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Mathematical Methods for Physics and Engineering - Matematica.NET

Mathematical Methods for Physics and Engineering - Matematica.NET

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PARTIAL DIFFERENTIATIONit exact. Consider the general differential containing two variables,df = A(x, y) dx + B(x, y) dy.We see that∂f∂f= A(x, y),∂x= B(x, y)∂y<strong>and</strong>, using the property f xy = f yx , we there<strong>for</strong>e require∂A∂y = ∂B∂x . (5.9)This is in fact both a necessary <strong>and</strong> a sufficient condition <strong>for</strong> the differential tobe exact.◮Using (5.9) show that xdy+3ydx is inexact.In the above notation, A(x, y) =3y <strong>and</strong> B(x, y) =x <strong>and</strong> so∂A∂y =3,∂B∂x =1.As these are not equal it follows that the differential is inexact. ◭Determining whether a differential containing many variable x 1 ,x 2 ,...,x n isexact is a simple extension of the above. A differential containing many variablescanbewritteningeneralas<strong>and</strong> will be exact ifdf =n∑g i (x 1 ,x 2 ,...,x n ) dx ii=1∂g i∂x j= ∂g j∂x i<strong>for</strong> all pairs i, j. (5.10)There will be 1 2n(n − 1) such relationships to be satisfied.◮Show thatis an exact differential.(y + z) dx + xdy+ xdzIn this case, g 1 (x, y, z) =y + z, g 2 (x, y, z) =x, g 3 (x, y, z) =x <strong>and</strong> hence ∂g 1 /∂y =1=∂g 2 /∂x, ∂g 3 /∂x =1=∂g 1 /∂z, ∂g 2 /∂z =0=∂g 3 /∂y; there<strong>for</strong>e, from (5.10), the differentialis exact. As mentioned above, it is sometimes possible to show that a differential is exactsimply by finding by inspection the function from which it originates. In this example, itcan be seen easily that f(x, y, z) =x(y + z)+c. ◭156

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