SSM-Ch12 - Chap. 12 Sec. 12.1 Partial Differential...

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Chap . 12 Partial Differential Equations ( PDEs ) Sec . 12 . 1 Basic Concepts A PDE has a much greater variety of solutions than an ODE. Indeed, whereas the solution of an ODE of second order contains 2 arbitrary constants , the solution of a PDE of second order generally contains 2 arbitrary functions . You determine the latter from intial conditions (from given initial functions, for instance, given initial displacement and velocity of a mechanical system). This section introduces basic concepts, lists the most important PDEs to be considered, and shows that some PDEs can be solved as ODEs. Problem Set 12 . 1 . Page 537 1 . PDE solvable as ODE . x does not occur explicitly. As an ODE it would be u uu U 16 u ± 0 with a general solution u ± A cos 4 y U B sin 4 y involving arbitrary constants A and B . Since you are dealing with a PDE and look for a solution u ± u u x , y U , you can have arbitrary functions A ± A u x U and B ± B u x U , so that u u x , y U ± A u x U cos 4 y U B u x U sin 4 y . 5 . PDE solvable as a linear ODE . u u U u ± e xy can be solved by (4) in Sec. 1.5 with x ± y and the present x as a parameter, that is, u ± e u y U e y e xy dy U c u x U ± c u x U e u y U e u y e u x U 1 U y x U 1 ± c u x U e u y U e xy x U 1 . 11 . Substitution u x ± v gives from u xy ± u x the PDE u y ± v solvable as an ODE. Separate variables and integrate with respect to y to get v ± c 1 u x U e y and integrate, this time with respect to x , to get the answer u ± U c 1 u x U e y dx U c 2 u y U ± e y U c 1 u x U dx U c 2 u y U , which on p. A33 in Appendix 2 of the book is denoted by c u x U e y U h u y U . This is the same, except for notation, as you can see by noting the dependence of the two summands on the variables x and y . 27 . Boundary value problem . Verify the solution. Observing the chain rule, you obtain by differentiation u x ± 2 ax x 2 U y 2 and by another differentiation, with the product rule applied to 2 ax and 1/ u x 2 U y 2 U u xx ± 2 a x 2 U y 2 U u 1 ± 2 ax ± 2 x u x 2 U y 2 U 2 . (A) Similarly, with y instead of x , u yy ± 2 a x 2 U y 2 u 4 ay 2 u x 2 U y 2 U 2 . (B) Taking the common denominator u x 2 U y 2 U 2 , you obtain in (A) the numerator 2 a u x 2 U y 2 U u 4 ax 2 ± u 2 ax 2 U 2 ay 2 and in (B)
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Chap. 12 Partial Differential Equations (PDEs) 119 2 a u x 2 u y 2 U u 4 ay 2 U u 2 ay 2 u 2 ax 2 . Addition of the two expressions on the right gives 0 and completes the verification. Now determine a and b in u u x , y U U a ln u x 2 u y 2 U u b from the boundary conditions. For x 2 u y 2 U 1 you have ln1 U 0, so that b U 110 from the first boundary condition. From this and the second boundary condition 0 U a ln 100 u b you obtain a ln 100 U u 110. Hence a U u 110/ln 100, in agreement with the answer on p. A33. Sec
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SSM-Ch12 - Chap. 12 Sec. 12.1 Partial Differential...

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