Method of Characteristics

[member 428835]

Homework Statement

$$\frac{\partial u}{\partial t} + x^2 + t+\left(\frac{\partial u}{\partial x}\right)^2 = 0\\
u(x,0)=0$$

Homework Equations

$$
\dot{x} = 2 u_x ;\,\,t=0,\,\,x=\xi\\
\dot{u}=2(u_x)^2+u_t ;\,\,t=0,\,\,u=0\\
\dot{u_x}=-2x ;\,\,t=0,\,\,p=0\\
\dot{u_t}=-1 ;\,\,t=0,\,\,u_t=-\xi^2.
$$
where $\dot{f}$ is the total derivative of $f$ with respect to $t$, or $\dot{f} \equiv \frac{df}{dt}$ where $x$ is a function of $t$.

The Attempt at a Solution

Write equation 1 as $\ddot{x} = 2 \dot{u_x}$. Next substitute equation 3 into arrive at $$\ddot{x}=-4x^2 \implies\\ x = A \sin(2t) + B\cos(2t)$$ The first BC associated with equation 1 implies $B = \xi$, but now I'm stuck. Any ideas how to proceed?

 

[mighty2000]

Your approach seems to be on the right track. To find the constant A, we can use the second boundary condition, which implies that at t=0, x=ξ. So we have ξ=Asin(0)+ξcos(0) which simplifies to A=0. Therefore, the solution for x is x=ξcos(2t).

To find the solution for u, we can use the third boundary condition, which implies that at t=0, u_x=0. Substituting this into equation 2, we get u_t=0. This means that u is a constant with respect to time. Using the first boundary condition, we can find this constant to be u=-(ξ^3)/3. Therefore, the solution for u is u=-(ξ^3)/3.

So the final solution is u=-(ξ^3)/3 and x=ξcos(2t).