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Please provide step by step solution.

(a) Express the Falkner-Skan equation 1 2 = -m, 11 + 3(m +1)yy – m (4) with initial conditions y(0) = 0, y(0) = 0, y(0)

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Answer:-

a) Falkner. skan equation is

{y}^{'''}+\frac{1}{2}(m+1)y{y}''-m({y}')^{2}=-m...................................(1)

with  y(0)=0,y{}'(0)=0,   y{}''(0)=G,

Where m,G are constants.

Let

x_{1}=y

x_{2}=y{}'

x_{3}=y{}''

So from (1) , we get

x_{1}{}'=x_{2}

x_{2}{}'=x_{3}

x_{3}{}'=-m-\frac{1}{2}(m+1)x_{1}x_{3}+mx_{2}^{2}

With  (x_{1}(0),x_{2}(0),x_{3}(0))=(0,0,G)

i.e . if we take   X=(x_{1},x_{2},x_{3}). Then

X{}'=f(x)

X(0)=(0,0,G)..........................................(2)

Where f(x)=f(x_{1},x_{2},x_{3})

  =(x_{2},x_{3},-m-\frac{1}{2}(m+1)x_{1}x_{3}+mx_{2}^{2})

\therefore (2) is the required system of , 1st order \triangle DE, equivalent to (1)

b) Clearly, as  f(x_{1},x_{2},x_{3})=(x_{2},x_{3},-m-\frac{1}{2}(m+1)x_{1}x_{3}+mx_{2}^{2}) has each component is smooth (being polnomial in (x_{1},x_{2},x_{3}) . so with any given initial condition , (2) ( or associated (1) ) has locally unique solution around O. (By existence uniqueness theorem ). and because of the same reason, there are no such intervals for which, the conditions of existence unidueness theorom are violated.

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