By Yehuda Pinchover and Jacob Rubinstein

ISBN-10: 0511111576

ISBN-13: 9780511111570

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**Additional info for An introduction to partial differential equations**

**Example text**

Substituting the initial condition into the solution above leads to the parametric integral surface (x(t, s), y(t, s), u(t, s)) = (s cos t, s sin t, et ψ(s)). 34 First-order equations y projection of Γ x char. 6. Isolating s and t we obtain the explicit representation y . 4). Therefore, each one of them intersects the projection of the initial curve (the x axis) twice. We also saw that the Jacobian vanishes at the origin. So how is it that we seem to have obtained a unique solution? The mystery is easily resolved by observing that in choosing the positive sign for the square root in the argument of ψ, we effectively reduced the solution to the ray {x > 0}.

Char. 5 Self-intersection of characteristics. for y. We obtain y = (x + 1 − s)3 , and, thus, for each ﬁxed s this is an equation for a characteristic. 5. While the picture indicates no problems, we were not careful enough in solving the characteristic equations, since the function y 2/3 is not Lipschitz continuous at the origin. Thus the characteristic equations might not have a unique solution there! In fact, it can be easily veriﬁed that y = 0 is also a solution of yt = 3y 2/3 . 5, the well behaved characteristics near the projection of the initial curve y = 1 intersect at some point the extra characteristic y = 0.

26) Thus we anticipate a unique solution at each point where s = 0. Since we are limited to the regime x > 0 we indeed expect a unique solution. The parametric integral surface is given by (x(t, s), y(t, s), u(t, s)) = (s + t, s + s 2 + t, 1 − (1 − sin s)e−t ). In order to invert the mapping (x(t, s), y(t, s)), we substitute the equation for x into the equation for y to obtain s = (y − x)1/2 . The sign of the square root was selected 1 according to the condition x > 0. Now it is easy to ﬁnd t = x − (y − x) 2 , whence the explicit representation of the integral surface 1 u(x, y) = 1 − [1 − sin(y − x) 2 ]e−x+(y−x) 2 .

### An introduction to partial differential equations by Yehuda Pinchover and Jacob Rubinstein

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