Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2018-10-29 15:01:38

Emma22
Guest

Quasilinear 2nd order PDEs with inital data

Hi, I'm having trouble with applying initial data to a partial differential equation I've solved.

The general solution obtained is u(x,t) =F(x^2-t^2*exp(u)) and the initial condition is u(x,0)=2ln(x)

What I tired was plugging in the data to the general solution to get:
2ln(x) = F(x^2)
From there i let x^2 = z and hence x=sqrt(z), which gives me:
2*ln(sqrt(z)) = F(z)

and then I tried taking this and subbing it through to the original solution I had, which gives:

u(x,t) = 2*ln(sqrt(x^2-t^2 *exp(U))

The solution to the problem is u(x,t) = ln((x^2)/(1+t^2)), was wondering if anyone could help point out what i did wrong

#2 2018-10-29 22:32:14

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,432
Website

Re: Quasilinear 2nd order PDEs with inital data

Hi Emma22,

Welcome to the forum. Have you considered registering an account with us?

Emma22 wrote:

The general solution obtained is u(x,t) =F(x^2-t^2*exp(u)) and the initial condition is u(x,0)=2ln(x)

What is
here? (You have later called this
.)

Offline

#3 2018-10-30 09:44:53

Grantingriver
Member
Registered: 2016-02-01
Posts: 129

Re: Quasilinear 2nd order PDEs with inital data

Hi Emman22, the solution is as follow:


and hence

which is the required solution. Your solution is also correct, but it required some manipulation to reach the desired form.

Offline

Board footer

Powered by FluxBB