Partial Diff. Eqn 1st order

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!!!

Reply 1

thebadgeroverlord

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.

Reply 2

Darkprince

But I have to deduce u(x,y) not u(x/3)! How do I do that?

Reply 3

thebadgeroverlord

It's your notation which is confusing you.

Let

P(x,y,u)\frac{\partial u}{\partial x} + Q(x,y,u)\frac{\partial u}{\partial y} = R(x,y,u)

We parameterise along characteristics with

s

and our intial data with

t

so our initial data is given by

\gamma (t) = (x(t),y(t),u(t))

. We set

s=0

wlog on the characteristics so

x(0,t)=x(t), y(0,t)=y(t), u(0,t)=u(t)

We find

x=x(s,t), y=y(s,t), z=z(s,t)

by solving

\frac{\partial x}{\partial s} = P

\frac{\partial y}{\partial s} = Q

\frac{\partial u}{\partial s} = R

For this problem

\gamma (t) = (0,t,t^2)

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

x(s,t) = 3s + A(t)[br]y(s,t) = -4s + B(t)

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

A(t) = 0, B(t)=t

x = 3s[br]y = -4s + t [br]so t = y + \frac{4x}{3}

so now solve for u

u(s,t) = s + C(t)e^{-s}

using our initial data

u(0,t) = C(t) = t^2

u(s,t) = s + t^2e^{-s}

but

s = \frac{x}{3}

and

t = y + \frac{4x}{3}

then sub in to u(s,t) to get

u(x,y) = \frac{x}{3} + (y + \frac{4x}{3})^2e^{-\frac{x}{3}}

which is the answer you have got.

(edited 12 years ago)

Reply 4

Darkprince

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 :smile:

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? :smile:

(edited 12 years ago)

Reply 5

thebadgeroverlord