Solving the Diff.Eq. to Find v(x,p) and u(0,t)

[Gekko]

u(x,t) defined for x>0 t>0
d2u/dx2=du/dt

Conditions:
u(x,0)=0
d/dx u(0,t)=-1
u(x,t) tends to 0 as x tends to infinity
d2v/dx2 =pv
d/dx v(0,p) =-1/p

v(x,p)=integral(0 to inf) exp(-pt) u(x,t) dt

How do we use this to find v(x,p) and u(0,t)?

For u(0,t) can we simply integrate wrt x d/dx u(0,t) = -x?


Thanks in advance

 

[jegues]

I could be wrong, but isn't there something wrong with the conditions and the way you defined,

$$u(x,t)$$

?

You said,

$$u(x,t)$$ $$\exists \forall x > 0, t>0$$

but in the conditions you are somehow evaluating at,

$$t = 0$$

$$u(x,0) = 0$$

 

[Gekko]

You are right. I saw this too but assumed something specific to the heat equation allows it. Or maybe we assume tending to? In any case this is how the question was stated