28 Beam PDE

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28.1 Beam PDE with zero initial velocity

problem number 175

Added January 20, 2018.

Solve \[ \frac {\partial ^2 u}{\partial t^2} + \frac {\partial ^4 u}{\partial x^4} =0 \]

With boundary conditions

\begin {align*} u(0,t) &= -12 t^2\\ f(1,t) &=1-12 t^2\\ \frac {\partial ^2 u}{\partial x^2}u(0,t) &=0 \\ \frac {\partial ^2 u}{\partial x^2}u(1,t) &=12 \end {align*}

And initial conditions

\begin {align*} u(x,0) &= x^4\\ \frac {\partial u}{\partial t}u(x,0) &=0 \end {align*}

Mathematica ✓

ClearAll[u, x, t]; 
 pde = D[u[x, t], {t, 2}] + D[u[x, t], {x, 4}] == 0; 
 bc = {u[0, t] == -12*t^2, u[1, t] == 1 - 12*t^2, Derivative[2, 0][u][0, t] == 0, Derivative[2, 0][u][1, t] == 12}; 
 ic = {u[x, 0] == x^4, Derivative[0, 1][u][x, 0] == 0}; 
 sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], x, t], 60*10]];
 

\[ \left \{\left \{u(x,t)\to x^4-12 t^2\right \}\right \} \]

Maple ✓

 
x:='x'; t:='t'; L:='L'; c:='c';u:='u'; 
interface(showassumed=0); 
pde:=diff(u(x,t),t$2)+diff(u(x,t),x$4)=0; 
bc:=u(0,t)=-12*t^2, 
    u(1,t)=1-12*t^2,D[1,1](u)(0,t)=0, 
    D[1,1](u)(1,t)=12; 
ic:=u(x,0)=x^4,D[2](u)(x,0)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t),HINT=`+`)),output='realtime'));
 

\[ u \left ( x,t \right ) ={x}^{4}-12\,{t}^{2} \]