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I know a PDE is linear when the dependent variable $u$ and its derivatives appear only to the first power. So, $u_t + u_x +5u = 1$ would be linear.

However, I do not quite understand the other two.

My professor described

  • "semilinear" PDE's as PDE's whose highest order terms are linear, and

  • "quasilinear" PDE's as PDE's whose highest order terms appear only as individual terms multiplied by lower order terms.

No examples were provided; only equivalent statements involving sums and multiindices were shown, which I do not think I could decipher by tomorrow.

Can someone provide some examples of "semilinear" and "quasilinear" PDE's?

311411
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Ninosław Brzostowiecki
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  • Another good reference book is from the book of Zachmanoglou"introduction to PDE with applications" – dmtri Jul 29 '18 at 09:32
  • You can find a formal definition in 'Partial Differential Equations': Second Edition written by Evans. It is on the second page of chapter 1. – esmo Jul 29 '18 at 09:11

4 Answers4

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I think this will help you to understand the PDE $:$

Linear PDE: $a(x,y)u_x+b(x,y)u_y+c(x,y)u=f(x,y)$

Semi-linear PDE: $a(x,y)u_x+b(x,y)u_y=f(x,y,u)$

Quasi-linear PDE: $a(x,y,u)u_x+b(x,y,u)u_y=f(x,y,u)$

MatheMagic
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    could you please give a similar explanation for the case of second order pde ? – Nizar Sep 30 '18 at 12:38
  • @MatheMagic thanks for your answer.I have one query.You consider first order PDEs in your answer,can I know how coefficients will change if it is of second order means suppose in case of quasi-linear $u_{xx}$ is present then coefficient will remain a(x,y,u) or will get modify? – ogirkar Jan 16 '19 at 13:19
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Using Einstein convention for summation and omitting dependence on $x$ of $u$ and its derivatives,

  1. linear: $a^{ij}(x)D_{ij}u+b^i(x)D_iu+c(x)u+d(x)=0$
  2. semi-linear: $a^{ij}(x)D_{ij}u+b(x,u,Du)=0$
  3. quasi-linear: $a^{ij}(x,u,Du)D_{ij}u+b(x,u,Du)=0$.
No-one
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I hope these examples will help you.

Semilinear/Almost Linear PDE:

1) $a(x,y)u_x+b(x,y)u_y+c(x,y,u)=0$

2) $U_{tt}-U_{xx}+U^3=0$

Quasi Linear PDE:

1) $a(x,y,u)u_x+b(x,y,u)u_y-c(x,y,u)=0$

2) $U_x+UV_y=0$

3) $U_{tt}-UU_{xx}+U^3=0$

4) $U_{tt}-UU_{xx}+U=0$

5) Navier Stokes equation is also Quasi Linear Equation

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Hammad Khalid
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    (Just because the question has a lot of views) I am fairly sure that Navier Stokes is considered as being semilinear (see e.g. the definition in Evan's book, second page of the first chapter). – Lorenzo Pompili Nov 08 '21 at 08:07
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Linear. The degree for the unknown function is one through out. And no functions of the Unknown function. Semilinear. The derivatives are linear but the unknown function is not likear. Quasilinear. Derivatives of the order are not linear. Once the whole eqn is not linear then it becomes non linear.

MUGISHA FREDRICK
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Apr 26 '22 at 00:15