Terminology — Question 1

Question 1 – Solution

Strategy. Classify the differential expression on arbitrary admissible fields, rather than on a special solution.

Step 1: Count the derivatives that remain. The independent variables are $x,t$ and the dependent variable is $u$. Since $1+x^2>0$, the term $u_{xx}$ never disappears. The order is therefore $\boxed{2}$ for every $a$. When $a=0$, the equation is first order in time but still second order as a PDE. When $a\ne 0$, it is also second order in time.

Step 2: Separate coefficients from unknowns. The expression is linear exactly when $\boxed{b=c=0}$. Then $1+x^2$ is an allowed variable coefficient, so the PDE is linear with variable coefficients. It is nonhomogeneous because $\sin t$ is not identically zero on the domain. A forcing term may vanish at some times without making the equation homogeneous.

Step 3: Test scaling on informative fields. Write the nonlinear part as $N[u]=b u u_x+c(u_x)^2$. It satisfies $N[2u]-2N[u]=2N[u]$, whereas a linear operator would give zero. For the smooth test field $u=x$, this difference is $$2(bx+c).$$ It vanishes for every $x$ only when $b=c=0$. In particular, if $b\ne 0$ it is not the zero function; if $b=0$ and $c\ne 0$ it is the nonzero constant $2c$. The restricted tests $u=h(t)$ have $u_x=0$ and conceal both nonlinear terms. Linearity must hold on all fields in the operator’s domain.

Step 4: Classify the equation after changing the forcing. With zero right-hand side and $b=c=0$, the equation is linear homogeneous. With $(b,c)\ne(0,0)$, the same scaling obstruction remains, regardless of the right-hand side. Under the stated convention, “homogeneous linear PDE” is not a label for an arbitrary nonlinear equation written with zero on one side.