General solution to partial differential equation (PDE)

manchester20
Messages
1
Reaction score
0
Hi,
I have the following PDE-S\frac{\partial\vartheta}{\partial\tau}+\frac{1}{2}\sigma^2\frac{X^2}{S}\frac{\partial^2\vartheta}{\partial\xi^{2}} + [\frac{S}{T} + (r-D)X]\frac{\partial\vartheta}{\partial\xi}I am asked to seek a solution of the form \vartheta=\alpha_1(\tau)\xi + \alpha_0(\tau) and give a general solution for \alpha_1(\tau) and \alpha_0(\tau)

where we have
\tau=T-t
and
\xi=\frac{t}{T}-\frac{X}{S}

I have tried doing the partial differentials of \vartheta with respect to τ and ε, but the answer doesn't allow me to get a general solution for the two unknown functions of τ.
If anyone could help i would be really grateful.
Thanks

NOTE: the word 'partial' in the equation should be a symbol for the partial derivative.
 
Last edited:
Physics news on Phys.org
manchester20 said:
Hi,
I have the following PDE


-S\frac{\partial\vartheta}{\partial\tau}+\frac{1}{2}\sigma^2\frac{X^2}{S}\frac{\partial^2\vartheta}{\partial\xi^{2}} + [\frac{S}{T} + (r-D)X]\frac{\partial\vartheta}{\partial\xi}


I am asked to seek a solution of the form \vartheta=\alpha_1(\tau)\xi + \alpha_0(\tau) and give a general solution for \alpha_1(\tau) and \alpha_0(\tau)

where we have
\tau=T-t
and
\xi=\frac{t}{T}-\frac{X}{S}

I have tried doing the partial differentials of \vartheta with respect to τ and ε, but the answer doesn't allow me to get a general solution for the two unknown functions of τ.
If anyone could help i would be really grateful.
Thanks

NOTE: the word 'partial' in the equation should be a symbol for the partial derivative.

Hi !

Sorry to say, but the wording of the problem seems very fishy (see attachment)
 

Attachments

  • Ambiguous PDE.JPG
    Ambiguous PDE.JPG
    41.6 KB · Views: 594
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Replies
36
Views
4K
Replies
1
Views
3K
Replies
3
Views
4K
Replies
1
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
Back
Top