Verify directly from (3) that the solution has derivatives of all orders in {z > 0}. Asume that h(x, y) is a continuous function that vanishes outside some circle. (Hint: See Section A.3 for differentiation under an integral sign.) on z = 0. Therefore, the solution of (2) is Z0 u(xo, yo, Zo) /| Ix – xo)° + (y – yo)² + (zo)°]=3/²h(x, y) dx dy, (3) where both integrals run over (-o, ), noting that z = 0 in the integrand.
Verify directly from (3) that the solution has derivatives of all orders in {z > 0}. Asume that h(x, y) is a continuous function that vanishes outside some circle. (Hint: See Section A.3 for differentiation under an integral sign.) on z = 0. Therefore, the solution of (2) is Z0 u(xo, yo, Zo) /| Ix – xo)° + (y – yo)² + (zo)°]=3/²h(x, y) dx dy, (3) where both integrals run over (-o, ), noting that z = 0 in the integrand.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
M3
![Verify directly from (3) that the solution has
derivatives of all orders
in {z > 0}. Assume that h(x, y) is a continuous
function that vanishes
outside some circle. (Hint: See Section A.3 for
differentiation under an
integral sign.)
on z = 0. Therefore, the solution of (2) is
u(x0, yo, zo)
* || (x – xo)° + (y – yo)² + (zo)°]¯3/²h(x, y) dx dy,
(3)
where both integrals run over (-o, ), noting that z = 0 in the integrand.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5fb4bc67-5d6a-40d6-a872-72e7a1893aba%2F2abd4620-0ea3-4571-a0de-2bd0330f28b4%2Fmplwsyb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Verify directly from (3) that the solution has
derivatives of all orders
in {z > 0}. Assume that h(x, y) is a continuous
function that vanishes
outside some circle. (Hint: See Section A.3 for
differentiation under an
integral sign.)
on z = 0. Therefore, the solution of (2) is
u(x0, yo, zo)
* || (x – xo)° + (y – yo)² + (zo)°]¯3/²h(x, y) dx dy,
(3)
where both integrals run over (-o, ), noting that z = 0 in the integrand.
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