# First order PDE

• Nov 28th 2011, 03:06 PM
Darkprince
First order PDE
I have to solve 3 * du/dx - 4*du/dy + u = 1 + x/3
with initial condition u(0,y)= y ^2 using the method of characteristics

The above derivatives are partial derivatives

My working until now:
Writing the characteristic equations: dx/ds = 3 => x=3s+x(0)
dy/ds = -4 => y=-4s + y(0)

and du/ds = 1- u + x/3

I am not sure how to proceed with the equation du/ds and so to proceed the exercise.

Thank you!
• Nov 28th 2011, 03:15 PM
TheEmptySet
Re: First order PDE
Quote:

Originally Posted by Darkprince
I have to solve 3 * du/dx - 4*du/dy + u = 1 + x/3
with initial condition u(0,y)= y ^2 using the method of characteristics

The above derivatives are partial derivatives

My working until now:
Writing the characteristic equations: dx/ds = 3 => x=3s+x(0)
dy/ds = -4 => y=-4s + y(0)

and du/ds = 1- u + x/3

I am not sure how to proceed with the equation du/ds and so to proceed the exercise.

Thank you!

You have already solved for what x is. $x=3s+x(0)$

Just sub this in and you have a first order equation in u and s

$\frac{du}{ds}=1-u+\frac{1}{3}(3s+x(0))=1-u+s+\frac{x(0)}{3}$

So now you have the first order ODE

$\frac{du}{ds}+u=s+1+\frac{x(0)}{3}$

This can be solve using an integrating factor or a number of other ways.
• Nov 28th 2011, 03:31 PM
Darkprince
Re: First order PDE
Quote:

Originally Posted by TheEmptySet
You have already solved for what x is. $x=3s+x(0)$

Just sub this in and you have a first order equation in u and s

$\frac{du}{ds}=1-u+\frac{1}{3}(3s+x(0))=1-u+s+\frac{x(0)}{3}$

So now you have the first order ODE

$\frac{du}{ds}+u=s+1+\frac{x(0)}{3}$

This can be solve using an integrating factor or a number of other ways.

So say that our integrating factor is e^integral(1ds) = e^s

Then d/ds(e^s * u) = e^s + se^s + s*x(0)/3

So e^s * u = e^s + se^s - e^s + ( x(0)/3 ) * 1/2 * s^2 + c, where c is constant

So finally I find u(s) = s + x(0)/3 * 1/2 * s^2 * e^(-s) + ce^(-s)

Am I correct? Then how do I proceed? I am new to characteristics that's why I am trying to solve this exercise!
• Nov 28th 2011, 03:53 PM
Jester
Re: First order PDE
You really need to bring in your boundary conditions to find these functions of integration (yes, they are functions of integration). If we integrate your first two characteristic equations (assuming that $x =x(r,s), y = y(r,s)$) then we have to solve

$x_s = 3, y_s = -4$ (as you've said)

If we associate that the boundary $x = 0$ (in the $x-y$ plane) corrresponds to say $s = 0$ (in the $s-r$ plane) and that $y = r$ on this boundary then $u = y^2 = r^2$ on this boundary.

Integrating the first two gives

$x = 3s + a(r), \;\;y = -4s + b(r)$.

Using your BC gives what $a(r)$ and $b(r)$ are. Then use this and the remaining BC in

$u_s = - u + 1 + \frac{x}{3}$ (as you've said)

to find $u$. Then eliminate $r$ and $s$.
• Nov 28th 2011, 04:02 PM
Darkprince
Re: First order PDE
If I want to proceed with my last post how do I use the initial conditions? We have x=3s+x(0) y= -4s+y(0) and that u(s) = s + x(0)/3 * 1/2 * s^2 * e^(-s) + ce^(-s) [if I am correct when I used the integrating factor method]

Then I have to use that u(0,y) = y^2 but I don't know how to use it!
• Nov 28th 2011, 04:24 PM
Jester
Re: First order PDE
If you use what I said then

if $x = 3s + a(r)$ then when $s = 0,\; x = 0$. This gives $a(r) = 0$

If $y = -4s + b(r),$ then when $s=0$ then $y = r$ gives $b(r) = r$.

So far you have $x = 3r, y = -4s + r$.

Now solve

$u_s = -u + 1 + \frac{x}{3} = -u + 1 + r$ subject to $u(0,r)=r^2$.
• Nov 28th 2011, 05:25 PM
Darkprince
Re: First order PDE
Your way confuses me a bit since we have been taught a different approach (maybe is the same but with different terms). May I tell you what I did and please help me complete it using my way? Thank you!

So let's start again.

We have dx/ds = 3 =>x=3s+x(0) => s=(x-x(0))/3

dy/ds = -4 => y= -4s + y(0) => y= -4(x-x(0))/3 + y(0) {using that s=(x-x(0))/3}

du/ds = 1+ x/3 - u

But since s=(x-x(0))/3 => du/ds = 1+s+x(0)/3 - u

From our initial conditions now, x(0)=0 so x=3s

Also y= -4(x-x(0))/3 + y(0) from above becomes y+ 4/3 * x = y(0)

Furthermore du/ds = 1+s - u {since x(0)=0}

Solving using the integrating factor yields that u(s) = s + c*e^(-s), where c is an arbitrary constant!

Moreover u(s=0) = u(0) = y(0) ^2 {using the initial condition u(0,y) = y^2}

Finally, since x = 3s from above, we deduce u(x/3) = x/3 + (y + 4/3 * x)^2 * e^(-x/3)

Then I don't know how to proceed to find u(x,y)

May you review my work with my way, since I know how to justify this way ans tell me how do I proceed to finish the exercise?

Thank you again and sorry for not following your way!

Kind regards!

I have just realised that u(x/3) = x/3 + (y + 4/3 * x)^2 * e^(-x/3) is the solution of my PDE (I did direct differentiation and substitution and I get what I needed)
but how do I go from u(x/3) to u(x,y)?

Something goes wrong there with my notation but I don't know why!

With a bit of reconsideration I think I am correct!

If I use my notation and just say that u=s+c*e^(-s)
then say s=x/3
then say u(s=0) = u(0) = c = y(0)^2 and then that y(0)=y+4/3 * x, so c= (y+4/3 * x)^2

And then I just say u(x,y)=..... {without saying before u(x/3)=..... etc }

If I use my notation and the way I just used (in all my post), am I correct? :)