Meaning of quasi-linear PDE (Where is linearity in quasi-linear PDE?)











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My book says, "PDE is said to be quasi-linear if it is linear in
the highest-ordered derivative of the unknown function." For example,
$$u_xu_{xx} + xuu_y = sin y$$ is quasi-linear PDE. Now, what is the meaning of 'linear in highest ordered derivative of unknown function'? I read first answer of this question: Linear vs nonlinear differential equation. But it doesn't answer my question. How can we find out linear operator in this case (or where is linearity here)? On which variable should we apply linear operator here?
Some sources defines quasilinear PDE as the PDE in which the coefficients of the highest derivatives of $u$ depend only on lower derivatives of $u$. How does this make PDE linear in highest order derivative? Any help would be appreciated. Thank you.










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    down vote

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    My book says, "PDE is said to be quasi-linear if it is linear in
    the highest-ordered derivative of the unknown function." For example,
    $$u_xu_{xx} + xuu_y = sin y$$ is quasi-linear PDE. Now, what is the meaning of 'linear in highest ordered derivative of unknown function'? I read first answer of this question: Linear vs nonlinear differential equation. But it doesn't answer my question. How can we find out linear operator in this case (or where is linearity here)? On which variable should we apply linear operator here?
    Some sources defines quasilinear PDE as the PDE in which the coefficients of the highest derivatives of $u$ depend only on lower derivatives of $u$. How does this make PDE linear in highest order derivative? Any help would be appreciated. Thank you.










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      down vote

      favorite









      up vote
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      down vote

      favorite











      My book says, "PDE is said to be quasi-linear if it is linear in
      the highest-ordered derivative of the unknown function." For example,
      $$u_xu_{xx} + xuu_y = sin y$$ is quasi-linear PDE. Now, what is the meaning of 'linear in highest ordered derivative of unknown function'? I read first answer of this question: Linear vs nonlinear differential equation. But it doesn't answer my question. How can we find out linear operator in this case (or where is linearity here)? On which variable should we apply linear operator here?
      Some sources defines quasilinear PDE as the PDE in which the coefficients of the highest derivatives of $u$ depend only on lower derivatives of $u$. How does this make PDE linear in highest order derivative? Any help would be appreciated. Thank you.










      share|cite|improve this question













      My book says, "PDE is said to be quasi-linear if it is linear in
      the highest-ordered derivative of the unknown function." For example,
      $$u_xu_{xx} + xuu_y = sin y$$ is quasi-linear PDE. Now, what is the meaning of 'linear in highest ordered derivative of unknown function'? I read first answer of this question: Linear vs nonlinear differential equation. But it doesn't answer my question. How can we find out linear operator in this case (or where is linearity here)? On which variable should we apply linear operator here?
      Some sources defines quasilinear PDE as the PDE in which the coefficients of the highest derivatives of $u$ depend only on lower derivatives of $u$. How does this make PDE linear in highest order derivative? Any help would be appreciated. Thank you.







      differential-equations pde






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      asked Nov 20 at 21:10









      ramanujan

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          In short, you have
          begin{align}
          u_{xx}+xufrac{u_y}{u_x}=frac{sin y}{u_x}
          end{align}

          which is linear in the highest derivative.



          On the other hand
          begin{align}
          u_{xx}^2+xufrac{u_y}{u_x}=frac{sin y}{u_x}
          end{align}

          is not quasi-linear.






          share|cite|improve this answer





















          • But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
            – ramanujan
            Nov 20 at 21:20






          • 2




            @ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
            – Jacky Chong
            Nov 20 at 21:22






          • 2




            @ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
            – AlexanderJ93
            Nov 20 at 21:25










          • @Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
            – ramanujan
            Nov 20 at 21:29










          • Yes. You are correct.
            – Jacky Chong
            Nov 20 at 21:30











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          In short, you have
          begin{align}
          u_{xx}+xufrac{u_y}{u_x}=frac{sin y}{u_x}
          end{align}

          which is linear in the highest derivative.



          On the other hand
          begin{align}
          u_{xx}^2+xufrac{u_y}{u_x}=frac{sin y}{u_x}
          end{align}

          is not quasi-linear.






          share|cite|improve this answer





















          • But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
            – ramanujan
            Nov 20 at 21:20






          • 2




            @ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
            – Jacky Chong
            Nov 20 at 21:22






          • 2




            @ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
            – AlexanderJ93
            Nov 20 at 21:25










          • @Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
            – ramanujan
            Nov 20 at 21:29










          • Yes. You are correct.
            – Jacky Chong
            Nov 20 at 21:30















          up vote
          1
          down vote



          accepted










          In short, you have
          begin{align}
          u_{xx}+xufrac{u_y}{u_x}=frac{sin y}{u_x}
          end{align}

          which is linear in the highest derivative.



          On the other hand
          begin{align}
          u_{xx}^2+xufrac{u_y}{u_x}=frac{sin y}{u_x}
          end{align}

          is not quasi-linear.






          share|cite|improve this answer





















          • But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
            – ramanujan
            Nov 20 at 21:20






          • 2




            @ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
            – Jacky Chong
            Nov 20 at 21:22






          • 2




            @ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
            – AlexanderJ93
            Nov 20 at 21:25










          • @Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
            – ramanujan
            Nov 20 at 21:29










          • Yes. You are correct.
            – Jacky Chong
            Nov 20 at 21:30













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          In short, you have
          begin{align}
          u_{xx}+xufrac{u_y}{u_x}=frac{sin y}{u_x}
          end{align}

          which is linear in the highest derivative.



          On the other hand
          begin{align}
          u_{xx}^2+xufrac{u_y}{u_x}=frac{sin y}{u_x}
          end{align}

          is not quasi-linear.






          share|cite|improve this answer












          In short, you have
          begin{align}
          u_{xx}+xufrac{u_y}{u_x}=frac{sin y}{u_x}
          end{align}

          which is linear in the highest derivative.



          On the other hand
          begin{align}
          u_{xx}^2+xufrac{u_y}{u_x}=frac{sin y}{u_x}
          end{align}

          is not quasi-linear.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 21:16









          Jacky Chong

          17.4k21128




          17.4k21128












          • But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
            – ramanujan
            Nov 20 at 21:20






          • 2




            @ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
            – Jacky Chong
            Nov 20 at 21:22






          • 2




            @ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
            – AlexanderJ93
            Nov 20 at 21:25










          • @Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
            – ramanujan
            Nov 20 at 21:29










          • Yes. You are correct.
            – Jacky Chong
            Nov 20 at 21:30


















          • But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
            – ramanujan
            Nov 20 at 21:20






          • 2




            @ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
            – Jacky Chong
            Nov 20 at 21:22






          • 2




            @ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
            – AlexanderJ93
            Nov 20 at 21:25










          • @Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
            – ramanujan
            Nov 20 at 21:29










          • Yes. You are correct.
            – Jacky Chong
            Nov 20 at 21:30
















          But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
          – ramanujan
          Nov 20 at 21:20




          But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here?
          – ramanujan
          Nov 20 at 21:20




          2




          2




          @ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
          – Jacky Chong
          Nov 20 at 21:22




          @ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms.
          – Jacky Chong
          Nov 20 at 21:22




          2




          2




          @ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
          – AlexanderJ93
          Nov 20 at 21:25




          @ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case.
          – AlexanderJ93
          Nov 20 at 21:25












          @Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
          – ramanujan
          Nov 20 at 21:29




          @Jacky Chong In $u_xu_{xx} + xuu_y = sin y$, $L(u)= u_{xx}$ and $F= frac{sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true?
          – ramanujan
          Nov 20 at 21:29












          Yes. You are correct.
          – Jacky Chong
          Nov 20 at 21:30




          Yes. You are correct.
          – Jacky Chong
          Nov 20 at 21:30


















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