itex]\frac{\pi}{2})) = x^2 + x - \frac{\pi}{2}[/itex] for [itex]-1 \leq x \leq 3[/itex]

determine the characteristic curves in the xy plane and draw 3 of them
determine the general solution of this pde
determine the particular solution and show that u(0,0) = e
sketch the region of influence of the initial conditions

Homework Equations

The Attempt at a Solution

dy/dx = 1/siny this gives K = x + cosy
if i draw this out is it a cos wave along the y-axis and to draw 3 of them i let k = 0,1,2?
what happens as i add k?

 

[gtfitzpatrick]

to get the general solution [itex] \frac{ du}{ dx} [/itex] = [itex] \frac{xcos(y) - sin^2(y))u}{ sin y}[/itex]

for which i get ln(u) = [itex] x^2 \frac{cosy}{2siny} -xsiny +F(x-cosy)[/itex]

so then i transferred in the given value[itex] y= \pi/2 [/itex] just to check if it works out but it doesnt. what am i doing wrong here. Am i even going about this the right way...

 

[hunt_mat]

Write the PDE as:
[tex]
\dot{x}=\sin y,\quad\dot{y}=1,\quad\dot{u}=(x\cos y-\sin^{2}y)u
[/tex]
Integrate the y equation to obtain [itex]y=s+\pi /2[/itex], substitute this into the equation for x and integrate to obtain [itex]x=\sin s+r[/itex], where we parametrised the initial data as x=r, and [itex]u(r)=r^{2}+r-\pi /2[/itex], insert this into the final equation and integrate as a function of s, the characteristic variable. This will be your general solution.

 

[gtfitzpatrick]

Write the PDE as:
[tex]
Integrate the y equation to obtain [itex]y=s+\pi /2[/itex],

Hi Mat,
Thanks a million.
I don't understand where the pi/2 came from?the constant of integration..

 

[hunt_mat]

If s is the characteristic variable, then [itex]y(0)=\pi /2[/itex].

Do you have my notes on first order PDEs?

 

[gtfitzpatrick]

Thanks, that makes sense.

are your notes online?

 

[hunt_mat]

As the initial conditions (at s=0) are paramatrised by [itex]x=r[/itex], [itex]y=\pi /2[/itex] and [itex]\log u=r^{2}+r-\pi /2[/itex].

My notes are on this site somewhere, but if you can't find them, drop me a PM a with you e-mail and I will e-mail them to you.

I have noticed that you do ask quite a few questions about 1st order PDEs.

 

[gtfitzpatrick]

we use t as s here so I'm just going to leave it as t

[itex]\frac{dx}{dt} = sin(y)[/itex]

[itex]\frac{dy}{dt} = 1 [/itex]
y=t+k
@t=0 y=[itex]\frac{\pi}{2}[/itex]
=>y=t+[itex]\frac{\pi}{2}[/itex]

then
[itex]\frac{dx}{dt} = sin(t+ \frac{\pi}{2})[/itex]
Simplifies to [itex]\frac{dx}{dt} = cos(t)[/itex]
so x = sint + k

you parameterised x=r and subed into eq 3.

do i sub in the values I've got for x and y into eq3? no mater what i sub in it doesn't work out.

 

[hunt_mat]

You have dx/dt=sin(t-pi/2) and using the initial values you also have x+sin(y)=r. So what you do is substitute x and y for their respective functions and integrate.

 

[gtfitzpatrick]

You have dx/dt=sin(t-pi/2)

should that be dx/dt=sin(t+pi/2)

 

[hunt_mat]

Sorry, you're right.

 

[gtfitzpatrick]

[itex]\frac{du}{dt} = xcos(t+\frac{\pi}{2}) -sin^2(t+\frac{\pi}{2})[/itex]

[itex]\frac{du}{dt} = -xsin(t) -sin^2(t+\frac{\pi}{2})[/itex]

[itex]\frac{du}{dt} = -(sin(t) + k)sin(t) -sin^2(t+\frac{\pi}{2})[/itex]

[itex]\frac{du}{dt} = -(sin^2(t) + ksin(t) -sin^2(t+\frac{\pi}{2})[/itex]

and then integrate?

 

[gtfitzpatrick]

sorry there should be a u at the end of all them eqs

 

[hunt_mat]

Expand the trig function and then integrate.

 

[gtfitzpatrick]

i do it all out and i get ln(u) = t-ksin(t)

 

[hunt_mat]

We have [itex]y=t+\pi /2[/itex] and [itex]x=\sin t+k[/itex], inserting these in the equation for u to obtain:
[tex]
\frac{\dot{u}}{u}=(\sin t+k)(-\sin t)-\sin^{2}t
[/tex]
Now to integrate this.

 

[gtfitzpatrick]

is that not the same as [itex]\frac{\dot{u}}{u}=(-2sin^2 t + ksin t) [/itex]

for which i get [itex] -(t-\frac{1}{2}Sin 2t) + kcos t [/itex] and then i throw in the values for k and t but...

 

[hunt_mat]

That is, you should get the solution:
[tex]
\log u=\frac{1}{2}\sin 2t-t-k\cos t+k^{2}+k-\frac{\pi}{2}
[/tex]
Upon using the initial condition for u. Now insert [itex]t=y-\pi /2[/itex] and [itex]k=x+\cos y[/itex] and you have your solution.

 

[gtfitzpatrick]

is the intregral of [tex]-2sin^2 t [/tex] not [itex]\frac{1}{2}(t-\frac{1}{2}sin(2t) [/itex] and ksin(t) =kcos(t)...i don't follow where you got the k^2 from...

 

[hunt_mat]

Oh yes, you're correct my mistake. The k^2 term comes from the initial condition for u which is:
[tex]
u\left( k,\frac{\pi}{2}\right) =k^{2}+k-\frac{\pi}{2}
[/tex]

 

[gtfitzpatrick]

right so we've integrated but i still don't get why you just added k^2+k-pi/2 on to the end of the integral...

 

[gtfitzpatrick]

oh feck, sorry its obvious. Thanks a million

 

[hunt_mat]

From here it should be easy to compute u(0,0) to get the answer required. My notes have notes of examples like this.

 

[gtfitzpatrick]

u(0,0) is grand, no bothers there.
The region of influence though, is there anything in your notes about it?

 

[hunt_mat]

Sadly no, you will have to do a bit of googling for that.