# Lecture5 - ESE 502 Mathematics of Modern Engineering II...

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ESE 502 Mathematics of Modern Engineering II Lecture 5 ”Some general facts about PDEs. The wave equation”. Let u ( t, x ) : [0 , ) × S IR be a given real-valued function where t [0 , ) and S IR n is a subset of IR n . The variable t is interpreted here as a time parameter and x as a space parameter. Since u is a function of more than one variable, any derivative of u is a partial derivative. Under a partial differential equation (PDE) we understand any equation involving the unknown function u and some of its partial derivatives. Classification of PDEs is similar to that of ODEs. Any expression of the form f ( t, x, u, u t , u x ) = 0 determines a PDE of 1st order . Here f is used as a function that can depend on the variables indicated. Any expression of the form f ( t, x, u, u t , u x , u tx , u tt , u xx ) = 0 determines a PDE of 2nd order . If the function f is linear in u and all partial derivatives of u , then the PDE is called linear . Any equation of the form f ( u, u t , u x ) = 0 is called 1st order homogeneous PDE . The 2nd order homogeneous PDE is given then by f ( u, u t , u x , u tx , u tt , u xx ) = 0 . Any function u that satisfies the given PDE is called a solution . As for any equation, the following questions arise: Does a PDE have a solution? (Existence question) 1

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If yes, then is the solution unique? (Uniqueness question) Can a solution (unique solution) of a PDE be found explicitely or only numerically ? Usually a PDE has (if at all) infinitely many solutions. It is clear why: since to find a solution, we have to integrate the equation in order to find u (getting rid of the derivatives involved). The integration procedure will generate arbitrary constants and even functions. Therfore, it must be clear that in order to find a unique solution, one needs to have additional assumptions to be satisfied by the function u . The assumptions could be of two kinds: boundary conditions which are specified by the behavior of the function on the boundary of S . For example, if u is sought on [0 , L ], then the conditions could be of the form u ( t, 0) = g 1 ( t ) , u ( t, L ) = g 2 ( t );
• Fall '09
• Partial differential equation

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