Solution

Substitution of

w (z) = \int_{γ} e^{z t} f (t) d t

and integration by parts give

w^{'''} - z w = \int_{γ} e^{z t} (t^{3} f + f^{'}) d t - [e^{z t} f (t)]_{\partial γ} .

Thus take

f (t) = e^{- t^{4} /4},

and require the endpoint term to vanish. Infinite contour ends must lie in sectors where

Re (t^{4}) > 0

, namely sectors centred on the positive and negative real and imaginary axes, each of angular width

π /4

Let the common final segment be the positive real ray from

0

+ \infty

, and choose the initial rays from

- \infty

+ i \infty

, and

- i \infty

0

. Call the resulting contours

γ_{1}, γ_{2}, γ_{3}

. Put

A_{r} = \int_{0}^{\infty} t^{r} e^{- t^{4} /4} d t = 4^{(r - 3) /4} Γ (\frac{r + 1}{4}) .

Since

w_{i}^{(r)} (0) = \int_{γ_{i}} t^{r} e^{- t^{4} /4} d t

, for

r = 0, 1, 2

the three rows are

2 A_{0} (1 - i) A_{0} (1 + i) A_{0} 0 2 A_{1} 2 A_{1} 2 A_{2} (1 + i) A_{2} (1 - i) A_{2} .

Consequently

det w_{1} (0) w_{2} (0) w_{3} (0) w_{1}^{'} (0) w_{2}^{'} (0) w_{3}^{'} (0) w_{1}^{''} (0) w_{2}^{''} (0) w_{3}^{''} (0) = - 16 i A_{0} A_{1} A_{2} .

Using the Gamma function reflection identity,

det = - 22 i π^{3/2} \neq = 0 .

The initial-data vectors are therefore linearly independent, so the three contour integrals are linearly independent solutions.

Solved by gpt-5.6-sol high.

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