Thread: PDE

**PvtBillPilgrim** · February 4th 2009, 02:56 PM

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.

**Danny** · February 4th 2009, 03:02 PM

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?

**PvtBillPilgrim** · February 4th 2009, 03:54 PM

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

**ThePerfectHacker** · February 4th 2009, 05:00 PM

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 $u(x,t) = \sum_{k=0}^n a_k e^{-4(\pi k t/2)^2} \sin \tfrac{\pi k x}{2}$ .

**Danny** · February 4th 2009, 07:09 PM

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.

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