PDE for Heat Diffusion Equation

Smythe

The one-dimensional heat diffusion equation is given by :

∂t(x,t)/∂t = α[∂^2T(x,t) / ∂x^2]

where α is positive.

Is the following a possible solution?

Assume that the constants a and b can take any positive value.

T(x,t) = exp(at)cos(bx)


SOLUTION:
T(x,t) = exp(at)cos(bx)

∂T/∂t = a exp(at) cos(bx)
∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

As a, b and α are all positive this cannot be a solution.

A friend and I were working on this and got the answer above. Though as it was about a week ago I can't exactly remember what we've done here.

In the first step I think we've differentiated exp(at) with respect to t, treating the cos(bx) as a constant multiplier, which become a exp(at) cos(bx).

Is that all correct above?

Then we differentiated again with respect to t.

So wouldn't that become;

∂^2 T/∂x^2 = a^2 exp(at) cos(bx)

Why did we previously get ∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

Thanks!

Michael Collins

Looks like you have the right answer anyway.

The left hand side is

[latex] \displaystyle \frac{\partial T(x,y)}{\partial t} = a \exp(\hbox{a}t) \cos(\hbox{b}x).[/latex]

The right hand side is

[latex] \displaystyle \alpha \frac{\partial^2 T(x,y)}{\partial x^2} = \alpha \frac{\partial}{\partial x } \Big(-\hbox{b} \exp(\hbox{a}t) \sin(\hbox{b}x)\Big) = (\alpha)(-\hbox{b}^2) \exp(\hbox{a}t) \cos(\hbox{b}x), [/latex]

i.e. to get the last line in your post you differentiate with respect to x, twice. Just as you are 'instructed' to do according to the right hand side of your PDE.

Clearly, as you said, since all constants are constrained to be positive, there can be no equality here between the left and right hand side using the [latex] T(x,y) [/latex] function you are given.

Smythe

Cheers for that! I see what I was doing wrong - getting x and t confused...

Smythe

Another possible solution we investigated for the same heat diffusion equation was;

T(x,t) = exp(-at + ibx)

The PDEs being;

∂T/∂t = -a exp(-at + ibx)
∂^2 T / ∂x^2 = - b^2 exp(-at + ibx)

So the possible solution is possible if a = b^2 α

I understand all the above now.

Though apparently T(x, t) is complex in this instance. How is this apparent?

Michael Collins

T(x,t) contains the imaginary unit, i.

This is how the second derivative of T(x,t) is negative - you end up multiplying i by itself, which by definition is -1.

Smythe

Michael Collins wrote: »

T(x,t) contains the imaginary unit, i.

This is how the second derivative of T(x,t) is negative - you end up multiplying i by itself, which by definition is -1.

Thanks again for the help!