# Need help for solving a set of differential eqns

1. Aug 27, 2010

### ygrl

Hi, I'm not good in math but I need to learn how to solve these differential equations.
It was really hard to write in TeX form, sorry.

d a / d t = d^2 a / d y^2 + d / d y (a * d x / d y)

d b / d t = d^2 b / d y^2 - d / d y (b * d x / d y)

d^2 x / d y^2 = - (a - b)

I need to solve for a, b and x.

What type of DE is this, a PDE? Which methods should I use?

2. Aug 28, 2010

### JJacquelin

It's difficult to answer since the typography of the equations is ambiguous.
Since symbol "d" is used everywhere instead of "d" in some places and "" in other places, it is impossible to know if each derivative is a partial or a total derivative.
Morover, in the third equation, since a and b are functions of t, then x might be a function of t. So x=x(y,t) instead of x=x(y) has a different meaning . [ except if a/t = b/t ]

3. Aug 28, 2010

### jackmell

Ok. I had some problems understanding it as well but did not look at it as a coupled system of PDEs. Perhaps then it should have been written as:

$$\frac{\partial a}{\partial t}=\frac{\partial^2 a}{\partial y^2}+a \frac{\partial^2 x}{\partial y^2}+\frac{\partial x}{\partial y}\frac{\partial a}{\partial y}$$

$$\frac{\partial b}{\partial t}=\frac{\partial^2 b}{\partial y^2}-b \frac{\partial^2 x}{\partial y^2}-\frac{\partial x}{\partial y}\frac{\partial b}{\partial y}$$

$$\frac{\partial^2 x}{\partial y^2}=b-a$$

for a,b, and x as functions of t and y. If that's what you want, then it's a coupled system of non-linear PDEs. If it were mine, I would resort to numerical methods. See:

Ortega, J.M. and Rheinholdt, W.C., "Iterative solutions of nonlinear equations in several variables"

Last edited: Aug 28, 2010
4. Aug 28, 2010

### JJacquelin

OK, jackmell, I agree.
Nevertheless, I am not sure that x is function of y AND t or if x is function of y only.
ygrl should state if there are dx/dy or x/y in the equations.
It is not yet clear if y is function of t or not.
If x was function of y only and if y was not function of t, it should be much simpler to solve.

5. Aug 28, 2010

### jackmell

Personally, I think my system is way more fun to work on although I'm pretty sure it would take me at least til' October to get a good handle on it. Would be a nice substitute for the final exam though. :)

6. Aug 28, 2010

### ygrl

Hi, jackmann's notation is correct. Instead of d there should be $$\partial$$

Thanks for helping me out. I'm gonna read the book that you've mentioned.

7. Aug 29, 2010

### JJacquelin

Well, in this case, analytical solving will be very arduous, probably impossible in present state of knowledge.
It is possible to reduce the set of three DPE to one DPE only with one unknown function only X(y,t) , as shown in attachment.
But this DPE is a fourth order non-linear DPE. So, numerical solving is probably the only realistic way.
Numerical treatment would probably be better directly with the original set of three PDE than with only one transformed DPE, because further integrations in ordrer to compute the functions a(y,t) and b(y,t) would be rather complicated and because it would need to transform the boundaries conditions as well.

#### Attached Files:

• ###### Setof3PDE.JPG
File size:
41.7 KB
Views:
132
8. Aug 29, 2010

### jackmell

Hi. Anyone know how to set up the proper initial and boundary values to solve the original system numerically in Mathematica? For example, how about solve it in a region 0<t<1 and 0<y<1 with these conditions:

a[0, y] == y, a[t, 0] == 0, a[t, 1] == 1,
b[0, y] == y, b[t, 0] == 0, b[t, 1] == 1,
x[0, y] == y, x[t, 0] == 0, x[t, 1] == 1

However when I attempt to set up NDSolve with the following code it returns several errors including:

"Warning: An insufficient number of boundary conditions \
have been specified for the direction of independent variable
Artificial boundary effects may be present in the solution.
"
and:

" step size is effectively zero;
singularity or stiff system suspected"

as well as other errors.

Code (Text):

mysol = NDSolve[{D[a[t, y], t] ==
D[a[t, y], {y, 2}] + a[t, y] D[x[t, y], {y, 2}] +
D[x[t, y], y] D[a[t, y], y],
D[b[t, y], t] ==
D[b[t, y], {y, 2}] - b[t, y] D[x[t, y], {y, 2}] -
D[x[t, y], y] D[b[t, y], y],
D[x[t, y], {y, 2}] == b[t, y] - a[t, y],
a[0, y] == y, a[t, 0] == 0, a[t, 1] == 1,
b[0, y] == y, b[t, 0] == 0, b[t, 1] == 1,
x[0, y] == y, x[t, 0] == 0, x[t, 1] == 1}, {a, b, x}, {t, 0,
1}, {y, 0, 1}]

9. Aug 30, 2010

### jackmell

Hi guys. I received some help with this and would like to document it as I think it's a beautiful equation. Here is the code to numerically solve it in the domain above:

Code (Text):
eqn = {D[a[t, y], t] ==
D[a[t, y], {y, 2}] + a[t, y] D[x[t, y], {y, 2}] +
D[x[t, y], y] D[a[t, y], y],
D[b[t, y], t] ==
D[b[t, y], {y, 2}] - b[t, y] D[x[t, y], {y, 2}] -
D[x[t, y], y] D[b[t, y], y],
D[D[x[t, y], {y, 2}] == b[t, y] - a[t, y], t]}

mysol = NDSolve[{{eqn}, a[0, y] == y, a[t, 0] == 0, a[t, 1] == 1,
b[0, y] == y, b[t, 0] == 0, b[t, 1] == 1, x[t, 0] == 0,
x[t, 1] == 1, x[0, y] == y}, {a[t, y], b[t, y], x[t, y]}, {t, 0,
1}, {y, 0, 1}, Method -> {"MethodOfLines", "TemporalVariable" -> t}]

Note how the method of lines is being specified and that x has been diffferentiated with respect to t. I don't understand why and that's why I said October. Anyway, NDSolve returns a solution and now, I'd like to back-substitute the solution into the three equations with the following code:

Code (Text):

In[135]:= myx[t_, y_] = Evaluate[x[t, y] /. mysol];
mya[t_, y_] = Evaluate[a[t, y] /. mysol];
myb[t_, y_] = Evaluate[b[t, y] /. mysol];
myxd1y[t_, y_] = D[myx[t, y], {y, 1}];
myxd2y[t_, y_] = D[myx[t, y], {y, 2}];
myad1t[t_, y_] = D[mya[t, y], {t, 1}];
myad1y[t_, y_] = D[mya[t, y], {y, 1}];
myad2y[t_, y_] = D[mya[t, y], {y, 2}];
mybd1t[t_, y_] = D[myb[t, y], {t, 1}];
mybd1y[t_, y_] = D[myb[t, y], {y, 1}];
mybd2y[t_, y_] = D[myb[t, y], {y, 2}];

tval = 1/3;
yval = 7/8;
(* first equation *)
rightside =
N[myad2y[tval, yval] + mya[tval, yval] myxd2y[tval, yval] +
(* second equation *)
leftside = N[mybd1t[tval, yval]]
rightside =
N[mybd2y[tval, yval] - myb[tval, yval] myxd2y[tval, yval] -
myxd1y[tval, yval] mybd1y[tval, yval]]
(* third equation *)
leftside = N[myxd2y[tval, yval]]
rightside = N[myb[tval, yval] - mya[tval, yval]]

Out[148]= {0.0084824}

Out[149]= {0.00838773}

Out[150]= {-0.0149216}

Out[151]= {-0.0148}

Out[152]= {-0.0935406}

Out[153]= {-0.0935406}

and note the three pairs of leftside and right side are quite close. I don't know about you guys, but I think it's pretty amazing to have something like Mathematica to help solve what a very long time ago I considered to be near the summit of mathematical difficulty. :)