Solution for PDE

∂u/∂t=k ∂/∂x (∂u/∂x)

I.C.→ u(x,0)=12 cos(9πx⁄L)-7 sin(4πx⁄L)

B.C.→ u(-L,t)=u(L,t) & ∂u/∂x (-L,t)=∂u/∂x(L,t)

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- August 6th 2012, 05:03 AMtykimSolution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)
Solution for PDE

∂u/∂t=k ∂/∂x (∂u/∂x)

I.C.→ u(x,0)=12 cos(9πx⁄L)-7 sin(4πx⁄L)

B.C.→ u(-L,t)=u(L,t) & ∂u/∂x (-L,t)=∂u/∂x(L,t)

- August 6th 2012, 05:06 AMProve ItRe: Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)
- August 6th 2012, 05:10 AMtykimRe: Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)
Dear Prove IT

Correct.

Function u and its derivatives as u(x, t) - August 6th 2012, 05:36 AMProve ItRe: Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)
I expect you can solve this with separation of variables.

Assume you can write , then we have and .

Substituting into the DE gives

Now notice that the LHS is only a function of t, and the RHS is only a function of x, so for them to be equal, they must be equal to the same constant value .

We also have

Now see what you can do with the Boundary and Initial conditions to evaluate .

This might help you further... - August 6th 2012, 07:26 AMtykimRe: Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)
Dear Prove IT,

Thanks for your kind guidance.

There may be type error, and I corrected as below. Sorry for that.

I.C.→ u(x,0)=12**cos**(9πx⁄L)-7 sin(4πx⁄L)

Could you let me know how to get**root-square of (lambda)**

Thanks