06.09.2021 Views

Basics of Fluid Mechanics, 2014a

Basics of Fluid Mechanics, 2014a

Basics of Fluid Mechanics, 2014a

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

582 APPENDIX A. MATHEMATICS FOR FLUID MECHANICS<br />

Solution<br />

The transformation is<br />

v = u 2<br />

Using the definition (1.III.b) equation (1.III.a) becomes<br />

dv<br />

dt<br />

1−p<br />

{}}{<br />

−2 t 2 v =<br />

The homogeneous solution <strong>of</strong> equation (1.III.c) is<br />

1−p<br />

{}}{<br />

−2 sin(t)<br />

(1.III.b)<br />

(1.III.c)<br />

u(t) =ce −t3<br />

3 (1.III.d)<br />

And the general solution is<br />

u = e −t3 3<br />

⎛<br />

⎞<br />

private solution<br />

{ }} {<br />

⎜∫<br />

⎝<br />

e t3 ⎟<br />

⎠<br />

3 sin (t) dt +c<br />

(1.III.e)<br />

End Solution<br />

A.2.3.1<br />

Homogeneous Equations<br />

Homogeneous function is given as<br />

du<br />

= f(u, t) =f(au,at)<br />

dt (A.63)<br />

for any real positive a. For this case, the transformation <strong>of</strong> u = vttransforms equation<br />

(A.63) into<br />

t dv<br />

dt + v = f(1,v)<br />

(A.64)<br />

In another words if the substitution u = vtis inserted the function f become a function<br />

<strong>of</strong> only v it is homogeneous function. Example <strong>of</strong> such case u′ = ( u 3 − t 3) /t 3 becomes<br />

u′ = ( v 3 +1 ) . The solution is then<br />

∫<br />

dv<br />

ln |t| =<br />

f(1,v) − v + c<br />

(A.65)<br />

Example A.4:<br />

Solve the equation<br />

du<br />

( u<br />

) ( u 4<br />

dt = sin − t 4 )<br />

+<br />

t t 4<br />

(1.IV.a)

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!