# Solving an initial value problem for a PDE

조회 수: 2(최근 30일)
Salma fathi 2022년 4월 26일
편집: Torsten 2022년 4월 26일
Having the following initial value problem
with some mathematical computations we reach to an end that an implicit general solution of this pde can have the following form
if we had phi=e^(-x^2) for example,
I have been able to solve a similar problem to this but the genral solution was only a function of x and t, but here we have also u, so how can we possibly do that.

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### 답변(1개)

Torsten 2022년 4월 26일
편집: Torsten 2022년 4월 26일
The method of characteristics gives the equations
dt/ds = 1, t(0) = 0
dx/ds = u, x(0) = x0
du/ds = 0, u(0) = phi(x0)
with solution
x = x0 + phi(x0) * t
Thus to get the solution u(x,t) in (x,t), you will have to solve
x - x0 - phi(x0)*t = 0
for x0.
The solution u(x,t) in (x,t) is then given by u(x,t) = phi(x0).
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Torsten 2022년 4월 26일
편집: Torsten 2022년 4월 26일
You can't set
x0 = x - exp(-x0.^2).*t
For each pair (x(i),t(j)) of your linspace, you have to find x0 (if it exists) for that
x0-x(i)+exp(-x0.^2)*t(j) = 0.
Use "fzero" to do that in a double loop over i and j.

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