MHB How Can You Solve the Scaled Transport Equation in the First Quadrant?

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To solve the scaled transport equation ut(x,t) + ux(x,t) = 0 in the first quadrant, the method of separation of variables is recommended. By introducing auxiliary variables, the equation can be transformed into a simpler form, leading to the general solution u = f(η) = f(x - ct). In this case, the specific solution is u(x,t) = -sin(x - t). This approach effectively addresses the initial and boundary conditions provided.
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am given a scaled transport equation
ut(x,t) + ux(x,t)=0 x>0; t>0
u(x,0)=0 x>0
u(0,t)= sint t>0

how can I begin to find a solution in the quadrant {x.0,t>0} to this problem, am really struglling:(
 
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onie mti said:
am given a scaled transport equation
ut(x,t) + ux(x,t)=0 x>0; t>0
u(x,0)=0 x>0
u(0,t)= sint t>0

how can I begin to find a solution in the quadrant {x.0,t>0} to this problem, am really struglling:(

You have to use the method: separation of variables.
 
onie mti said:
am given a scaled transport equation
ut(x,t) + ux(x,t)=0 x>0; t>0
u(x,0)=0 x>0
u(0,t)= sint t>0

how can I begin to find a solution in the quadrant {x.0,t>0} to this problem, am really struglling:(

A PDE of the form...

$\displaystyle u_{t} + c\ u_{x} = 0\ (1)$

... can be solved with the auxiliary variables $\xi= x + c\ t$ and $\eta = x - c\ t$. Applying the chain rule You arrive to the equivalent PDE...

$\displaystyle 2\ c\ \frac{\partial{u}}{\partial{\xi}} = 0\ (2)$

... the solution of which is...

$\displaystyle u = f(\eta) = f(x - c\ t)\ (3)$

... where $f(*,*) \in C^{1}$ is arbitrary. In Your case is...

$\displaystyle u(x,t)= - \sin (x - t)\ (4)$

Kind regards

$\chi$ $\sigma$
 

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