2.14.1 \(u_t + u_{xxx} = 0\)

problem number 109

Added May 30, 2019.

Airy PDE

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

\[ u_t+u_{xxx} =0 \]

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] + D[u[x, t], {x,3}] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, t], {x, t}], 60*10]];
 

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

Maple ✓

restart; 
pde := diff(u(x,t),t)+diff(u(x,t),x$3)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,t),'build')),output='realtime'));
 

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

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

\[ u\left ( x,t\right ) =C_{4}e^{-\lambda t}\left ( C_{1}e^{\left ( -\frac {\lambda ^{\frac {1}{3}}}{2}-\frac {i}{2}\lambda ^{\frac {1}{3}}\sqrt {3}\right ) x}+C_{2}e^{\left ( -\frac {\lambda ^{\frac {1}{3}}}{2}+\frac {i}{2}\lambda ^{\frac {1}{3}}\sqrt {3}\right ) x}+C_{3}e^{\lambda ^{\frac {1}{3}x}}\right ) \]