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

**tykim** · August 6th 2012, 04:03 AM

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)

**Prove It** · August 6th 2012, 04:06 AM

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 $\displaystyle \begin{align*} u(t, x) \end{align*}$ ?

**tykim** · August 6th 2012, 04:10 AM

Dear Prove IT

Correct.

Function u and its derivatives as u(x, t)

**Prove It** · August 6th 2012, 04:36 AM

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

This might help you further...

**tykim** · August 6th 2012, 06:26 AM

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

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