Question

trying to solve this first order pde

\[u_t+xtu_x=0\]\[u(x,0)=f(x)\]\[-\infty 12 years ago

Answer 1

my prof really skimmed over this, I have no idea where to start here. I can't find any characteristic equations...

Answer 2

\[\dot{x}=xt\]\[\ln x=\frac{t^2}{2}+C\]\[x(t)=Ce^{t^2/2}\]

Answer 3

How about Separation of Variables?

Answer 4

I don't see how SOV would work?

Answer 5

Try now. Let me see your work,

Answer 6

\[\frac{u_t}{t}=-xu_x\]

Answer 7

... and i'm stuck haha.

Answer 8

That's a simple DE to solve.

Answer 9

And I meant Separation of Variables by assuming U is a product of x and t functions.

Answer 10

Meaning, let U = R(x)*G(t).

Answer 11

oh. you mean like every other SOV in this course? oops. just a second.

Answer 12

\[u(x,t)=R(x)G(t)\]\[R(x)G'(t)=-xtR'(x)G(t)\]\[\frac{G'(t)}{tG(t)}=-\frac{xR'(x)}{x}=\text{constant}\equiv C\]
\[G'(t)=CtG(t)\]\[G(t)=Ce^{kt^2/2}\]
\[\frac{R'(t)}{R(t)}=-\frac{k}{x}\]\[\ln R(t)=-k \ln x+C\]\[R(t)=Cx^{-k}\]
\[u(x,t)=C\frac{e^{kt^2/2}}{x^k}\]

Answer 13

I didn't go through your work, but it looks like you get the point.

Answer 14

oops. I messed up copying it.

Answer 15

This is the first method you should always attempt.

Answer 16

I still don't have a solution that works. not even close.

Answer 17

well I do. but I have no idea how to "get" it.

Answer 18

What's the problem?

Answer 19

Your answer will be in terms of f(x).

Answer 20

if I could get to this\[u(x,t)=\frac{t^2}{2}-\ln x\]I would be stoked.

Answer 21

Is that the final answer?

Answer 22

I don't know. but that satisfies the pde. just from inspection

Answer 23

Dude/Dudette, there will be many solutions to the PDE.

Answer 24

Use the initial condition given.

Answer 25

u(x,0)=f(x) is also true with the solution I have?

Answer 26

I dunno honestly I am getting nowhere with separation of variables.

Answer 27

No, because nobody said what f(x) is. Your answer should be in terms of f(x). Said that already.