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Example Problems:

1) Examples of Non-linear PDEs Non linear PDEs are linear PDEs with non-linear terms. These are usually much harder to study or solve than simple PDEs. The basic problems of PDEs like singularities, uniqueness and existence still exist but are much harder to deal with in the case of non-linear PDEs. Famous examples of non-linear PDEs are :

Burgers Equation: \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} = \vartheta \frac{\partial^2 u}{\partial x^2}

Eulers' Equations: \frac{\partial \rho }{\partial t} + \nabla. (\rho u) = 0 \frac{\partial \rho u}{\partial t} + \nabla. (u \times \rho u)+ \nabla p = 0 \frac{\partial E}{\partial t}+ \nabla.( u(E p))=0

Some other examples are the  Navier-Stokes Equations  and the Schrodinger Equation.

2) The Hessian Operator: Consider a function f= f(x,y,z). The Hessian  H  of function f is given by: \begin{bmatrix} \frac{\partial^2 f}{\partial x^2}  \frac{\partial^2 f}{\partial x \partial y}  \frac{\partial^2 f}{\partial x \partial z}  \\  \frac{\partial^2 f}{\partial y \partial x}   \frac{\partial^2 f}{\partial y^2}  \frac{\partial^2 f}{\partial y \partial z} \\  \frac{\partial^2 f}{\partial z \partial x}   \frac{\partial^2 f}{\partial z \partial y}  \frac{\partial^2 f}{\partial z^2}  \end{bmatrix} The term 'hessian' is used to refer sometimes to both the matrix above and its determinant. It is also used in a number of calculus tests such as determining critical points of the function f or points where f attains a local minima or maxima.  If the derivative of f w.r.t x is zero at a certain x, it is said that this x is a critical point. If the determinant of the hessian at x is zero, then x is called a ''' degenerate critical point. Similarly if the Hessian has positive eigenvalues at x, then f has a local minima at x. Otherwise f has a local maxima at x. If the hessian has both positive and negative eigenvalues, then x is defined as a saddle-point for the function f.

3)The operator D(u) is also denoted by D(.) Hence if D(.)= 6(.)^3+ 2(.)^2+ \sqrt u then D(u)= 6u^3+2u^2+\sqrt u

4) In order to prove linearity, we can use this simple test. For a function f(x) to be linear we need  f(\alpha x)= \alpha f(x) , where \alpha is a constant. Expanding  D(\alpha u) = 6(\alpha u)^3+ 2(\alpha u)^2+ \sqrt (\alpha u)  which is  NOT  equal to \alpha D(u) .  Hence D(.) is not a linear operator. .