Solving an initial value problem for a PDE

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.

Answer 1

[Torsten]

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).

[Salma fathi]

Thank you for the quick reply, when solving this out I have got x0 to be a function of u, i.e, x0=x-ut, so my confussion is how can we implement x0 when it is a function of u? or is my understanding to the problem is wrong?

[Torsten]

x0 depends on x, t and phi, and all three are known.

[Salma fathi]

would the following lines be a correct implementationt o the problem's solution ?

x=linspace(-10,10,41)';
t=linspace(0,1,41)'; 
[x,t] = meshgrid(x,t);
x0=x-exp(-x.^2).*t;
u= exp(-x0.^2) ;
s=surf (x,t,u,'FaceAlpha','1');
s.FaceColor = 'texturemap';
xlabel('x'),ylabel('t'),zlabel('u')
view(-40,25) 

[Torsten]

x0=x-exp(-x0.^2).*t;

instead of

x0=x-exp(-x.^2).*t;

Thus for each (x,t) combination, you have to iterate to get the correct x0 value.

[Salma fathi]

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]

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.