Tutorial 11 solution

# Tutorial 11 solution - NATIONAL UNIVERSITY OF SINGAPORE...

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NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA 1505 Mathematics I Tutorial 11 (Solution Notes) 1. Show that u ( x, y )= F ( y 3 x ), where F is an arbitrary single variable function, is a solution of the partial diFerential equation u x +3 u y =0 . ±ind the particular solution which satis²es each of the following conditions separately : (a) u (0 ,y )=4s in y ; (b) u ( x, 0) = e x +1 . Solution ±or veri²cation is quite easy, we just need to substitute the candidate (the solution given in the question) to the partial diFerential equation. So, here comes the veri²cation: We substitute the solution u ( x, y F ( y 3 x )( ) to the partial diFerential equation. ±irst, we get the partial derivatives: u x = ∂x u = F ( y 3 x ) (Substitute the solution ( )) = d dx F ( y 3 x ) (Since the function F is single-variable) = F ± ( y 3 x ) · d dx ( y 3 x C h a i n R u l e ) = 3 F ± ( y 3 x ) u y = ∂y u = F ( y 3 x ) (Substitute the solution ( )) = d dy F ( y 3 x ) (Since the function F is single-variable) = F ± ( y 3 x ) · d dy ( y 3 x C h a i n R u l e ) = F ± ( y 3 x ) Therefore, u x u y = ± 3 F ± ( y 3 x ) ² ± F ± ( y 3 x ) ² , i.e. the P.D.E. is satis²ed and ( ) is a general solution. 224

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To fnd the particular solution to the partial diFerential equation u x +3 u y = 0, we need to fnd the specifc expression o± the single-variable ±unction F ( t ) that satisfes the given initial condition. (a) The initial condition is u (0 ,y )=4s in y ,sowesubst itute x =0in( ): u ( x, y )= F ( y 3 x )( ) , we get u (0 F ( y 3 · 0) = F ( y ) . Comparing with the given condition u (0 y ,wehave F ( y )=4 s i n y ( ) . There±ore, substituting ( )into( ), we get the particular solution with initial condition u (0 y is given by u ( x, y F ( y 3 x s i n ( y 3 x ) . (b) The initial condition is u ( x, 0) = e x +1 , so we substitute y ): u ( x, y F ( y 3 x ) , we get u ( x, 0) = F (0 3 · x F ( 3 x ) . Comparing with the given condition u ( x, 0) = e x +1 F ( 3 x e x +1 ( ± ) . We set t = 3 x to get x = t/ 3. So equation ( ± ) can be rewritten as F ( t e ( t/ 3+1) ( ) . There±ore, substituting ( ), we get the particular solution with initial condition u ( x, 0) = e x +1 is given by u ( x, y F ( y 3 x e ± ( y 3 x ) 3 +1 ² . ± 225
2. Solve the following partial diFerential equations: (a) u xy = u x ; (b) u x =2 xyu ; Preliminaries Separable Ordinary Diferential Equation Defnition. A frst-order ordinary diFerential equation is called separable i± the equation can be written as the ±orm A ( y ) · dy dx = B ( x ) . Solution Scheme. Multiply dx on both sides (in the way, we separate the variables x and y ), and get: A ( y ) dy = B ( x ) dx. Now, the ordinary diFerential equation can by solved by integration: ± A ( y ) dy = ± B ( x ) dx + c. Applications 1. Of course, the most useful application is use the scheme to solve ordinary diFerential equations.

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## This note was uploaded on 05/05/2009 for the course MA MA1505 taught by Professor Ma1505 during the Spring '09 term at National University of Juridical Sciences.

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Tutorial 11 solution - NATIONAL UNIVERSITY OF SINGAPORE...

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