J4James

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These are questions asked by J4James

I have the following fuction in Laplace domain,

restart:with(plots):with(inttrans):

u:=Pi^4*s3^(alpha-1)/((s1^2+Pi^2)*(s2^2+Pi^2)*(-s1^2+Pi^2*s3^alpha-s2^2))-Pi*s1*s2^(alpha-1)/(s3*(1+s2^alpha)*(-s1^2+Pi^2*s3^alpha-s2^2))-Pi*s1*s2^(alpha-1)/(-s1^2+Pi^2*s3^alpha-s2^2);

Where, s1, s2, s3 are the Laplace variables.

x1:=invlaplace(u, s1, x);

This worked. But the next two doesn't work.

y1:=invlaplace(x1, s2, y);

uu:=invlaplace(y1, s3,t);

Even, I tried to plot the unevaluated invlaplace but no luck.

alpha:=1:t:=1:
plot3d(uu,x =1..2, y=1..2);
 

Am I missing something?

 

I have a system of ode's

restart:with(plots):
eq1:=((diff(f(x),x$3)))+f(x)*diff(f(x),x$2)-a*diff(f(x),x$1)^2=0;
eq2:=(diff(g(x),x$2))+b*f(x)*diff(g(x),x$1)=0;
bc1:=f(0)=0,D(f)(0)=1,D(f)(5)=0,g(0)=0.5,g(5)=0;

dsolve can solve it easily.

sol:=dsolve(subs(a=0.5,b=0.5,{bc1,eq1,eq2}), numeric):

Now I was to use the numerical data from the sol in the following expression

Expr:=a*f(x)+b*g'(x)+c*f'(x)*g(x) # at x=1:

to make a table of this form a  b  c  Expr, where a, b, c vary simultaneously between 0 and 1 by 0.2 producing data for Expr.

Any suggestions? 

 

I have this H(alpha,beta,eta) con ‡uent hypergeometric function, which I can easily adopt in Mathematica like this

Hypergeometric1F1[( (\[Alpha] + S \[Beta]))/\[Beta]^2, 
 1 + ((\[Alpha] + S \[Beta]))/\[Beta]^2, -\[Alpha] ((
   E^(-\[Beta] \[Eta]))   )/\[Beta]^2]

But when I try to input this in maple using the built-in `hypergeom` it gives an error.

hypergeom((S*lambda2+alpha)/lambda2^2, 1+(S*lambda2+alpha)/lambda2^2, -alpha*exp(-lambda2*eta)/lambda2^2)

Any suggestions?

 

I have solved a BVP using finite difference method in maple. But the comparsion with the dsolve solution shows big difference,

Why there is too much difference? Is there an elegant way to implement FDM?

FDM_Dbvp_sheet_1.mw

I am having difficulty to solve the following ODE

ode:=diff(R(x),x$6)=Dirac(x-y);

with initial and boundary conditions

bcs := ((D@@3)(R))(1) = 0, R(0)-((D@@5)(R))(0) = 0, ((D@@2)(R))(0)-((D@@3)(R))(0) = 0, ((D@@5)(R))(1) = 0, (D(R))(0)+((D@@4)(R))(0) = 0, ((D@@4)(R))(1) = 0;

I want to solve the above ODE for two cases 1) x>y and 2) x<=y

Thanks

 

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