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I've been asked to solve the diffusion equation with variable dissipation, I have given the start of my answer but can't seem to proceed. Would appreciate a full solution as the work is due soon, thank you for any help though:

$$ \frac{∂u}{∂t} - D \frac{∂^2u}{∂x^2} + e^{-pt}u = 0 , x∈(-∞,∞), t∈[0,∞] $$

subject to

$u(x,0) = φ(x),$

where $D>0$ and $p>0$ are given constants, and φ(x) is a given function.

I've managed to come up with this so far: Let $u(x,t) = h(t)v(x,t)$ then

$$ h_tv + hv_t - Dhv_{xx} + e^{-pt}hv = 0 $$

$$ v_t - Dv_{xx} + \left(\frac{h_t}{h}+e^{-pt}\right)v = 0 $$

$v$ will satisfy the heat equation if the third term on the LHS is zero, i.e.

$$ \frac{h_t}{h} + e^{-pt} = 0 $$

$$ v_t - Dv_{xx} = 0 $$

with initial condition $$ v(x,0) = \frac{u(x,0)}{h(0)} = \frac{\varphi(x)}{h(0)} $$

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    $\begingroup$ I already told you to solve for $h$ first. Have you done that? What are you still having trouble with? $\endgroup$ – Dylan Mar 4 at 4:21
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@DPH24 You can just choose your $h$ to be a solution of the first differential equation (the one for $h$ above). You can choose, for example, $$ h(t)=e^{e^{-pt}/p} $$ and then you solve for $v$ a standard IVP for the heat equation with $v(x,0)=e^{1/p}\phi(x)$. You get $$ v(x,t)=\frac{e^{1/p}}{2\sqrt{\pi Dt}}\int\limits_{-\infty}^{+\infty}\phi(y)e^{-\frac{(x-y)^2}{4Dt}}\,dy $$ And then $u=hv$.

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  • $\begingroup$ This is what I thought but I wasn't sure on what to choose for φ(y) $\endgroup$ – DPH24 Mar 4 at 8:22

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