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# Partial Diff. Eqn 1st order watch

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1. 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:

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

The above were our characteristic equations.

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 someone review my work and help me please?

Thanks!!!
2. (Original post by Darkprince)

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 someone review my work and help me please?

Thanks!!!
This is the answer! You've got u only in terms of x and y.
3. But I have to deduce u(x,y) not u(x/3)! How do I do that?
4. It's your notation which is confusing you.

Let

We parameterise along characteristics with and our intial data with so our initial data is given by
. We set wlog on the characteristics so

We find by solving

For this problem

we solve for x,y, denoting our constants of intergration A(t) and B(t)

Using in our initial data (ie setting s = 0) we find

so now solve for u

using our initial data so

but and then sub in to u(s,t) to get which is the answer you have got.
5. To be honest your notation confuses me even more, may you review my answer using my notation please? At the stage
u(x/3) = x/3 + (y + 4/3 * x)^2 * e^(-x/3)
since s=x/3

what should I say?

Thanks again for your time, I appreciate it

With a bit of reconsideration 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, am I correct?
6. When finding the characteristic equation for u you are solving

where t is the parameter along your initial data curve, so , not just a function of s!

You've written u as a function of x and y which is what you want. You do not need to do anything else!
7. So if I just use my notation and the exact way I used in my first post am I correct? Nothing more?
8. (Original post by Darkprince)
So if I just use my notation and the exact way I used in my first post am I correct? Nothing more?
You are correct.
9. Thanks for your time and help!

Kind regards

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Updated: November 30, 2011
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