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[?] Solve the following PDE?

vishal nandigana

Posted 7 years ago

Hi I want to solve the following PDE heqn = D[u[x, t], t] == cD[u[x, t], {x, 2}] -A u[x, t]- B; // A, B and c are constants ic = u[x, 0] == 2*A/b; // b is a constant bc = u[0,t] ==0, u[500,t]==0; sol = DSolveValue[{heqn, ic,bc}, u[x, t], {x, t}] The solver is not working, can you please let me know where I am doing a mistake. Thanks, Vishal

Hi I want to solve the following PDE

heqn = D[u[x, t], t] == c*D[u[x, t], {x, 2}] -A* u[x, t]- B; // A, B and c are constants

ic = u[x, 0] == 2*A/b; // b is a constant
bc = u[0,t] ==0, u[500,t]==0;
sol = DSolveValue[{heqn, ic,bc}, u[x, t], {x, t}]

The solver is not working, can you please let me know where I am doing a mistake.

Thanks, Vishal

POSTED BY: vishal nandigana

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Vishal Nandigana

Posted 7 years ago

Thanks, Mariusz. The code works now. Can you post the maple analytical solution for the above initial and boundary conditions. Sorry for troubling you for this. Regards, Vishal

POSTED BY: Vishal Nandigana

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 7 years ago

POSTED BY: Mariusz Iwaniuk

vishal nandigana

Posted 7 years ago

Thanks, Mariusz for solving the equation. Can you tell me what is the solution when I have initial conditions which are as follows, u(x,0) = exp(-Ax) + aexp(Ax) -d I tried using NDSolve, but it did not fetch me any results again. I used A = 0.005; B = 2; c = 0.002; b = 3; a = 0.003; d = 0.004; heqn = D[u[x, t], t] == cD[u[x, t], {x, 2}] - Au[x, t] - B; ic = u[x, 0] ==exp(-Ax) + aexp(Ax)-d; bc = {u[0, t] == 0, u[500, t] == 0}; sol = NDSolve[{heqn, ic, bc}, u, {x, 0, 1}, {t, 0, 1}] Plot3D[u[x, t] /. sol, {x, 0, 1}, {t, 0, 1}, AxesLabel -> Automatic] Thanks, Vishal

Thanks, Mariusz for solving the equation. Can you tell me what is the solution when I have initial conditions which are as follows,

u(x,0) = exp(-Ax) + a*exp(Ax) -d

I tried using NDSolve, but it did not fetch me any results again.

I used

 A = 0.005;
      B = 2;
      c = 0.002;
      b = 3;
    a = 0.003;
    d = 0.004;
      heqn = D[u[x, t], t] == c*D[u[x, t], {x, 2}] - A*u[x, t] - B;
      ic = u[x, 0] ==exp(-Ax) + a*exp(Ax)-d;
      bc = {u[0, t] == 0, u[500, t] == 0};
      sol = NDSolve[{heqn, ic, bc}, u, {x, 0, 1}, {t, 0, 1}]
      Plot3D[u[x, t] /. sol, {x, 0, 1}, {t, 0, 1}, AxesLabel -> Automatic]

Thanks, Vishal

POSTED BY: vishal nandigana

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 7 years ago

Hello. You make syntax mistakes: u[x, 0] ==exp(-Ax) + aexp(Ax)-d;(its Wrong - Check Documentation Center or Help ) ( it should be:) u[x, 0] == Exp[-A x] + aExp[A x] - d; Corrected code: A = 0.005; B = 2; c = 0.002; b = 3; a = 0.003; d = 0.004; heqn = D[u[x, t], t] == cD[u[x, t], {x, 2}] - Au[x, t] - B; ic = u[x, 0] == Exp[-A x] + a*Exp[A x] - d; bc = {u[0, t] == 0, u[500, t] == 0}; sol = NDSolve[{heqn, ic, bc}, u, {x, 0, 1}, {t, 0, 1}] Plot3D[u[x, t] /. sol, {x, 0, 1}, {t, 0, 1}, AxesLabel -> Automatic] Regards,Mariusz

Hello.

You make syntax mistakes:

 u[x, 0] ==exp(-Ax) + a*exp(Ax)-d;(*its Wrong - Check Documentation Center or Help *)
 (* it should be:*)
 u[x, 0] == Exp[-A x] + a*Exp[A x] - d;

Corrected code:

 A = 0.005;
 B = 2;
 c = 0.002;
 b = 3;
 a = 0.003;
 d = 0.004;
 heqn = D[u[x, t], t] == c*D[u[x, t], {x, 2}] - A*u[x, t] - B;
 ic = u[x, 0] == Exp[-A x] + a*Exp[A x] - d;
 bc = {u[0, t] == 0, u[500, t] == 0};
 sol = NDSolve[{heqn, ic, bc}, u, {x, 0, 1}, {t, 0, 1}]
 Plot3D[u[x, t] /. sol, {x, 0, 1}, {t, 0, 1}, AxesLabel -> Automatic]

Regards,Mariusz

POSTED BY: Mariusz Iwaniuk

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 7 years ago

POSTED BY: Mariusz Iwaniuk

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