Solution to heat equation

**charikaar** · December 10th 2009, 02:26 AM

$u_t = 2u_{xx} \quad$

$T'(t) = -2 \lambda T(t) \quad$
and
$X''(x) = - \lambda X(x). \quad$

Does it make any difference if choose constant to be $\lambda$ and not $-\lambda$ as above

Thanks

**TheEmptySet** · December 10th 2009, 06:05 AM

Originally Posted by charikaar

$u_t = 2u_{xx} \quad$

$T'(t) = -2 \lambda T(t) \quad$
and
$X''(x) = - \lambda X(x). \quad$

Does it make any difference if choose constant to be $\lambda$ and not $-\lambda$ as above

Thanks

It can be shown that the seperation constant must be negative or you will only get the trivial solution.

Here is the proof for homogenoious Dirichlet or Neumann boundry contdtions

$u_t=au_{xx}$ after seperating we get

$\frac{T'}{aT}=\frac{X''}{X}=\beta$

So we get

$X''-\beta X=0$

Now multiply this equation by $X$ and integrate the equation.

$XX''-\beta X^2=0 \iff \int_{0}^{L}XX''dx -\int_{0}^{L}\beta X^2dx=0$

Now integrate the first term by parts with $u=X \implies du=X'$ and $dv=X'' \implies v=X'$ to get

$XX'\bigg|_{0}^{L}-\int_{0}^{L}(X')^2dx-\int_{0}^{L}X^2dx =0$

Now if we have the homogenious condtions mentioned above either

$X(0)=X'(L)=0 \text{ or } X'(0)=X'(L)=0$ in either case the first term above is equal to zero when evaluated at the end points. So we get

$-\int_{0}^{L} [(X')^2+\beta (X)^2]dx=0$

Since both $(X')^2 \text{ and } (X)^2$ are always non negative the only when this integral can be zero is if $\beta < 0$

So the seperation constant must be negative in the heat equation.

This can be shown for the mixed and other boundry conditions but it is not as clean. I hope this helps.

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