# Math Help - PDE

1. ## PDE

I'm in an elementary PDE's class and I'm trying to do problems in my textbook. I'm having trouble starting. I was wondering if someone could show me how to do this example, so I can model similar solutions after it.

Find the solution and give a physical interpretation of the problem uxx = 4ut (i.e. second partial derivative of u with respect to x = 4 times the first partial derivative of u with respect to t)
for 0 < x < 2, t > 0,
with the conditions
u(0,t) = 0
u(2,t) = 0
u(x,0) = 2sin(pi*x/2) - sin(pi*x) + 4sin(2pi*x)

Thanks.

2. Originally Posted by PvtBillPilgrim
I'm in an elementary PDE's class and I'm trying to do problems in my textbook. I'm having trouble starting. I was wondering if someone could show me how to do this example, so I can model similar solutions after it.

Find the solution and give a physical interpretation of the problem uxx = 4ut (i.e. second partial derivative of u with respect to x = 4 times the first partial derivative of u with respect to t)
for 0 < x < 2, t > 0,
with the conditions
u(0,t) = 0
u(2,t) = 0
u(x,0) = 2sin(pi*x/2) - sin(pi*x) + 4sin(2pi*x)

Thanks.
Do you know the technique of separation of variables?

3. I remember it from ODE's, but I'm not sure how to apply it here.

4. Originally Posted by PvtBillPilgrim
I'm in an elementary PDE's class and I'm trying to do problems in my textbook. I'm having trouble starting. I was wondering if someone could show me how to do this example, so I can model similar solutions after it.

Find the solution and give a physical interpretation of the problem uxx = 4ut (i.e. second partial derivative of u with respect to x = 4 times the first partial derivative of u with respect to t)
for 0 < x < 2, t > 0,
with the conditions
u(0,t) = 0
u(2,t) = 0
u(x,0) = 2sin(pi*x/2) - sin(pi*x) + 4sin(2pi*x)

Thanks.
The solution to $u_{xx} = 4u_t$ on $[0,2]\times [0,\infty)$ with initial conditions $u(x,0) = \sum_{k=0}^n a_k \sin \tfrac{\pi k x}{2}$ and boundary conditions $u(0,t)=u(2,t)=0$ is given by $u(x,t) = \sum_{k=0}^n a_k e^{-4(\pi k t/2)^2} \sin \tfrac{\pi k x}{2}$.

5. Originally Posted by ThePerfectHacker
The solution to $u_{xx} = 4u_t$ on $[0,2]\times [0,\infty)$ with initial conditions $u(x,0) = \sum_{k=0}^n a_k \sin \tfrac{\pi k x}{2}$ and boundary conditions $u(0,t)=u(2,t)=0$ is given by $u(x,t) = \sum_{k=0}^n a_k e^{-4(\pi k t/2)^2} \sin \tfrac{\pi k x}{2}$.
Not for all n - right, just those to recover the initial condition.