Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2011/9/solution

Introduce a characteristic parameter and label each point of the initial segment by . The method of characteristics gives
Integrating gives the parametric solution
The projected characteristic curves satisfy , so each lies on a straight line through . Its branch from the data always has ; the common vertex is approached only as .
Eliminating using and yields
Its derivatives give , and substitution of gives the prescribed value one.
The Jacobian determinant of is , and the inverse just displayed is unique. Hence the data determine the solution exactly on
This is the domain determined by a finite segment under a radial characteristic flow. Its interior is , ; the two boundary rays come from the endpoints of the initial segment. Every point there lies on exactly one characteristic from the data. Outside this sector, a characteristic either misses the initial segment or lies on the branch , which cannot reach it. Thus the formula may be extended beyond this sector, but such an extension is not uniquely specified by the given data.

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