Generalized linear transport equation











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I stumbled upon a transport equation of the form
$$u_t(x,t)=u_x(x,t) + u_x(1,t).$$
Since I can write it in the form $u_t(x,t) = Lu(x,t)$ where L is some linear operator I thought that there must be some theory behind this type of equations. Unfortunately, so far my research was unsuccessful.
Is there a way to solve the problem explicitly, e.g., by using method of characteristics in some way?



Thank you in advance!










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    up vote
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    down vote

    favorite












    I stumbled upon a transport equation of the form
    $$u_t(x,t)=u_x(x,t) + u_x(1,t).$$
    Since I can write it in the form $u_t(x,t) = Lu(x,t)$ where L is some linear operator I thought that there must be some theory behind this type of equations. Unfortunately, so far my research was unsuccessful.
    Is there a way to solve the problem explicitly, e.g., by using method of characteristics in some way?



    Thank you in advance!










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I stumbled upon a transport equation of the form
      $$u_t(x,t)=u_x(x,t) + u_x(1,t).$$
      Since I can write it in the form $u_t(x,t) = Lu(x,t)$ where L is some linear operator I thought that there must be some theory behind this type of equations. Unfortunately, so far my research was unsuccessful.
      Is there a way to solve the problem explicitly, e.g., by using method of characteristics in some way?



      Thank you in advance!










      share|cite|improve this question













      I stumbled upon a transport equation of the form
      $$u_t(x,t)=u_x(x,t) + u_x(1,t).$$
      Since I can write it in the form $u_t(x,t) = Lu(x,t)$ where L is some linear operator I thought that there must be some theory behind this type of equations. Unfortunately, so far my research was unsuccessful.
      Is there a way to solve the problem explicitly, e.g., by using method of characteristics in some way?



      Thank you in advance!







      pde differential-operators transport-equation






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      share|cite|improve this question











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      asked Nov 12 at 12:20









      Jfischer

      31




      31






















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          $$u_t(x,t)-u_x(x,t) = u_x(1,t).$$
          $u_x(1+t)$ is a function of $t$ only. Let $u_x(1+t)=f(t)$. Of course, this is an unknown function.
          $$u_t(x,t)-u_x(x,t) = f(t).$$
          Solving for the general solution. Charpit-Lagrange equations :
          $$frac{dt}{1}=frac{dx}{-1}=frac{du}{f(t)}$$
          A first family of characteristic curves comes from $frac{dt}{1}=frac{dx}{-1}$
          $$x+t=c_1$$
          A second family of characteristic curves comes from $frac{dt}{1}=frac{du}{f(t)}$
          $$u-int f(t)dt=c_2$$
          General solution of the PDE : $u-int f(t)dt=Phi(x+t)$



          $Phi(X)$ is an arbitrary function of one variable only. In the above equation $X=(x+t)$.
          $$u(x,t)= Phi(x+t)+int f(t)dt$$
          $u_x(x,t)=left(frac{dPhi(X)}{dX}right)_{(X=x+t)}=Phi'(x+t)$



          $u_x(1,t)=left(frac{dPhi(X)}{dX}right)_{(X=1+t)}=Phi'(1+t)$.
          $$f(t)=u_x(1,t)=Phi'(1+t).$$
          $int f(t)dt=int_{t_0}^t f(zeta)dzeta+C= int_{t_0}^t Phi'(1+zeta)dzeta +C =Phi(1+t)+C$.



          Finally the solution is :
          $$u(x,t)= Phi(x+t)+Phi(1+t)+C$$
          $Phi$ is an arbitrary function. $C$ is an arbitrary constant.
          They have to be determined to fit some boundary and initial conditions. One cannot go further since those conditions are not specified in the wording of the question.



          CHECKING :
          $$u_x(x,t)=Phi'(x+t)$$
          $$u_t(x,t)=Phi'(x,t)+Phi'(1+t)$$
          $$u_t(x,t)-u_x(x,t)=Phi'(1+t)=u_x(1,t)$$
          The PDE is satisfied. The above result is correct.



          IN ADDITION after the comments :



          Case of condition $u(x,0)=g(x)=Phi(x)+Phi(1)+C=Phi(x)+C_2$



          $Phi(x)=g(x)-C_2$



          $u(x,t)= (g(x+t)-C_2)+(g(1+t)-C_2)+C$



          Thus $quad -2C_2+C+g(1)=0quad;quad C=2C_2-g(1)$



          $u(x,t)=(g(x+t)-C_2)+(g(1+t)-C_2)+2C_2-g(1)$



          $$u(x,t)=g(x+t)+g(1+t)-g(1)$$



          Example :



          $$g(x)=x^mqquad:qquad u(x,t)=(x+t)^m+(1+t)^m-1$$






          share|cite|improve this answer























          • Thank you for your response. That was really helpful. In general my initial condition will just look like this: $u(x,0)=g(x)$ where g is some continuous function. For the sake of my example I choose $g(x) = x^m$ for some natural number $m$. Setting $C=-g(1)$ and $Phi=g$ would yield a solution, since $$u(x,0)=Phi(x+0) + Phi(1) - g(1) = Phi(x) = g(x)$$ and by your calculation. Is that right? But this choice is always possible independent of what my initial condition looks like as long as $g(1)$ is well defined and that is what puzzles me a bit.
            – Jfischer
            Nov 15 at 8:13












          • I hoped to get the general concept from that toy example, i.e., how to solve an equation of type $$u_t(x,t) = w(x)u_x(x,t) + v(x)u_x(1,t)$$. I struggle with the step from $c_1 = x+t$ and $c_2 = u-int f(t) dt$ to $u-int f(t) dt = Phi(x+t)$. Does that mean that $c_2$ is represented as a function along the characteristic? I would also be very thankful for a source for the case where a term of type $u_x(1,t)$ appears to read it up.
            – Jfischer
            Nov 18 at 8:10











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          accepted










          $$u_t(x,t)-u_x(x,t) = u_x(1,t).$$
          $u_x(1+t)$ is a function of $t$ only. Let $u_x(1+t)=f(t)$. Of course, this is an unknown function.
          $$u_t(x,t)-u_x(x,t) = f(t).$$
          Solving for the general solution. Charpit-Lagrange equations :
          $$frac{dt}{1}=frac{dx}{-1}=frac{du}{f(t)}$$
          A first family of characteristic curves comes from $frac{dt}{1}=frac{dx}{-1}$
          $$x+t=c_1$$
          A second family of characteristic curves comes from $frac{dt}{1}=frac{du}{f(t)}$
          $$u-int f(t)dt=c_2$$
          General solution of the PDE : $u-int f(t)dt=Phi(x+t)$



          $Phi(X)$ is an arbitrary function of one variable only. In the above equation $X=(x+t)$.
          $$u(x,t)= Phi(x+t)+int f(t)dt$$
          $u_x(x,t)=left(frac{dPhi(X)}{dX}right)_{(X=x+t)}=Phi'(x+t)$



          $u_x(1,t)=left(frac{dPhi(X)}{dX}right)_{(X=1+t)}=Phi'(1+t)$.
          $$f(t)=u_x(1,t)=Phi'(1+t).$$
          $int f(t)dt=int_{t_0}^t f(zeta)dzeta+C= int_{t_0}^t Phi'(1+zeta)dzeta +C =Phi(1+t)+C$.



          Finally the solution is :
          $$u(x,t)= Phi(x+t)+Phi(1+t)+C$$
          $Phi$ is an arbitrary function. $C$ is an arbitrary constant.
          They have to be determined to fit some boundary and initial conditions. One cannot go further since those conditions are not specified in the wording of the question.



          CHECKING :
          $$u_x(x,t)=Phi'(x+t)$$
          $$u_t(x,t)=Phi'(x,t)+Phi'(1+t)$$
          $$u_t(x,t)-u_x(x,t)=Phi'(1+t)=u_x(1,t)$$
          The PDE is satisfied. The above result is correct.



          IN ADDITION after the comments :



          Case of condition $u(x,0)=g(x)=Phi(x)+Phi(1)+C=Phi(x)+C_2$



          $Phi(x)=g(x)-C_2$



          $u(x,t)= (g(x+t)-C_2)+(g(1+t)-C_2)+C$



          Thus $quad -2C_2+C+g(1)=0quad;quad C=2C_2-g(1)$



          $u(x,t)=(g(x+t)-C_2)+(g(1+t)-C_2)+2C_2-g(1)$



          $$u(x,t)=g(x+t)+g(1+t)-g(1)$$



          Example :



          $$g(x)=x^mqquad:qquad u(x,t)=(x+t)^m+(1+t)^m-1$$






          share|cite|improve this answer























          • Thank you for your response. That was really helpful. In general my initial condition will just look like this: $u(x,0)=g(x)$ where g is some continuous function. For the sake of my example I choose $g(x) = x^m$ for some natural number $m$. Setting $C=-g(1)$ and $Phi=g$ would yield a solution, since $$u(x,0)=Phi(x+0) + Phi(1) - g(1) = Phi(x) = g(x)$$ and by your calculation. Is that right? But this choice is always possible independent of what my initial condition looks like as long as $g(1)$ is well defined and that is what puzzles me a bit.
            – Jfischer
            Nov 15 at 8:13












          • I hoped to get the general concept from that toy example, i.e., how to solve an equation of type $$u_t(x,t) = w(x)u_x(x,t) + v(x)u_x(1,t)$$. I struggle with the step from $c_1 = x+t$ and $c_2 = u-int f(t) dt$ to $u-int f(t) dt = Phi(x+t)$. Does that mean that $c_2$ is represented as a function along the characteristic? I would also be very thankful for a source for the case where a term of type $u_x(1,t)$ appears to read it up.
            – Jfischer
            Nov 18 at 8:10















          up vote
          1
          down vote



          accepted










          $$u_t(x,t)-u_x(x,t) = u_x(1,t).$$
          $u_x(1+t)$ is a function of $t$ only. Let $u_x(1+t)=f(t)$. Of course, this is an unknown function.
          $$u_t(x,t)-u_x(x,t) = f(t).$$
          Solving for the general solution. Charpit-Lagrange equations :
          $$frac{dt}{1}=frac{dx}{-1}=frac{du}{f(t)}$$
          A first family of characteristic curves comes from $frac{dt}{1}=frac{dx}{-1}$
          $$x+t=c_1$$
          A second family of characteristic curves comes from $frac{dt}{1}=frac{du}{f(t)}$
          $$u-int f(t)dt=c_2$$
          General solution of the PDE : $u-int f(t)dt=Phi(x+t)$



          $Phi(X)$ is an arbitrary function of one variable only. In the above equation $X=(x+t)$.
          $$u(x,t)= Phi(x+t)+int f(t)dt$$
          $u_x(x,t)=left(frac{dPhi(X)}{dX}right)_{(X=x+t)}=Phi'(x+t)$



          $u_x(1,t)=left(frac{dPhi(X)}{dX}right)_{(X=1+t)}=Phi'(1+t)$.
          $$f(t)=u_x(1,t)=Phi'(1+t).$$
          $int f(t)dt=int_{t_0}^t f(zeta)dzeta+C= int_{t_0}^t Phi'(1+zeta)dzeta +C =Phi(1+t)+C$.



          Finally the solution is :
          $$u(x,t)= Phi(x+t)+Phi(1+t)+C$$
          $Phi$ is an arbitrary function. $C$ is an arbitrary constant.
          They have to be determined to fit some boundary and initial conditions. One cannot go further since those conditions are not specified in the wording of the question.



          CHECKING :
          $$u_x(x,t)=Phi'(x+t)$$
          $$u_t(x,t)=Phi'(x,t)+Phi'(1+t)$$
          $$u_t(x,t)-u_x(x,t)=Phi'(1+t)=u_x(1,t)$$
          The PDE is satisfied. The above result is correct.



          IN ADDITION after the comments :



          Case of condition $u(x,0)=g(x)=Phi(x)+Phi(1)+C=Phi(x)+C_2$



          $Phi(x)=g(x)-C_2$



          $u(x,t)= (g(x+t)-C_2)+(g(1+t)-C_2)+C$



          Thus $quad -2C_2+C+g(1)=0quad;quad C=2C_2-g(1)$



          $u(x,t)=(g(x+t)-C_2)+(g(1+t)-C_2)+2C_2-g(1)$



          $$u(x,t)=g(x+t)+g(1+t)-g(1)$$



          Example :



          $$g(x)=x^mqquad:qquad u(x,t)=(x+t)^m+(1+t)^m-1$$






          share|cite|improve this answer























          • Thank you for your response. That was really helpful. In general my initial condition will just look like this: $u(x,0)=g(x)$ where g is some continuous function. For the sake of my example I choose $g(x) = x^m$ for some natural number $m$. Setting $C=-g(1)$ and $Phi=g$ would yield a solution, since $$u(x,0)=Phi(x+0) + Phi(1) - g(1) = Phi(x) = g(x)$$ and by your calculation. Is that right? But this choice is always possible independent of what my initial condition looks like as long as $g(1)$ is well defined and that is what puzzles me a bit.
            – Jfischer
            Nov 15 at 8:13












          • I hoped to get the general concept from that toy example, i.e., how to solve an equation of type $$u_t(x,t) = w(x)u_x(x,t) + v(x)u_x(1,t)$$. I struggle with the step from $c_1 = x+t$ and $c_2 = u-int f(t) dt$ to $u-int f(t) dt = Phi(x+t)$. Does that mean that $c_2$ is represented as a function along the characteristic? I would also be very thankful for a source for the case where a term of type $u_x(1,t)$ appears to read it up.
            – Jfischer
            Nov 18 at 8:10













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          $$u_t(x,t)-u_x(x,t) = u_x(1,t).$$
          $u_x(1+t)$ is a function of $t$ only. Let $u_x(1+t)=f(t)$. Of course, this is an unknown function.
          $$u_t(x,t)-u_x(x,t) = f(t).$$
          Solving for the general solution. Charpit-Lagrange equations :
          $$frac{dt}{1}=frac{dx}{-1}=frac{du}{f(t)}$$
          A first family of characteristic curves comes from $frac{dt}{1}=frac{dx}{-1}$
          $$x+t=c_1$$
          A second family of characteristic curves comes from $frac{dt}{1}=frac{du}{f(t)}$
          $$u-int f(t)dt=c_2$$
          General solution of the PDE : $u-int f(t)dt=Phi(x+t)$



          $Phi(X)$ is an arbitrary function of one variable only. In the above equation $X=(x+t)$.
          $$u(x,t)= Phi(x+t)+int f(t)dt$$
          $u_x(x,t)=left(frac{dPhi(X)}{dX}right)_{(X=x+t)}=Phi'(x+t)$



          $u_x(1,t)=left(frac{dPhi(X)}{dX}right)_{(X=1+t)}=Phi'(1+t)$.
          $$f(t)=u_x(1,t)=Phi'(1+t).$$
          $int f(t)dt=int_{t_0}^t f(zeta)dzeta+C= int_{t_0}^t Phi'(1+zeta)dzeta +C =Phi(1+t)+C$.



          Finally the solution is :
          $$u(x,t)= Phi(x+t)+Phi(1+t)+C$$
          $Phi$ is an arbitrary function. $C$ is an arbitrary constant.
          They have to be determined to fit some boundary and initial conditions. One cannot go further since those conditions are not specified in the wording of the question.



          CHECKING :
          $$u_x(x,t)=Phi'(x+t)$$
          $$u_t(x,t)=Phi'(x,t)+Phi'(1+t)$$
          $$u_t(x,t)-u_x(x,t)=Phi'(1+t)=u_x(1,t)$$
          The PDE is satisfied. The above result is correct.



          IN ADDITION after the comments :



          Case of condition $u(x,0)=g(x)=Phi(x)+Phi(1)+C=Phi(x)+C_2$



          $Phi(x)=g(x)-C_2$



          $u(x,t)= (g(x+t)-C_2)+(g(1+t)-C_2)+C$



          Thus $quad -2C_2+C+g(1)=0quad;quad C=2C_2-g(1)$



          $u(x,t)=(g(x+t)-C_2)+(g(1+t)-C_2)+2C_2-g(1)$



          $$u(x,t)=g(x+t)+g(1+t)-g(1)$$



          Example :



          $$g(x)=x^mqquad:qquad u(x,t)=(x+t)^m+(1+t)^m-1$$






          share|cite|improve this answer














          $$u_t(x,t)-u_x(x,t) = u_x(1,t).$$
          $u_x(1+t)$ is a function of $t$ only. Let $u_x(1+t)=f(t)$. Of course, this is an unknown function.
          $$u_t(x,t)-u_x(x,t) = f(t).$$
          Solving for the general solution. Charpit-Lagrange equations :
          $$frac{dt}{1}=frac{dx}{-1}=frac{du}{f(t)}$$
          A first family of characteristic curves comes from $frac{dt}{1}=frac{dx}{-1}$
          $$x+t=c_1$$
          A second family of characteristic curves comes from $frac{dt}{1}=frac{du}{f(t)}$
          $$u-int f(t)dt=c_2$$
          General solution of the PDE : $u-int f(t)dt=Phi(x+t)$



          $Phi(X)$ is an arbitrary function of one variable only. In the above equation $X=(x+t)$.
          $$u(x,t)= Phi(x+t)+int f(t)dt$$
          $u_x(x,t)=left(frac{dPhi(X)}{dX}right)_{(X=x+t)}=Phi'(x+t)$



          $u_x(1,t)=left(frac{dPhi(X)}{dX}right)_{(X=1+t)}=Phi'(1+t)$.
          $$f(t)=u_x(1,t)=Phi'(1+t).$$
          $int f(t)dt=int_{t_0}^t f(zeta)dzeta+C= int_{t_0}^t Phi'(1+zeta)dzeta +C =Phi(1+t)+C$.



          Finally the solution is :
          $$u(x,t)= Phi(x+t)+Phi(1+t)+C$$
          $Phi$ is an arbitrary function. $C$ is an arbitrary constant.
          They have to be determined to fit some boundary and initial conditions. One cannot go further since those conditions are not specified in the wording of the question.



          CHECKING :
          $$u_x(x,t)=Phi'(x+t)$$
          $$u_t(x,t)=Phi'(x,t)+Phi'(1+t)$$
          $$u_t(x,t)-u_x(x,t)=Phi'(1+t)=u_x(1,t)$$
          The PDE is satisfied. The above result is correct.



          IN ADDITION after the comments :



          Case of condition $u(x,0)=g(x)=Phi(x)+Phi(1)+C=Phi(x)+C_2$



          $Phi(x)=g(x)-C_2$



          $u(x,t)= (g(x+t)-C_2)+(g(1+t)-C_2)+C$



          Thus $quad -2C_2+C+g(1)=0quad;quad C=2C_2-g(1)$



          $u(x,t)=(g(x+t)-C_2)+(g(1+t)-C_2)+2C_2-g(1)$



          $$u(x,t)=g(x+t)+g(1+t)-g(1)$$



          Example :



          $$g(x)=x^mqquad:qquad u(x,t)=(x+t)^m+(1+t)^m-1$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 15 at 9:09

























          answered Nov 12 at 19:14









          JJacquelin

          42k21750




          42k21750












          • Thank you for your response. That was really helpful. In general my initial condition will just look like this: $u(x,0)=g(x)$ where g is some continuous function. For the sake of my example I choose $g(x) = x^m$ for some natural number $m$. Setting $C=-g(1)$ and $Phi=g$ would yield a solution, since $$u(x,0)=Phi(x+0) + Phi(1) - g(1) = Phi(x) = g(x)$$ and by your calculation. Is that right? But this choice is always possible independent of what my initial condition looks like as long as $g(1)$ is well defined and that is what puzzles me a bit.
            – Jfischer
            Nov 15 at 8:13












          • I hoped to get the general concept from that toy example, i.e., how to solve an equation of type $$u_t(x,t) = w(x)u_x(x,t) + v(x)u_x(1,t)$$. I struggle with the step from $c_1 = x+t$ and $c_2 = u-int f(t) dt$ to $u-int f(t) dt = Phi(x+t)$. Does that mean that $c_2$ is represented as a function along the characteristic? I would also be very thankful for a source for the case where a term of type $u_x(1,t)$ appears to read it up.
            – Jfischer
            Nov 18 at 8:10


















          • Thank you for your response. That was really helpful. In general my initial condition will just look like this: $u(x,0)=g(x)$ where g is some continuous function. For the sake of my example I choose $g(x) = x^m$ for some natural number $m$. Setting $C=-g(1)$ and $Phi=g$ would yield a solution, since $$u(x,0)=Phi(x+0) + Phi(1) - g(1) = Phi(x) = g(x)$$ and by your calculation. Is that right? But this choice is always possible independent of what my initial condition looks like as long as $g(1)$ is well defined and that is what puzzles me a bit.
            – Jfischer
            Nov 15 at 8:13












          • I hoped to get the general concept from that toy example, i.e., how to solve an equation of type $$u_t(x,t) = w(x)u_x(x,t) + v(x)u_x(1,t)$$. I struggle with the step from $c_1 = x+t$ and $c_2 = u-int f(t) dt$ to $u-int f(t) dt = Phi(x+t)$. Does that mean that $c_2$ is represented as a function along the characteristic? I would also be very thankful for a source for the case where a term of type $u_x(1,t)$ appears to read it up.
            – Jfischer
            Nov 18 at 8:10
















          Thank you for your response. That was really helpful. In general my initial condition will just look like this: $u(x,0)=g(x)$ where g is some continuous function. For the sake of my example I choose $g(x) = x^m$ for some natural number $m$. Setting $C=-g(1)$ and $Phi=g$ would yield a solution, since $$u(x,0)=Phi(x+0) + Phi(1) - g(1) = Phi(x) = g(x)$$ and by your calculation. Is that right? But this choice is always possible independent of what my initial condition looks like as long as $g(1)$ is well defined and that is what puzzles me a bit.
          – Jfischer
          Nov 15 at 8:13






          Thank you for your response. That was really helpful. In general my initial condition will just look like this: $u(x,0)=g(x)$ where g is some continuous function. For the sake of my example I choose $g(x) = x^m$ for some natural number $m$. Setting $C=-g(1)$ and $Phi=g$ would yield a solution, since $$u(x,0)=Phi(x+0) + Phi(1) - g(1) = Phi(x) = g(x)$$ and by your calculation. Is that right? But this choice is always possible independent of what my initial condition looks like as long as $g(1)$ is well defined and that is what puzzles me a bit.
          – Jfischer
          Nov 15 at 8:13














          I hoped to get the general concept from that toy example, i.e., how to solve an equation of type $$u_t(x,t) = w(x)u_x(x,t) + v(x)u_x(1,t)$$. I struggle with the step from $c_1 = x+t$ and $c_2 = u-int f(t) dt$ to $u-int f(t) dt = Phi(x+t)$. Does that mean that $c_2$ is represented as a function along the characteristic? I would also be very thankful for a source for the case where a term of type $u_x(1,t)$ appears to read it up.
          – Jfischer
          Nov 18 at 8:10




          I hoped to get the general concept from that toy example, i.e., how to solve an equation of type $$u_t(x,t) = w(x)u_x(x,t) + v(x)u_x(1,t)$$. I struggle with the step from $c_1 = x+t$ and $c_2 = u-int f(t) dt$ to $u-int f(t) dt = Phi(x+t)$. Does that mean that $c_2$ is represented as a function along the characteristic? I would also be very thankful for a source for the case where a term of type $u_x(1,t)$ appears to read it up.
          – Jfischer
          Nov 18 at 8:10


















           

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