# PDE Solving in mathematica

I'd like to ask how I might go about solving this equation:

# $\frac{\partial Ci}{\partial t} + \frac{1}{r^2}\frac{\partial (r^2 Ci)}{\partial r} = D\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial Ci}{\partial r})$

as suggested a simplification

# $\frac{\partial Ci}{\partial t} + \frac{1}{r}2Ci +\frac{1}{r^2}\frac{\partial Ci}{\partial r} = D(\frac{2}{r}\frac{\partial Ci}{\partial r} +\frac{\partial^2 Ci}{\partial r^2})$

$r = 0, t = 0 \rightarrow Ci = 0$
$r = 1, t = 0 \rightarrow Ci = 0$

I am fairly new to Mathematica and I don't really know how to go about this; I've also tried the steps outlined in this post here but while I understood what to do I get recursion limit reached errors (using the code provided).

Could you please help me out on how to solve this equation?

a partial and first attempt on solving this taking in account the notes of the posted link is the following:

Clear[y];
f1[r_ /; r > 0] := 1;
f1[r_ /; r == 0] := 2;
f2[r_ /; r > 0] := 1/r^2;
f2[r_ /; r == 0] := 0;
f3[r_ /; r > 0] := 1/r;
f3[r_ /; r == 0] := 0;

eqns = {
Derivative[0, 0, 1][y][t, r, z] +
f3[r]*2*y[t, r, z] +
f2*Derivative[0, 1, 0][y][t, r, z] -
2*f3[r]*Derivative[0, 1, 0][y][t, r, z] -
Derivative[0, 2, 0][y][t, r, z] == Derivative[1, 0, 0][y][t, r, z],
y[t, 0, z] == 0,
y[t, r, 0] == 0,
Derivative[0, 1, 0][y][t, 0, z] == 0,
Derivative[0, 0, 1][y][t, r, 0] == 0,
y[0, r, z] == (1 - r^2)*(1 - z^2)}

$RecursionLimit = 1536 y[t_, r_, z_] = y[t, r, z] /. First[NDSolve[eqns, y[t, r, z], {t, 0, 1}, {r, 0, 1}, {z, 0, 1}, Method -> {"MethodOfLines", Method -> "StiffnessSwitching", "DifferentiateBoundaryConditions" -> {True, "ScaleFactor" -> 1}}]];  This gives me a lot of errors but the first one is the following: - You do not provide any code. – b.gatessucks Oct 18 '13 at 21:16 I said that I tried the code (tailored for my equation) that was supplied in the post I linked, do you need a more? – jtimz Oct 18 '13 at 21:23 Yes, at least the equation. – Sektor Oct 18 '13 at 21:34 Well, I have a problem with that I don't know how to type in the second partial; that is I don' know how to write the$\frac{\partial r^2Ci}{\partial r}$. This has both r and Ci inside which I don't how to type them. – jtimz Oct 18 '13 at 22:05 You mean D[ r^2 Ci[r,t], r] ? (I suppose Ci is a function of r and t). By the way, Ci and D are not the best names for variables in Mathematica. I'd switch to lowercase if I were you. At least for D. – Peltio Oct 18 '13 at 22:47 show 2 more comments ## 1 Answer Here is a solution using the method of characteristics. But the conditions as given above will cause a problem at$r=0$. There is no Cauchy data to use. So I kept the solution in terms of the constants of integrations. Let$u\left( t,r\right) be the solution \begin{align*} \frac{\partial u}{\partial t}+\frac{1}{r^{2}}\frac{\partial\left( r^{2}u\right) }{\partial r} & =D\frac{1}{r^{2}}\frac{\partial\left( r^{2}u\right) }{\partial r}\\ \frac{\partial u}{\partial t}+\frac{1}{r^{2}}\left( 2ru+r^{2}\frac{\partial u}{\partial r}\right) & =D\frac{1}{r^{2}}\left( 2ru+r^{2}\frac{\partial u}{\partial r}\right) \\ \frac{\partial u}{\partial t}+\left( 2\frac{u}{r}+\frac{\partial u}{\partial r}\right) & =D\left( 2\frac{u}{r}+\frac{\partial u}{\partial r}\right) \\ \frac{\partial u}{\partial t}+\frac{\partial u}{\partial r}\left( 1-D\right) & =-2\frac{u}{r}\left( 1-D\right) \end{align*} Let1-D=k, hence \begin{align*} \frac{\partial u}{\partial t}+k\frac{\partial u}{\partial r} & =-2k\frac {u}{r}\\ \frac{1}{k}\frac{\partial u}{\partial t}+\frac{\partial u}{\partial r} & =-2\frac{u}{r} \end{align*} To find the solution using the characteristics method, we write the above in the standard form $$\frac{dt}{k^{-1}}=\frac{dr}{1}=\frac{du}{-2\frac{u}{r}}%$$ Hencedt=\frac{dr}{k}$or$t=\frac{r}{k}+r_{0}$or $$r_{0}=t-\frac{r}{k}$$ Also,$\frac{du}{-2\frac{u}{r}}=dt$or$\frac{du}{u}=-\frac{2}{r}dt$, hence$\ln u=-\frac{2}{r}t+c_{2}$, or$u=Ae^{-\frac{2t}{r}}$. At$t=0$, to satisfy initial conditions at$r_{0}$, then$u\left( 0,r_{0}\right) =f\left( r_{0}\right) =A$. So the solution is $$u\left( t,r\right) =A\left( t-\frac{r}{k}\right) e^{-\frac{2kt}{r}}%$$ Mathematica gives this btw:  ClearAll[u, r, t, k]; ode = D[u[r, t], t] + k D[u[r, t], r] + 2 k u[r, t]/r sol = u[r, t] /. First@DSolve[ode == 0, u[r, t], {r, t}]  Again, I think the initial conditions given are not correct. for more information. - Thanks for the reply @Nasser; if that is the case then you can assume since we cannot solve for all$D$that$D$is$\frac{1}{2}$for that$k=\frac{1}{2}$. That way I get an answer but I don't think it is a correct one. Also the limits can be$0 \rightarrow 1$and if I do (sol /. {r -> 1, t -> 1}) == 0 I get an answer but that is a differential equation as well. – jtimz Oct 19 '13 at 1:16 This is a first order linear PDE, but it is inhomogeneous PDE. But you can still solve it analytically using separation of variables. You need to use eigenfunction expansion method. – Nasser Oct 19 '13 at 1:23 can you give me an example? The post I read had a similar equation for solving but I can't figure how to adapt my code to that code. – jtimz Oct 19 '13 at 1:32 @jtimz You can actually use the method of characteristics also to solve it? start by writing$\frac{dt}{1}=\frac{dr}{k}=\frac{du}{-2k\frac{u}{r}}$and then$\frac{dr}{dt}=k$or$r=tk+c_{1}$or$r-tk=c_{1}$and$\frac{du}{dt}=-2k\frac{u}{r}$or$\frac{du}{u}=-2\frac{k}{r}dt$, hence$\ln u=-2\frac{k}{r}t+c_{2}$, hence$u=c_{2}e^{-2\frac{k}{r}t}$So the the solution is function$f\left( r-tk,ue^{2\frac{k}{r}t}\right)\$. I just forgot more of the details now. Need to look it more...separation of variables will also work.... –  Nasser Oct 19 '13 at 2:13
thanks will look it a bit more and will let you know! Thanks a ton! –  jtimz Oct 19 '13 at 2:27