2.14.1 $$u_t + u_{xxx} = 0$$

problem number 109

Airy PDE

Solve for $$u(x,t)$$

$u_t+u_{xxx} =0$

Mathematica

$\left \{\left \{u(x,t)\to x^3 (c_{11} t+c_4)+x^2 (c_3-60 c_6 t)+x (c_2-24 c_5 t)-3 t (c_{11} t+2 c_4)-\frac {c_{11} x^6}{120}+c_6 x^5+c_5 x^4+c_1\right \}\right \}$

Maple

$u \left (x , t\right ) = c_{4} \left (c_{1} {\mathrm e}^{-\frac {i \sqrt {3}\, x \mathit {\_c}_{1}^{\frac {1}{3}}}{2}} {\mathrm e}^{-\frac {x \mathit {\_c}_{1}^{\frac {1}{3}}}{2}}+c_{2} {\mathrm e}^{\frac {i \sqrt {3}\, x \mathit {\_c}_{1}^{\frac {1}{3}}}{2}} {\mathrm e}^{-\frac {x \mathit {\_c}_{1}^{\frac {1}{3}}}{2}}+c_{3} {\mathrm e}^{x \mathit {\_c}_{1}^{\frac {1}{3}}}\right ) {\mathrm e}^{-t \mathit {\_c}_{1}}$

Hand solution

Solve for $$u_{t}+u_{xxx}=0$$ on the real line for $$t>0$$. Let $$u=T\left ( t\right ) X\left ( t\right )$$. The pde becomes\begin {align*} T^{\prime }X+X^{\prime \prime \prime }T & =0\\ \frac {T^{\prime }}{T} & =-\frac {X^{\prime \prime \prime }}{X}=-\lambda \end {align*}

Hence $$X^{\prime \prime \prime }+\lambda X=0$$. This ODE has solution $X\left ( x\right ) =C_{1}e^{\left ( -\frac {\lambda ^{\frac {1}{3}}}{2}-\frac {i}{2}\lambda ^{\frac {1}{3}}\sqrt {3}\right ) x}+C_{2}e^{\left ( -\frac {\lambda ^{\frac {1}{3}}}{2}+\frac {i}{2}\lambda ^{\frac {1}{3}}\sqrt {3}\right ) x}+C_{3}e^{\lambda ^{\frac {1}{3}x}}$ The ODE $$T^{\prime }+\lambda T=0$$ has the solution $$T\left ( t\right ) =C_{4}e^{-\lambda t}$$. Therefore the solution to the PDE is $$T\left ( t\right ) X\left ( t\right )$$ given by$u\left ( x,t\right ) =C_{4}e^{-\lambda t}\left ( C_{1}e^{\left ( -\frac {\lambda ^{\frac {1}{3}}}{2}-\frac {i}{2}\lambda ^{\frac {1}{3}}\sqrt {3}\right ) x}+C_{2}e^{\left ( -\frac {\lambda ^{\frac {1}{3}}}{2}+\frac {i}{2}\lambda ^{\frac {1}{3}}\sqrt {3}\right ) x}+C_{3}e^{\lambda ^{\frac {1}{3}x}}\right )$