Will separation of variables work in solving this PDE?

[AxiomOfChoice]

Homework Statement

As part of the solution to a HW problem of mine, I have to solve the PDE

[tex]
p_t = -vk^2 p - k \delta p_k,
[/tex]

where [itex]p = p(k,t)[/itex] and [itex]v,\delta[/itex] are known constants.

Homework Equations

I tried to look for a solution of the form [itex]p(k,t) = K(k)T(t)[/itex] and found one, but I'm not sure if I need to sum/integrate over my arbitrary constant.

The Attempt at a Solution

Separation of variables gave me the solution

[tex]
p(k,t) = A_0 e^{-ct} k^{c/\delta} e^{-vk^2/2\delta},
[/tex]

where [itex]c[/itex] is the constant one gets from the separation of variables and [itex]A_0[/itex] is a constant of integration. But I tried solving the PDE in Mathematica, and got a different solution:

[tex]
p(k,t) = A_0(t-(\log k)/\delta) e^{-vk^2/2\delta}.
[/tex]

Can someone explain why Mathematica's answer differs from mine? Also, do I need to perform an integration/sum in [itex]c[/itex] to get the most general solution? I've plugged both my solution and Mathematica's in, and they both work, so I'm thinking I have to go another step to get the general solution.

 

[SammyS]

...

I tried to look for a solution of the form [itex]p(x,t) = X(x)T(t)[/itex] and found one, but I'm not sure if I need to sum/integrate over my arbitrary constant.

The Attempt at a Solution

Separation of variables gave me the solution

[tex]
p(k,t) = A_0 e^{-ct} k^{c/\delta} e^{-vk^2/2\delta},
[/tex]
...

Can someone explain why Mathematica's answer differs from mine? Also, do I need to perform an integration/sum in [itex]c[/itex] to get the most general solution? I've plugged both my solution and Mathematica's in, and they both work, so I'm thinking I have to go another step to get the general solution.

Show us some details about how you arrived at your solution.

One thing I noticed:

[tex]
\log\left(e^{-ct} k^{c/\delta}\right)=-c\left(t-\frac{\log(k)}{\delta}\right)
[/tex]

 

[AxiomOfChoice]

I'll be happy to. I have a function to go to right now, but I'll be back in a few hours and will type up how I arrived at my separation of variables solution.

 

[AxiomOfChoice]

Ok. So, on to my separation of variables calculations. Note that I have also corrected a typo in my original post, replacing some [itex]x[/itex]'s with [itex]k[/itex]'s. Writing [itex]p(k,t) = K(k)T(t)[/itex] and plugging into the PDE, we find (after dividing through on both sides by [itex]KT[/itex])

[tex]
\frac{1}{T} \frac{\partial T}{\partial t} = -v^2 k^2 - k \delta \frac{1}{K} \frac{\partial K}{\partial k}.
[/tex]

The LHS is a function of t only; the RHS a function of k only. So we can (I think) set them both equal to a constant, which I call [itex]-c[/itex] (anticipating that it will be negative). The equation for T becomes

[tex]
\frac{d T}{d t} = -c T.
[/tex]

This has the obvious solution [itex]T(t) = C_0 e^{-ct}[/itex]. The equation for K, however, becomes (after canceling the minus sign)

[tex]
v^2 k^2 + k \delta \frac{1}{K} \frac{d K}{d k} = c.
[/tex]

This can be rearranged and rewritten as

[tex]
\frac{dK}{K} = \left( \frac{c}{k \delta} - v^2 k \right) dk.
[/tex]

Integrating on both sides, we obtain (after combining the two constants we get into one, C^0_1, on the RHS)

[tex]
\log K = \frac{c}{\delta} \log k - \frac{v^2 k^2}{2\delta} + C^0_1.
[/tex]

Hence, after exponentiating both sides, we obtain

[tex]
K(k) = C_1 k^{c/\delta} \exp \left( -\frac{v^2 k^2}{2\delta} \right).
[/tex]

Now we put them together:

[tex]
p(k,t) = K(k)T(t) = A_0 e^{-ct} k^{c/\delta} \exp \left( -\frac{v^2 k^2}{2\delta} \right),
[/tex]

which is what I have listed above. Does anyone see anything wrong with this?

 

[SammyS]

It looks good to me.