# Solving Integro-Differential equations

I need help solving this equation. Is there a built in function that solves this type of equations? DSolve wouldn't work.

Updated equation:

DSolve[D[p[u, t], t]*(1/M + β*(b^2/Kp)) - (b/(3*Kp))*
D[pa[t], t] + ((2*b^2)/Kp)*(1 - β)*
Integrate[D[p[u, t], t]*u, {u, 0, 1}] == (1/u)*D[k[u]*u*D[p[u, t], u], u],
p[u, 0] = po, p[1, t] = pa[t], p, {u, t}]


The error I get is:

Input is not an ordinary differential equation.

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• Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need.
– user9660
Mar 2, 2015 at 6:26
• @Raven: your equation for p[u,t] is linear (I guess pa'(t) means D[p[u,t],u]/.u->a) and can therefore be solved by standard mathematical methods once you provide the boundary and initial conditions. Physically it decribes Diffusion in a cylinder. A necessary condition can be obtained by differentiating the equation with respect to u. The resulting equation for q = D[p,u] is solved by MMA exactly in terms of Bessel functions. Mar 2, 2015 at 9:15

Outline

As you didn't provide boundary and initial conditions and the function pa'[t] this solution must be generic.

Your equation for p[u,t] is linear (I guess pa'(t) means D[p[u,t],u]/.u->a) and can therefore be solved by standard mathematical methods once you provide the boundary and initial conditions. Physically it describes diffusion in a cylinder. A necessary condition can be obtained by differentiating the equation with respect to u. The resulting equation for q = D[p,u] is solved by Mathematica exactly in terms of Bessel functions.

Derivation

The integral differential equation in Mathematica terms is (pap is pa'(t))

D[p[u, t], t] + Integrate[u D[p[u, t], t], {u, 0, 1}] - pap[t] == (1/
u) (D[p[u, t], u] + u D[ p[u, t], {u, 2}]) (* equation 1 *)


Differentiating the equation with respect to u lets the integral and the term pa'[t] vanish and gives for

q[u,t] = D[p[u,t],u]


the equation

D[q[u, t], t] == D[(1/u D[r q[u, t]]), u] (* equation 2 *)


This linear equation can be solved by standard methods. In fact separating of variables writing q = T[t] * U[u] gives

T'[t]/T[t] = - k^2 = U''[u] + 1/u U'[u] - 1/u^2 U[u]


where -k^2 is the separation parameter. The ODEs for T[t] and U[u] and their respective solutions are

solT = DSolve[T'[t] == -k^2 T[t], T[t], t]

(*
Out= {{T[t] -> E^(-k^2 t) C}}
*)

solU = DSolve[U''[u] + 1/u U'[u] - 1/u^2 U[u] == -k^2 U[u], U[u], u]

(*
Out= {{U[u] -> BesselJ[1, k u] C + BesselY[1, k u] C}}
*)


The separation parameter must be determined by the boundary conditions of U[u] at some u=u1 and u=u2.

Most frequently u1 = 0 and it is requested that U remains finite. This forces the BesselY term to vanish (which we assume in what follows). The other condition might be U[u=1] = 0 which leads to discrete values of k, called k[i] here.

The general solution of equation 2 is therefore of the form

q[u, t] = Sum[ a[i] Exp[- k[i]^2 t] BesselJ[1, k[i] u], {i, 1, ∞}]


where the coefficients a[i] have to be determined by the initial condition q[u,t=0].

Now we turn to the original function p[u,t].

Integrating q[u,t] from u=0 to u gives (notice J instead of J, and letting b[i] = a[i]/k[i])

p[u, t] = Sum[
b[i] Exp[- k[i]^2 t] BesselJ[0, k[i] u], {i, 1, ∞}] + g[t]


where g[t] is an arbitrary function of t alone.

Inserting this solution into equation 1 we have with

pk[u_, t_] := Exp[- k[i]^2 t] BesselJ[0, k[i] u]


on the one hand

(1/u) (D[pk[u, t], u] + u D[ pk[u, t], {u, 2}]) // FullSimplify

(*
Out= -E^(-t k[i]^2) BesselJ[0, u k[i]] k[i]^2
*)

D[pk[u, t], t] // FullSimplify

(*
-E^(-t k[i]^2) BesselJ[0, u k[i]] k[i]^2 + g'[t]
*)


Hence we are left with

Integrate[u D[p[u, t], t], {u, 0, 1}] - pap[t] == g'[t]


The integral can be done giving

Integrate[u BesselJ[0, k[i] u], {u, 0, 1}]

(*
Out= BesselJ[1, k[i]]/k[i]
*)


and the final equation is then

Sum[ b[i]/k[i] Exp[- k[i]^2 t] BesselJ[1, k[i] u], {i, 1, ∞}] +
g[t] == g'[t] + pap[t] (* equation 3 *)


Summary

Apart from the time dependence through the unknown function pap[t], the solution can be expressed by standard functions within Mathematica.

• The function pa[t] is the pressure applied on the surface of the cylinder. p[1,t]=pa[t]. I have also updated my complete equation. Mar 2, 2015 at 16:33
• @Raven: I have described the solution procedure completely in my answer. You should now go through it step by step provide the boundary and initial conditions and the external funtion pf[t]. In the end you'll obtain p[u,t]. Mar 2, 2015 at 17:28