APM346-2015F > Quiz 1

Quiz1-P1

(1/1)

Yumeng Wang:
(a)  ux + xuy = 0   non-linear linear homogeneous. V.I.
(b)  ux + uuy = 0  quasilinear

I am not sure of expressions of this two equations, but my answer for the first one is non-linear and the second one is quasilinear.

Emily Deibert:
I think that the first one should be considered linear homogeneous. We see in section 1.3 Classification of Equations in the textbook [http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.3.html] that a linear equation may have variable coefficients. In this case, the x is a variable coefficient, but the equation is still linear.

At least, this is what I think, but I might be wrong. Does anyone else have any idea?

You are right. V.I.

Yumeng Wang:
I think the first one is semilinear because it depends on definition 2 in Chapter 2.1 in textbook. In this case, b=b (x, y) and I don't think similinear is type of linear. So I don't get why it is linear.

Bruce Wu:

--- Quote from: Yumeng Wang on October 02, 2015, 11:13:39 PM ---I think the first one is semilinear because it depends on definition 2 in Chapter 2.1 in textbook. In this case, b=b (x, y) and I don't think similinear is type of linear.

--- End quote ---

But there is no right hand expression that is a non-zero function of $u$. If there was, say $u_{x}+xu_{y}=u^{2}$ then it would be semilinear.

Victor Ivrii:
\begin{align}
&a (x,t) u_t + b(x,t) u_x + c(x,t)u=0 && \text{linear homogeneous},\\[3pt]
&a (x,t) u_t + b(x,t) u_x + c(x,t)u=f(x,t) && \text{linear inhomogeneous},\\[3pt]
&a (x,t) u_t + b(x,t) u_x + c(x,t)u=f(x,t,u) && \text{seminilear},\\[3pt]
&a (x,t,u) u_t + b(x,t,u) u_x = f(x,t,u) && \text{quasilinear},\\[3pt]
&F(x,t,u,u_t,u_x)=0 && \text{nonlinear}
\end{align}
and each type is more general than the previous one.

Writing corresponding ODE we get respectively
\begin{align}
&\frac{dx}{a}=\frac{dy}{b}=-\frac{du}{cu} \implies &&\phi(x,t)=C_1,   u=C_2\psi(x,t)\tag{1},\\
&\frac{dx}{a}=\frac{dy}{b}=-\frac{du}{cu} \implies &&\phi(x,t)=C_1,   u=C_2\psi(x,t)+g(x,t)\tag{2},\\
&\frac{dx}{a}=\frac{dy}{b}=-\frac{du}{c} \implies &&\phi(x,t)=C_1, \psi(x,t,u)=C_2\tag{3},\\
&\frac{dx}{a}=\frac{dy}{b}=-\frac{du}{c} \implies &&\phi(x,t,u)=C_1, \psi(x,t,u)=C_2\tag{4},\\
\end{align}
For nonlinear see http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.2.html#sect-2.2.2, not covered in the course