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Burgers Equation Question -- cannot satisfy initial conditions
Use characteristics to solve [tex]u_t+uu_x=0[/tex] on half line x≥0 with [tex]u(x,0)=x^2[/tex]
NA
I think I have an issue with the initial condition. So solving via characteristics gives:
[tex] \frac{dx}{dt}=u \Rightarrow x=ut+x_0 \Rightarrow u=f(x-ut)[/tex]. Then we plug in initial values and get:
[tex]u(x,0)=f(x)=x^2 \Rightarrow u=(x-ut)^2 \Rightarrow u^2t^2-u(1+2xt)+x^2=0[/tex]
Then by quadratic formula
[tex] u=\dfrac{(1+2xt) \pm \sqrt{(1+2xt)^2+4x^2t^2}}{2t^2} [/tex]
Now here I have a problem. So, in order to pic which sign solution to use, I need to use the initial condition. However, at t=0, I have an obvious problem. If anyone could offer help, or whether there is a typo in the problem, that would be much appreciated!
Homework Statement
Use characteristics to solve [tex]u_t+uu_x=0[/tex] on half line x≥0 with [tex]u(x,0)=x^2[/tex]
Homework Equations
NA
The Attempt at a Solution
I think I have an issue with the initial condition. So solving via characteristics gives:
[tex] \frac{dx}{dt}=u \Rightarrow x=ut+x_0 \Rightarrow u=f(x-ut)[/tex]. Then we plug in initial values and get:
[tex]u(x,0)=f(x)=x^2 \Rightarrow u=(x-ut)^2 \Rightarrow u^2t^2-u(1+2xt)+x^2=0[/tex]
Then by quadratic formula
[tex] u=\dfrac{(1+2xt) \pm \sqrt{(1+2xt)^2+4x^2t^2}}{2t^2} [/tex]
Now here I have a problem. So, in order to pic which sign solution to use, I need to use the initial condition. However, at t=0, I have an obvious problem. If anyone could offer help, or whether there is a typo in the problem, that would be much appreciated!