PDE: Method of characteristics question

[mike1111]

Homework Statement
x u_t +u_x =0
intial condition
u(x,0)=f(x)

1. Find the characteristics curves
2. What area of the xt-plane do u expect a solution
3. Find solution when f(x)=cos x
4.Now u(x,0)=f(x) (again), Find the level curves of u i.e for each c find the set L_c={(x,t):u(x,t)=f(c)}
5. Show there is not solution for u(x,0)=sin x
6. For what function is there a solution for u(x,0)=f(x), Then what the soution for u(x,y)?

The attempt at a solution
Can anyone check my solution to tell me if this is right and advise me on how to do this correctly. I think it wrong leading to incorrect answers everywhere else.
1. dt/dx = x
y=x² +C

2. only on the x=0 since it can't apss throught the charactistics curve more than once

3. From initial
x(0,s) =s x(k,s) =s+k
t(0,s)=0 t(k,s) =xk
u(0,s)= cos s u(k,s) =cos s

u= cos (x/(x-y))
4. Not sure what 4 is asking can anyone head point me in the right direction?
5. I don't see how this is any different from the cos equation in part 3,
I end up getting u= sin(x/(x-y)) by the same method, but there not meant to be a solution
6. Any help is appreciated

 

[mighty2000]

1. Your solution for the characteristics curves is correct. However, you could also write it as y = x^2 + C, which is the same as x = √(y - C). This form may be more useful for finding the solution later on.

2. The area in the xt-plane where we would expect a solution is the entire plane except for the line x = 0. This is because the characteristics curves are not defined for x = 0, so the solution cannot pass through this line.

3. To find the solution when f(x) = cos x, we can use the method of characteristics. We have the initial condition u(x,0) = cos x, which gives us the initial characteristic curve y = x^2 + C, as we found in part 1. Plugging this into the equation xut + ux = 0, we get:

(x^2 + C)ut + (x^2 + C) = 0

We can solve this using separation of variables:

∫(1/u) du = -∫(x^2 + C) dt

ln(u) = -x^3/3 - Ct + D

where D is a constant of integration. Solving for u, we get:

u(x,t) = e^(-x^3/3 - Ct + D)

Substituting in our initial condition u(x,0) = cos x, we get:

cos x = e^(-x^3/3 + D)

Solving for D, we get D = ln(cos x), so our solution is:

u(x,t) = e^(-x^3/3 - Ct + ln(cos x)) = cos(x - x^3/3 - Ct)

4. The level curves of u are the curves in the xt-plane where u is constant. In other words, they are the curves where u(x,t) = f(c) for some constant c. To find these curves, we can use the solution we found in part 3:

u(x,t) = cos(x - x^3/3 - Ct)

Setting u(x,t) = f(c), we get:

cos(x - x^3/3 - Ct) = f(c)

This equation defines a curve in the xt-plane. To find the set of all points (x,t) that satisfy this equation, we can use inverse trigonometric functions:

x -