 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-86/3/a/solution

Let

s_{j} = sin β_{j}

c_{j} = cos β_{j}

. Inverting the two orthogonal derivative systems gives, on the vertical axis,

q_{x} (0, y) = - s_{1} h_{1} (y) - c_{1} j_{1} (y), q_{y} (0, y) = c_{1} h_{1} (y) - s_{1} j_{1} (y),

and, on the horizontal axis,

q_{x} (x, 0) = c_{2} h_{2} (x) + s_{2} j_{2} (x), q_{y} (x, 0) = - s_{2} h_{2} (x) + c_{2} j_{2} (x) .

q_{z} = (q_{x} - i q_{y}) /2

, these become

q_{z} (i y) = - \frac{e ^{- i β_{1}}}{2} [j_{1} (y) + i h_{1} (y)], q_{z} (x) = \frac{e ^{i β_{2}}}{2} [h_{2} (x) - i j_{2} (x)] .

Define the appropriate boundary transforms by

H_{1} (k) H_{2} (k) = \int_{0}^{\infty} e^{k y} h_{1} (y) d y, = \int_{0}^{\infty} e^{- ik x} h_{2} (x) d x, J_{1} (k) J_{2} (k) = \int_{0}^{\infty} e^{k y} j_{1} (y) d y, = \int_{0}^{\infty} e^{- ik x} j_{2} (x) d x .

Re k < 0

Im k < 0

. Parametrizing the vertical boundary by

z = i y

gives

d z = i d y

, which must be retained. Substitution yields the quarter-plane oblique boundary spectral functions:

q_{1} (k) = \frac{e ^{- i β_{1}}}{2} [H_{1} (k) - i J_{1} (k)], q_{2} (k) = - \frac{e ^{i β_{2}}}{2} [H_{2} (k) - i J_{2} (k)] .

Their signs follow from the stated clockwise boundary orientations: upward on the left and inward on the bottom.

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