Solving an initial value problem for a PDE
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Having the following initial value problem
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/979870/image.png)
with some mathematical computations we reach to an end that an implicit general solution of this pde can have the following form
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/979875/image.png)
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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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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