# Second derivative test failed

I have $$H(\psi_1(t),\psi_2(t),x_1(t),x_2(t),u(t))= -1 + \psi_1(3x_1+x_2) + \psi_2(4x_1+3x_2 +u)$$

Note: $\psi_1 = -2Ae^t + 2Be^{5t}$, $\psi_2 = Ae^t + Be^{5t}$

Hence we have

$$\frac{\partial H}{\partial u}=\psi_2$$

$$\frac{\partial^2 H}{\partial u^2}= 0$$

What... I always get some $\frac{\partial^2 H}{\partial u^2}= f(u,\psi_i)$

The second derivative test failed,now what??

• It seems to me that $\dfrac{\partial H}{\partial u}=\psi_2'$, not $\psi_2$. Then $\dfrac{\partial^2 H}{\partial u^2}=\psi_2''=Ae^{4x_1+3x_2+u}+25Be^{5(4x_1+3x_2+u)}$. What are you trying to show? – robjohn Oct 28 '14 at 14:33
• @robjohn $\psi_2' = -\frac{\partial H}{\partial x_2}$. Where did $\frac{\partial H}{\partial u} = \psi_2'$ come from? – Partly Putrid Pile of Pus Oct 29 '14 at 6:38
• Your notation is very confusing. You write $\psi_1(t)=-2Ae^t+2Be^{5t}$ and $\psi_2(t)=Ae^t+Be^{5t}$, yet $H(\psi_1,\psi_2,x_1,x_2,u)=-1+\color{#C00000}{\psi_1}(3x_1+x_2) +\color{#C00000}{\psi_2}(4x_1+3x_2+u)$. Perhaps you meant the red $\psi_k$'s not to be functional applications. Would it be accurate to say instead $H(\psi_1,\psi_2,x_1,x_2,u)=-1+(3x_1+x_2)\psi_1+(4x_1+3x_2+u)\psi_2$? – robjohn Oct 29 '14 at 7:24
• @robjohn Oh my god wow, I am sorry, I didn't even think of it that way. The problem came in that I derived $\psi_1$ & $\psi_2$ after trying to find that $\frac{\partial H}{\partial u}$ – Partly Putrid Pile of Pus Oct 29 '14 at 11:23
• @robjohn Note a friend is working on this with me, here: meta.math.stackexchange.com/questions/4666/… . We are solving a question by another user, and it is similar content to in class exam soon – Partly Putrid Pile of Pus Oct 29 '14 at 11:24

Base on your calculation, I guess $\psi_1$, $\psi_2$ are constants. In that case, $H$ is linear (or rather affine). Hence, $H$ doesn't have any critical points. That said, you shouldn't be applying second derivative test here.
• All of $x_1,x_2,u$ are functions of time, and I want to maximise $H$ as a function of $u$ – Partly Putrid Pile of Pus Oct 28 '14 at 7:54
• and $\psi_1 = -2Ae^t + 2Be^{5t}$ and $\psi_2 = Ae^t + Be^{5t}$ – Partly Putrid Pile of Pus Oct 28 '14 at 7:58
• Still, $H$ is linear in $u$, no critical points. It doesn't work unless you are maximizing $H$ as a function of $t$ and $u$ (which you should, as you are using partial derivatives here). – Quang Hoang Oct 28 '14 at 8:02