# Separate and Solve $u_x+2u_{tx}-10u_{tt}=0$

I came across a question requesting that the PDE below along with its boundary condition be separated:

$$u_x+2u_{tx}-10u_{tt}=0$$ $$u(0,t)=0, u_x(L,t)=0$$

Which I was able to do using the method described below:

$$u(x,t)=X(x)T(t)$$

Thus: $$u_x=X'T$$ $$u_{tx}=X'T'$$ $$u_{tt}=XT''$$ Substituting these into the original PDE and simplifying gave me: $$\frac{X'}{X}=\frac{10T''}{T+2T'}=-\lambda$$ This thus permitted me to split the PDE into 2 ODEs; $$X'+\lambda X=0$$ $$10T''+2\lambda T'+\lambda T=0$$

I was also able to split the boundary conditions: $$u(0,t)=X(0)T(t)=0$$ $$u_x(L,t)=X'(L)T(t)=0$$

This gives me: $$X(0)=0$$ $$X'(L)=0$$ Since $T(t)=0$ would be trivial (thus not desirable)

Now my problem is how to solve the BVP below without getting a trivial equation: $$X'+\lambda X=0$$ $$X(0)=0$$ $$X'(L)=0$$

• The last BVP has the only trivial solution (since the general solution to $X'+\lambda X=0$ is $X(x) = C e^{-\lambda x}$, $C$ is a constant). Maybe there is a misprint in the initial problem? – Voliar Oct 31 '16 at 13:39
• You might have swapped $x$ and $t$ in the original problem. If $X$ obeys the second order ODE you found for $T$, then you obtain a nontrivial solution to your boundary value problem. – Frits Veerman Oct 31 '16 at 13:58
• Could it possibly have been $u_{xx}$ rather than $u_x$ in the PDE? This would make the separated ODE as second-order and solvable. – Semiclassical Oct 31 '16 at 13:59
• @Semiclassical separation of variables for the case $u_{xx}$ instead of $u_x$ is less trivial and definitely wont produce the same ODEs. – Voliar Oct 31 '16 at 15:53
• @Voliar et al (FritsVeerman & Semiclassical) you guys just confirmed my suspicion. Thanks – Obinoscopy Oct 31 '16 at 19:27