Thx acer,

Your approach is one of the possibilty.

I am trying to combine the follwoing two plots as a line plot

plot([subs(R3=-5,k=4,R1=0.0006,PL=1,Q=1.2,P),subs(R3=-5,k=4,R1=0.0006,PL=1,Q=2,P),

subs(R3=-5,k=4,R1=0.0006,PL=1,Q=4,P)],x=0..1,numpoints=200);

plot([subs(R3=-5,k=4,R1=0.0006,x=1,Q=1.2,P),subs(R3=-5,k=4,R1=0.0006,x=1,Q=2,P),

subs(R3=-5,k=4,R1=0.0006,x=1,Q=4,P)],PL=0..1,numpoints=200);

because, I also need to chose three different values for Q.

Thx acer,

Your approach is one of the possibilty.

I am trying to combine the follwoing two plots as a line plot

plot([subs(R3=-5,k=4,R1=0.0006,PL=1,Q=1.2,P),subs(R3=-5,k=4,R1=0.0006,PL=1,Q=2,P),

subs(R3=-5,k=4,R1=0.0006,PL=1,Q=4,P)],x=0..1,numpoints=200);

plot([subs(R3=-5,k=4,R1=0.0006,x=1,Q=1.2,P),subs(R3=-5,k=4,R1=0.0006,x=1,Q=2,P),

subs(R3=-5,k=4,R1=0.0006,x=1,Q=4,P)],PL=0..1,numpoints=200);

because, I also need to chose three different values for Q.

@Carl Love 

I tried to plot the residual but it doesnt look good

restart:with(plots):

Digits:=10:
fw:=0.1:Pe:=1:R1:=0.5:R:=0:d1:=1:n:=10:
eq1 := diff(F1(eta),`$`(eta,3))+R1*d1*(F1(eta)*diff(F1(eta),`$`(eta,2))-(diff(F1(eta),eta))^2)-diff(F1(eta),eta)=0;

eq2 := (1+4*R/3)*diff(theta(eta),`$`(eta,2))+Pe*d1*F1(eta)*diff(theta(eta),eta)=0;

bc:=F1(0)=fw,D(F1)(0)=1,D(F1)(n)=0,theta(0)=1,theta(n)=0;

A1:=dsolve({eq1,eq2,bc},numeric,output= listprocedure):

Exact:=.906216927-exp(-eta)+(1/20*(-4.749257710-4.749257710*eta))*exp(-eta)-(1/800*(-344.9967669-344.9967669*eta+22.55544879*eta^2))*exp(-eta);

Res:= eval(F1(eta), A1) - unapply(eval(Exact), eta):

plot(Res, 0..10);

Please have look.

Thanks

@Carl Love 

I tried to plot the residual but it doesnt look good

restart:with(plots):

Digits:=10:
fw:=0.1:Pe:=1:R1:=0.5:R:=0:d1:=1:n:=10:
eq1 := diff(F1(eta),`$`(eta,3))+R1*d1*(F1(eta)*diff(F1(eta),`$`(eta,2))-(diff(F1(eta),eta))^2)-diff(F1(eta),eta)=0;

eq2 := (1+4*R/3)*diff(theta(eta),`$`(eta,2))+Pe*d1*F1(eta)*diff(theta(eta),eta)=0;

bc:=F1(0)=fw,D(F1)(0)=1,D(F1)(n)=0,theta(0)=1,theta(n)=0;

A1:=dsolve({eq1,eq2,bc},numeric,output= listprocedure):

Exact:=.906216927-exp(-eta)+(1/20*(-4.749257710-4.749257710*eta))*exp(-eta)-(1/800*(-344.9967669-344.9967669*eta+22.55544879*eta^2))*exp(-eta);

Res:= eval(F1(eta), A1) - unapply(eval(Exact), eta):

plot(Res, 0..10);

Please have look.

Thanks

@Preben Alsholm

Thanks for the detailed response. I wasn't expecting it to be so complicated.

I thought maybe there is something wrong with my PC. Anyway, I have no idea how to

proceed with it.

@Preben Alsholm

Thanks for the detailed response. I wasn't expecting it to be so complicated.

I thought maybe there is something wrong with my PC. Anyway, I have no idea how to

proceed with it.

@Carl Love @acer

I have tried to plot an integral but couldn't

here is my try

restart:with(plots):
N:=1/sqrt(Pi*t)-1/(2*sqrt(Pi))*int((t-s)*sin(t-s)/(s*sqrt(s)),s=0..t)+

1/sqrt(Pi)*int((t-s)*sin(t-s)/(s*sqrt(s)),s=0..t);

plot(N,t=0..10);

THanks

@Carl Love @acer

I have tried to plot an integral but couldn't

here is my try

restart:with(plots):
N:=1/sqrt(Pi*t)-1/(2*sqrt(Pi))*int((t-s)*sin(t-s)/(s*sqrt(s)),s=0..t)+

1/sqrt(Pi)*int((t-s)*sin(t-s)/(s*sqrt(s)),s=0..t);

plot(N,t=0..10);

THanks

Thanks @Preben Alsholm

I agree with you that there is some issue with the bcs. But, I want to see the problem in more detailand 

have found the exact expression for diff(T(0,t),x) which is

N := 1/sqrt(Pi*t)+(1/2)*(int(-(t-s)*sin(-t+s)/s^(3/2), s = 0 .. t))/sqrt(Pi);

If I am not wrong the integral has finite values only for multiple of Pi?

Thanks

Thanks @Preben Alsholm

I agree with you that there is some issue with the bcs. But, I want to see the problem in more detailand 

have found the exact expression for diff(T(0,t),x) which is

N := 1/sqrt(Pi*t)+(1/2)*(int(-(t-s)*sin(-t+s)/s^(3/2), s = 0 .. t))/sqrt(Pi);

If I am not wrong the integral has finite values only for multiple of Pi?

Thanks

@Markiyan Hirnyk @Preben Alsholm

Thanks for your interest.

I have tried three different ways to find out whats going on,

1. This one was suggested by Lopez for finding diff(u(eta,t),eta,eta) in one of my ealier posts

          W:=rhs(pds:-value(output=listprocedure)[4]);

           p1:=plot(W(0,t), t= 0..1,color=blue,thickness=2,legend = ["p1"]):

           I am not sure whether [4] is the position of the diff(T(eta,t)) or not

           but I thought its worth a try.

    2.   The second one was suggested by Preben

           T2Val:=op(subs(VAL,[T2(eta,t)]));

           p2:=plot(T2Val(0,t),t=0..1,legend = ["p2"]);

    3.  The third one  is the same as 2 but here I collect the values of T2Val(0,t) point by point

         for each t and then plot.

Here is the output I got 

pdsolveplot.mw

Thanks

Thanks @Preben Alsholm

I tried to apply your suggested approach like this

restart:with(PDEtools):with(plots):

sys := {diff(T(x, t), t) = diff(T(x, t), x, x)+(diff(u(x, t), x))^2, diff(u(x, t), t) = diff(u(x, t), x, x)};

PDE:={diff(u(x,t),t)=diff(u(x,t),x$2),diff(T(x, t), t) = diff(T2(x,t), x)+(diff(u(x, t), x))^2,diff(T(x, t), x)=T2(x,t)};

BCs := {u(0,t)=sin(t), u(10,t)=0,T(0,t)=1, T(10,t)=0,u(x,0)=0,T(x,0)=0};

pds2 := pdsolve(PDE, BCs, numeric, spacestep=1/50);

but I have a question, why we are suppose to have 3dplot instead of plot?

I just want to plot T2(x,t) vs t for x=0. Note for this plot x must be zero.

THanks

Thanks @Preben Alsholm

I tried to apply your suggested approach like this

restart:with(PDEtools):with(plots):

sys := {diff(T(x, t), t) = diff(T(x, t), x, x)+(diff(u(x, t), x))^2, diff(u(x, t), t) = diff(u(x, t), x, x)};

PDE:={diff(u(x,t),t)=diff(u(x,t),x$2),diff(T(x, t), t) = diff(T2(x,t), x)+(diff(u(x, t), x))^2,diff(T(x, t), x)=T2(x,t)};

BCs := {u(0,t)=sin(t), u(10,t)=0,T(0,t)=1, T(10,t)=0,u(x,0)=0,T(x,0)=0};

pds2 := pdsolve(PDE, BCs, numeric, spacestep=1/50);

but I have a question, why we are suppose to have 3dplot instead of plot?

I just want to plot T2(x,t) vs t for x=0. Note for this plot x must be zero.

THanks

@Carl Love 

Thanks for your kind response. Well, if we don't know the exact solution then how we can prove that the numerical scheme offers high accuracy?

@Carl Love 

Thanks for your kind response. Well, if we don't know the exact solution then how we can prove that the numerical scheme offers high accuracy?