# Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)

• Aug 6th 2012, 05:03 AM
tykim
Solution 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)

• Aug 6th 2012, 05:06 AM
Prove It
Re: Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)
Quote:

Originally Posted by tykim
Solution for PDE

u/t=k ∂/∂x (∂u/∂x)

I.C.
→ u(x,0)=12 sin(9πx⁄L)-7 sin(4πx⁄L)
B.C.→
u(-L,t)=u(L,t) & ∂u/∂x (-L,t)=∂u/∂x(L,t)

Just so we're clear, is this your DE? \displaystyle \begin{align*} \frac{\partial u}{\partial t} = k\,\frac{\partial ^2 u}{\partial x^2} \end{align*}

Also in your boundary conditions, are you writing the function \displaystyle \begin{align*} u \end{align*} and its derivatives as \displaystyle \begin{align*} u(t, x) \end{align*}?
• Aug 6th 2012, 05:10 AM
tykim
Re: Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)
Dear Prove IT

Correct.

Function u and its derivatives as u(x, t)
• Aug 6th 2012, 05:36 AM
Prove It
Re: Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)
I expect you can solve this with separation of variables.

Assume you can write \displaystyle \begin{align*} u = T(t)\,X(x) \end{align*}, then we have \displaystyle \begin{align*} \frac{\partial u}{\partial t} &= \frac{dT}{dt}\,X \end{align*} and \displaystyle \begin{align*} \frac{\partial^2 u}{\partial x^2} = T\,\frac{d^2X}{dx^2} \end{align*}.

Substituting into the DE gives

\displaystyle \begin{align*} \frac{\partial u}{\partial t} &= k\,\frac{\partial ^2 u}{\partial x^2} \\ \frac{dT}{dt}\,X &= k\,T\,\frac{d^2X}{dx^2} \\ \frac{1}{k\,T}\,\frac{dT}{dt} &= \frac{1}{X}\,\frac{d^2X}{dx^2}\end{align*}

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 \displaystyle \begin{align*} -\lambda \end{align*}.

\displaystyle \begin{align*} \frac{1}{k\,T}\,\frac{dT}{dt} &= -\lambda \\ \int{\frac{1}{k\,T}\,\frac{dT}{dt}\,dt} &= \int{-\lambda\,dt} \\ \frac{1}{k}\int{\frac{1}{T}\,dT} &= -\lambda \,t + C_1 \\ \frac{1}{k}\ln{|T|} + C_2 &= -\lambda\,t + C_1 \\ \frac{1}{k}\ln{|T|} &= -\lambda \, t + C_1 - C_2 \\ \ln{|T|} &= -k\,\lambda\,t + k\left(C_1 - C_2\right) \\ |T| &= e^{-k\,\lambda\,t + k\left(C_1 - C_2\right)} \\ |T| &= e^{k\left(C_1 - C_2\right)}e^{-k\,\lambda\,t} \\ T &= A\,e^{-k\,\lambda \, t} \textrm{ where } A = \pm e^{k\left(C_1 - C_2\right)} \end{align*}

We also have

\displaystyle \begin{align*} \frac{1}{X}\,\frac{d^2X}{dx^2} &= -\lambda \\ \frac{d^2X}{dx^2} &= -\lambda\,X \\ \frac{d^2X}{dx^2} + \lambda\,X &= 0 \\ \\ m^2 + \lambda &= 0 \\ m^2 &= -\lambda \\ m &= \pm i\sqrt{\lambda} \\ \\ X &= B\sin{\left(\sqrt{\lambda}\,x\right)} + C\cos{\left(\sqrt{\lambda}\,x\right)} \end{align*}

Now see what you can do with the Boundary and Initial conditions to evaluate \displaystyle \begin{align*} \lambda \end{align*}.

• Aug 6th 2012, 07:26 AM
tykim
Re: Solution for PDE, ∂u/∂t=k ∂/∂x (∂u/∂x)
Dear Prove IT,

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