Changes between Version 2 and Version 3 of u/johannjc/scratchpad24


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Timestamp:
03/21/17 17:19:38 (8 years ago)
Author:
Jonathan
Comment:

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  • u/johannjc/scratchpad24

    v2 v3  
    3636and then we can linearize the flux function about $q$
    3737
    38 $\frac{\partial f(q(x,t))}{\partial x} = \frac{\partial f}{\partial q} \frac{\partial q(x,t)}{\partial x} = A \frac{\partial q}{\partial x}$
     38$\frac{\partial f(q(x,t))}{\partial x} = \frac{\partial f}{\partial q} \frac{\partial q(x,t)}{\partial x} = \lambda \frac{\partial q(x,t)}{\partial x}  $
     39
     40
     41$q_L(\Delta t/2) =  q_L(0) + \displaystyle \int_0^{t/2}  \lambda \frac{\partial (q(x_L,t'))}{\partial x} dt'+ \int_0^{t/2} s(q(x_L,t')) dt' $
     42
     43Now if we ignore the source term, we can directly solve for
     44
     45$q(x,t)=q(x-\lambda t, 0)$
     46
     47and under the change of variables $x'=x_L-\lambda t'$ we can rewrite the integrals as
     48
     49$q_L(\Delta t/2) =  q_L(0) + \displaystyle \int_{x_L}^{x_L-\lambda t/2}  \frac{\partial (q(x',0))}{\partial x} dx' + \int_{x_L}^{x_L-\lambda t/2} s(q(x',0)) dx'$
     50
     51which just gives us
     52
     53$q_L(\Delta t/2) =  q_L(0) - (q(x_L,0) - q(x_L-\lambda \Delta t/2)) + s(\bar{q})\Delta t/2$
     54
     55where $\bar{q} = \displaystyle \int_{x_L-\lambda \Delta t/2}^{x_L}q(x',0) dx'$
     56
     57
     58
     59$\frac{\partial f(q(x,t))}{\partial x} = \frac{\partial f}{\partial q} \frac{\partial q(x,t)}{\partial x} = A \frac{\partial q(x,t)}{\partial x} = R \Lambda L \frac{\partial q(x,t)}{\partial x} = \displaystyle \sum_i R_i \lambda_i L_i \frac{\partial q(x,t)}{\partial x} \approx \displaystyle \sum_i R_i \lambda_i L_i \frac{\partial q(x-\lambda_i t,0)}{\partial x} $
     60
    3961
    4062and use the characteristics (eigenvectors of the Jacobian of the flux) to linearize the flux
    4163
    42 $q_L(\Delta t/2) =  q_L(0) + \displaystyle \int_0^{t/2} \left [ A \frac{\partial (q(x_L,t'))}{\partial x} + s(q(x_L,t')) \right ] $
     64$q_L(\Delta t/2) =  q_L(0) + \displaystyle \int_0^{t/2} \left [ R \Lambda L \frac{\partial (q(x_L,t'))}{\partial x} + s(q(x_L,t')) \right ] $
    4365
    44 and then we can use the
     66$q_L(\Delta t/2) =  q_L(0) + \displaystyle  \int_0^{t/2} \left [ \sum R_i \lambda_i L_i \frac{\partial (q(x_L,t'))}{\partial x} + s(q(x_L,t')) \right ] $
    4567
    46 $q_L(\Delta t/2) =  q_L(0) + A \displaystyle \int_0^{t/2} \left [ \frac{\partial (q())}{\partial x} + s(q_L(t')) \right ] $
     68$q_L(\Delta t/2) =  q_L(0) + \displaystyle  \int_0^{t/2} \left [ \sum R_i \lambda_i \delta l (L_i \frac{\partial (q(x_L,t'))}{\partial x} + s(q(x_L,t')) \right ] $
     69
     70
     71$q_L(\Delta t/2) =  q_L(0) + \displaystyle  \int_0^{t/2} \left [ \sum R_i \lambda_i \frac{\partial (q(x_L,t'))}{\partial x} + s(q(x_L,t')) \right ] $
     72
     73
     74
     75and then we can use the solution for the characteristics $q(x_L,t')=q(x_L-\lambda t',0)$ to turn the time integral into a spatial integral
     76
     77$q_L(\Delta t/2) =  q_L(0) + A \displaystyle \int_0^{t/2} \left [ \frac{\partial (q(x_L-\lambda t',0))}{\partial x} + s(q(x_L-\lambda t',0)) \right ] $
     78
     79which we can do a variable substitution $x' = \lambda t'$