At 1:18 the very first eqn of 1D heat conduction seems to be doubtfull (LHS side may be wrong as per my understanding Bcoz if convection is not considered then temperature gradient term should be wrt time)
Professor how would I solve this system of first order edos numerically by plotting the graph for the different values of (n). the derivatives are in relation to ha (r). a'/r = -e^2*v^2*(g^2 - 1) g' = - a*g/r given the boundary conditions a(0) = n a(inf)=0 g(0) = 0 g(inf)=1 where (e)=0.5 and (v)=1 are constant. please give a helping hand there, I looked for and did not find any problems like this on the matlab website.
If you look at about the 7 minute mark of the video, you'll see that time is provided to the boundary condition function. So you can use time in your function.
Hi Professor Blanchard, I'm trying to use pdepe to solve a pde based on Fick's Second Law for the evolution of a solution concentration within a spherical bead. The general equation is == du/dt = D*d²u/dx² + 2D/x*du/dx. My problem right now is defining the right boundary condition, which in my system is == dudx = (V/D/A/p)*dudt (V, D, A, p are constants). How can I evoke dudt in setting a boundary condition within pdepe? Thank you very much for your kind attention!
@@jakeblanchard Thank you very much for your quick response. I understand your suggestion. If you don't mind, I'd like to ask your opinion on how I can proceed to model my current set of pdes. To be more precise, the model is for estimating the evolution of a drug's concentration in an external finite volume (dCb/dt), that initially comes from a spherical bead by diffusion (i.e., the drug is initially inside the bead, then diffuses inside to reach the bead's surface, where it diffuses to the outside external finite volume). The general equation that describes the concentration change within the spherical bead is dependent on both time t and the bead coordinate r: dC/dt = D*(d²C/dr²) + 2D/r*(dC/dr) The initial condition is: C(r, t = 0) = Co (constant, initial concentration inside the spherical bead) The boundary conditions are: (t > 0, r = 0), dCdr = 0 (t > 0, r = a), D*dC/dr = V/A*(dCb/dt), where dCb/dt = 1/p*(dC/dt). By combining these two, this right boundary condition then becomes == dC/dr = (V/(D*A*p))*dCdt The term "Cb" represents the drug's concentration in the external finite volume V (V, p, D, A are constants) after being diffused from the spherical bead. Thus, while the concentration within the bead (C) is dependent on both t and r, the external concentration (Cb) is only dependent on time. Do you think this can be solved using pdepe?
@@lucasdanda I don't think you have access to dC/dt from within pdepe's boundary condition subroutine, so I don't think you can do this with pdepe. I could be wrong. I would start with a Crank-Nicholson finite difference approach and see if I could work out how to handle this boundary condition that way. There is a lot of stuff on the internet about how to do Crank-Nicholson.
Is there any way to get the solution as an equation? Doing it with pdepe I get a 3 by 3 matrix which allows me to plot the solution, but for checking answers sake I would like to view the data in equation form.
+william leach That requires symbolic algebra. Matlab has a symbolic toolkit, but I don't use it much. I would suggest you try Maple or Mathematica for this type of problem. Keep in mind that only the simplest of partial differential equations will have a clean, closed-form solution. You might be able to get a series solution, depending on the equation.
This page discusses how to solve a system of PDE's. www.mathworks.com/help/matlab/math/solve-system-of-pdes.html You might want to take a look. Are you sure you can fit your differential equation into the standard format for pdepe?
@@jakeblanchard After i corrected my equation, i narrowed the bc to two. Having problems with identifying my second bc as matlab's function (p and q). My general equation is εdC/dt= D(d2C/dr2 +2/r*dC/dr) with boundary conditions of 1) D(dC/dr)=k(Cao-Cas) 2) -D(dC/dr)= aCso(dr/dt)
Hi! Is there a way to implement a time dependent heat source? In the case shown the heating is constant, so your heat source is only acting in terms of a stationary boundary condition.....
Yes. Go to about the 7 minute mark of the video. You'll see that for the boundary conditions the pdexlbc function passes time as "t" so you can use that in defining your boundary conditions.
Dear Sir, thanks for the informative video. I am writing a report about opinion dynamic borrowed from the kinetic equation. Is it possible to use this method to solve a large time behaviour ( t --> \infty)of the Fokker Planck equation (PDE) , in other words the steady state profile?
Vincent Winsion I don't know much about Fokker Planck equations, but, from the sound of it, it isn't going to be a parabolic equation. Hence, pdepe is not your answer. Feel free to correct me if I'm wrong.
xl and xr are the positions of the boundaries, so I'm guessing it gets them from your definition of x. In the video, I define x=linspace(0,L,200), so xl=0 and xr=L. ul and ur are the solution vectors at the boundary and are calculated each iteration. So your job is to define the desired values for ul and ur and let Matlab take care of the rest.
u r amazing jake... i would really appreciate if u can upload more videos on Matlab... Also some introductory videos on Fortran will be greatly appreciated....
Hi Jake, thank you so much for putting in the effort in this video. It is very clear and helpful! I have a small side note: I think you have a typo in your first PDE statement where it says rho*cp*dT/dx
hi, I have a doubt. When using the pdepe function, it is not necessary to make the discretization in space ? it is possible to solve such a problem by using the tool ode45 and when to use, do the discretization of space in the algorithm and the program integrates in time. Because I am with doubts. Furthermore, the pdepe function solves 2D problems ?
+Bryan Silveira pdepe solves systems of partial differential equations in 1 space variable and time. It will not solve problems involving two spatial dimensions. Ode45 solves initial value problems. I'm guessing you could difference a pde in space and then solve the resulting system of initial value problems using ode45, but why bother? pdepe is easier. It will handle the discretization/differencing for you.
+Jake Blanchard It's because I just know how to solve this problems using ode45 and do the discretization a pde in space haha. I never used pdepe before, but now I will try to use this function. Do you know a function used to solve problems involving a pde involving two spatial dimensions ?? In my research, I have a boundary moving problem involving a pde with two spatial dimensions.
Sure, but I'm not aware of any pre-built Matlab tools for solving that problem. Many would use finite elements for a 2-D, time-dependent problem. You could do finite differences, but you'd have to write up your own script. You can find descriptions of the algorithms in many places on the internet.
+Yuhang Cai ur is passed in by the pdepe solver as the temperature on the right hand side. By setting p=ur, we are asking the solver to iterate until T=0. Similarly, DuDx is provided by the solver and passed into the pdex1pde routine as an argument.
+Yuhang Cai No, you are supposed to define the functions p and q at both boundaries for any value of x, t, and u. Then pdepe will solve the equations in such a way that those boundary conditions are met.
Hi! how can i implement a heat source that depends on the time. In your example q= const. How do i have to change the boundary conditions in the code, if q changes with the time? Thanks for helping!
No. It uses a method from Skeel, R. D. and M. Berzins, "A Method for the Spatial Discretization of Parabolic Equations in One Space Variable," SIAM Journal on Scientific and Statistical Computing, Vol. 11, 1990, pp.1-32.
Hi Jake, thanks for the information. Do you know how to solve N=2 coupled parabolic PDE? ie Du1/Dt = d2u1/Dx2 - D2u2/Dx2 Du2/Dt = d2u2/Dx2 - D2u1/Dx2 Thanks!
It looks to me like pdepe will solve this. You just need to compare your equations to the general equation for pdepe and then build the script. I can try to help if you send me a script.
Jake Blanchard Jake, thanks for the response. someone already helped me on an internet fourm. I'm so appreciative that so many ppl like yourself are willing to help with MATLAB problems. Thanks
It's a linear equation, so you can solve the whole problem using Celcius if you want. One boundary condition was T=0, so if that's 0 Celcius, then all the temperatures will be in Celcius.
The BC is p+q*f=0. I've defined f to be f=k*dT/dx and the BC we want is -k*dT/dx=C at x=0, where C is the prescribed heat flux (a constant). Hence, we are just looking for -f=q or f+q=0. In other words, k is already included in f.
It is a differential partial ecuation that can be expressed as the ine required for the method. The only problem i have is that my boundary conditions are du/dx( x=a) =somenthing and u(x=a) =something
Alexandra Urbano You have to have something to define the other boundary. So if you only have an equation which is second order in the spatial variable, then I don't think this is going to work.
+Nikhil Yenumula I'm not aware of any built-in Matlab functions that will solve this directly. The PDE toolbox might have something. Or you might be able to find something in Matlab Central.
Thank you sir for this video. One of my boundary condition is a∂T/∂t=-K∂T/∂x. In this case my f is K∂T/∂x so q will be 1/a. How should I do with ∂T/∂t ? Can I put ∂T/∂t as p? Thanks for any help
Sir,very informative tutorial I want to ask,I have these two boundaries how i will put in the above form u mentioned.I have these two boundary conditions. λrad ∂Tp/ ∂rp@ rp=Rp =αf(Tf −Tp); (∂Tp /∂rp)@rp=0 = 0 I need your guidance.
OK. It looks like you'll want to choose f=lambda/rho/cp*dT/drp as you define your equation, so the BC will use f to get the first derivative. Try it and send me a script if you can't get it working.
I don't think the pdepe routine, which is built into Matlab, can solve this set of equations. I actually think you can solve this analytically, so you might give that a shot.
I don't know of a built-in tool for this type of problem. There may be something in the PDE Toolbox, but I'm not sure. If that is not of use, you'd be left with developing your own tool.
Hello, first of all thank you for this useful video, I'm trying to write a little code to solve a partial differential equation with the command "pdepe". With this equation I want to calculate the velocity "u" over time of a liquid flowing on an inclined plane. I decided to put the boundary conditions so that at time t = 0 and the initial position x = 0 I have u = 0 (the liquid is stopped) while at time t = t0 and x = x0 (at the end of the inclined plane) I have that velocity is u = u0 (x, t) then, the flow is free to slide and will have a certain velocity (which is my incognita). I do not know how to define the coefficients ql pl pr qr to define my boundary conditions in the form requested by Matlab. the equation that I developed is this ρ ∂u/∂t=D (∂^2 u)/(∂x)^2 -u ∂u/∂x. thanks i advance
Send me what you have written so far and a detailed description of what you want for boundary conditions. I'll take a look at it. Email is blanchard@engr.wisc.edu.
Hi Jake .. Can i solve a differential eq. using Method of Characteristics in MATLAB. Actually I want to solve transient flow equations. So pls help me on this matter...
Sure, you can do it, but I'm not aware of any built-in tools that would facilitate using characteristics. You might be able to find something in the File Exchange on Matlab Central.
Jake Blanchard Excuse me, Could you give me a hint as to solve this proble,: du/dt = alpha^2 *(d^2u/dx^2 + d^2u/dy^2) - (Q/A)*(du/dx) + (V^2/(rho*rho1*c*L^2))
Sankarshan Acharya I'm not aware of any built-in tools in Matlab that will solve that equation. It's possible there is something in the PDE toolbox, but I'm not sure. Try finite differences.
Hi Sir , I'm working now in my project of graduation in Master Applied Mathematics , our goal is to show that the solution of parabolic problem go to the solution of the associated elliptic problem (is called the stationary solution ) for two spaces in competition : du/dt-\sigma1 \Delta u =u(a-bu-cv) dv/dt- \sigma2 \Delta v= v(e-gv-du) (this is the parabolic problem ) ... (1) -\sigma1 \Delta u =u(a-bu-cv) - \sigma2 \Delta v= v(e-gv-du) (this is the elliptic problem ) ... (2) Can you help me Sir to program (1) and (2 ) and show that the solution of (1) converge to the solution of (2) when t go to \infty ?????
I understand your explanation, but why not demonstrate it in MATLAB？...I‘ve spent more than 16hr to find where to type those commands rather than use PDEtool box.
It helps when you actually show MATLAB, execute it and show the result
It's almost unbelievable that there are videos covering MATLAB's pdepe solver. You are awesome.
Among the most useful videos on Matlab out there. Thanks from Purdue U.
At 1:18 the very first eqn of 1D heat conduction seems to be doubtfull (LHS side may be wrong as per my understanding Bcoz if convection is not considered then temperature gradient term should be wrt time)
Yes, that's a typo. The first term should be rho*cp*dTdt.
Professor how would I solve this system of first order edos numerically by plotting the graph for the different values of (n). the derivatives are in relation to ha (r).
a'/r = -e^2*v^2*(g^2 - 1)
g' = - a*g/r
given the boundary conditions
a(0) = n a(inf)=0
g(0) = 0 g(inf)=1
where (e)=0.5 and (v)=1 are constant. please give a helping hand there, I looked for and did not find any problems like this on the matlab website.
1:22 surely the left hand side is dT/dt and not dT/dx?
Yes, that's a typo. Good catch.
If you look at about the 7 minute mark of the video, you'll see that time is provided to the boundary condition function. So you can use time in your function.
Hi Professor Blanchard,
I'm trying to use pdepe to solve a pde based on Fick's Second Law for the evolution of a solution concentration within a spherical bead. The general equation is == du/dt = D*d²u/dx² + 2D/x*du/dx. My problem right now is defining the right boundary condition, which in my system is == dudx = (V/D/A/p)*dudt (V, D, A, p are constants). How can I evoke dudt in setting a boundary condition within pdepe? Thank you very much for your kind attention!
I don't think pdepe allows you to use the time derivative in the boundary conditions. You might have to write your own solver.
@@jakeblanchard Thank you very much for your quick response. I understand your suggestion. If you don't mind, I'd like to ask your opinion on how I can proceed to model my current set of pdes.
To be more precise, the model is for estimating the evolution of a drug's concentration in an external finite volume (dCb/dt), that initially comes from a spherical bead by diffusion (i.e., the drug is initially inside the bead, then diffuses inside to reach the bead's surface, where it diffuses to the outside external finite volume). The general equation that describes the concentration change within the spherical bead is dependent on both time t and the bead coordinate r:
dC/dt = D*(d²C/dr²) + 2D/r*(dC/dr)
The initial condition is:
C(r, t = 0) = Co (constant, initial concentration inside the spherical bead)
The boundary conditions are:
(t > 0, r = 0), dCdr = 0
(t > 0, r = a), D*dC/dr = V/A*(dCb/dt), where dCb/dt = 1/p*(dC/dt). By combining these two, this right boundary condition then becomes == dC/dr = (V/(D*A*p))*dCdt
The term "Cb" represents the drug's concentration in the external finite volume V (V, p, D, A are constants) after being diffused from the spherical bead.
Thus, while the concentration within the bead (C) is dependent on both t and r, the external concentration (Cb) is only dependent on time.
Do you think this can be solved using pdepe?
@@lucasdanda I don't think you have access to dC/dt from within pdepe's boundary condition subroutine, so I don't think you can do this with pdepe. I could be wrong. I would start with a Crank-Nicholson finite difference approach and see if I could work out how to handle this boundary condition that way. There is a lot of stuff on the internet about how to do Crank-Nicholson.
Is there any way to get the solution as an equation? Doing it with pdepe I get a 3 by 3 matrix which allows me to plot the solution, but for checking answers sake I would like to view the data in equation form.
+william leach
That requires symbolic algebra. Matlab has a symbolic toolkit, but I don't use it much. I would suggest you try Maple or Mathematica for this type of problem. Keep in mind that only the simplest of partial differential equations will have a clean, closed-form solution. You might be able to get a series solution, depending on the equation.
Thank you Ms.J.Blanchard , it's a great work and helpful .
I can't believe this video was made when I was a first grader and now I learned from it
Makes me feel very old!
you made such a high-quality lecture that it is still being watched today and has learners from all generations@@jakeblanchard
Hi sir, im trying to solve pde for a shrinking core model which has 3 boundary conditions. How do i input the coding for 3 BC?
This page discusses how to solve a system of PDE's. www.mathworks.com/help/matlab/math/solve-system-of-pdes.html You might want to take a look. Are you sure you can fit your differential equation into the standard format for pdepe?
@@jakeblanchard After i corrected my equation, i narrowed the bc to two. Having problems with identifying my second bc as matlab's function (p and q). My general equation is εdC/dt= D(d2C/dr2 +2/r*dC/dr) with boundary conditions of 1) D(dC/dr)=k(Cao-Cas) 2) -D(dC/dr)= aCso(dr/dt)
@@faudziahismail6726 I don't understand the dr/dt term. Can you explain?
Hello Prof. Jake. Thank you very much for this video. It is really useful. I have a question.
Is it possible to solve any time-delayed PDE in matlab?
I don't think pdepe will do that.
Okay. Thank you Prof. Jake for your reply.
Hello Jake,thanks you very much fot this video, is posible work with the source term (s) for exemple a matriz o only a column[1 2 3 4... n]?
I don't understand the question.
Hi! Is there a way to implement a time dependent heat source? In the case shown the heating is constant, so your heat source is only acting in terms of a stationary boundary condition.....
Yes. Go to about the 7 minute mark of the video. You'll see that for the boundary conditions the pdexlbc function passes time as "t" so you can use that in defining your boundary conditions.
Dear Sir,
thanks for the informative video. I am writing a report about opinion dynamic borrowed from the kinetic equation. Is it possible to use this method to solve a large time behaviour ( t --> \infty)of the Fokker Planck equation (PDE) , in other words the steady state profile?
Vincent Winsion
I don't know much about Fokker Planck equations, but, from the sound of it, it isn't going to be a parabolic equation. Hence, pdepe is not your answer. Feel free to correct me if I'm wrong.
Thank you prof, but I have one doubt, in the boundary condition how does matlab identify what the ul,xl,ur,xr are?
xl and xr are the positions of the boundaries, so I'm guessing it gets them from your definition of x. In the video, I define x=linspace(0,L,200), so xl=0 and xr=L. ul and ur are the solution vectors at the boundary and are calculated each iteration. So your job is to define the desired values for ul and ur and let Matlab take care of the rest.
can i get programs of solving partial differential equations in FORTRAN particularly of heat transfer and fluid flow in porous medium.
u r amazing jake... i would really appreciate if u can upload more videos on Matlab... Also some introductory videos on Fortran will be greatly appreciated....
Hi Jake, thank you so much for putting in the effort in this video. It is very clear and helpful!
I have a small side note: I think you have a typo in your first PDE statement where it says rho*cp*dT/dx
Good catch on the typo. Nobody ever mentioned that before.
hi, I have a doubt.
When using the pdepe function, it is not necessary to make the discretization in space ?
it is possible to solve such a problem by using the tool ode45 and when to use, do the discretization of space in the algorithm and the program integrates in time. Because I am with doubts.
Furthermore, the pdepe function solves 2D problems ?
+Bryan Silveira
pdepe solves systems of partial differential equations in 1 space variable and time. It will not solve problems involving two spatial dimensions. Ode45 solves initial value problems. I'm guessing you could difference a pde in space and then solve the resulting system of initial value problems using ode45, but why bother? pdepe is easier. It will handle the discretization/differencing for you.
+Jake Blanchard
It's because I just know how to solve this problems using ode45 and do the discretization a pde in space haha.
I never used pdepe before, but now I will try to use this function.
Do you know a function used to solve problems involving a pde involving two spatial dimensions ??
In my research, I have a boundary moving problem involving a pde with two spatial dimensions.
+Bryan Silveira
There is nothing in standard Matlab for problems in two spatial dimensions. The PDE toolbox has some stuff.
+Jake Blanchard
Thank you
Is it possible to generalise the same concept for the 2D plane ?
Sure, but I'm not aware of any pre-built Matlab tools for solving that problem. Many would use finite elements for a 2-D, time-dependent problem. You could do finite differences, but you'd have to write up your own script. You can find descriptions of the algorithms in many places on the internet.
@@jakeblanchard, I have used 'pdetool'. It is effective, but it does not show you the solution.
That must be in the PDE toolbox. I've never used it. But surely there is a way to display the results.
@@jakeblanchard Thank you very much, Sir.
Can you explain why pr=ur? Since I found ur haven't been defined yet? Also don't know how matlab recognize "DuDx", which also not defined.
+Yuhang Cai
ur is passed in by the pdepe solver as the temperature on the right hand side. By setting p=ur, we are asking the solver to iterate until T=0.
Similarly, DuDx is provided by the solver and passed into the pdex1pde routine as an argument.
+Jake Blanchard thank you! But I thought "ur" should be a pre-defined term. Is it not necessary to define "ur" and "DuDx" in this pdepe solver?
+Yuhang Cai
No, you are supposed to define the functions p and q at both boundaries for any value of x, t, and u. Then pdepe will solve the equations in such a way that those boundary conditions are met.
Hi!
how can i implement a heat source that depends on the time.
In your example q= const. How do i have to change the boundary conditions in the code, if q changes with the time?
Thanks for helping!
THANKS SO MUCH, realy its useful....
I would to ask you...does the solution methodology applying Crank-Nikolson method?
No. It uses a method from
Skeel, R. D. and M. Berzins, "A Method for the Spatial Discretization of Parabolic Equations in One Space Variable," SIAM Journal on Scientific and Statistical Computing, Vol. 11, 1990, pp.1-32.
Hi Jake, thanks for the information.
Do you know how to solve N=2 coupled parabolic PDE? ie
Du1/Dt = d2u1/Dx2 - D2u2/Dx2
Du2/Dt = d2u2/Dx2 - D2u1/Dx2
Thanks!
It looks to me like pdepe will solve this. You just need to compare your equations to the general equation for pdepe and then build the script. I can try to help if you send me a script.
Jake Blanchard Jake, thanks for the response. someone already helped me on an internet fourm. I'm so appreciative that so many ppl like yourself are willing to help with MATLAB problems. Thanks
Excellent lesson.
Hello Jake, Is there any way I could use this approach to solve 2-D heat equation like this : T_yy + Txx - a*T_y -b* Tx + c = d*T_t ?
Thanks
No, pdepe is not set up for these. You might try the pde toolbox, if you have access to it.
Sir, how can we solve 3rd order nonlinear pde (like KdV equation) in matlab
If you send me questions, I can try to respond.
In which part you convert Kelvin to Celcius? Due to you plot as Celcius the temperature results
It's a linear equation, so you can solve the whole problem using Celcius if you want. One boundary condition was T=0, so if that's 0 Celcius, then all the temperatures will be in Celcius.
Thank you for these videos they are a great help!
For the first boundary condition, isn't q supposed to be k? Don't quite get why q=1
The BC is p+q*f=0. I've defined f to be f=k*dT/dx and the BC we want is -k*dT/dx=C at x=0, where C is the prescribed heat flux (a constant). Hence, we are just looking for -f=q or f+q=0. In other words, k is already included in f.
Should've listened better, you even said that it's the same f as in the PDE itself :) Thank you for your quick response and great tutorial!
Hi, if i have two boundary contions at x=0, can I use this method? Thanks in advance!
What is the equation?
It is a differential partial ecuation that can be expressed as the ine required for the method. The only problem i have is that my boundary conditions are du/dx( x=a) =somenthing and u(x=a) =something
Alexandra Urbano You have to have something to define the other boundary. So if you only have an equation which is second order in the spatial variable, then I don't think this is going to work.
Ok, thank you for your answer :)
Thank you for the video, it was really helpful. On a separate note, I can't help but notice an eerie resemblance of the voice with Morgan Freeman.
can we solve for (x,y,z,t) coordinates using matlab? Is it possible? Can u please tell me?
+Nikhil Yenumula
What kind of equation are you trying to solve?
+Jake Blanchard Heat equation involving all cartesian coordinates
+Nikhil Yenumula
I'm not aware of any built-in Matlab functions that will solve this directly. The PDE toolbox might have something. Or you might be able to find something in Matlab Central.
Thank you sir for this video. One of my boundary condition is a∂T/∂t=-K∂T/∂x. In this case my f is K∂T/∂x so q will be 1/a. How should I do with ∂T/∂t ? Can I put ∂T/∂t as p? Thanks for any help
I don't think pdepe can do what you are trying to do. That is, I don't think it will allow the use of dT/dt in the boundary conditions.
Excuse me, sir ! Can I learn and ask some helps from you about Matlab ??
Sir,very informative tutorial I want to ask,I have these two boundaries how i will put in the above form u mentioned.I have these two boundary conditions. λrad ∂Tp/ ∂rp@ rp=Rp =αf(Tf −Tp); (∂Tp /∂rp)@rp=0 = 0
I need your guidance.
It depends on how you set up f, which depends on how you set up your differential equations. So you need to show me the pde as well.
Ok sir. ∂Tp/ ∂t = 1 / rp2 ∂/∂rp(rp2 λp /ρCp ∂Tp /∂rp)+ rA∆Hr,A* ρCp
OK. It looks like you'll want to choose f=lambda/rho/cp*dT/drp as you define your equation, so the BC will use f to get the first derivative. Try it and send me a script if you can't get it working.
Ok sir.
I don't have any code that employs finite volume techniques. You'll have to look elsewhere.
can we solve three dependent pde equations with pdepe function ?
Yes, pdepe solves systems of parabolic pde's, so it will handle three (or more).
Jake Blanchard Ok sir thanks
I don't think the pdepe routine, which is built into Matlab, can solve this set of equations. I actually think you can solve this analytically, so you might give that a shot.
How to solve the parabolic system of pde over a 2d space?
I don't know of a built-in tool for this type of problem. There may be something in the PDE Toolbox, but I'm not sure. If that is not of use, you'd be left with developing your own tool.
How can we solve KdV equations (3rd order pde) in matlab
Hey wow its a very well defined .. Very happy Thank you +Jake Blanchard. .
How to solve a system of PDE, say for example Navier-Stokes equations?
Actually how can we solve KdV equation by pde in matlab
Now I am interested by solving kdv by matlab if you found
Thanks for this video. it's rally helpful.
sir i need the code in FORTRAN for governing equations of heat transfer and fluid flow FOR POROUS MATERIAL/POROUS MEDIA USING FINITE VOLUME APPROACH.
HOW CAN WE SOLVE NONLINEAR 3RD ORDER PARTIAL DIFFERENTIAL EQUATION IN MATLAB
I'm not sure how to do this. I think you might have to just finite difference it manually and find a way to solve the resulting equations.
Any topics you could suggest? I'm sort of short on ideas. I'm short on time, as well, but some ideas might help.
How about solving level set function in Matlab for moving boundary problems?
Hello, first of all thank you for this useful video, I'm trying to write a little code to solve a partial differential equation with the command "pdepe". With this equation I want to calculate the velocity "u" over time of a liquid flowing on an inclined plane. I decided to put the boundary conditions so that at time t = 0 and the initial position x = 0 I have u = 0 (the liquid is stopped) while at time t = t0 and x = x0 (at the end of the inclined plane) I have that velocity is u = u0 (x, t) then, the flow is free to slide and will have a certain velocity (which is my incognita). I do not know how to define the coefficients ql pl pr qr to define my boundary conditions in the form requested by Matlab. the equation that I developed is this ρ ∂u/∂t=D (∂^2 u)/(∂x)^2 -u ∂u/∂x. thanks i advance
Send me what you have written so far and a detailed description of what you want for boundary conditions. I'll take a look at it. Email is blanchard@engr.wisc.edu.
ur voice is like THE GREAT MORGAN FREEMAN>>>>
How can we solve du/dx+A(d3u/dT3)+B udu/dT+Cu=0
Hi Jake ..
Can i solve a differential eq. using Method of Characteristics in MATLAB. Actually I want to solve transient flow equations. So pls help me on this matter...
Sure, you can do it, but I'm not aware of any built-in tools that would facilitate using characteristics. You might be able to find something in the File Exchange on Matlab Central.
Jake Blanchard
Excuse me,
Could you give me a hint as to solve this proble,:
du/dt = alpha^2 *(d^2u/dx^2 + d^2u/dy^2) - (Q/A)*(du/dx) + (V^2/(rho*rho1*c*L^2))
Sankarshan Acharya
I'm not aware of any built-in tools in Matlab that will solve that equation. It's possible there is something in the PDE toolbox, but I'm not sure. Try finite differences.
Hi Sir , I'm working now in my project of graduation in Master Applied Mathematics , our goal is to show that the solution of parabolic problem go to the solution of the associated elliptic problem (is called the stationary solution ) for two spaces in competition :
du/dt-\sigma1 \Delta u =u(a-bu-cv)
dv/dt- \sigma2 \Delta v= v(e-gv-du) (this is the parabolic problem ) ... (1)
-\sigma1 \Delta u =u(a-bu-cv)
- \sigma2 \Delta v= v(e-gv-du) (this is the elliptic problem ) ... (2)
Can you help me Sir to program (1) and (2 ) and show that the solution of (1) converge to the solution of (2) when t go to \infty ?????
Qui peut m'aider ???? Qui me donner une rapport sur la bibliothèque pde toolbox du matleb svp ?????
It sounds like you might need a finite element code.
I've never seen an equation like that. You might have to difference it yourself and try to come up with an algorithm for solving it.
I understand your explanation, but why not demonstrate it in MATLAB？...I‘ve spent more than 16hr to find where to type those commands rather than use PDEtool box.
+Yiran Lin
You can download a working script from here: blanchard.ep.wisc.edu/PublicMatlab/index.html#PPDE
+Jake Blanchard Thank you
Fantastic
temp=temp(z,r)
1/r*D/Dr(r*Dtemp/Dr)+f(temp,r)=0 here c is zero but its giving me an error jake need your help
Send me a script and I can try to take a look at it.
Jake Blanchard sir your email please
Jake Blanchard the equation is (1/r*D/Dr(r*Dtemp/Dr))+(C1-temp)*C2*f(r)f(r)=(r*(5*r - 600))/2 + 5000)
say r
blanchard "at" engr.wisc.edu
Jake Blanchard i sent u the mail with the attached programs sir...looking forward to your response
I'm not aware of any tools devoted to this. You'll likely have to develop a solver on your own.
very good!
at 1.20 it should be DT/Dt instead of DT/Dx
You are correct. Good catch.
Oh ok thank you sir.
Thanx, dude!
thanks
MOAR!
morgan freeman?
da paiura! :D