Solving an initial value problem for a PDE

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Salma fathi
Salma fathi 2022-4-26
编辑： Torsten 2022-4-26
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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回答（1 个）

Torsten
Torsten 2022-4-26
编辑：Torsten 2022-4-26
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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Salma fathi
Salma fathi 2022-4-26
but this will result in the follwong error
Arrays have incompatible sizes for this operation.
Error in gradproject (line 28)
x0=x-exp(-x0.^2).*t;
Torsten
Torsten 2022-4-26
编辑：Torsten 2022-4-26
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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