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Hi, I need help in PDE homework. There are 10 problems. Page 71, Problem 4.1 (a, b, c) For 4.1-c, What happens if you try to solve it (Choose one method and try to solve it). Page 75, Problems 5.1,...

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Hi,
I need help in PDE homework. There are 10 problems.

Page 71, Problem 4.1 (a, b, c)

For 4.1-c, What happens if you try to solve it (Choose one method and try to solve it).

Page 75, Problems 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, (5.8 for 5.6 only)

Please Read the lecture notes to know how to solve the problem like the teacher style.
I need step by step answer for integral and differential.
Also check your solution for u1 and u2
Also grad(u1)Xgrad(u2)
And the check z
Take a look at the attached file.
Answered Same Day Dec 24, 2021

Solution

Robert answered on Dec 24 2021
126 Votes
(a) For the given equation and the initial curve C: x=t, y=t, t>0 we can write the ordinary differential
equation can be written as,
Given that z = 2t on Curve C , so the above equation will become
1/z = 1/y = 2/x = μ
So from theorem 4.2 we have infinitely many solutions
So, x = 2 μ , y = μ, z = μ, since μ is a constant number then it can have multiple values so infinite
solutions exits.
(b) For z = 1 on C
Given that z = t
1/z = 1/y = 1/x = μ
So from theorem 4.2 we have infinitely many solutions
So, x = μ, y = μ, z = μ , since μ is a constant number then it can have multiple values so infinite solutions
exits.
(c) z = sin(pi*t/2 ) on Curve C
Given that z = sin(pi*t/2 )
1/z = 1/y = cos(pi*t/2 )/x = μ
1/z = 1/y = 0/x = μ
So, here μ = 0
So from theorem 4.2 we have unique solution.
So, x = 0, y = 0 , z = 0
Hence unique solution exits.
Answer
Given partial differential equation is,
Where f and a are constant.
If we need to find a solution to the problem we can consider the case of ordinary differential equation and
the associated equation can be written as,
To solve this system of equation we need to take two independent functions u1 and u2 as,
u1 = z and u2 = x –a(z)y
So on solving the above differential equation for these two systems we get,
z = F(x-a(z)y)
If we apply the boundary condition then we get,
F(x) = f(x)
Now if we use the implicit function theorem for the given equation we see that at point(x0,y0) and the
oundary condition z(x,0) the two lines will intersect at a single point which shows that unique solution
exits for such kind of partial differential...
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