## 355 Reputation

11 years, 361 days

## Sorry again!...

I didn't got you at first.

## Maple 18...

Sorry for the issue with the header. I am using Maple 18.

## seq in contour...

What's the main purpose of seq in the contour command here?

@tejolson I totally miss understood the question.

## dsolve...

Same type of question has been answered here using FDM.

## Missing stuff...

You need to specify the missing 8 ics and value for N.

## I = sqrt(-1)...

You are right. I is equal to sqrt(-1). And I need to know whhether, Maple can simplify this expression or not.

## S!...

"S" is the end point of y ( y is [0, S]), being just another numerical constant.

For example, We take S = 0.5 then f0(S=0.5)=1 having f0(0)=0.

Hope this make sense.

P.S. we need ''S'' to be symbolica as mentioned by @Preben Alsholm answering this question.

Thank You @Axel Vogt and @Preben Alsholm

## Worked...

Thanks dear. It worked the way I wanted.

## Diffusion equation...

I already uploaded the complete sheet, where I'm tring to solve the diffusion equation.

 > restart;
 > with(PDEtools):with(plots):
 > with(linalg):with(inttrans):
 >
 > eq:=diff(u(x,t),t)=phi*diff(u(x,t),x\$2);
 (1)
 > u(x,0):=0;
 (2)
 > bc1:=u(0,t)=sin(t);
 (3)
 > bc2:=u(infinity,t)=0;
 (4)
 > eqs:=laplace(eq,t,s);
 (5)
 > eqs:=subs(laplace(u(x,t),t,s)=U(x),eqs);
 (6)
 > bc1:=laplace(bc1,t,s);
 (7)
 > bc1:=subs(laplace(u(0,t),t,s)=U(0),bc1);
 (8)
 > U(x):=rhs(dsolve({eqs,bc1},U(x)));
 (9)
 > eqc:=coeff(U(x),exp(sqrt(s)*x/sqrt(phi)));
 (10)
 > _C2:=solve(eqc,_C2);
 (11)
 > U(x):=eval(U(x));
 (12)
 > u:=invlaplace(U(x),s,t);
 (13)
 > U(x):=subs(x=x1,phi=phi1,U(x));
 (14)
 > assume(x1>0,phi1>0);
 > u:=invlaplace(U(x),s,t);
 (15)
 > u:=subs(x1=x,phi1=phi,u);
 (16)
 >

Thanks

## M vs M...

As was pointed out by @tomleslie, I used the simplified version and tried to find the invlapalce using both Maple and Mathemaica.

Maple is able to give numerical results but no analytical expression.

u := invlaplace(exp(-sqrt(s)*x/sqrt(phi))/(s^2+1), s, t);

plot([subs(phi=1,t=Pi,u)],x=0..10,axes=boxed);

Mathematica gives me an analytical expression

g[s_] = Exp[-Sqrt[s] x/Sqrt[1]]/(s^2 + 1)

a = InverseLaplaceTransform[g[s], s, t] // Simplify[#, x > 0] &

t = 3*Pi/2;

Plot[a, {x, 0, 15}]

## Background of U(x)...

Thanks all for your valuable comments. Here is the backgroud of the function U(x).

ss.mw

Hand calculations gives me (tau=t)

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